internal index. When $f_{S}=1$ and $f_{P}=0$, we see that $\mathcal{T}^{(i)}$
reduces to $-i\,\mathbb{I}$, which expresses the usual imaginary-anti-self-
duality constraint.

Similarly, the general currents

$\displaystyle\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}_{w}=\frac{i}{4}\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu^{\overline{i}\overline{j}}\left(f_{S}-f_{P}-1\right)\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu_{w}^{\overline{j}}\,,\qquad\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}_{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}=\frac{i}{4}\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu^{\overline{i}\overline{j}}\left(f_{S}-f_{P}+1\right)\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}^{\overline{j}}\,,$
(3.46)

satisfy the non-linear self-duality condition

$\displaystyle\left(\ast\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu\right)^{\overline{i}}_{\alpha}$ $\displaystyle=\tensor{\left(\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern
1.5mu^{(\overline{i})}\right)}{{}_{\alpha}^{\beta}}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}_{\beta}\,,$ $\displaystyle\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern
1.5mu^{(\overline{i})}$
$\displaystyle=i\begin{bmatrix}1&-\frac{2\tensor{\overline{k}}{{}^{\overline{i}}_{\overline{j}}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu_{w}^{\overline{j}}(1+f_{P}-f_{S})}{\tensor{\overline{k}}{{}^{\overline{i}}_{\overline{k}}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}^{\overline{k}}(-1+f_{P}-f_{S})}\\\ 0&1\end{bmatrix}\,.$ (3.47)

Likewise, when $f_{S}=1$ and $f_{P}=0$, we see that $\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern
1.5mu^{(\overline{i})}=i\,\mathbb{I}$ so this reduces to the usual imaginary-
self-duality condition $\ast\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}=i\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}$.

We should point out that, in other studies of twisted self-duality in Chern-
Simons theories such as [76, 22], the twisting operator $\mathcal{T}$ commutes
with the Hodge star operation. As a result, acting with the Hodge star
operator on each side of the twisted self-duality constraint $\ast
J=\mathcal{T}J$, one has

$\displaystyle\ast\ast J=\ast\mathcal{T}J=\mathcal{T}\ast
J=\mathcal{T}^{2}J\,.$ (3.48)

Since the Hodge star is an anti-involution, $\ast\ast=-\mathbb{I}$, in two
Euclidean dimensions, one therefore arrives at the constraint

$\displaystyle\mathcal{T}^{2}=-\mathbb{I}\,.$ (3.49)

In Lorentzian signature, this would instead give the constraint
$\mathcal{T}^{2}=\mathbb{I}$.

However, in our case the twisting operators $\mathcal{T}^{(i)}$ and $\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern
1.5mu^{(\overline{i})}$ have non-trivial structure in their Lorentz indices
and therefore do not commute with the Hodge star. This is why, in our case,
these twisting operators do not satisfy an anti-involutive constraint like
(3.49).

One can now proceed as in the linear case and define projection operators

$\displaystyle P_{+}^{(i)}$ $\displaystyle=\begin{bmatrix}0&0\\\
-\frac{\tensor{k}{{}^{i}_{j}}A_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}^{j}}{\tensor{k}{{}^{i}_{k}}A_{w}^{k}}\frac{f_{S}+f_{P}-1}{f_{S}+f_{P}+1}&1\end{bmatrix}\,,\qquad\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{+}^{(\overline{i})}=\begin{bmatrix}0&\frac{\tensor{\overline{k}}{{}^{\overline{i}}_{\overline{j}}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu_{w}^{\overline{j}}(1+f_{P}-f_{S})}{\tensor{\overline{k}}{{}^{\overline{i}}_{\overline{k}}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}^{\overline{k}}(-1+f_{P}-f_{S})}\\\ 0&1\end{bmatrix}\,,$ $\displaystyle
P_{-}^{(i)}$ $\displaystyle=\begin{bmatrix}1&0\\\
\frac{\tensor{k}{{}^{i}_{j}}A_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}^{j}}{\tensor{k}{{}^{i}_{k}}A_{w}^{k}}\frac{f_{S}+f_{P}-1}{f_{S}+f_{P}+1}&0\end{bmatrix}\,,\qquad\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{-}^{(\overline{i})}=\begin{bmatrix}1&-\frac{\tensor{\overline{k}}{{}^{\overline{i}}_{\overline{j}}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu_{w}^{\overline{j}}(1+f_{P}-f_{S})}{\tensor{\overline{k}}{{}^{\overline{i}}_{\overline{k}}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}^{\overline{k}}(-1+f_{P}-f_{S})}\\\ 0&0\end{bmatrix}\,,$ (3.50)

which satisfy the expected properties of orthogonal projectors,

$\displaystyle\left(P_{\pm}^{(i)}\right)^{2}=P_{\pm}^{(i)}\,,\quad\left(\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{\pm}^{(\overline{i})}\right)^{2}=\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{\pm}^{(\overline{i})}\,,\quad P_{\pm}^{(i)}P_{\mp}^{(i)}=0=\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{\pm}^{(\overline{i})}\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{\mp}^{(\overline{i})}\,,$ (3.51)

along with the chirality conditions

$\displaystyle P_{-}^{(i)}J^{i}=J^{i}\,,\quad P_{+}^{(i)}J^{i}=0\,,\quad\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{+}^{(\overline{i})}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}=\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}\,,\quad\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{-}^{(\overline{i})}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}=0\,.$
(3.52)

Therefore, even in the interacting case, one can view the currents as
satisfying an appropriate non-linear self-duality constraint. This expresses,
in Chern-Simons language, the equations of motion (2.23) for interacting
Floreanini-Jackiw bosons.

We should point out that this construction has now produced two separate pairs
of projection operators $P_{\pm}^{(i)}$, $\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu_{\pm}^{(\overline{i})}$
for each fixed choice of indices $i,\overline{i}$, or equivalently, two
separate twist operators $\mathcal{T}^{(i)}$ and $\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern 1.5mu^{(i)}$. This
is in contrast with the linear-self duality constraint, which is described by
only two projectors $P_{\pm}=\frac{1}{2}\left(1\mp i\ast\right)$, where

$\displaystyle P_{+}=\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{+}\,\qquad P_{-}=\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu_{-}\,.$ (3.53)

In the linear case, there are relations that cause these four operators to
collapse to just two independent projectors, and it is clear that these
operators project onto one-dimensional eigenspaces which represent physically
opposite chiralities.

In the non-linear case, there are also relations (albeit more complicated
ones) between the two twist operators. For instance, one can see that
$\mathcal{T}^{(i)}$ can be obtained from $\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern 1.5mu^{(i)}$ by
simultaneously transposing the matrix in its Lorentz indices and interchanging
all barred and unbarred quantities. That is, one exchanges

$\displaystyle k^{ij}\longleftrightarrow\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu^{\overline{i}\overline{j}}\,,\quad A^{i}\longleftrightarrow\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}\,\quad
w\longleftrightarrow{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}\,,$ (3.54)

which also has the effect of sending $P\to-P$ (and thus $f_{P}\to-f_{P}$).
This relation holds regardless of the choice of boundary term. When the
function $f(S,P)$ satisfies the self-duality condition (3.35) necessary to
describe chiral modes, there are further constraints between the twist
operators. To see one such constraint, we can rewrite (3.35) as

$\displaystyle J^{i}\wedge\ast J^{i}=\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}\wedge\ast\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}\,.$
(3.55)

Since $\ast J^{i}=\mathcal{T}^{(i)}J^{i}$ and $\ast\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\overline{i}}=\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern
1.5mu^{(\overline{i})}\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}$, this relation can also be expressed as

$\displaystyle J^{i}\wedge\mathcal{T}^{(i)}J^{i}=\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}\wedge\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern
1.5mu^{(\overline{i})}\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}\,.$ (3.56)

Equation (3.56) is a consequence of the fact that, when the boundary term
obeys the self-duality constraint, the chiral and anti-chiral twist operators
are ``compatible'' in a sense which generalizes the statements that
$\mathcal{T}^{(i)}=-\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{T}\mkern-1.5mu}\mkern
1.5mu^{(\overline{i})}$, or that the projection operators satisfy (3.53), in
the linear case.

###  3.3 Current deformations of boundary terms

We will now consider flow equations which modify the boundary term
$\mathcal{L}_{\text{bdry}}$ of a bulk Chern-Simons theory.131313Although we
focus on $U(1)$ Chern-Simons theories in this work, stress tensor deformations
of the boundary term for $SL(2)\times SL(2)$ Chern-Simons have been considered
in [77, 78, 79]. In particular, we are interested in differential equations
for $\mathcal{L}_{\text{bdry}}$ which are driven by conserved quantities. We
will refer to any such flow equation as a ``current deformation'' regardless
of whether the conserved currents driving the flow are the objects
$J_{\alpha}^{i}$ and $\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu_{\alpha}^{\overline{i}}$ defined in equation (3.14), or the energy-
momentum tensor $T_{\alpha\beta}$, which is another type of conserved current
in the theory.

Let us first study deformations which involve the spin-$1$ currents
$J_{\alpha}^{i}$ and $\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu_{\alpha}^{\overline{i}}$. A general flow equation in this class takes
the form

$\displaystyle\frac{\partial\mathcal{L}_{\text{bdry}}}{\partial\lambda}=\mathcal{O}\left(J_{\alpha}^{i},\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu_{\alpha}^{\overline{i}},\lambda\right)\,,$ (3.57)

where $\mathcal{O}$ is a Lorentz scalar and $O(N)\times O(\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)$ singlet constructed
from the currents. Within this class, there are fewer interesting
possibilities. The most natural deformation to consider is to begin with the
conventional boundary term $\mathcal{L}_{\text{bdry}}=S$ and deform by a
marginal combination of the form

$\displaystyle\mathcal{O}=k_{ij}J_{\alpha}^{i}J^{\alpha j}\,,\;\text{ or
}\quad\mathcal{O}=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu_{\overline{i}\overline{j}}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu_{\alpha}^{\overline{i}}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\alpha\overline{j}}\,.$
(3.58)

However, by virtue of the chirality of the currents given in equation (3.26),
both of these operators vanish. One might instead construct a deforming
operator which mixes the currents on the two sides, such as

$\displaystyle\mathcal{O}=C_{i\overline{j}}J_{\alpha}^{i}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\alpha\overline{j}}\,,$
(3.59)

where $C_{i\overline{j}}$ is a constant tensor with mixed indices. For
instance, in the case $N=\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$, we do not need to
distinguish between barred and unbarred indices, and can choose
$C_{i\overline{j}}=\delta_{i\overline{j}}\equiv\delta_{ij}$.141414Of course,
when $N\neq\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$, a
deformation of this form does not preserve $O(N)\times O(\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)$ symmetry. For
instance, a deformation by $\sum_{i=1}^{M}J^{i}_{\alpha}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{i\alpha}$, where
$M=\min(N,\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)$,
treats the currents asymmetrically. Let us consider the effect of this
deformation with the simplifying assumption $k^{ij}=\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu^{ij}=\delta^{ij}$. In
this case, at leading order in the deformation parameter, one finds a deformed
boundary term

$\displaystyle\mathcal{L}_{\text{bdry}}^{(1)}=\frac{1}{2}\left(A_{i}^{\alpha}A^{i}_{\alpha}+\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{i}^{\alpha}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu^{i}_{\alpha}\right)+\lambda A_{i}^{\alpha}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu^{i}_{\alpha}\,,$ (3.60)

up to the normalization of $\lambda$. That is, such an operator has introduced
an off-diagonal mixing between the barred and unbarred gauge fields. Ignoring
possible subtleties about quantization of the Chern-Simons levels, such a
quadratic mixing can always be undone by performing a Bogoliubov-like field
redefinition. Indeed, note that beginning with the undeformed boundary term

$\displaystyle\mathcal{L}_{\text{bdry}}^{(0)}=\frac{1}{2}\left(A_{i}^{\alpha}A^{i}_{\alpha}+\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{i}^{\alpha}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu^{i}_{\alpha}\right)$
(3.61)

and then performing a change of variables to

$\displaystyle A^{i}_{\alpha}=\cosh(\mu)B^{i}_{\alpha}+\sinh(\mu)\mkern
1.5mu\overline{\mkern-1.5muB\mkern-1.5mu}\mkern
1.5mu^{i}_{\alpha}\,,\qquad\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu^{i}_{\alpha}=\cosh(\mu)\mkern
1.5mu\overline{\mkern-1.5muB\mkern-1.5mu}\mkern
1.5mu^{i}_{\alpha}+\sinh(\mu)B^{i}_{\alpha}\,,$ (3.62)

gives the transformed boundary term

$\displaystyle\mathcal{L}_{\text{bdry}}^{(0)}=\cosh(2\mu)\left[\frac{1}{2}\left(B_{i}^{\alpha}B^{i}_{\alpha}+\mkern
1.5mu\overline{\mkern-1.5muB\mkern-1.5mu}\mkern 1.5mu_{i}^{\alpha}\mkern
1.5mu\overline{\mkern-1.5muB\mkern-1.5mu}\mkern
1.5mu^{i}_{\alpha}\right)+\tanh(2\mu)B_{i}^{\alpha}\mkern
1.5mu\overline{\mkern-1.5muB\mkern-1.5mu}\mkern 1.5mu^{i}_{\alpha}\right]\,.$
(3.63)

Up to an overall rescaling, this is equivalent to the deformed boundary term
(3.60) if we identify $\tanh(2\mu)=\lambda$. Therefore, the marginal $J\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu$ deformation of equation
(3.59) can be viewed as inducing a rotation between the fields
$A_{\alpha}^{i}$ and $\mkern 1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu_{\alpha}^{\overline{i}}$. We will see later that the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation, in the
case $N=\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=1$, is
qualitatively similar to this $J\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu$ deformation.

In principle, one could consider more general operators constructed from the
currents $J$ and $\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu$, such as powers of the form $\mathcal{O}=\left(J_{\alpha}^{i}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu^{\alpha i}\right)^{n}$
or other structures such as
$\mathcal{O}=\left(J_{\alpha}^{i}J_{\beta}^{i}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\alpha\overline{j}}\mkern
1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern
1.5mu^{\beta\overline{j}}\right)^{m}$, both of which preserve $O(N)\times
O(\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)$ symmetry.
These operators are irrelevant for $n>1$ and $m>\frac{1}{2}$, respectively.
However, we will now instead turn our attention to deformations which are
constructed from the energy-momentum tensor,

$\displaystyle\frac{\partial\mathcal{L}_{\text{bdry}}}{\partial\lambda}=\mathcal{O}\left(T_{\alpha\beta}^{(\lambda)},\lambda\right)\,.$
(3.64)

The first choice that one must make in defining such a flow is _which_ stress
tensor to use. There are generally many definitions of the energy-momentum
tensor which are all conserved but which differ by improvement
transformations. One natural choice is the Hilbert stress tensor defined by
varying the metric. Of course, neither the Chern-Simons action (3.5) nor the
boundary action (3.6) depend on the _bulk_ metric, but the term
$I_{\text{bdry}}$ does depend on the boundary metric. One can therefore define
a boundary stress tensor,

$\displaystyle T_{\alpha\beta}=-\frac{2}{\sqrt{g}}\frac{\delta I}{\delta
g^{\alpha\beta}}=-\frac{2}{\sqrt{g}}\frac{\delta I_{\text{bdry}}}{\delta
g^{\alpha\beta}}\,.$ (3.65)

However, this stress tensor is qualitatively different from the one obtained
in equation (2.2) by coupling a chiral boson theory to vielbeins. In that
context, the coupling to vielbeins treated chiral and anti-chiral modes
differently, and as a result the stress tensor component $T_{t\theta}=-P$ is
sensitive to the difference between chiral and anti-chiral fields. Exchanging
the fields $\phi$ with $\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu$, and vice-versa,
reverses the sign of $P$ and therefore changes $T_{t\theta}$.

In contrast, since both the barred gauge fields and unbarred gauge fields
couple to the boundary metric in the same way, the Hilbert stress tensor
(3.65) treats the fields $A_{\alpha}^{i}$ and $\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{\overline{i}}^{\alpha}$
on equal footing. Unlike (2.2), the Hilbert stress tensor associated with the
standard boundary Lagrangian $\mathcal{L}_{\text{bdry}}=S$ is unchanged under
the process of exchanging barred and unbarred gauge fields. To make this point
explicit, let us write this boundary term as

$\displaystyle\mathcal{L}_{\text{bdry}}=\frac{1}{2}\tensor{S}{{}_{\alpha}^{\alpha}}\,,\qquad
S_{\alpha\beta}=k^{ij}A_{i\alpha}A_{j\beta}+\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu^{\overline{i}\overline{j}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu_{\overline{i}\alpha}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{\overline{j}\beta}\,.$
(3.66)

With this definition, one has $\tensor{S}{{}_{\alpha}^{\alpha}}=2S$. The
Hilbert stress tensor computed from (3.66), after rescaling to eliminate the
overall prefactor of $-\frac{1}{16\pi}$ in $I_{\text{bdry}}$, is

$\displaystyle T_{\alpha\beta}=-S_{\alpha\beta}+g_{\alpha\beta}S\,.$ (3.67)

Deforming the standard boundary term by a generic function of the stress
tensor (3.67), which necessarily involves the single independent non-vanishing
Lorentz invariant $T^{\alpha\beta}T_{\alpha\beta}$, will introduce dependence
on the new variable

$\displaystyle S_{2}=S_{\alpha\beta}S^{\alpha\beta}\,.$ (3.68)

Note that $S_{2}$ is functionally independent from the invariant
$P=\frac{1}{2}\left(k^{ij}A_{i}^{\alpha}A^{j}_{\alpha}-\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu^{\overline{i}\overline{j}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern
1.5mu_{\overline{i}}^{\alpha}\overline{A}^{\overline{j}}_{\alpha}\right)$.
Therefore, the class of boundary terms that can be described by functions
$f(S,P)$ is not closed under deformations by the Hilbert stress tensor.
Instead, to describe flows driven by this choice of stress tensor, we should
instead parameterize the boundary term as a function of different invariants:

$\displaystyle\mathcal{L}_{\text{bdry}}=f(S_{1},S_{2})\,,$ (3.69)

where

$\displaystyle S_{1}=\Tr(S)=\tensor{S}{{}_{\alpha}^{\alpha}}=2S\,,\qquad
S_{2}=\Tr(S^{2})=S_{\alpha\beta}S^{\alpha\beta}\,.$ (3.70)

The structure of Hilbert stress tensor deformations of the class of functions
(3.69) is identical to the structure of such flows for a collection of _non-
chiral_ bosons. Indeed, as was worked out in [30], a general Lagrangian for a
collection of $N$ non-chiral bosons $\varphi^{i}$ with target space metric
$G_{ij}$ is a function of the matrix

$\displaystyle\tensor{X}{{}_{\alpha}^{\beta}}=G_{ij}\partial_{\alpha}\varphi^{i}\partial^{\beta}\varphi^{j}\,,$
(3.71)

which has two independent traces,

$\displaystyle x_{1}=\Tr(X)=\tensor{X}{{}_{\alpha}^{\alpha}}\,,\qquad
x_{2}=\Tr(X^{2})=\tensor{X}{{}_{\alpha}^{\beta}}\tensor{X}{{}_{\beta}^{\alpha}}\,.$
(3.72)

All higher traces can be expressed in terms of $x_{1}$ and $x_{2}$ using
identities derived from the Cayley-Hamilton theorem for $2\times 2$ matrices.
Precisely the same results apply in the Chern-Simons context, except replacing
the matrix $\tensor{X}{{}_{\alpha}^{\beta}}$ with
$\tensor{S}{{}_{\alpha}^{\beta}}$ and thus replacing the invariants $x_{1}$,
$x_{2}$ with $S_{1}$, $S_{2}$. For instance, the Hilbert stress tensor
associated with a general boundary term (3.69) is

$\displaystyle T_{\alpha\beta}=-2\frac{\partial f}{\partial
S_{1}}S_{\alpha\beta}-4\frac{\partial f}{\partial
S_{2}}S_{\alpha\gamma}\tensor{S}{{}^{\gamma}_{\beta}}+g_{\alpha\beta}f\,.$
(3.73)

One can then construct deformations of the boundary term which depend on the
two independent traces of the stress tensor, which can be written as

$\displaystyle T^{\alpha\beta}T_{\alpha\beta}$
$\displaystyle=2\left(f+2S_{1}^{2}\frac{\partial f}{\partial
S_{2}}\right)\left(f-2S_{1}\left(\frac{\partial f}{\partial
S_{1}}+S_{1}\frac{\partial f}{\partial
S_{2}}\right)\right)+8S_{2}^{2}\left(\frac{\partial f}{\partial
S_{2}}\right)^{2}$ $\displaystyle\quad+4S_{2}\left(\left(\frac{\partial
f}{\partial S_{1}}\right)^{2}+6S_{1}\frac{\partial f}{\partial
S_{1}}\frac{\partial f}{\partial S_{2}}-2\frac{\partial f}{\partial
S_{2}}\left(f-2S_{1}^{2}\frac{\partial f}{\partial S_{2}}\right)\right)\,,$
(3.74) $\displaystyle\tensor{T}{{}^{\alpha}_{\alpha}}$
$\displaystyle=-2S_{1}\frac{\partial f}{\partial S_{1}}-4S_{2}\frac{\partial
f}{\partial S_{2}}+2f\,.$ (3.75)

All of the results concerning stress tensor flows for non-chiral bosons in two
dimensions (see, for instance, [30] and section 4 of [41]) therefore
immediately apply to deformations of Chern-Simons boundary terms which take
the form (3.69).

One way to think about this class of deformations, using the parameterization
(3.69) and the Hilbert stress tensor, is the following. In the case $N=\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$ – when the unbarred
gauge fields $A^{i}_{\alpha}$ and barred gauge fields $\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu^{i}_{\alpha}$ are dual
to equal numbers of chiral bosons $\phi^{i}$ and anti-chiral bosons $\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{i}$, respectively –
one can collect these fields into a collection of non-chiral bosons
$\varphi^{i}$ as

$\displaystyle\varphi^{i}=\frac{1}{\sqrt{2}}\left(\phi^{i}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{i}\right)\,.$ (3.76)

We will revisit the quantization of the boundary theory after performing this
repackaging of the field content into non-chiral fields in section 5. We claim
that deformations using the Hilbert stress tensor and the parameterization
(3.69) are appropriate for understanding flows in which the bosons are
assembled into non-chiral fields in this way. This is why such flows are
naturally studied using the invariants $(S_{1},S_{2})$, which have the same
structure as the ones appearing in ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-like deformations of
non-chiral bosons, rather than the invariants $(S,P)$, which we have used in
section 2 to understand stress tensor flows for chiral bosons.

One might ask whether there is a different presentation of stress tensor
deformations for the boundary term whose structure is more similar to that of
flows in the Floreanini-Jackiw description of section 2. This brings us to the
second natural choice of stress tensor, besides the Hilbert definition in
equation (3.65). Rather than coupling the boundary theory to a metric on
$\partial\mathcal{M}_{3}$, one could instead couple to vielbeins in the same
way as we did in equation (2.49) for chiral boson theories. To do this, we
again introduce frame fields $\tensor{E}{{}^{a}_{\alpha}}$, although now the
flat indices will be raised or lowered with the _Euclidean_ tangent-space
metric $\eta_{ab}=\left[\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right]$. In this case, the appropriate flat-space values
for the vielbeins are

$\displaystyle E^{+}_{w}=E^{-}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}=\frac{1}{\sqrt{2}}\,,\qquad E^{+}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}=E^{-}_{w}=0\,,$ (3.77)

whose inverses produce the desired spacetime metric $ds^{2}=dw\,d{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}$,

$\displaystyle
E^{a}_{\alpha}E^{b}_{\beta}\eta_{ab}=g_{\alpha\beta}=\begin{bmatrix}0&\frac{1}{2}\\\
\frac{1}{2}&0\end{bmatrix}\,.$ (3.78)

One can then couple the Chern-Simons boundary term $I_{\text{bdry}}$ to
vielbeins as

$I_{\text{bdry}}=-\frac{i}{16\pi}\int_{\partial\mathcal{M}_{3}}d^{2}x\,\left(2\left(E^{+}_{w}E^{-}_{w}-E^{+}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}E^{-}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}\right)P+2Ef(S,P)\right)\,,$ (3.79)

where we include factors of $2$ since, in the conventions of this section,
$E=\frac{1}{2}$. Likewise, the overall factor of $i$ in (3.79) arises because
$\sqrt{g}=\frac{i}{2}$ but $E=\frac{1}{2}$. To compare with equation (2.49),
note that in the conventions of section 2, we instead had $E=1$. Now $S$ and
$P$ are coupled to vielbeins as

$\displaystyle S$ $\displaystyle=\frac{1}{4\left(E^{+}_{w}E^{-}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}+E^{-}_{w}E^{+}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}+E^{+}_{w}E^{-}_{w}+E^{+}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}E^{-}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}\right)}\left(k^{ij}A_{iw}A^{j}_{j{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}+\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu^{\overline{i}\overline{j}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{\overline{i}w}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu^{\overline{j}}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}\right)\,,$
$\displaystyle P$ $\displaystyle=\frac{1}{4\left(E^{+}_{w}E^{-}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}+E^{-}_{w}E^{+}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}+E^{+}_{w}E^{-}_{w}+E^{+}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}E^{-}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}\right)}\left(k^{ij}A_{iw}A^{j}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}-\mkern
1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern
1.5mu^{\overline{i}\overline{j}}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{\overline{i}w}\mkern
1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu^{\overline{j}}_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}\right)\,,$ (3.80)

in such a way that they reduce to their flat-space values when the vielbeins
are given by (3.77). Because these expressions are written with explicit
$(w,{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu})$ indices,
the resulting coupling to gravity is not manifestly Lorentz-invariant.
However, this is to be expected since we are performing the equivalent of the
procedure used in equation (2.49) for coupling Floreanini-Jackiw bosons to
gravity, which is also not manifestly Lorentz-invariant.

We now compute the stress tensor (2.6) using this coupling to the frame
fields. In order to make comparison with the results of section 2 easier, we
will re-scale the stress tensor by an overall factor to absorb the
multiplicative constant of $-\frac{i}{16\pi}$ in the boundary term (3.79), as
well as the relative factor of $2$ due to the conventions for the vielbein in
this section. Therefore we instead compute

$\displaystyle\tensor{T}{{}_{\beta}^{a}}=-\frac{8\pi i}{E}\frac{\delta
S}{\delta\tensor{E}{{}^{\beta}_{a}}}\,,$ (3.81)

and convert to spacetime indices to find

$\displaystyle T_{ww}$
$\displaystyle=-\frac{1}{4}\left(2SV_{S}+2P\left(1+V_{P}\right)\right)\,,$
$\displaystyle T_{w{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}}$ $\displaystyle=T_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern
1.5mu}w}=\frac{1}{2}\left(V-SV_{S}-PV_{P}\right)\,,$ $\displaystyle T_{{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}{\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu}}$
$\displaystyle=\frac{1}{2}\left(P-PV_{P}-SV_{S}\right)\,.$ (3.82)

The two Lorentz scalars that we use for constructing flows are therefore

$\displaystyle\tensor{T}{{}^{\alpha}_{\alpha}}$
$\displaystyle=2\left(V-SV_{S}-PV_{P}\right)\,,$ $\displaystyle
T^{\alpha\beta}T_{\alpha\beta}$
$\displaystyle=V^{2}-2P^{2}+\left(V-2\left(SV_{S}+PV_{P}\right)\right)^{2}\,,$
(3.83)

which exactly matches equations (2.55) and (2.56).

It now follows that all of our comments about stress tensor flows in section 2
immediately apply to deformations of Chern-Simons boundary terms which are
constructed using the stress tensor (3.81) obtained from coupling to
vielbeins, as opposed to the standard Hilbert stress tensor. For instance, any
deformation by a function of the vielbein stress tensor (3.81) necessarily
preserves the condition (3.35). This means that, if one begins with a seed
Chern-Simons boundary term which is invariant under the symmetry (3.32) that
guarantees the chirality (or self-duality) of the theory, and then deforms
this seed by any function of the energy-momentum tensor, the resulting
deformed boundary term will also be invariant under the same symmetry.
Furthermore, any one-parameter family of Chern-Simons boundary terms which are
all invariant under the duality rotation (3.32) must satisfy a differential
equation driven by a function of the vielbein stress tensor.

It also follows that the closed-form solutions to flow equations driven by
functions of the stress tensor discussed in section 2 – such as the two-
parameter family of solutions (2.40) to the commuting ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ and root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow equations – also
have obvious analogs for deformations of Chern-Simons boundary terms. Besides
solving these differential equations directly, a complementary way to analyze
stress tensor deformations is by performing a perturbative expansion which
computes the deformed action order-by-order in the flow parameter. This
approach is discussed in appendix A for deformations by various functions of
the energy-momentum tensor, using the version of $T_{\alpha\beta}$ defined by
coupling to vielbeins.

To conclude this section, let us summarize and mention some applications. We
have seen that the boundary term of a bulk $U(1)$ Chern-Simons theory can be
deformed either by functions of the Hilbert stress tensor or by functions of
the vielbein stress tensor (3.81). The former deformations lead to a class of
modified boundary terms $\mathcal{L}_{\text{bdry}}(S_{1},S_{2})$ with the same
properties as Lagrangians obtained by stress tensor deformations of non-chiral
boson theories. Conversely, the latter flows generate a family of boundary
terms $\mathcal{L}_{\text{bdry}}(S,P)$ with the same structure as the
Lagrangians in section 2 arising from stress tensor deformations of chiral
boson theories. We have thus described two complementary ways to view
deformations of Chern-Simons boundary terms by functions of the energy-
momentum tensor.

These results provide a general framework for studying three-dimensional
$U(1)$ Chern-Simons theories subject to boundary deformations. Throughout our
discussion, we have been agnostic as to the specific setting in which such
Chern-Simons terms arise, but let us briefly mention two specific applications
of the formalism we have developed. One context in which these results could
be useful is when considering $\mathrm{AdS}_{3}/\mathrm{CFT}_{2}$ holography
with $U(1)$ gauge fields. One could use our machinery to derive flow equations
for various observables under stress tensor deformations, just as [79] found
expressions for ${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern
1.5mu}$-deformed Wilson lines and loops, and [32] obtained formulas for the
masses of BTZ black holes under a boundary root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation. For
instance, one could use the results of this section to analyze the dependence
of the $U(1)$ charges of charged BTZ black holes as a function of the
deformation parameter for boundary ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ or root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformations. Another
possible application of these results is to study quantum Hall systems subject
to boundary deformations, which we will briefly describe in the conclusion of
this paper.

##  4 Quantization Along Classical Flows

In this section, we will consider the quantization of a member of the general
class of interacting chiral boson models. We will work purely within the
Floreanini-Jackiw description, described by an action of the form (2.22),
rather than in the Chern-Simons formulation of section 3. We will also work in
Lorentzian signature with spacetime coordinates $(t,\theta)$. Although in the
preceding discussion we have been agnostic as to the spacetime topology,
within this section we will assume that $\theta$ is compact and subject to the
identification $\theta\sim\theta+2\pi$. We focus on the case of a compact
spatial manifold because our primary observable of interest is the finite-
volume spectrum of energy levels $E_{n}$, and in particular how these energies
depend on a deformation parameter along a stress tensor flow.

The most well-studied example of a stress tensor deformation for which the
deformed cylinder spectrum can be determined is the ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation. Under the
${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow, the
energy levels of the deformed theory obey the inviscid Burgers' equation,

$\displaystyle\frac{\partial E_{n}}{\partial\lambda}=E_{n}\frac{\partial
E_{n}}{\partial R}+\frac{P_{n}^{2}}{R}\,,$ (4.1)

where $R$ is the radius of the cylinder and $E_{n},P_{n}$ are the energy and
momentum of the eigenstate under consideration [1, 2, 3].151515One can also
study various generalizations of this flow for the spectrum, such as the
energy levels of tensor product theories where the factors are sequentially
deformed by multiple ${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern
1.5mu}$ flows [80].

This example is remarkable because the flow equation (4.1) can be proven
directly at the quantum level using the properties of the local ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ operator, which is
defined by

$\displaystyle\mathcal{O}_{{T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}}(x)=\lim_{y\to
x}\left(T^{\alpha\beta}(x)T_{\alpha\beta}(y)-\tensor{T}{{}^{\alpha}_{\alpha}}(x)\tensor{T}{{}^{\beta}_{\beta}}(y)\right)\,.$
(4.2)

It was demonstrated in [1] that the coincident point limit on the right side
of (4.2) actually gives rise to a well-defined local operator, up to total
derivative ambiguities which can be ignored. One can therefore prove results
about a ${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern
1.5mu}$-deformed quantum field theory at the quantum level using the
properties of this operator; for instance, an argument involving a certain
factorization property of $\mathcal{O}_{{T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}}$ and the
interpretation of the components of the stress tensor in terms of energy and
momentum lead to the flow equation (4.1).

This is in contrast with a different method for attempting to learn about the
quantum mechanical properties of a stress tensor deformation, which we refer
to as _quantization along a classical flow_. In this case, one first finds the
solution to a differential equation of the form (2.1) for the Lagrangian of a
deformed theory, and then attempts to quantize this deformed Lagrangian
directly.

Assuming that a given classical deformation can be rigorously defined at the
quantum level, we do not expect that quantization along the classical flow
will give accurate information about _all_ aspects of the deformed quantum
field theory. Indeed, this is already true for the ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation. For
instance, it can be shown that the S-matrix of a ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed quantum field
theory is equal to the S-matrix of the undeformed theory multiplied by a
certain momentum-dependent phase known as a CDD factor [81, 82]. However, if
one studies scattering using quantization along the classical ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow, one finds that
this CDD factor is not reproduced unless one adds specific counter-terms which
are engineered to obtain the expected scattering behavior [83, 84, 18].
Therefore, quantization along the classical flow is not sufficient to fully
characterize the properties of the ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed theory
without additional input from the quantum definition.161616Another argument
for this conclusion is that quantization of theories with fermions along the
classical ${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$
flow can give different Hilbert spaces depending on which definition of the
stress tensor one uses [85, 86].

Despite this, one may hope that quantization of a classical deformed
Lagrangian will still give _some_ information about the corresponding
deformation at the quantum level, at least in particular limiting cases. For
instance, the solution to the classical ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow equation
beginning from a seed theory of free scalars is the Nambu-Goto action of
string theory, and one generically expects that string theories exhibit a
high-energy density of states which is Hagedorn rather than Cardy. This
predicted Hagedorn scaling agrees with an analysis of the high-energy behavior
of a ${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern
1.5mu}$-deformed CFT at the quantum level, which can be seen either from the
energies [3] or the partition function [87, 88]. Thus certain limiting
features of the quantum theory can still be inferred from the ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed classical
Lagrangian.

For other stress tensor deformations, like the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow, it is not yet
known whether one can give a rigorous definition of the deforming operator at
the quantum level. Therefore, we do not yet have any exact data about the
deformed quantum theory against which to compare results obtained by other
methods. However, extrapolating from the ${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ case, one might
perform quantization along a classical root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow in the hope that
this procedure will still give useful information in certain limits. Our goal
in this section is to carry out this quantization procedure for root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed theories of
chiral bosons and examine the behavior of the deformed spectrum in such
limiting cases.

One regime for which we have additional data about the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed spectrum is
the limit of a large-$c$ holographic CFT which admits a bulk
$\mathrm{AdS}_{3}$ dual. When restricting to states for which the stress
tensor is approximately constant (which are dual to BTZ black holes), one
obtains the formula (1.6) for the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformed spectrum
[32]. We will see that our analysis using quantization along the classical
flow agrees with this ``zero mode formula'' for states that correspond to
constant stress tensor backgrounds. However, we will also be able to probe
other limits of a root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed theory, such
as a large-momentum limit which is _not_ close to a constant stress tensor
configuration for which the zero mode formula is expected to apply. This
result may therefore give novel information about the behavior of a putative
root-${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern
1.5mu}$-deformed field theory in a different regime.

###  4.1 Generalities on quantization

Let us now study the quantum mechanics of interacting chiral boson models such
as (2.22). This Floreanini-Jackiw form of the Lagrangian, although it is not
manifestly Lorentz-invariant, is nonetheless convenient for quantization
because it is first-order in time derivatives. This allows us to perform
canonical quantization in a uniform way which does not depend on the details
of the interaction function $V(S,P)$.

We begin by reviewing some basic features of quantization of first-order
systems in the simpler setting of $(0+1)$-dimensional theories, i.e. particle
mechanics.

#### Quantization of first-order particle mechanics

We will first consider a collection of $(0+1)$-dimensional fields $q_{i}(t)$,
whose time derivatives will be denoted $\dot{q}_{i}(t)$. A general first-order
Lagrangian for such a system takes the form

$\displaystyle L=\frac{1}{2}C^{ij}q_{i}\dot{q}_{j}-V(q)\,,$ (4.3)

where $C^{ij}$ is a non-singular constant matrix. Without loss of generality,
we may assume that $C^{ij}$ is antisymmetric. Indeed, if we instead split
$C^{ij}=C^{[ij]}+C^{(ij)}$ into symmetric and anti-symmetric parts, the
Lagrangian would be

$\displaystyle
L=\frac{1}{2}C^{[ij]}q_{i}\dot{q}_{j}+\frac{1}{2}C^{(ij)}\frac{d}{dt}\left(q_{j}q_{i}\right)-V(q)\,,$
(4.4)

where the second term is a total time derivative that can be ignored.

The canonical momentum which is conjugate to $q_{j}(t)$ is

$\displaystyle p^{j}=\frac{\partial
L}{\partial\dot{q}_{j}}=\frac{1}{2}C^{ij}q_{i}\,,$ (4.5)

and thus the Hamiltonian associated with (4.3) is

$\displaystyle H(q,p)=\frac{\partial
L}{\partial\dot{q}_{i}}\dot{q}_{i}-L=V(q)\,.$ (4.6)

The Hamiltonian (4.6) appears to depend only on the position variables but not
on the momenta, but this is misleading, since equation (4.5) implies that some
combinations of the $q_{i}$ _are_ momenta. The Euler-Lagrange equations
arising from the Lagrangian (4.3) are

$\displaystyle C^{ij}\dot{q}_{i}=\frac{\partial H}{\partial q_{j}}\,,$ (4.7)

where we now use the symbols $V$ and $H$ interchangeably. Alternatively, by
defining $C_{ij}$ to be the inverse matrix $\left(C^{-1}\right)_{ij}$ of
$C^{ij}$, the equations of motion can be written as

$\displaystyle\dot{q}_{i}=C_{ij}\frac{\partial H}{\partial q_{j}}\,.$ (4.8)

Next we consider the quantization of this model. Ordinarily, for Lagrangians
which are quadratic in time derivatives, one would impose the canonical
commutation relations

$\displaystyle[x_{i},p_{j}]=i\delta_{ij}\,.$ (4.9)

However, imposing the relations (4.9) for a first-order system like (4.3)
gives results that differ from the correct commutation relations by a factor
of $2$. To arrive at the correct relations, we follow the prescription
outlined in appendix A of [7], and further justified in [89], which is to
define commutators so that the Heisenberg-picture time evolution of operators
in the quantum theory takes the same form as the classical Euler-Lagrange
equations.171717In conventional quantum systems with second-order Lagrangians,
the fact that these two equations should take the same form is the content of
the Ehrenfest theorem. We demand that the same is true here.

In general, the Heisenberg equation of motion for an operator $\mathcal{O}$
reads $\dot{\mathcal{O}}=i[H,\mathcal{O}]$. In the case of the operator
$\mathcal{O}=q_{i}$, we have

$\displaystyle\dot{q}_{i}=i[H,q_{i}]=i\frac{\partial H}{\partial
q_{j}}[q_{j},q_{i}]\,.$ (4.10)

Comparing (4.10) to (4.8), we find that the two take the same form if we
identify

$\displaystyle[q_{i},q_{j}]=iC_{ij}\,.$ (4.11)

As we mentioned, since $p^{j}=\frac{1}{2}C^{ij}q_{i}$, this differs from the
canonical prescription (4.9) which would give $[q_{i},q_{j}]=2iC_{ij}$. The
errant factor of $2$ is due to the fact that, in a first-order system, there
is a constraint on the phase space.

#### Quantization of first-order field theories

Having reviewed the quantum mechanics of first-order $(0+1)$-dimensional
systems, we now turn to the quantization of first-order $(1+1)$-dimensional
field theories, and in particular the theories of chiral bosons which are the
focus of this work.

As a simple example to set the stage, we will first consider a single chiral
boson described by the Floreanini-Jackiw Lagrangian (1.1) which we repeat
here:

$\mathcal{L}=\frac{1}{2}\left(\phi^{\prime}\dot{\phi}-\phi^{\prime}\phi^{\prime}\right)\,.$
(4.12)

As usual, we write $\dot{\phi}$ for the time derivative of $\phi$ and
$\phi^{\prime}$ for the spatial derivative of $\phi$. The quantization of this
system in infinite volume, i.e. with a spatial coordinate $x\in\mathbb{R}$,
was first studied in [7]. In short, one can view $x$ as a continuous
generalization of the discrete labels $i$, $j$ in (4.3) and rewrite the first
term as

$\displaystyle\frac{1}{2}\int dx\,\partial_{x}\phi(x,t)\dot{\phi}(x,t)$
$\displaystyle=\frac{1}{2}\int dx\int
dy\,\delta(x-y)\partial_{x}\phi(x,t)\dot{\phi}(y,t)$ (4.13)
$\displaystyle=-\frac{1}{2}\int dx\int
dy\,[\partial_{x}\delta(x-y)]\phi(x,t)\dot{\phi}(y,t)\,.$

The role of the constant antisymmetric matrix $C^{ij}$ in the particle
mechanics example is now played by the function

$\displaystyle C(x-y)=-\partial_{x}\delta(x-y)\,,$ (4.14)

and the role of the inverse matrix $C_{ij}$ is played by the Green's function
of $C(x-y)$. This suggests that we impose the commutation relations

$[\phi(x),\phi(y)]=-\frac{i}{2}\text{sgn}(x-y)~{},$ (4.15)

which is the field theory analog of (4.11) and which matches the result in
[7]. It is then straightforward to use the above equal-time commutation
relations to confirm the Heisenberg equations of motion are indeed equivalent
to the Euler-Lagrange equations of the Lagrangian (4.12), which describe a
chiral boson:

$\frac{\partial\mathcal{O}}{\partial
t}=-i[\mathcal{O},H]\quad\implies\quad\dot{\phi}=\phi^{\prime}\,.$ (4.16)

Next we will study this theory in finite volume. We now replace the spatial
coordinate $x\in\mathbb{R}$ with an angular coordinate $\theta$ labeling a
position on $S^{1}$, and subject to the identification
$\theta\sim\theta+2\pi$. We will also assume that the _target space_ is
compact, which means that $\phi$ likewise takes values in a circle so that
$\phi\sim\phi+2\pi$. As we will see, the structure of this theory on a
cylinder is closely related to the particle mechanics example considered
above.

First let us write the function $\phi(t,\theta)$ using a mode expansion:

$\phi(t,\theta)=\frac{1}{\pi}x(t)+p(t)\theta+\frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\left(a_{n}(t)e^{in\theta}+a^{\dagger}_{n}(t)e^{-in\theta}\right)\,.$
(4.17)

We have included a zero-mode term $x(t)$ in addition to a momentum
contribution which is linear in $\theta$; the latter is permissible, despite
not being periodic in $\theta$, since both $\theta\sim\theta+2\pi$ and
$\phi\sim\phi+2\pi$, so such a term is compatible with our identifications if
$p\in\mathbb{Z}$. The remaining sum is the standard Fourier expansion of the
periodic part of $\phi$ in the $\theta$ direction.

It is now necessary to distinguish between the Lagrangian density
$\mathcal{L}$ and the Lagrangian $L=\int d\theta\,\mathcal{L}$. Substituting
the mode expansion (4.17) into the Lagrangian density (4.12) and performing
the integral over the $\theta$ coordinate gives

$L=\int_{0}^{2\pi}d\theta\,\mathcal{L}=p\dot{x}-\pi
p^{2}+\frac{i}{2}\left(\sum_{n=1}^{\infty}(\dot{a}^{\dagger}_{n}a_{n}-\dot{a}_{n}a_{n}^{\dagger})\right)-\left(\frac{1}{2}\sum_{n=1}^{\infty}n(a_{n}a^{\dagger}_{n}+a^{\dagger}_{n}a_{n})\right)~{},$
(4.18)

where we have dropped a term that is a total derivative in time. Because $p$
is integer-quantized, as we mentioned above, the first term describes the
well-known quantum system which is a particle on a ring. The Hilbert space is
generated by states $|p\rangle$ labeled by integer $p\in\mathbb{Z}$ with
energy $E_{p}=\pi p^{2}$. The remaining terms are nothing but the familiar
first-order particle mechanics system discussed previously. To make this
analogy clearer, it is convenient to define

$\displaystyle a_{-n}=a_{n}^{\dagger}\,,$ (4.19)

so that the Lagrangian can be written as

$L=p\dot{x}-\pi
p^{2}+\frac{i}{2}\left(\sum_{n=1}^{\infty}(\dot{a}_{-n}a_{n}-\dot{a}_{n}a_{-n})\right)-\left(\frac{1}{2}\sum_{n=1}^{\infty}n(a_{n}a_{-n}+a_{-n}a_{n})\right)\,.$
(4.20)

The $a_{n}$'s now play the role of $q_{i}$'s, except the modes are labeled by
$n\in\mathbb{Z}$ so the phase space is infinite-dimensional. Comparing the two
sums in the Lagrangian (4.20) with the general form (4.3), we find that the
two agree if we identify

$C^{n,m}=i\,\text{sgn}(n)\delta_{n,-m}\,.$ (4.21)

Therefore, when we promote the $a_{n}$ from functions appearing in the
expansion of the classical field $\phi$ to quantum operators, the appropriate
commutation relations (4.11) are

$\displaystyle[a_{n},a_{m}]=\text{sgn}(n)\delta_{n,-m}\,.$ (4.22)

When expressed in terms of $a^{\dagger}_{m}$, this is the familiar commutation
relation of ladder operators:

$[a_{n},a_{m}^{\dagger}]=\delta_{n,m}\,.$ (4.23)

It is perhaps surprising that, if we had worked with the Fourier modes $a_{n}$
of the field $\phi$ from the beginning (rather than with the field $\phi$
itself), then imposing the standard commutation relations (4.23) gives the
correct result, without the errant factor of $2$ which we mentioned around
equation (4.11) that occurs due to the phase space constraint on first-order
systems. The reason for this is that, after performing the mode expansion, the
positive Fourier modes $a_{n}$ with $n>0$ act as the position variables and
the negative modes $a_{n}$ with $n<0$ (or equivalently $a^{\dagger}_{n}$) act
as the conjugate momentum variables. Therefore, in Fourier space, the
separation between coordinates and momenta is automatic, and we need not
impose phase space constraints or consider commutation relations like (4.11)
which naïvely appear to involve two position variables.181818See section 6.1.3
of [90] for a pedagogical review of the quantization of the chiral boson from
this momentum-space perspective, and later sections of this reference for
applications to quantum Hall physics.

The Hamiltonian obtained from the Legendre transform of the Lagrangian (4.20),
written in terms of $a_{n}^{\dagger}$ rather than $a_{-n}$, is

$\displaystyle H$ $\displaystyle=\pi
p^{2}+\frac{1}{2}\sum_{n=1}^{\infty}n(a_{n}a_{n}^{\dagger}+a_{n}^{\dagger}a_{n})$
$\displaystyle=-\frac{1}{24}+\pi
p^{2}+\sum_{n=1}^{\infty}na^{\dagger}_{n}a_{n}\,,$ (4.24)

where we have used $a_{n}a_{n}^{\dagger}=a_{n}^{\dagger}a_{n}+1$ and the well-
known $\zeta$-function regularization

$\sum_{n=1}^{\infty}n=-\frac{1}{12}\,.$ (4.25)

It is straightforward to generalize the above discussion to the case of
multiple chiral and anti-chiral bosons. We work with a Lagrangian density for
$N$ chiral bosons $\phi^{i}$, $i=1,\ldots,n$, and $\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$ anti-chiral bosons
$\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\overline{i}}$, of the form (1.2) which we have been considering in
section 2. For simplicity we take trivial target-space metrics for the bosons,
$G_{ij}=\delta_{ij}$ and $\mkern
1.5mu\overline{\mkern-1.5muG\mkern-1.5mu}\mkern
1.5mu_{\overline{i}\overline{j}}=\delta_{\overline{i}\overline{j}}$. The
Lagrangian density for this system is then

$\mathcal{L}=\frac{1}{2}\left(\phi^{\prime}_{i}\dot{\phi}^{i}-\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu_{\overline{i}}^{\prime}\dot{\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu}{}^{\overline{i}}\right)-V(\phi_{i}^{\prime},\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu_{\overline{i}}^{\prime})~{}.$ (4.26)

We expand both the chiral and anit-chiral fields in modes as

$\displaystyle\phi_{i}(t,\theta)$
$\displaystyle=\frac{1}{\pi}x_{i}(t)+p_{i}(t)\theta+\frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\left(a_{i,n}(t)e^{in\theta}+a_{i,n}^{\dagger}(t)e^{-in\theta}\right)~{},$
(4.27) $\displaystyle\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu_{\overline{i}}(t,\theta)$ $\displaystyle=-\frac{1}{\pi}\mkern
1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_{\overline{i}}(t)+\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu_{\overline{i}}(t)\theta+\frac{1}{\sqrt{2\pi}}\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}}\left(b_{\overline{i},n}(t)e^{-in\theta}+b_{\overline{i},n}^{\dagger}(t)e^{in\theta}\right)~{}.$

The non-zero commutation relations between the various expansion coefficients
are

$[x_{i},p_{j}]=i\delta_{ij},\quad[\mkern
1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_{\overline{i}},\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu_{\overline{j}}]=i\delta_{\overline{i}\overline{j}},\quad[a_{i,n},a_{j,m}^{\dagger}]=\delta_{ij}\delta_{nm},\quad[b_{\overline{i},n},b_{\overline{j},m}^{\dagger}]=\delta_{\overline{i}\overline{j}}\delta_{nm}\,,$
(4.28)

with all other commutators vanishing.

Note that here we take all $\phi_{i}$ and $\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu_{\overline{i}}$ to be
compact with radius $2\pi$. Therefore, the eigenvalues of $p_{j}$ and $\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_{\overline{j}}$ must be
integers. The Hamiltonian is given by

$H=\int_{0}^{2\pi}d\theta\,V(\phi^{\prime}_{i},\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}_{\overline{j}})~{}.$ (4.29)

The commutation relations (4.28) allow us to build the Hilbert space of the
quantum theory for any potential $V$. In the next subsection, we will use this
to study the spectrum of the ``Modified Scalar'' theory, that is, the theory
obtained by applying a root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation to a seed
theory of free chiral and anti-chiral bosons.

###  4.2 Root-${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern
1.5mu}$-deformed spectrum

We will now use the formalism reviewed in section 4.1 to study root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed free boson
theories. In principle, this can be done for any numbers $(N,\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)$ of chiral and anti-
chiral bosons, respectively. However, there is a sharp distinction between the
case $N=\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=1$, for
which the deformation is comparatively simple and can be interpreted as a re-
scaling of the target space radius for the boson, and all other cases with
$N\geq 1$ and $\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu\geq 1$, where the deformation is more non-trivial.191919Note that, if
either $N=0$ or $\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu=0$, then the theory is a fixed point of stress tensor flows so the
root-${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$
deformation is trivial. We will therefore first discuss the simpler case
$N=\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=1$ in detail,
and then as an illustrative example of the latter class, we will study the
example with $N=2$ and $\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu=1$. We expect that the qualitative features of the deformed $(N,\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)=(2,1)$ model will be
similar to those of theories with larger $N$ and $\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$.

#### One compact boson

Let us begin by studying the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation of a
single (non-chiral) $c=1$ compact boson, or equivalently, a pair of $N=1$
left-moving and $\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu=1$ right-moving chiral bosons. It was already mentioned in the initial
work [30] that, in this case, the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow simply rescales
the kinetic term for the boson, which corresponds to a change in the radius if
the scalar is compact. We will revisit this claim by describing the deformed
model in terms of chiral bosons and determining the quantum spectrum exactly
to confirm that the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation of a
compact boson is just a change of radius.

This formalism also provides a way to realize a compact boson at an arbitrary
radius – even at irrational points where the theory does not factorize into
the chiral part and anti-chiral part – using a Lagrangian for one chiral and
one anti-chiral boson with a quadratic mixing term. Furthermore, treating this
example in detail will allow us to test the ``zero mode formula'' given in
equation (1.6) that is expected, due to evidence from holography [32], to
describe the energies of states in root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed CFTs for
which the energy-momentum tensor is constant in space. We will see explicitly
that this zero mode formula fails to give the energies of deformed states for
which this assumption is violated.

The Lagrangian for a root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed seed theory
of one left-moving and one right-moving chiral boson takes the form (1.2) with
an interaction function $V(S,P,\gamma)$ given by the $\lambda\to 0$ limit of
equation (2.40). To be pedantic, the resulting Lagrangian is technically

$\mathcal{L}^{(\gamma)}=\frac{1}{2}\left(\phi^{\prime}\dot{\phi}-\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}\dot{\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu}\right)-\frac{\cosh(\gamma)}{2}\left(\phi^{\prime 2}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime
2}\right)-\sinh(\gamma)\sqrt{\left(\phi^{\prime}\right)^{2}\left(\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}\right)^{2}}\,.$ (4.30)

That is, because $\phi^{\prime}$ and $\dot{\phi}$ can take both positive and
negative values the final term is really proportional to
$|\phi^{\prime}|\cdot|\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}|$. However,
we will ignore this subtlety and simply replace
$\sqrt{\left(\phi^{\prime}\right)^{2}\left(\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}\right)^{2}}$
with $\phi^{\prime}\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}$. This can be justified, for instance, by restricting attention
to small fluctuations of the fields around a background for which the
gradients are large and positive, so that both $\phi^{\prime}$ and $\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}$ have fixed
positive sign. This corresponds to a solution with large positive values of
$p_{i}$ and $\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_{i}$
in the expansion of equation (4.27). We will take a similar large-momentum
limit in the analysis with several bosons below, again resolving the square
root, which is more non-trivial in that setting because of an additional term
under the root.

After making this simplification, the Lagrangian we wish to study becomes

$\mathcal{L}^{(\gamma)}=\frac{1}{2}\left(\phi^{\prime}\dot{\phi}-\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}\dot{\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu}\right)-\frac{\cosh(\gamma)}{2}\left(\phi^{\prime 2}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime
2}\right)-\sinh(\gamma)\phi^{\prime}\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}\,.$ (4.31)

As discussed previously, the Hilbert space factorizes into two parts: the
particles on a ring and the infinite tower of harmonic oscillators. Due to the
special form of (4.31), the Hamiltonian does not mix the two parts. Therefore,
we can study them separately.

Let us first consider the sector of the Hilbert space which describes the
particles on a ring. We write the states in this Hilbert space as
$|p,\overline{p}\rangle$, which are labeled by two quantized momenta
$p,\overline{p}\in\mathbb{Z}$. The corresponding Hamiltonian and the momentum
operator are

$H_{\text{PR}}^{(\gamma)}=\pi(p^{2}+\overline{p}^{2})\cosh(\gamma)+2\pi
p\overline{p}\sinh(\gamma)\,,\quad
P^{(\gamma)}_{\text{PR}}=\pi(p^{2}-\overline{p}^{2})=P^{(0)}_{\text{PR}}\,,$
(4.32)

where we use the subscript PR to denote particles on a ring.

Because the corresponding undeformed states at $\gamma=0$ have energies

$\displaystyle H_{\text{PR}}^{(0)}=\pi(p^{2}+\overline{p}^{2})\,,$ (4.33)

we see that the prediction for the deformed energies from the zero mode
formula (1.6) is

$\displaystyle E_{\text{PR}}^{(\gamma)}$
$\displaystyle=H_{\text{PR}}^{(0)}\cosh(\gamma)+\sqrt{\left(H_{\text{PR}}^{(0)}\right)^{2}-\left(P^{(0)}_{\text{PR}}\right)^{2}}\sinh(\gamma)$
$\displaystyle=\pi(p^{2}+\overline{p}^{2})\cosh(\gamma)+\sqrt{\left(\pi(p^{2}+\overline{p}^{2})\right)^{2}-\left(\pi(p^{2}-\overline{p}^{2})\right)^{2}}\sinh(\gamma)$
$\displaystyle=\pi(p^{2}+\overline{p}^{2})\cosh(\gamma)+2\pi
p\overline{p}\sinh(\gamma)\,,$ (4.34)

which indeed agrees with the true deformed energies $H_{\text{PR}}^{(\gamma)}$
of equation (4.32), subject to the usual caveat that we have used the
assumption $\sqrt{p^{2}\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu^{2}}=p\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu$.

It is not too surprising that these states have deformed energies which agree
with the zero-mode formula, since the corresponding saddle points have
constant stress-energy tensors, and this is the assumption under which the
formula (1.6) was derived in holography.

To see this explicitly, we look for solutions to the equations of motion
associated with the deformed Lagrangian $\mathcal{L}^{(\gamma)}$ in equation
(4.31), which are

$\displaystyle\dot{\phi}^{\prime}-\phi^{\prime\prime}\cosh(\gamma)-\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime\prime}\sinh(\gamma)=0\,,\quad\dot{\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu}{}^{\prime}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime\prime}\cosh(\gamma)+\phi^{\prime\prime}\sinh(\gamma)=0\,.$
(4.35)

One can integrate these equations with respect to the spatial coordinate
$\theta$, up to an undetermined integration constant $h(t)$ which is an
arbitrary function of $t$. As in the discussion around equation (2.18), one
can always set $h(t)=0$ by a gauge transformation. Specializing to this $h=0$
gauge, the equations of motion become

$\dot{\phi}-\phi^{\prime}\cosh(\gamma)-\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}\sinh(\gamma)=0,\quad\dot{\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}\cosh(\gamma)+\phi^{\prime}\sinh(\gamma)=0\,.$ (4.36)

We wish to solve the equations of motion (4.36) subject to the boundary
conditions

$\phi(\theta+2\pi,t)-\phi(\theta,t)=2\pi p\,,\quad\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu(\theta+2\pi,t)-\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu(\theta,t)=2\pi\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu\,,$ (4.37)

where $p,\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu\in\mathbb{Z}$. The desired solutions with such periodic boundary
conditions are

$\displaystyle\phi_{p}^{(\gamma)}$
$\displaystyle=p\theta+\left(p\cosh(\gamma)+\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu\sinh(\gamma)\right)t\,,\quad\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu_{\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu}^{(\gamma)}=\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu\theta-\left(\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu\cosh(\gamma)+p\sinh(\gamma)\right)t\,.$ (4.38)

Since these solutions $\phi_{p}^{(\gamma)}$ and $\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu_{\overline{p}}^{(\gamma)}$ depend on $t,\theta$ linearly, the
corresponding stress-energy tensor is constant. Therefore, it is reasonable
that the energies of these states are indeed governed by the energy formula
derived via AdS3/CFT2 holography for constant stress tensor backgrounds, as we
found around equation (4.2).202020Strictly speaking, the derivation of this
zero mode formula also assumes that the boundary theory is a large-$c$
holographic CFT for which we can trust semiclassical bulk gravity. However,
this assumption does not seem strictly necessary for the zero mode formula to
hold, since the theory we study here has $c=1$.

We would also like to point out that the energies of these states agree with
the energies of momentum states for a compact boson with a different radius.
To see this, it is convenient to change variables as

$w=\sqrt{\pi}\left(p+\bar{p}\right)\,,\quad\bar{w}=\sqrt{\pi}\left(p-\bar{p}\right)\,,\quad
R=\exp\left(-\frac{\gamma}{2}\right)\,,$ (4.39)

so that the deformed Hamiltonian (4.32) can be written as

$H^{(\gamma)}_{\text{PR}}=\frac{1}{2}\left(\frac{w^{2}}{R^{2}}+R^{2}\bar{w}^{2}\right)\,.$
(4.40)

This supports the claim that the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation, in this
case, corresponds to a rescaling of the target-space radius for the compact
boson. However, to verify this conclusion, we should also study the effect of
the deformation in the other sector of the Hilbert space, which describes an
infinite tower of harmonic oscillators.

We turn to this task now. Expanding the field $\phi$ and $\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu$ as in (4.27), we
find the Hamiltonian operator and the momentum operator for this oscillator
sector are given by

$\displaystyle H^{(\gamma)}_{\text{OS}}$
$\displaystyle=\sum_{n=1}^{\infty}n(a_{n}^{\dagger}a_{n}+b^{\dagger}_{n}b_{n})\cosh(\gamma)+\sum_{n=1}^{\infty}n(a_{n}^{\dagger}b_{n}^{\dagger}+a_{n}b_{n})\sinh(\gamma)-\frac{1}{12}\cosh(\gamma)\,,$
(4.41) $\displaystyle P^{(\gamma)}_{\text{OS}}$
$\displaystyle=\sum_{n=1}^{\infty}n(a_{n}^{\dagger}a_{n}-b_{n}^{\dagger}b_{n})\,,$

where we have performed normal ordering as before and where OS stands for
oscillators. This Hamiltonian has exactly the same spectrum as its undeformed
counterpart, which can be made manifest by the following Bogoliubov
transformation:

$\displaystyle
a_{n}=\tilde{a}_{n}\cosh\left(\frac{\gamma}{2}\right)-\tilde{b}^{\dagger}_{n}\sinh\left(\frac{\gamma}{2}\right)\,,\quad
b_{n}=\tilde{b}_{n}\cosh\left(\frac{\gamma}{2}\right)-\tilde{a}^{\dagger}_{n}\sinh\left(\frac{\gamma}{2}\right)\,.$
(4.42)

We note that this has the same structure as the change of variables which
diagonalized the mixing term between the two Chern-Simons gauge fields in
equation (3.62). This transformation preserves the commutation relation, i.e.

$[\tilde{a}_{n},\tilde{a}^{\dagger}_{n}]=[\tilde{b}_{n},\tilde{b}^{\dagger}_{n}]=1~{}.$
(4.43)

In terms of the new oscillators, the Hamiltonian then reduces to the
undeformed one,

$H^{(\gamma)}_{\text{OS}}=-\frac{1}{12}+\sum^{\infty}_{n=1}n\left(\tilde{a}^{\dagger}_{n}\tilde{a}_{n}+\tilde{b}^{\dagger}_{n}\tilde{b}_{n}\right)\,,$
(4.44)

while the momentum operator is unchanged,

$P_{\text{OS}}^{(\gamma)}=\sum_{n=1}^{\infty}n\left(\tilde{a}_{n}^{\dagger}\tilde{a}_{n}-\tilde{b}_{n}^{\dagger}\tilde{b}_{n}\right)\,.$
(4.45)

Hence, we conclude that the energies in the oscillator sector of the Hilbert
space do not flow under the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation. This
agrees with the effect of changing the radius for a compact boson, which
likewise does not change the energies of oscillator excitations.

Therefore, combining this result with the flow of $H_{\text{PR}}^{(\gamma)}$,
we conclude that indeed the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation
corresponds to a change of radius for a single compact boson.

We have also verified that the zero-mode energy formula (1.6) proposed in [32]
does not apply to generic states in a root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed CFT. For
instance, any state with $p=\overline{p}=0$ but with oscillator excitations
will have an energy that is unchanged by the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ flow, whereas the
formula (1.6) would predict that the energy flows with $\gamma$. This is
because such oscillator states have non-constant stress tensors and therefore
violate the assumptions under which the zero-mode formula was derived.
However, we reiterate that the states which _do_ have constant stress tensors
– namely, states with general $p$ and $\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu$ but no oscillator
excitations – indeed have energies which flow according to the zero mode
formula.

#### Multiple compact bosons

Next we aim to study the spectrum for the theory of root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$-deformed free bosons
when there are more fields, rather than just a single left-mover and a single
right-mover. All of these cases are qualitatively similar, in the sense that
the argument of the square root appearing in the Lagrangian is no longer a
perfect square, and thus cannot be resolved to a simple product of fields as
in the $N=\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=1$ case
above. For simplicity, we will therefore focus on the first non-trivial case,
which has $N=2$ left-movers and $\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=1$ right-movers (the
case with $N=1$ and $\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu=2$ is identical, after exchanging chiral and anti-chiral fields).

The Hamiltonian for the deformed $(N,\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)=(2,1)$ theory is

$\displaystyle H^{(\gamma)}$ $\displaystyle=\int
d\theta\left[\frac{1}{2}\left(\phi^{\prime 2}_{1}+\phi^{\prime 2}_{2}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime
2}_{1}\right)\cosh(\gamma)+\sqrt{\phi^{\prime 2}_{1}+\phi^{\prime
2}_{2}}\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}_{1}\sinh(\gamma)\right]\,.$ (4.46)

To resolve the square root, our strategy will be to expand in large positive
momenta and compute the energies perturbatively. The mode expansion for the
fields takes the form

$\phi_{j}=p_{j}\theta+\frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\left(a_{j,n}e^{in\theta}+a_{j,n}^{\dagger}e^{-in\theta}\right)\,,\quad\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu_{1}=\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu_{1}\theta+\frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\left(b_{1,n}^{\dagger}e^{in\theta}+b_{1,n}e^{-in\theta}\right)\,,$
(4.47)

where $j=1,2$ and periodicity requires $\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_{1}$, $p_{1}$,
$p_{2}\in\mathbb{Z}$. Substituting the expansion (4.47) into our Hamiltonian
(4.46) and expanding in large $p_{1}$ and $\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_{1}$, to leading order
we find

$\displaystyle H^{(\gamma)}$
$\displaystyle=\left(\pi(p_{1}^{2}+p_{2}^{2}+\bar{p}_{1}^{2})-\frac{1}{8}+\sum^{\infty}_{n=1}n\left(N_{1,n}+N_{2,n}+\bar{N}_{1,n}\right)\right)\cosh(\gamma)$
(4.48) $\displaystyle\qquad+\left(2\pi
p_{1}\bar{p}_{1}+\sum^{\infty}_{n=1}n\left(a_{1,n}b_{1,n}+a^{\dagger}_{1,n}b^{\dagger}_{1,n}\right)\right)\sinh(\gamma)+\cdots\,,$

where $N_{i,n}=a_{i,n}^{\dagger}a_{i,n}$ and $\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu_{1,n}=b_{1,n}^{\dagger}b_{1,n}$ are number operators at level $n$ for
left- and right-movers respectively.

We would now like to compare the spectrum of the true large-momentum
Hamiltonian (4.48) to the zero-mode formula (1.6) predicted from holography
for states with constant stress tensors. The undeformed Hamiltonian and
momentum are

$\displaystyle H^{(0)}$ $\displaystyle=\pi\left(p_{1}^{2}+p_{2}^{2}+\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu_{1}^{2}\right)-\frac{1}{8}+\sum^{\infty}_{n=1}n\left(N_{1,n}+N_{2,n}+\bar{N}_{1,n}\right)\,,$
(4.49) $\displaystyle P^{(0)}$
$\displaystyle=\pi\left(p_{1}^{2}+p_{2}^{2}-\mkern
1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern
1.5mu_{1}^{2}\right)-\frac{1}{24}+\sum^{\infty}_{n=1}n\left(N_{1,n}+N_{2,n}-\bar{N}_{1,n}\right)\,.$

Let us restrict to an eigenstate of both the momentum operators $p_{1}$,
$p_{2}$, $\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_{1}$
and the number operators $N_{1,n}$, $N_{2,n}$, $\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu_{1,n}$, in the
undeformed theory. The energy and momentum of such a state are also given by
the expressions (4.49), if we simply re-interpret each symbol representing an
operator as instead representing the corresponding eigenvalue.212121We have
chosen not to denote operators by decorating them with hats, which would
distinguish between operators $\widehat{N}_{1}$ and their corresponding
eigenvalues $N_{1}$, to avoid cluttering the formulas.

Substituting the energy and momentum eigenvalues for this state into the zero-
mode formula (1.6) then gives a predicted value for a deformed energy:

$\displaystyle E^{(\gamma)}_{\text{zero mode}}$
$\displaystyle=\left(\pi\left(p_{1}^{2}+p_{2}^{2}+\bar{p}_{1}^{2}\right)-\frac{1}{8}+\sum^{\infty}_{n=1}n\left(N_{1,n}+N_{2,n}+\bar{N}_{1,n}\right)\right)\cosh(\gamma)$
$\displaystyle\qquad+2\pi p_{1}\bar{p}_{1}\sinh(\gamma)+\cdots\,.$ (4.50)

We should stress that equation (1.6) is a prediction for the deformed
_spectrum_ and not for the deformed _eigenstates_. Therefore, even if equation
(4.2) were correct, this would simply mean that there exists _some_ state in
the deformed theory whose energy is $E^{(\gamma)}_{\text{zero mode}}$.

However, with this caveat aside, it is now easy to see why the formula (4.2)
is incorrect, and what effect it fails to take into account. Were it not for
the final term in the true Hamiltonian (4.48), which involves
$\sum^{\infty}_{n=1}n\left(a_{1,n}b_{1,n}+a^{\dagger}_{1,n}b^{\dagger}_{1,n}\right)$,
then any eigenstate of the undeformed theory would remain an eigenstate of the
deformed theory at this order in the momentum expansion, and its energy would
indeed be given by (4.2). This is simply because the first several terms of
the true Hamiltonian (4.48) agree with the zero-mode prediction (4.2), after
replacing operators with their eigenvalues. However, the presence of this
final term in (4.48) means that an eigenstate of the undeformed Hamiltonian
will _not_ remain an eigenstate in the deformed theory, since terms like
$a_{1,n}b_{1,n}$ will mix such a state into other states with different
oscillator numbers. We conclude that the zero-mode energy formula (4.2) is not
correct for the deformed spectrum, even in this large-momentum limit.

#### Possible interpretation of root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation for higher
$N$, $\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$

We have seen that, in the special case $N=\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=1$, the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformations of chiral
bosons admits a simple interpretation as a rescaling of the target-space
radius. This can also be understood from the observation that, for this case,
the oscillator sector of the deformed theory is equivalent to that of the
undeformed theory due to the Bogoliubov transformation (4.42). To conclude
this section, we would like to make some speculative remarks about possible
generalizations of this interpretation to cases with higher $N$ and $\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$, which seem
considerably more complicated.

First let us point out that, for the case $(N,\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)=(1,1)$, the Bogoliubov
transformation which returns the oscillator sector of the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformed theory to its
undeformed form also has an analog at the level of the Lagrangian and
Hamiltonian densities. Indeed, for the quadratic theory (4.31), one can write
the Lagrangian and Hamiltonian densities as

$\displaystyle\mathcal{L}$
$\displaystyle=\frac{1}{2}\left(\Phi^{\prime}\dot{\Phi}-\mkern
1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\mkern 1.5mu^{\prime}\dot{\mkern
1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\mkern
1.5mu}\right)-\frac{1}{2}\left(\Phi^{\prime 2}+\mkern
1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\mkern 1.5mu^{\prime 2}\right)\,,$
(4.51) $\displaystyle\mathcal{H}$ $\displaystyle=\frac{1}{2}\left(\Phi^{\prime
2}+\mkern 1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\mkern 1.5mu^{\prime
2}\right)\,,$

where we have made a field redefinition

$\left(\begin{array}[]{c}\Phi\\\ \mkern
1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\end{array}\right)=\left(\begin{array}[]{cc}\cosh\left(\frac{\gamma}{2}\right)&\quad\sinh\left(\frac{\gamma}{2}\right)\\\
\sinh\frac{\gamma}{2}&\quad\cosh\left(\frac{\gamma}{2}\right)\end{array}\right)\left(\begin{array}[]{c}\phi\\\
\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\end{array}\right)\,,\quad\left(\begin{array}[]{c}\phi\\\
\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\end{array}\right)=\left(\begin{array}[]{cc}\cosh\left(\frac{\gamma}{2}\right)&\quad-\sinh\left(\frac{\gamma}{2}\right)\\\
-\sinh\left(\frac{\gamma}{2}\right)&\quad\cosh\left(\frac{\gamma}{2}\right)\end{array}\right)\left(\begin{array}[]{c}\Phi\\\
\mkern 1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\end{array}\right)\,.$
(4.52)

The deformed equations of motion, written in terms of the new fields $\Phi$
and $\mkern 1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\mkern 1.5mu$, are

$\Phi^{\prime\prime}=\dot{\Phi}^{\prime},\quad\mkern
1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\mkern
1.5mu^{\prime\prime}=-\dot{\mkern
1.5mu\overline{\mkern-1.5mu\Phi\mkern-1.5mu}\mkern 1.5mu}^{\prime}\,,$ (4.53)

which take the same form as those in the undeformed theory. Again, this is
analogous to the field redefinition (3.62) in the Chern-Simons setting, which
undoes a similar quadratic mixing between the barred and unbarred fields
induced by a $J\mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu$
deformation.

Next let us consider how this observation might extend to multiple bosons. We
focus on the case of $N=\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu$ for simplicity. The deformed Hamiltonian density for an equal number of
left- and right-moving chiral bosons is

$\mathcal{H}^{(\mu)}=\frac{1}{2}\left(\phi^{\prime}_{j}\phi^{\prime}_{j}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}_{\mkern
1.5mu\overline{\mkern-1.5muj\mkern-1.5mu}\mkern 1.5mu}\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}_{\mkern
1.5mu\overline{\mkern-1.5muj\mkern-1.5mu}\mkern
1.5mu}\right)\cosh(\gamma)+\sqrt{\phi^{\prime}_{j}\phi^{\prime}_{j}\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}_{\mkern
1.5mu\overline{\mkern-1.5muj\mkern-1.5mu}\mkern 1.5mu}\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime}_{\mkern
1.5mu\overline{\mkern-1.5muj\mkern-1.5mu}\mkern 1.5mu}}\sinh(\gamma)\,.$
(4.54)

We now ask whether some more complicated field redefinition might return this
Hamiltonian to a quadratic one, as in the case of (4.51). When $N=\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=2$, at least formally,
one can attempt to perform a change of variables that resembles a
transformation to polar coordinates in a $2d$ target space:

$\displaystyle\left(\begin{array}[]{c}\phi^{\prime}_{1}(\theta,t)\\\
\phi^{\prime}_{2}(\theta,t)\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}r^{\prime}(\theta,t)\cos\left(\Theta^{\prime}(\theta,t)\right)\\\
r^{\prime}(\theta,t)\sin\left(\Theta^{\prime}(\theta,t)\right)\end{array}\right)\,,\quad\left(\begin{array}[]{c}\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}_{1}(\theta,t)\\\ \mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern
1.5mu^{\prime}_{2}(\theta,t)\end{array}\right)=\left(\begin{array}[]{cc}\mkern
1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern
1.5mu^{\prime}(\theta,t)\cos\left(\mkern
1.5mu\overline{\mkern-1.5mu\Theta\mkern-1.5mu}\mkern
1.5mu^{\prime}(\theta,t)\right)\\\ \mkern
1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern
1.5mu^{\prime}(\theta,t)\sin\left(\mkern
1.5mu\overline{\mkern-1.5mu\Theta\mkern-1.5mu}\mkern
1.5mu^{\prime}(\theta,t)\right)\end{array}\right)\,.$ (4.55)

Here we interpret $\Theta^{\prime}(\theta,t)$ and $r^{\prime}(\theta,t)$ as
spatial derivatives of new fields which depend on the derivatives
$\phi^{\prime}(\theta,t)$ in a nonlinear way. In terms of these quantities,
the Hamiltonian density (4.54) with $N=\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu=2$ takes the form

$\displaystyle\mathcal{H}^{(\mu)}$
$\displaystyle=\frac{1}{2}\left(\phi^{\prime 2}_{1}+\phi^{\prime 2}_{2}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime 2}_{1}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime
2}_{2}\right)\cosh\left(\gamma\right)+\sqrt{\left(\phi^{\prime
2}_{1}+\phi^{\prime 2}_{2}\right)\left(\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime 2}_{1}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{\prime
2}_{2}\right)}\sinh\left(\gamma\right)$ (4.56)
$\displaystyle=\frac{1}{2}\left(r^{\prime 2}+\mkern
1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern 1.5mu^{\prime
2}\right)\cosh\left(\gamma\right)+r^{\prime}\mkern
1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern
1.5mu^{\prime}\sinh\left(\gamma\right)\,,$

where we assumed $r^{\prime}\mkern
1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern 1.5mu^{\prime}>0$ in order to
simplify the square root. Now we perform a second field redefinition, just as
in (4.52), to a new field $\rho$:

$\left(\begin{array}[]{c}r(\theta,t)\\\ \mkern
1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern
1.5mu(\theta,t)\end{array}\right)=\left(\begin{array}[]{cc}\cosh\left(\frac{\gamma}{2}\right)&\quad-\sinh\left(\frac{\gamma}{2}\right)\\\
-\sinh\left(\frac{\gamma}{2}\right)&\quad\cosh\left(\frac{\gamma}{2}\right)\end{array}\right)\left(\begin{array}[]{c}\rho(\theta,t)\\\
\mkern 1.5mu\overline{\mkern-1.5mu\rho\mkern-1.5mu}\mkern
1.5mu(\theta,t)\end{array}\right)\,.$ (4.57)

Expressing the Hamiltonian density (4.56) in terms of the $\rho$ variables
rather than the $r$ variables, we conclude

$\mathcal{H}^{(\mu)}=\frac{1}{2}\left(\rho^{\prime 2}+\mkern
1.5mu\overline{\mkern-1.5mu\rho\mkern-1.5mu}\mkern 1.5mu^{\prime 2}\right)\,.$
(4.58)

Therefore, again at a formal classical level, it appears that this series of
field redefinitions has returned the Hamiltonian density to that of the free
theory. Furthermore, the latter change of variables (4.57) can be interpreted
as rescaling the overall target space radius $r$, much as in the $(N,\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu)=(1,1)$ case. For a
larger number of bosons $N=\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu>2$, one can perform a
similar series of manipulations using higher-dimensional spherical
coordinates.

Several technical issues preclude us from taking this series of field
redefinitions seriously, at least without further investigation. First, the
change of variables (4.55) was at the level of _derivatives_ of the fields,
and it is not clear that this corresponds to a sensible change of variables
for the fields themselves. Second, all of these manipulations have been purely
classical, and it is not guaranteed that one could make sense of these field
redefinitions within a path integral (which would produce Jacobian factors
from each change of variables). And third, we have not been careful about the
identifications that each field is subject to. For instance, if indeed the
field $\Theta$ can be interpreted as a target-space angle in polar
coordinates, then it should be subject to the identification
$\Theta\sim\Theta+2\pi$.

Nonetheless, it would be very interesting if an argument of this form could be
used to endow the root-${T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$ deformation of $N$
chiral and anti-chiral bosons with a geometrical target-space interpretation.

##  5 Perturbative Quantization Using Background Field Method

In the preceding sections, we have considered interacting theories with
arbitrary numbers $N,\mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern
1.5mu$ of chiral and anti-chiral bosons, respectively, and sacrificed manifest
Lorentz invariance in order to use a first-order formulation which is
convenient for canonical quantization. In the special case $N=\mkern
1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu$, however, we also have
the option of assembling the field content of our theory into $N$ _non-chiral_
bosons by summing the left-movers and right-movers:

$\displaystyle\varphi^{i}=\frac{1}{\sqrt{2}}\left(\phi^{i}+\mkern
1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{i}\right)\,.$ (5.1)

Here we now use the same index $i=1,\ldots,N$ for both the chiral and anti-
chiral fields, rather than distinct indices $i$ and $\overline{i}$. As this
change of variables is merely a field redefinition, stress tensor deformations
of such a theory of $N$ bosons must be equivalent, regardless of whether the
theory is presented in terms of left-movers and right-movers $\phi^{i}$,
$\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu^{i}$, or in
terms of their non-chiral counterparts $\varphi^{i}$. Indeed, for the case of
the ${T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu}$
deformation of a free seed theory, this equivalence was checked explicitly in
[78].

In this section, we will provide a complementary analysis of the perturbative
quantization of the Modified Scalar theory using this presentation in terms of
non-chiral fields $\varphi^{i}$. For concreteness, we will focus on the case
where both the fields $\varphi^{i}$ and the Lorentzian spacetime coordinates
$(t,x)$ are non-compact, and we will use middle Greek letters like $\mu$,
$\nu$ (rather than early Greek letters like $\alpha$, $\beta$, which were used
in sections 2 and 3) for spacetime indices in this section. We will write
$g_{\mu\nu}$ for the (Minkowski) spacetime metric.

In terms of the non-chiral fields $\varphi^{i}$, the Lagrangian for the
Modified Scalar theory can be written in the manifestly Lorentz-invariant form

$\displaystyle\mathcal{L}=\frac{1}{2}\left(\cosh\left(\gamma\right)\partial_{\mu}\varphi^{i}\partial^{\mu}\varphi^{i}+\sinh\left(\gamma\right)\sqrt{2\left(\partial_{\mu}\varphi^{i}\partial^{\nu}\varphi^{i}\right)\left(\partial_{\nu}\varphi^{j}\partial^{\mu}\varphi^{j}\right)-\left(\partial_{\mu}\varphi^{i}\partial^{\mu}\varphi^{i}\right)^{2}}\right)\,.$
(5.2)

The advantage of this representation is that one can more easily apply
standard diagrammatic techniques to compute loop corrections in the quantum
theory. Of course, the second term in the Lagrangian (5.2) is still non-
analytic around the vacuum of the theory, or around any field configuration
for which

$\displaystyle\partial_{\mu}\varphi^{i}=0\,.$ (5.3)

We will circumvent this issue by working in a background field expansion
around a field configuration $\varphi^{i}$ for the scalars which we assume has
non-zero gradients and which satisfies the classical equations of motion for
the theory, but which is otherwise arbitrary.

###  5.1 Background field expansion and Feynman rules

Throughout this section, we will use the notation

$\displaystyle\varphi^{i}=C^{i}+Q^{i}\,,$ (5.4)

where $C^{i}$ is a classical (background) field configuration around which we
perform our expansion, and $Q^{i}$ is a quantum field which is allowed to
fluctuate within the path integral. This classical background $C^{i}$ is the
analog of the large-momentum configuration around which we performed our
expansion in section 4.2. Our goal will be to investigate the terms which
contribute to the quantum effective action, as a function of the background
$C^{i}$.

To avoid cluttering the formulas, it will also be convenient to adopt the
following shorthand for spacetime derivatives of the various fields:

$\displaystyle\tensor{\varphi}{{}_{\mu}^{i}}=\partial_{\mu}\varphi^{i}\,,\quad\tensor{C}{{}_{\mu}^{i}}=\partial_{\mu}C^{i}\,,\quad\tensor{Q}{{}_{\mu}^{i}}=\partial_{\mu}Q^{i}\,.$
(5.5)

In our analysis of chiral boson theories, we introduced two useful quantities
$S$ and $P$ in equation (2.34) which were independent combinations of
derivatives of the scalar fields. In the present non-chiral analysis, let us
similarly introduce the quantities

$\displaystyle S=\tensor{\varphi}{{}_{\mu}^{i}}\tensor{\varphi}{{}^{\mu
i}}\,,\qquad P^{2}=\tensor{\varphi}{{}_{\mu}^{i}}\tensor{\varphi}{{}^{\nu
i}}\tensor{\varphi}{{}_{\nu}^{j}}\tensor{\varphi}{{}^{\mu j}}\,.$ (5.6)

We note that these are not the precise analogs of $S$ and $P$ in the chiral
setting; for instance, the role of the combination $S^{2}-P^{2}$ in section 2
is now played by $2P^{2}-S^{2}$. Therefore, in terms of these quantities
(5.6), the Modified Scalar Lagrangian (5.2) can be written as

$\displaystyle\mathcal{L}=\frac{1}{2}\left(\cosh(\gamma)S+\sinh(\gamma)\sqrt{2P^{2}-S^{2}}\right)\,.$
(5.7)

We decompose $S$ into a classical piece $S_{C}$ and a quantum piece $S_{Q}$,
along with a cross term:

$\displaystyle S$
$\displaystyle=\left(\tensor{C}{{}_{\mu}^{i}}+\tensor{Q}{{}_{\mu}^{i}}\right)\left(\tensor{C}{{}^{\mu}{}^{i}}+\tensor{Q}{{}^{\mu}{}^{i}}\right)$
$\displaystyle=\underbrace{\tensor{C}{{}_{\mu}^{i}}\tensor{C}{{}^{\mu
i}}}_{S_{C}}+2\tensor{C}{{}_{\mu}^{i}}\tensor{Q}{{}^{\mu
i}}+\underbrace{\tensor{Q}{{}_{\mu}^{i}}\tensor{Q}{{}^{\mu i}}}_{S_{Q}}\,.$
(5.8)

Next we will consider the splitting of $S^{2}$ and $P^{2}$ into classical and
quantum pieces. Because we assume that the field configuration $C^{i}$ is a
solution to the classical equations of motion, by definition the action is
stationary to linear order when expanding around such a solution. This means
that the effective action cannot contain any terms which are linear in the
fluctuation field $Q^{i}$, because the sum of all such contributions must
conspire to form an on-shell total derivative. We will therefore label all
terms linear in $Q^{\mu i}$ as ``on-shell deriv.'' and ignore them in what
follows, although with the caveat that _individual_ terms of this form need
not separately drop out; we are only guaranteed that the _combined_ effect of
all such terms is to form an on-shell total derivative.

With this in mind, the quantity $S^{2}$ can be expanded as

$\displaystyle S^{2}$
$\displaystyle=S_{C}^{2}+\underbrace{4S_{C}\tensor{C}{{}_{\mu}^{i}}\tensor{Q}{{}^{\mu
i}}}_{\text{on-shell deriv.}}+2S_{C}S_{Q}+4\tensor{C}{{}_{\mu}^{i}}Q^{\mu
i}\tensor{C}{{}_{\nu}^{j}}Q^{\nu
j}+\underbrace{4S_{Q}\tensor{C}{{}_{\mu}^{i}}Q^{\mu
i}}_{\mathcal{O}\left(Q^{3}\right)}+\underbrace{S_{Q}^{2}}_{\mathcal{O}\left(Q^{4}\right)}$
$\displaystyle\simeq S_{C}^{2}+2S_{C}S_{Q}+4\tensor{C}{{}_{\mu}^{i}}Q^{\mu
i}\tensor{C}{{}_{\nu}^{j}}Q^{\nu j}\,,$ (5.9)

where the symbol $\simeq$ means equal modulo all terms that are either linear
in $Q^{i}$ (which will form on-shell total derivatives) or that are of cubic
order or higher in $Q^{i}$ (which do not contribute to the one loop effective
action). A similar computation for $P^{2}$ gives

$\displaystyle P^{2}$
$\displaystyle=\tensor{C}{{}_{\mu}^{i}}\tensor{C}{{}^{\mu
j}}\tensor{C}{{}_{\nu}^{i}}\tensor{C}{{}^{\nu
j}}+\underbrace{4\tensor{C}{{}_{\mu}^{i}}\tensor{C}{{}^{\mu}{}^{j}}\tensor{C}{{}^{\nu}{}^{i}}\tensor{Q}{{}_{\nu}^{j}}}_{\text{on-
shell deriv.}}+2\tensor{C}{{}_{\mu}^{i}}C^{\nu
i}\tensor{Q}{{}_{\nu}^{j}}\tensor{Q}{{}^{\mu
j}}+2\tensor{Q}{{}_{\mu}^{i}}C^{\nu i}\tensor{Q}{{}_{\nu}^{j}}C^{\mu j}$
$\displaystyle\qquad+2\tensor{Q}{{}_{\mu}^{i}}C^{\nu
i}\tensor{C}{{}_{\nu}^{j}}Q^{\mu j}+\mathcal{O}\left(Q^{3}\right)\,$
$\displaystyle\simeq\tensor{C}{{}_{\mu}^{i}}\tensor{C}{{}^{\mu
j}}\tensor{C}{{}_{\nu}^{i}}\tensor{C}{{}^{\nu
j}}+2\tensor{C}{{}_{\mu}^{i}}C^{\nu
i}\tensor{Q}{{}_{\nu}^{j}}\tensor{Q}{{}^{\mu
j}}+2\tensor{Q}{{}_{\mu}^{i}}C^{\nu i}\tensor{Q}{{}_{\nu}^{j}}C^{\mu
j}+2\tensor{Q}{{}_{\mu}^{i}}C^{\nu i}\tensor{C}{{}_{\nu}^{j}}Q^{\mu j}\,.$
(5.10)

Therefore, the combination $2P^{2}-S^{2}$ under the square root in (5.7) has
an expansion

$\displaystyle 2P^{2}-S^{2}$ $\displaystyle\simeq
2P_{C}^{2}-S_{C}^{2}-2S_{C}S_{Q}-4\tensor{C}{{}_{\mu}^{i}}Q^{\mu
i}\tensor{C}{{}_{\nu}^{j}}Q^{\nu j}+4\tensor{C}{{}_{\mu}^{i}}C^{\nu
i}\tensor{Q}{{}_{\nu}^{j}}\tensor{Q}{{}^{\mu
j}}+4\tensor{Q}{{}_{\mu}^{i}}C^{\nu i}\tensor{Q}{{}_{\nu}^{j}}C^{\mu j}$
$\displaystyle\qquad+4\tensor{Q}{{}_{\mu}^{i}}C^{\nu
i}\tensor{C}{{}_{\nu}^{j}}Q^{\mu j}\,$ $\displaystyle\equiv
2P_{C}^{2}-S_{C}^{2}+2\mathcal{Q}_{1}\,.$ (5.11)

Here we introduce the shorthand $\mathcal{Q}_{1}$ which is proportional to the
correction to the classical part of (5.1) up to quadratic order in
fluctuations,

$\displaystyle\mathcal{Q}_{1}=-S_{C}S_{Q}-2\tensor{C}{{}_{\mu}^{i}}Q^{\mu
i}\tensor{C}{{}_{\nu}^{j}}Q^{\nu j}+2\tensor{C}{{}_{\mu}^{i}}C^{\nu
i}\tensor{Q}{{}_{\nu}^{j}}\tensor{Q}{{}^{\mu
j}}+2\tensor{Q}{{}_{\mu}^{i}}C^{\nu i}\tensor{Q}{{}_{\nu}^{j}}C^{\mu
j}+2\tensor{Q}{{}_{\mu}^{i}}C^{\nu i}\tensor{C}{{}_{\nu}^{j}}Q^{\mu j}\,,$
(5.12)

which is not to be confused with $Q^{i}$ or
$\tensor{Q}{{}_{\mu}^{i}}=\partial_{\mu}Q^{i}$. Let us also define
$\mathcal{Q}_{2}\simeq\mathcal{Q}_{1}^{2}$ to be the square of this quantity,
retaining terms only up to second order in $Q^{i}$, so that

$\displaystyle\mathcal{Q}_{2}=S_{C}^{2}\tensor{C}{{}_{\mu}^{i}}Q^{\mu
i}\tensor{C}{{}_{\nu}^{j}}Q^{\nu j}-8S_{C}\tensor{C}{{}_{\mu}^{i}}Q^{\mu
i}\tensor{C}{{}_{\nu}^{j}}C^{\nu k}C^{\rho
j}\tensor{Q}{{}_{\rho}^{k}}+16\left(\tensor{C}{{}_{\nu}^{j}}C^{\nu k}C^{\rho
j}\tensor{Q}{{}_{\rho}^{k}}\right)^{2}\,.$ (5.13)

In terms of these combinations, we can expand the square root appearing in
(5.7) as

$\displaystyle\sqrt{2P^{2}-S^{2}}\simeq\sqrt{2P_{C}^{2}-S_{C}^{2}}+\frac{\mathcal{Q}_{1}}{\sqrt{2P_{C}^{2}-S_{C}^{2}}}-\frac{\mathcal{Q}_{2}}{2\left(2P_{C}^{2}-S_{C}^{2}\right)^{3/2}}\,.$
(5.14)

Finally, we can express the Modified Scalar Lagrangian expanded to quadratic
order in fluctuations around a given classical solution as

$\displaystyle\mathcal{L}\simeq\mathcal{L}_{C}+\frac{1}{2}\left(\cosh(\gamma)S_{Q}+\sinh(\gamma)\left(\frac{\mathcal{Q}_{1}}{\sqrt{2P_{C}^{2}-S_{C}^{2}}}-\frac{\mathcal{Q}_{2}}{2\left(2P_{C}^{2}-S_{C}^{2}\right)^{3/2}}\right)\right)\,,$
(5.15)

where $\mathcal{L}_{C}$ represents the Lagrangian evaluated on the background
solution $C^{i}$, i.e.

$\displaystyle\mathcal{L}_{C}$
$\displaystyle\equiv\frac{\cosh(\gamma)}{2}S_{C}+\frac{\sinh(\gamma)}{2}\sqrt{2P_{C}^{2}-S_{C}^{2}}\,.$
(5.16)

It is also convenient to write the Lagrangian for the quantum field $Q^{i}$ in
terms of a bilinear form. Defining the tensor

$\displaystyle\tensor{P}{{}_{\mu\nu}^{ij}}$
$\displaystyle=-\bigg{(}\frac{-S_{C}g_{\mu\nu}\delta^{ij}-2\tensor{C}{{}_{\mu}^{i}}\tensor{C}{{}_{\nu}^{j}}+2\tensor{C}{{}_{\mu}^{k}}\tensor{C}{{}_{\nu}^{k}}\delta^{ij}+2\tensor{C}{{}_{\mu}^{j}}\tensor{C}{{}_{\nu}^{i}}+2\tensor{C}{{}_{\rho}^{i}}C^{\rho
j}g_{\mu\nu}}{2\sqrt{2P_{C}^{2}-S_{C}^{2}}}$
$\displaystyle\qquad\qquad-\frac{S_{C}^{2}\tensor{C}{{}_{\mu}^{i}}\tensor{C}{{}_{\nu}^{j}}-8S_{C}\tensor{C}{{}_{\mu}^{i}}\tensor{C}{{}_{\rho}^{k}}\tensor{C}{{}^{\rho}{}^{j}}\tensor{C}{{}_{\nu}^{k}}+16\tensor{C}{{}_{\rho}^{k}}\tensor{C}{{}^{\rho
i}}\tensor{C}{{}_{\mu}^{k}}\tensor{C}{{}_{\tau}^{m}}\tensor{C}{{}^{\tau
j}}\tensor{C}{{}_{\nu}^{m}}}{4\left(2P_{C}^{2}-S_{C}^{2}\right)^{\frac{3}{2}}}\bigg{)}\,,$
(5.17)

we can write the Lagrangian $\mathcal{L}_{Q}$ for the fluctuating field as

$\displaystyle\mathcal{L}_{Q}$ $\displaystyle=Q^{\mu
i}\left(\frac{\cosh(\gamma)}{2}g_{\mu\nu}\delta^{ij}+\sinh(\gamma)\tensor{P}{{}_{\mu\nu}}^{ij}\right)Q^{\nu
j}\,,$ (5.18)

or after integrating by parts to move the derivative acting on $Q^{\mu
i}=\partial^{\mu}Q^{i}$, as

$\displaystyle\mathcal{L}_{Q}$
$\displaystyle=-Q^{i}\left(\frac{\cosh(\gamma)}{2}\delta^{ij}\partial^{2}+\sinh(\gamma)\left(\partial^{\mu}\tensor{P}{{}_{\mu\nu}^{ij}}\right)\partial^{\nu}+\sinh(\gamma)\tensor{P}{{}_{\mu\nu}^{ij}}\partial^{\mu}\partial^{\nu}\right)Q^{j}\,.$
(5.19)

The first term in (5.19) is proportional to a conventional free kinetic term
for the fields $Q^{i}$. The second and third terms, involving
$\tensor{P}{{}_{\mu\nu}^{ij}}$ and its derivative, encode the interactions
which are induced by expanding around the classical field configuration
$C^{i}$.

#### Feynman rules

Now that we have obtained the Lagrangian (5.19), we may read off the Feynman
rules which we will need for computing diagrams. The propagator for the
quantum field is

$\displaystyle\tensor{D}{{}^{ij}}$
$\displaystyle=-\frac{i}{\cosh(\gamma)}\frac{\delta^{ij}}{k^{2}}\,.$ (5.20)

Next we must work out the vertex associated to the interaction between $Q^{i}$
and the classical field via the combination $\tensor{P}{{}_{\mu\nu}^{mn}}$. We
will draw quantum fields as solid lines and the cumulative effect of the
background fields as a single coiled line. Consider the trivalent interaction
between a field $Q^{i}$ with momentum $p$, a field $Q^{j}$ with momentum $q$,
and an insertion of the background $\tensor{P}{{}_{\mu\nu}^{mn}}$ with
momentum $r$. This vertex can be visualized as

$\displaystyle\raisebox{-0.5pt}{\includegraphics[width=164.77771pt]{interaction_vertex.pdf}}\;.$
(5.21)

Let the vertex factor for this interaction be $g_{ij}$.222222The vertex factor
$g_{ij}$ should not be confused with the target-space metric $G_{ij}(\phi)$
for the bosons which appears in equation (2.49). There are four ways that we
can get a contribution to this factor from the Lagrangian (5.19). First, there
is a piece arising from the term
$\sinh(\gamma)\left(\partial^{\mu}\tensor{P}{{}_{\mu\nu}^{mn}}\right)\partial^{\nu}$
when $m=j$ and $n=i$, which gives a term proportional to $r^{\mu}q^{\nu}$
because of the first derivative $\partial^{\mu}$ acting on
$\tensor{P}{{}_{\mu\nu}^{mn}}$ and the second derivative $\partial^{\nu}$
acting on $Q^{j}$. There is another term of the same form when $m=i$ and
$n=j$. Then there are two more contributions from the term
$\sinh(\gamma)\tensor{P}{{}_{\mu\nu}^{mn}}\partial^{\mu}\partial^{\nu}$, when
either $m=j$ and $n=i$, or when $m=i$ and $n=j$, which both come with a factor
of $q^{\mu}q^{\nu}$ from the two derivatives acting on $Q^{j}$. Altogether,
the value of this vertex is

$\displaystyle g_{ij}$
$\displaystyle=i\sinh(\gamma)\left(\delta^{mj}\delta^{ni}P_{\mu\nu}^{mn}r^{\mu}q^{\nu}+\delta^{mi}\delta^{nj}P_{\mu\nu}^{mn}r^{\mu}q^{\nu}+\delta^{mj}\delta^{ni}P_{\mu\nu}^{mn}q^{\mu}q^{\nu}+\delta^{mi}\delta^{nj}P_{\mu\nu}^{mn}q^{\mu}q^{\nu}\right)$
$\displaystyle=-i\sinh(\gamma)\left(\delta^{mj}\delta^{ni}P_{\mu\nu}^{mn}p^{\mu}q^{\nu}+\delta^{mi}\delta^{nj}P_{\mu\nu}^{mn}p^{\mu}q^{\nu}\right)\,$
$\displaystyle=-i\sinh(\gamma)\left(P^{ij}_{\mu\nu}+P^{ji}_{\mu\nu}\right)p^{\mu}q^{\nu}\,,$
(5.22)

where in the second step we have used $r^{\mu}=-q^{\mu}-p^{\mu}$ to cancel
terms. This gives the desired value of the trivalent vertex $g_{ij}$ between
$Q^{i}$, $Q^{j}$, and the classical background. However, in the calculations
that follow, it will be convenient to factor out the dependence on
$P_{\mu\nu}^{mn}$ and use an ``uncontracted'' vertex factor $\widetilde{g}$
defined by

$\displaystyle g_{ij}$
$\displaystyle=P_{\mu\nu}^{mn}\left(\widetilde{g}^{mn}_{ij}\right)^{\mu\nu}\,,$
$\displaystyle\left(\tilde{g}^{mn}_{ij}\right)^{\mu\nu}$
$\displaystyle=-i\sinh(\gamma)\left(\delta^{mj}\delta^{ni}p^{\mu}q^{\nu}+\delta^{mi}\delta^{nj}p^{\mu}q^{\nu}\right)\,.$
(5.23)

Let us emphasize that $\left(\tilde{g}^{mn}_{ij}\right)^{\mu\nu}$ is _not_ the
full value of the interaction vertex, but rather a useful intermediate
quantity which has removed all factors of $P_{\mu\nu}^{mn}$. After computing
Feynman diagrams using ``uncontracted'' vertices $\tilde{g}$, we must contract
the final result with one factor of $P_{\mu\nu}^{mn}$ for each vertex in order
to recover the true value of the diagram.

###  5.2 Quantum effective action

We are now ready to compute the leading quantum corrections to the Modified
Scalar Lagrangian. Most of our discussion will focus on the one-loop effective
action, defined by the first term beyond the classical contribution in the
expansion

$\displaystyle\Gamma[C^{i}]=S[C^{i}]+\frac{i}{2}\Tr\left[\log\left(\frac{\delta^{2}S}{\delta\varphi^{i}\,\delta\varphi^{j}}\right)\Bigg{|}_{\varphi^{k}=C^{k}}\right]+\cdots\,.$
(5.24)

Although we are primarily interested in the one-loop contribution to $\Gamma$,
we will also present some partial results concerning corrections at higher
loop order.

There are several techniques for computing the one-loop effective action
$\Gamma$. One way is to use heat kernel methods; we will not pursue this
strategy here, but we refer the reader to the thesis [53] for a discussion of
this approach in the related context of the $4d$ ModMax theory. Rather, we
will compute contributions to the effective action perturbatively, using the
Feynman rules derived in the preceding subsection. This amount to a
diagrammatic evaluation of the one-loop determinant of the operator
$\frac{\delta^{2}S}{\delta\varphi^{i}\,\delta\varphi^{j}}$, which is the
operator appearing in $\mathcal{L}_{Q}$ that we have computed in equation
(5.19).

In particular, our goal is to evaluate divergent Feynman diagrams in the
Modified Scalar theory using dimensional regularization, as a function of the
background configuration $C^{i}$. Each such divergent contribution
necessitates the addition of an appropriate counterterm to cancel the
divergence. The collection of all such counterterms which must be added to the
classical Lagrangian therefore reproduces the additional terms that appear in
the quantum effective action, giving a characterization of the corrections in
the expansion (5.24).

#### Constant background, one-loop diagrams

Let us begin by considering the simpler case in which the background field
configuration $C^{i}$ is linear in the spacetime coordinates, which means that
the classical field has constant gradients. That is, we assume that
$\tensor{C}{{}_{\mu}^{i}}=\partial_{\mu}C^{i}$ is constant for such
backgrounds, so that $\partial_{\mu}\tensor{C}{{}_{\nu}^{i}}=0$ for all
$\mu,\nu,i$. In this case, no momentum can flow through the classical fields
in the interaction vertex (5.21), which implies that $r=0$ and thus $p=-q$.

To obtain the one-loop effective action $\Gamma$, we must evaluate all Feynman
diagrams built from the quantum field propagator and interaction vertex (5.21)
which contain at most one loop. This corresponds to an infinite series of
diagrams given by

$\displaystyle\Gamma=\raisebox{-0.5pt}{\includegraphics[width=86.72267pt]{1loop_term_1.pdf}}\;+\;\raisebox{-0.5pt}{\includegraphics[width=108.405pt]{1loop_term_2.pdf}}\;+\;\raisebox{-0.5pt}{\includegraphics[width=86.72267pt]{1loop_term_3.pdf}}\;+\;\cdots\,.$
(5.25)

Let $\mathcal{D}_{n}$ represent the value of the diagram in the series (5.25)
which has $n$ insertions of the classical background. The first diagram in
this infinite series is

$\displaystyle\mathcal{D}_{1}=\raisebox{-0.5pt}{\includegraphics[width=108.405pt]{first_in_infinite_series.pdf}}\,.$
(5.26)

Following the comments around equation (5.1), we will evaluate this diagram –
and the others in this section – by using the Feynman rule associated with the
uncontracted vertex factor $\tilde{g}$, and then contracting with
$P_{\mu\nu}^{mn}$. Doing this and simplifying the resulting sum of Kronecker
delta functions using symmetry, one finds

$\displaystyle\mathcal{D}_{1}$
$\displaystyle=P_{\mu\nu}^{mn}\sinh(\gamma)\delta^{im}\delta^{jn}\int\frac{\differential{{}^{d}\ell}}{\left(2\pi\right)^{d}}\left(2i\ell^{\mu}\ell^{\nu}\right)D^{ij}\,.$
(5.27)

A term in the integrand which is proportional to $\ell^{\mu}\ell^{\nu}$ will
produce a result which scales like $\ell^{2}$ and which is a symmetric tensor
in $\mu$ and $\nu$. The only constant symmetric $2$-tensor in the problem is
the spacetime metric $g^{\mu\nu}$, so the integral of such a term must be
proportional to $\ell^{2}g^{\mu\nu}$. By taking the trace, one can fix the
dimensionless constant to be $\frac{1}{d}$. Thus, within the integral, we can
make the replacement

$\displaystyle\ell^{\mu}\ell^{\nu}$
$\displaystyle\to\frac{1}{d}\ell^{2}g^{\mu\nu}\,.$ (5.28)

Using this replacement and the propagator (5.20), we find

$\displaystyle\mathcal{D}_{1}$
$\displaystyle=-\frac{2\tanh(\gamma)}{d}P_{\mu\nu}^{mn}\delta^{mn}\int\frac{\differential{{}^{d}\ell}}{\left(2\pi\right)^{d}}g^{\mu\nu}\,.$
(5.29)

The integrand is now independent of $\ell$. Although this integral diverges as
$\Lambda^{d}$ with a naïve cutoff at momentum $\Lambda$, within dimensional
regularization it is exactly zero [91].

This result relies only on the momentum dependence of the integral. However,
note that the insertions of additional vertices appearing in the higher one-
loop diagrams $\mathcal{D}_{n}$ will not change the momentum dependence of the
integral. In general, we will have $n$ propagators $D_{ij}$ of the form
(5.20), each of which is proportional to $\frac{1}{\ell^{2}}$, and $n$ copies
of the vertex factor (5.1). Because the vertex factor contains products of
momenta like $\ell^{\mu}\ell^{\nu}$, the integrand of $\mathcal{D}_{n}$ will
involve a product of $2n$ momenta. We can replace such factors using a
generalization of the argument which led to the replacement rule (5.28). That
is, any integral involving a totally symmetric product of $2n$ momenta must
yield a result which is proportional to $\ell^{2n}$ multiplied by a totally
symmetrized combination of $n$ metric tensors, since the metric is the only
symmetric tensor in the problem. This leads to the replacement

$\displaystyle\prod_{i=1}^{n}\ell^{\mu_{2i-1}}\ell^{\mu_{2i}}\to\frac{\ell^{2n}\left(d-2\right)!!}{\left(d-2+2n\right)!!}g^{(\mu_{1}\mu_{2}}\cdots
g^{\mu_{2n-1}\mu_{2n})}\,,$ (5.30)

where we have used the double factorial $n!!=n\cdot(n-2)\ldots 4\cdot 2$. We
thus find an overall factor of $\ell^{2n}$ from the vertex factors, in
addition to a compensating factor of $\frac{1}{\ell^{2n}}$ from the $n$ copies
of the propagator, each of which scales like $\frac{1}{\ell^{2}}$. Note that
all of these momenta are equal due to momentum conservation around the loop,
as we assumed that no momentum can be carried by the classical fields, so the
powers of loop momentum precisely cancel. Therefore, every diagram
$\mathcal{D}_{n}$ involves an integrand which is independent of momentum, and
thus vanishes in dimensional regularization just as $\mathcal{D}_{1}$ does.

We conclude that the perturbative one-loop effective action $\Gamma[C^{i}]$,
with constant background field strength $\tensor{C}{{}_{\mu}^{i}}$, vanishes
in dimensional regularization. This implies that under these assumptions,
there are no 1-loop corrections to the classical theory.

#### Constant background, multi-loop diagrams

Proceeding to higher loops, more vertices in the perturbative expansion become
accessible, beginning at two loops with a vertex cubic in the quantum field.
The first of such diagrams that is not a tadpole, shown in equation (5.31),
emerges at order $\mathcal{O}(\gamma^{2})$, and one can show that it
nontrivially vanishes within dimensional regularization.

$\displaystyle\raisebox{-0.5pt}{\includegraphics[width=151.76964pt]{2_loop_diagram.pdf}}\,.$
(5.31)

The introduction of multiple loop momenta prevents the simple argument in the
one-loop case from generalizing immediately. However, since with constant
backgrounds there cannot be any external momenta and there is no
characteristic scale present in these integrals, it will always be possible to
iteratively symmetrize using (5.30) and integrate over each loop momentum,
leaving a symmetrizable integral that will vanish in dimensional
regularization. Therefore we expect that the argument presented above
generalizes to all loops, implying that the full effective action
$\Gamma[C^{i}]$ admits no corrections for constant background field strengths
$\tensor{C}{{}_{\mu}^{i}}$.

#### Background-varying, one-loop diagrams

We now study the more general case in which we do not assume that
$\partial_{\mu}\tensor{C}{{}_{\nu}^{i}}=0$, instead allowing the background
field to vary. Besides requiring that the field configuration $C^{i}$ is a
solution to the classical equations of motion, we make no further assumptions.

For this general background analysis, let us use the same notation
$\mathcal{D}_{n}$ for the diagrams appearing in the infinite sum (5.25). The
first diagram in this series, $\mathcal{D}_{1}$, is unchanged from the
constant background case, and thus it identically vanishes in dimensional
regularization.

The first nontrivial diagram is

$\displaystyle\mathcal{D}_{2}=\raisebox{-0.5pt}{\includegraphics[width=151.76964pt]{divergent_contribution.pdf}}\,.$
(5.32)

As usual, it will be convenient to strip off factors of
$\tensor{P}{{}_{\mu\nu}}^{ij}$ when computing the value of this diagram. This
corresponds to evaluating the diagram using the ``uncontracted'' vertex
$\tilde{g}$ of (5.1) and contracting the result with factors of
$\tensor{P}{{}_{\mu\nu}}^{ij}$. To this end, let us write the value of the
diagram as

$\displaystyle\mathcal{D}_{2}$
$\displaystyle=\frac{\tanh^{2}(\gamma)}{2}\int\frac{\differential{{}^{d}q}}{\left(2\pi\right)^{d}}\tensor{P}{{}_{\mu\nu}}^{ij}\left(-q\right)\Bigg{(}\int\frac{\differential{{}^{d}\ell}}{\left(2\pi\right)^{d}}\frac{1}{\ell^{2}\left(\ell+q\right)^{2}}\left(\delta^{ik}\delta^{jl}\ell^{\mu}\left(\ell+q\right)^{\nu}+\delta^{il}\delta^{jk}\left(\ell+q\right)^{\mu}\ell^{\nu}\right)$
$\displaystyle\hskip
180.0pt\cdot\left(\delta^{km}\delta^{ln}\ell^{\rho}\left(\ell+q\right)^{\tau}+\delta^{kn}\delta^{lm}\left(\ell+q\right)^{\rho}\ell^{\tau}\right)\Bigg{)}\tensor{P}{{}_{\rho\tau}^{mn}}\left(q\right)\,.$
(5.33)

Using the symmetry property
$\tensor{P}{{}_{\mu}{}_{\nu}^{i}{}^{j}}=\tensor{P}{{}_{\nu}{}_{\mu}^{j}{}^{i}}$,
this can also be expressed as

$\displaystyle\mathcal{D}_{2}$
$\displaystyle=2\tanh^{2}(\gamma)\int\frac{\differential{{}^{d}q}}{\left(2\pi\right)^{d}}\tensor{P}{{}_{\mu\nu}}^{ij}\left(-q\right){\mathcal{I}}^{(\mu\nu)(\rho\tau)}_{2}\tensor{P}{{}_{\rho\tau}^{ij}}\left(q\right)\,,$
(5.34)

where we have defined the simpler integral

$\displaystyle{\mathcal{I}}^{\mu\nu\rho\tau}_{2}=\int\frac{\differential{{}^{d}\ell}}{\left(2\pi\right)^{d}}\frac{(\ell+q)^{\nu}\ell^{\mu}(\ell+q)^{\tau}\ell^{\rho}}{\ell^{2}\left(\ell+q\right)^{2}}\,,$
(5.35)

and where our conventions for symmetrization are
$T^{(\mu\nu)}=\frac{1}{2}\left(T^{\mu\nu}+T^{\nu\mu}\right)$.

To study the divergence structure of the diagram $\mathcal{D}_{2}$, it
suffices to evaluate the quantity ${\mathcal{I}}^{\mu\nu\rho\tau}_{2}$ in
dimensional regularization, which is performed in appendix B.1. The resulting
divergent contribution is

$\displaystyle{\mathcal{I}}^{\mu\nu\rho\tau}_{2}$
$\displaystyle=\left(\frac{1}{\epsilon}\right)\frac{-i}{24\left(4\pi\right)}\bigg{[}q^{2}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)+2\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)$
$\displaystyle\quad+4\left(g^{\mu\rho}q^{\nu}q^{\tau}+g^{\nu\tau}q^{\mu}q^{\rho}\right)\bigg{]}\,.$
(5.36)

In order to cancel this $\frac{1}{\epsilon}$ divergence, one would introduce a
counter-term which involves two factors of $\tensor{P}{{}_{\mu\nu}^{ij}}$ in
the Lagrangian. Therefore, in the background-varying case, there is a non-
trivial contribution to the quantum effective action at one loop. Because the
higher diagrams $\mathcal{D}_{n}$ will involve higher powers of $\gamma$, the
result (5.2) represents the complete one-loop effective action at
$\mathcal{O}(\gamma^{2})$.

With the two-vertex diagram evaluated, to complete the computation of the one-
loop effective action, we seek to evaluate all remaining diagrams containing
one loop. Fortunately, there is only one diagram $\mathcal{D}_{n}$ for each
number of vertices $n$. The details of the evaluation of this diagram are
presented in appendix B.2. Here we merely summarize the results. The value of
$\mathcal{I}_{n}$ can be written as

$\displaystyle\left({\mathcal{I}}_{n}\right)^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle=\left(n-1\right)!\int_{0}^{1}\left(\prod_{i=0}^{n-1}\differential{x}_{i}\right)\delta\left(\sum_{i=0}^{n-1}x_{i}-1\right)\left(\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}+\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}\right)\,,$
(5.37)

where we have defined

$\displaystyle\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle=\frac{i\left(d-2\right)!!}{\left(d-2+2n\right)!!}g^{\mu_{1}\cdots\mu_{2n}}\frac{\Gamma\left(n+\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(n\right)\Gamma\left(\frac{d}{2}\right)}\Gamma\left(-\frac{d}{2}\right)\Delta^{d}\,\,$
$\displaystyle\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle=\frac{i\left(d-2\right)!!}{\left(d-4+2n\right)!!}\sum_{a=1}^{2n}\sum_{b>a}^{2n}g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}f^{\mu_{a}}\left(x,q,a\right)f^{\mu_{b}}\left(x,q,b\right)$
$\displaystyle\qquad\cdot\frac{\Gamma\left(n-1+\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(n\right)\Gamma\left(\frac{d}{2}\right)}\Gamma\left(1-\frac{d}{2}\right)\Delta^{d-2}\,.$
(5.38)

The notation $g^{\mu_{1}\cdots\mu_{2n}}$ refers to a symmetrized product of
metric tensor factors, which is defined in equation (B.34). Similarly,
$g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}$ is shorthand for such a symmetrized product
of metrics which which omits the two indices $\mu_{a}$ and $\mu_{b}$, which is
explained in more detail around equation (B.2). Finally, the function
$f^{\mu}(x,q,a)$ is defined in equation (B.33).

In dimensional regularization, with $d=2(1+\epsilon)$ and as $\epsilon\to 0$,
the overall momentum dependence and divergence structure of these terms is

$\displaystyle\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle\sim\frac{1}{\epsilon}q^{2}g^{\mu_{1}\cdots\mu_{2n}}\,,$
$\displaystyle\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle\sim\frac{1}{\epsilon}\sum_{a=1}^{2n}\sum_{b>a}^{2n}q^{\mu_{a}}q^{\mu_{b}}g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}\,,$
(5.39)

which is of the same qualitative form as the one-loop, two-vertex contribution
(5.2).

Therefore, the full one-loop effective action for the Modified Scalar theory
is obtained by introducing counterterms that cancel the divergent
contributions which we have described in equations (5.2) and (5.2). Because,
after Fourier transforms, only two derivatives arise acting on the external
background vertices, the counterterms are invariant under classical conformal
transformations.

#### Background varying, two vertex, $m$-loop diagrams

One could imagine computing the quantum effective action (5.24) using a double
expansion in both the number $n$ of vertices and the number $m$ of loops. The
preceding subsections have discussed the contributions at one loop but for any
number of vertices. We have also argued that higher loop corrections vanish
when expanding around constant backgrounds.

It is then natural to ask what one can say about the higher-loop contributions
in the general case of varying backgrounds. Although the structure of the
problem quickly becomes quite complicated, we can make some general remarks by
restricting to two vertices but any number of loops. For instance, we can
consider a diagram with $m+1$ internal quantum field lines, each of which runs
between two interaction vertices with a classical background field, thus
forming $m$ loops:

$\displaystyle\mathcal{D}_{m,2}\;=\;\vbox{\hbox{ \leavevmode\hbox
to32.67pt{\vbox to15.67pt{\pgfpicture\makeatletter\hbox{\hskip
16.33415pt\lower-7.83301pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\feynman \vertex(a); \vertex[below
right=of a] (b); \vertex[below left=of b] (c); \vertex[right=of b] (z);
\vertex[above left=of c] (d); \vertex[above right=of d] (e); \vertex[left=of
d] (x); \diagram* { (b) -- [quarter left] (d), (b) -- [quarter right] (d), (a)
[dot] -- [quarter left] (b) -- [quarter left] (c) [dot] -- [quarter left] (d)
-- [quarter left] (e), (z) [crossed dot] -- [rmomentum=$q$, gluon] (b), (d) --
[rmomentum=$q$, gluon] (x) [crossed dot], }; {}{{}}{}
{}{}{}{}{{{}{}}}{}{}{{{}{}}}{}{}{{{}{}}}\hbox{\hbox{\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-13.00114pt}{-2.80556pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$m+1$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-3.75pt}{-2.5pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\vdots$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-3.75pt}{-2.5pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\vdots$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}} }}\,.$ (5.40)

We use the notation $\mathcal{D}_{m,n}$ for a diagram which has $m$ loops and
$n$ vertices. In this notation, the one-loop diagrams which we called
$\mathcal{D}_{n}$ in the preceding subsections would be denoted
$\mathcal{D}_{1,n}$. For example, the diagram $\mathcal{D}_{2}$ of equation
(5.32) would be written as $\mathcal{D}_{1,2}$, since it is of the form in
equation (5.40) with $m=1$ because it has $2=1+1$ internal lines between two
vertices and thus one loop. Similarly, a diagram with $4$ internal lines
between two vertices would have three loops and be denoted
$\mathcal{D}_{3,2}$.

In order to study the diagrams $\mathcal{D}_{m,2}$, we will need to derive a
new Feynman rule for the $(m+2)$-valent vertex involving $(m+1)$ quantum
fields lines and one insertion of the classical background. These higher
vertex factors will come from further terms in the expansion of the square
root in equation (5.14),

$\displaystyle\sqrt{2P^{2}-S^{2}}$
$\displaystyle=\sqrt{2P_{C}^{2}-S_{C}^{2}}+\sum_{N=1}^{\infty}\binom{\frac{1}{2}}{N}\frac{2^{N}\mathcal{Q}_{N}}{\left(2P_{C}^{2}-S_{C}^{2}\right)^{N-\frac{1}{2}}}$
$\displaystyle=\sqrt{2P_{C}^{2}-S_{C}^{2}}+\sum_{M=2}^{\infty}\tensor{P}{{}^{i_{1}\cdots
i_{M}}_{\mu_{1}\cdots\mu_{M}}}\prod_{k=1}^{M}\partial^{\mu_{k}}Q_{i_{k}}\,.$
(5.41)

In the first line, the factor of $2^{N}$ is a choice of normalization which is
needed to match our conventions for $\mathcal{Q}_{1}$ and $\mathcal{Q}_{2}$
above. We will not compute the higher terms $\mathcal{Q}_{N}$ explicitly, but
we instead schematically denote the collection of all contributions from these
terms which involve a product of $M$ derivatives of the quantum fields by
writing the tensor $\tensor{P}{{}^{i_{1}\cdots
i_{M}}_{\mu_{1}\cdots\mu_{M}}}$. When $M=2$, this is precisely the tensor
$\tensor{P}{{}_{\mu}{}_{\nu}^{i}{}^{j}}$ of equation (5.1). We have changed
the summation variable to $M$ in the second line to emphasize that one must
collect contributions from several $\mathcal{Q}_{N}$ at each fixed order in
$M$. There are no linear vertices in $Q^{i}$, so the $M=1$ term is absent, but
both the $N=1$ term $\mathcal{Q}_{1}$ and the $N=2$ term $\mathcal{Q}_{2}$ of
the first sum contributes to the quadratic $M=2$ interaction of the second
sum, and so on.

In terms of the tensors $\tensor{P}{{}^{i_{1}\cdots
i_{M}}_{\mu_{1}\cdots\mu_{M}}}$ which are defined implicitly through the
expansion in equation (5.2), the Feynman rule for an $(M+1)$-valent
interaction with one classical field insertion is

$\displaystyle=\int\differential{{}^{d}q}\tensor{P}{{}^{i_{1}}{}^{\cdots}{}^{i_{M}}_{\nu_{1}\cdots\nu_{M}}}\left(q\right)\prod_{k=1}^{M}p^{\nu_{k}}_{i_{k}}\,.$
(5.42)

Using this Feynman rule, we can compute the value of the diagram
$\mathcal{D}_{m,2}$ in equation (5.40). Such a diagram has two vertices of the
form (5.42), each with $M=m+1$, along with $m$ loop momenta $\ell_{i}$. The
contribution from this diagram is given by the integral

$\displaystyle\mathcal{D}_{m,2}$
$\displaystyle=\frac{\sinh^{2}(\gamma)}{\cosh^{m+1}(\gamma)}\int\frac{\differential{{}^{d}q}}{(2\pi)^{d}}\left(\int\prod_{i=1}^{m}\frac{\differential{{}^{d}\ell_{i}}}{(2\pi)^{d}}\right)\left(\frac{1}{\prod_{j=1}^{m+1}p^{\mu}_{j}p_{\mu
j}}\right)$
$\displaystyle\qquad\qquad\qquad\cdot\tensor{P}{{}^{i_{1}}{}^{\cdots}{}^{i_{m+1}}_{\nu_{1}}{}_{\cdots}{}_{\nu_{m+1}}}\left(q\right)\left(\prod_{k=1}^{m+1}p_{i_{k}}^{\nu_{k}}p_{j_{k}}^{\mu_{k}}\right)\tensor{P}{{}^{j_{1}}{}^{\cdots}{}^{j_{m+1}}_{\mu_{1}}{}_{\cdots}{}_{\mu_{m+1}}}\left(-q\right)\,.$
(5.43)

Here the momenta of the internal lines are chosen to be $p_{1}=q-\ell_{1}$,
$p_{i}=\ell_{i-1}-\ell_{i}$ for $1<i<m+1$, and $p_{m+1}=\ell_{m}$, so that the
total momentum satisfies

$\displaystyle\sum_{i=1}^{m+1}p_{i}$ $\displaystyle=q\,.$ (5.44)

Besides the diagrams $\mathcal{D}_{m,2}$ drawn in equation (5.40), one might
ask whether we should account for additional diagrams where a loop begins and
ends on the same vertex. However, diagrams of this form do not contribute, as
they vanish in dimensional regularization. We can see this by noting that the
momentum $\ell$ running in such a loop will appear in the vertex factor only
in the combination $\ell^{\mu}\ell^{\nu}$, and in the propagator in the form
$\frac{1}{\ell^{2}}$. Therefore, the value of any diagram will be proportional
to

$\displaystyle\int\frac{\differential{{}^{d}\ell}}{\left(2\pi\right)^{d}}\frac{\ell^{\mu}\ell^{\nu}}{\ell^{2}}$
$\displaystyle=\frac{g^{\mu\nu}}{d}\int\frac{\differential{{}^{d}\ell}}{\left(2\pi\right)^{d}}1\
,$ (5.45)

which we have seen vanishes in dimensional regularization in the limit $d\to
2$, as desired.

Next let us consider the divergence structure of the diagram
$\mathcal{D}_{m,2}$. It is convenient to isolate the part of the integrand
which depends on the loop momenta and evaluate it separately. To do this, let
us define

$\displaystyle\left(\mathfrak{L}_{m,2}\right)_{\\{ij\\}}^{\\{\mu\nu\\}}$
$\displaystyle=\int\,\left(\prod_{i=1}^{m}\frac{\differential{{}^{d}\ell_{i}}}{(2\pi)^{d}}\right)\left(\frac{1}{\prod_{j=1}^{m+1}p^{\mu}_{j}p_{\mu
j}}\right)\left(\prod_{k=1}^{m+1}p_{i_{k}}^{\nu_{k}}p_{j_{k}}^{\mu_{k}}\right)\,.$
(5.46)

Here we use $\\{ij\\}$ as a shorthand for the multi-index $\\{i_{1}\ldots
i_{m+1}j_{1}\ldots j_{m+1}\\}$ and $\\{\mu\nu\\}$ for
$\\{\mu_{1}\ldots\mu_{m+1}\nu_{1}\ldots\nu_{m+1}\\}$. We will sometimes
suppress these multi-indices in writing $\mathfrak{L}_{m,2}$ for convenience.
The quantity $\mathfrak{L}_{m,2}$ determines the value of the diagram
$\mathcal{D}_{m,2}$ as

$\displaystyle\mathcal{D}_{m,2}$
$\displaystyle=\frac{\sinh^{2}(\gamma)}{\cosh^{m+1}(\gamma)}\int\frac{\differential{{}^{d}q}}{(2\pi)^{d}}\tensor{P}{{}^{i_{1}}{}^{\cdots}{}^{i_{m+1}}_{\nu_{1}}{}_{\cdots}{}_{\nu_{m+1}}}(q)\left(\mathfrak{L}_{m,2}\right)_{\\{ij\\}}^{\\{\mu\nu\\}}\tensor{P}{{}^{j_{1}}{}^{\cdots}{}^{j_{m+1}}_{\mu_{1}}{}_{\cdots}{}_{\mu_{m+1}}}\left(-q\right)\,,$
(5.47)

so to understand the divergences in $\mathfrak{D}_{m,2}$, it suffices to
understand those in $\mathfrak{L}_{m,2}$.

One can evaluate $\mathfrak{L}_{m,2}$ by performing the integral over each
loop momentum in succession. The details of one such integration, namely the
integral over the final variable $\ell_{m}$, are presented in appendix B.3.
After evaluating this single integral over $\ell_{m}$, one obtains a result
proportional to $\Gamma\left(-\frac{d}{2}\right)\ell^{d}_{m-1}$. One can then
apply the same argument recursively to conclude that performing all $m$ of the
integrals generates $m$ factors of this form. After evaluating all $m$
integrals, the final dependence on the momentum $q$ takes the form

$\displaystyle\left(\mathfrak{L}_{m,2}\right)_{\\{ij\\}}^{\\{\mu\nu\\}}\sim\Gamma\left(-\frac{d}{2}\right)^{m}q^{dm}\,,$
(5.48)

where we show only the dependence on $q$ and $d$ but suppress the tensor
structure in the $i,j,\mu,\nu$ indices.232323Each integral yields $6$
different symmetrizations of the external indices. Thus the exact form of an
$m$-loop diagram contains many different index structures and is challenging
to write explicitly in general.

It is also useful to translate the divergence structure of equation (5.48) in
dimensional regularization to an equivalent dependence on a momentum cutoff
$\Lambda$. For $d=2(1+\epsilon)$ we have the limiting behavior
$\Gamma\left(-\frac{d}{2}\right)\sim\frac{1}{\epsilon}$, and a divergence
proportional to $\frac{1}{\epsilon}$ in dimensional regularization corresponds