# Emission of twisted photons by a Dirac electron
in a strong magnetic field

I. Pavlov<EMAIL_ADDRESS>School of Physics and Engineering, ITMO
University,
197101 St. Petersburg, Russia D. Karlovets<EMAIL_ADDRESS>School of Physics and Engineering, ITMO University,
197101 St. Petersburg, Russia

###### Abstract

We study spontaneous emission of a photon during the transitions between
relativistic Landau states of an electron in a constant magnetic field that
can reach the Schwinger value of $H_{c}=4.4\times 10^{9}$ T. In contrast to
the conventional method in which detection of both the final electron and the
photon is implied in a certain basis, here we derive the photon state as it
evolves from the process itself. It is shown that the emitted photon state
represents a twisted Bessel beam propagating along the field axis with a total
angular momentum (TAM) projection onto this axis $\ell-\ell^{\prime}$ where
$\ell$ and $\ell^{\prime}$ are the TAM of the initial electron and of the
final one, respectively. Thus, the majority of the emitted photons turn out to
be twisted with $\ell-\ell^{\prime}\gtrsim 1$, even when the magnetic field
reaches the critical value of $H\sim H_{c}$. The transitions without a change
of the electron angular momentum, $\ell^{\prime}=\ell$, are possible, yet much
less probable. We also compare our findings with those for a spinless charged
particle and demonstrate their good agreement for the transitions without
change of the electron spin projection even in the critical fields, while the
spin-flip transitions are generally suppressed. In addition, we argue that
whereas the ambiguous choice of an electron spin operator affects the
differential probability of emission, this problem can partially be
circumvented for the photon evolved state because it is the electron TAM
rather than the spin alone that defines the TAM of the emitted twisted photon.

##  1 Introduction

According to classical electrodynamics, an electron moving along a helical
path in an external electromagnetic field – such as a magnetic field, that of
a helical undulator, etc. – emits electromagnetic waves possessing helical
phase structure and carrying orbital angular momentum (OAM) [1, 2, 3, 4]. In
quantum theory, the emitted field comprises the so-called twisted photons [5,
6, 7, 8]. The quantum description of this phenomenon is complicated by the
entanglement between the photon and the electron, which is rarely given
attention in theoretical works [9]. Furthermore, the conventional quantum
theory of synchrotron radiation [10, 11] deals with the so-called detected
states of the final particles silently implying that both the electron and the
photon are detected, and the photon detector state is usually chosen to be a
plane wave. More recently, the photon detector state was chosen to be a Bessel
beam with a definite projection of the total angular momentum (TAM) onto the
magnetic field axis [12, 13, 14]. However, even more general problem would be
to derive a quantum state of the photon as it has evolved from the emission
process itself, without specifying the detector. In particular, such
calculations in the scalar QED show that a spinless charged particle in the
magnetic field indeed emits a Bessel beam [15], the parameters of which can be
derived rather than postulated. Here, we present an analogous evolved state
analysis for a spin-$\frac{1}{2}$ Dirac electron and compare it to the scalar
case.

The magnetic field-induced Landau levels of charged particles are integral to
the quantum theory of synchrotron radiation, which is essential in describing
astrophysical objects like neutron stars, operating modern storage rings, and
producing spin-polarized electron beams. These states also have significant
implications in solid-state physics, particularly in graphene [16, 17, 18],
including quantum Hall effect [19], in diamagnetism of metals, in plasma [20],
and in the quantum dynamics of charged particles in Penning traps [21, 22].
Therefore, the fact that it is the twisted photons that are mostly emitted by
electrons in magnetic fields – both in the astrophysical and terrestrial
environments – could put a new twist on several fields. In this paper, we
study single photon emission in the first order of the perturbation theory by
an electron moving along the field lines of a constant and homogeneous
magnetic field $\bm{H}=\\{0,0,H\\}$ of an arbitrary strength $H>0$ [10, 11].
The magnetic field is taken into account exactly in the calculations and the
electron is described with the relativistic Landau states [10, 11, 23].

One of our main findings is that the evolved state of the emitted photon is
always twisted, i.e. its total angular momentum (TAM) projection onto the
magnetic field axis is well-defined, exactly like in the scalar QED [15]. The
distinction between the ”evolved” photon state and the traditional ”detected”
state (see Refs. [24, 25, 12, 13]) is similar to the difference between a
polarization state that arises from the process itself and a polarization
state that is measured by a detector (for instance, in Ref. [26]). The evolved
photon state is determined by a complex-valued S-matrix element $S_{fi}$,
whereas the emission probability and intensity only rely on its absolute value
$|S_{fi}|$. Therefore, the evolved state formalism clarifies the significance
of the S-matrix element phase and enriches the analysis of the emission
process.

The paper is organized as follows. In Sec. 2 the concept of the evolved states
is briefly introduced. In section 3 we derive the S-matrix element, which is
further used in the analysis of the photon evolved state in Sec. 4 and the
radiation probability and intensity in Sec.5. Finally, in Sec. 6 we draw a
conclusion. In Appendix A we explain a straightforward approach to solving the
Dirac equation in a uniform magnetic field and analyze the spin properties of
the obtained electron states.

We use natural system of units throughout the paper with $\hbar=c=1$ unless
stated otherwise, whereas the electron mass and charge are denoted as $m$ and
$e<0$, respectively. The Coulomb gauge is used for the four-vector potentials
of the background and the radiation fields.

##  2 Evolved photon state: definition

Let us consider emission $e\rightarrow e^{\prime}+\gamma$ of the photon by the
electron in a constant homogeneous magnetic field. The initial state of the
electron $\ket{i_{e}}$ and the evolved (pre-selected) state of the final
photon and electron $\ket{f}$ are connected via the first-order S-matrix in
the Furry picture $\hat{S}^{(1)}$, with the interaction of the electron with
the magnetic field being taken into account exactly:

$\ket{f}=\hat{S}^{(1)}\ket{i_{e}}.$ (1)

If the final electron is projected to a state $\ket{f_{e}}$, the state of the
whole system becomes

$\ket{f^{\prime}}=\sum_{\lambda=\pm
1}\int\frac{d^{3}k}{(2\pi)^{3}}\ket{f_{e};\bm{k},\lambda}S_{fi}^{(1)},$ (2)

where $S_{fi}^{(1)}=\bra{f_{e};\bm{k},\lambda}\hat{S}^{(1)}\ket{i_{e}}$ is the
transition matrix element. Plane waves with the momentum $\bm{k}$ and helicity
$\lambda=\pm 1$ are used here as a complete set of the one-particle photon
states. As the final electron is supposed to be measured, the evolved state
$\ket{f^{\prime}}$ is factorized into a product of the photon state and that
of the electron. Thus, the evolved state of the photon alone is

$\ket{\gamma}_{ev}=\sum_{\lambda=\pm
1}\int\frac{d^{3}k}{(2\pi)^{3}}\ket{\bm{k},\lambda}S_{fi}^{(1)},$ (3)

and the corresponding photon wave function [27] in the momentum representation
becomes [15, 9]

$\bm{A}^{(ev)}(\bm{k})=\sum_{\lambda=\pm
1}S_{fi}^{(1)}\bm{e}_{\lambda}(\bm{k}),$ (4)

where $\bm{e}_{\lambda}(\bm{k})$ are the polarization vectors of the plane-
wave photon. The first-order S-matrix element in the Furry picture is given by
the expression

$S_{fi}^{(1)}=-ie\int d^{4}x\>j^{\mu}_{fi}(x)A_{\mu}^{*}(x),$ (5)

where $A_{\mu}$ is the four-potential of the plane-wave photon and
$j^{\mu}_{fi}$ is the transition current. We emphasize that although the
emitted photon is generally not necessarily a plane wave, its vector potential
is expressed in terms of the plane-wave transition amplitudes $S_{fi}^{(1)}$.
This is simply due to the fact that we choose the plane waves as a convenient
basis for expansion of the photon states.

##  3 Matrix element for photon emission

Let us now derive the S-matrix amplitude with emission of a plane-wave photon
described by the state $\ket{\bm{k},\lambda}$. In the Furry picture of QED,
the first-order transition matrix element is given by Eq. (5) where the four-
potential of the plane-wave photon is

$A^{\mu}(x)=\frac{1}{\sqrt{2\omega V}}\,e^{\mu}_{\lambda}e^{-ik^{\nu}x_{\nu}}$
(6)

with the polarization four-vector in the Coulomb gauge

$e^{\mu}_{\lambda}(\bm{k})=\\{0,\bm{e}_{\lambda}(\bm{k})\\}$ (7)

and the wave vector

$k^{\nu}=\omega\\{1,\sin\theta\cos\varphi_{k},\sin\theta\sin\varphi_{k};\cos\theta\\}=\\{\omega,\bm{k}\\}.$
(8)

Note that in the cylindrical coordinates
$\bm{k}=k_{z}\bm{e}_{z}+\bm{k}_{\perp}$ and
$\bm{k}\cdot\bm{r}=k_{z}z+k_{\perp}\rho\cos(\phi_{k}-\phi)=k_{z}z+\omega
r\sin\theta\cos(\phi_{k}-\phi)$. The transition current reads

$j^{\mu}_{fi}(x)=\overline{\Psi}_{f}(x)\gamma^{\mu}\Psi_{i}(x),$ (9)

where $\Psi_{i}(x)$ and $\Psi_{f}(x)$ are the initial and the final (detected)
electron states, respectively.

In the Coulomb gauge with $\bm{k}\cdot\bm{e}_{\lambda}(\bm{k})=0$, one can
expand the photon polarization vectors in terms of the photon spin operator
eigenfunctions as [8]

$\displaystyle\bm{e}_{\lambda}(\bm{k})=\sum_{\sigma=0,\pm
1}e^{-i\sigma\varphi_{k}}d_{\sigma\lambda}^{(1)}(\theta)\bm{\chi}_{\sigma};$
(10) $\displaystyle\bm{\chi}_{0}=(0,0,1),\;\bm{\chi}_{\pm
1}=\mp\frac{1}{\sqrt{2}}(1,\pm i,0),$

where $d_{\lambda\lambda^{\prime}}^{(1)}(\theta)$ are the small Wigner
functions [28].

Let us now briefly discuss the electron states that we use to calculate the
transition amplitudes. We employ the stationary relativistic Landau states
obtained as an exact solution to the Dirac equation in the given magnetic
field in the so-called symmetric gauge with the following potential:
$\bm{A}=\frac{H}{2}\\{-y,x,0\\}$ (see Appendix A). As the QED in the Furry
picture is foremost used in problems with the very strong fields, $H\sim
H_{c}=m^{2}/|e|$, that can be found in astrophysical environments (say, in
neutron stars) and the final electron stays inside the same unperturbed field,
it is natural to describe its state in terms of the same basis as the initial
one.

Although spin is a fundamental property of electron, the corresponding
operator in relativistic quantum theory is defined ambiguously due to the
nonuniqueness of the way how the total angular momentum operator is split into
an external (orbital) part and an internal (spin) part [29, 30]. In some works
it is argued that the operator obtained with the Foldy-Wouthuysen
transformation is the most promising candidate for a proper relativistic spin
operator [29, 31]. However, the simplest spin operator – the so-called Pauli
operator – $\frac{1}{2}\widehat{\bm{\Sigma}}$, where

$\bm{\Sigma}=\begin{pmatrix}\bm{\sigma}&0\\\ 0&\bm{\sigma}\\\ \end{pmatrix}$
(11)

with $\bm{\sigma}$ being the Pauli matrices, is also not devoid of meaning in
relativistic theory [30, 29]. Since the discussion of the spin-related
phenomena is not the aim of this work, in the following derivations we use two
orthogonal solutions, $\Psi_{s,\ell}^{\uparrow}$ and
$\Psi_{s,\ell}^{\downarrow}$, which can be relatively easily obtained by the
procedure described in Appendix A. Here it is not especially significant
whether these states are the eigenstates of some specific spin operator (such
an operator, however, is introduced, for example, in Ref. [10]). The key point
is that they posses a definite value of the total angular momentum projection
onto the field axis. In the limit of a weak magnetic field, $H\ll H_{c}$, they
tend to the eigenstates of the spin operator $\frac{1}{2}\hat{\Sigma}_{z}$
with the eigenvalues $\pm\frac{1}{2}$, that is why we refer to these states as
“spin-up” and “spin-down”, respectively.

As an example, let us consider the transition between the “spin-up” states

$\Psi_{i}(x)=N_{i}^{\uparrow}\begin{pmatrix}(m+\varepsilon)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
0\\\ p_{z}\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
-ieH\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon+i\ell\varphi+ip_{z}z}$
(12)

and

$\Psi_{f}(x)=N_{f}^{\uparrow}\begin{pmatrix}(m+\varepsilon^{\prime})\Phi_{s^{\prime},\ell^{\prime}-1/2}(\rho)e^{-i\varphi/2}\\\
0\\\ p_{z}^{\prime}\Phi_{s,\ell^{\prime}-1/2}(\rho)e^{-i\varphi/2}\\\
-ieH\Phi_{s^{\prime},\ell^{\prime}+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon^{\prime}+i\ell^{\prime}\varphi+ip_{z}^{\prime}z},$
(13)

where the radial function is

$\Phi_{s,\ell}(\rho)\equiv\rho^{\ell}L_{s}^{\ell}(2\rho^{2}/\rho_{H}^{2})e^{-\rho^{2}/\rho_{H}^{2}}$
(14)

with $\rho_{H}=\sqrt{\frac{4}{\absolutevalue{e}H}}$ and $L_{s}^{\ell}$ being
the associated Laguerre polynomials. The normalization constants $N_{i,f}$ are
calculated in the Appendix A (see Eq. (61)). The energy $\varepsilon$ is given
by the equation (42). The independent quantum numbers of this state are the
continuous longitudinal momentum $p_{z}$, a radial quantum number
$s=0,1,2,...$ and the $z$-component of the total angular momentum $\ell=\pm
1/2,\pm 3/2,...$, provided $\ell\geq-s+\frac{1}{2}$.

The corresponding transition current components (Eqs. (78) - (81)) and the
details of the derivation are presented in the Appendix B. After the
integration we obtain the final expression for the transition amplitude (5):

$S_{\uparrow\uparrow}^{(1)}=ei^{-\ell+\ell^{\prime}+1}(2\pi)^{3}\frac{N_{i}^{\uparrow}N_{f}^{\uparrow}}{\sqrt{2\omega
V}}\delta(\omega+\varepsilon^{\prime}-\varepsilon)\delta(k_{z}+p_{z}^{\prime}-p_{z})e^{i(\ell-\ell^{\prime})\varphi_{k}}\times\\\
\times\Bigg{[}d_{0\lambda}^{(1)}(\theta)\rho_{H}^{\ell+\ell^{\prime}+1}\left[m(p_{z}+p_{z}^{\prime})+p_{z}^{\prime}\varepsilon+p_{z}\varepsilon^{\prime}\right]F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}-1/2}(y)+\\\
4\sqrt{2}\rho_{H}^{\ell+\ell^{\prime}}\left[(m+\varepsilon^{\prime})d_{-1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell+1/2,\ell^{\prime}-1/2}(y)-(m+\varepsilon)d_{1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}+1/2}(y)\right]\Bigg{]}.$
(15)

Here $y=k_{\perp}\rho_{H}$ and $F_{s,s^{\prime}}^{\ell,\ell^{\prime}}$ is
defined in Eq. (85). The remaining three S-matrix elements
$S_{\uparrow\downarrow}^{(1)}$, $S_{\downarrow\uparrow}^{(1)}$ and
$S_{\downarrow\downarrow}^{(1)}$, which describe the transitions
$\Psi_{s,\ell}^{\uparrow}\rightarrow\Psi_{s^{\prime},\ell^{\prime}}^{\downarrow}$,
$\Psi_{s,\ell}^{\downarrow}\rightarrow\Psi_{s^{\prime},\ell^{\prime}}^{\uparrow}$
and
$\Psi_{s,\ell}^{\downarrow}\rightarrow\Psi_{s^{\prime},\ell^{\prime}}^{\downarrow}$
respectively, are obtained in a similar fashion. Therefore, we skip the
details of the derivation and come straight to the final expressions (see
Appendix B).

##  4 Evolved photon state: analysis

Each of the four transition amplitudes obtained in the previous section can be
written in a form

$S_{fi}^{(1)}=(2\pi)^{3}\delta(\omega+\varepsilon^{\prime}-\varepsilon)\delta(k_{z}+p_{z}^{\prime}-p_{z})e^{i(\ell-\ell^{\prime})\varphi_{k}}\mathcal{F}(\varepsilon,\varepsilon^{\prime},p_{z},p_{z}^{\prime},s,s^{\prime},\ell,\ell^{\prime},k_{\perp},\theta),$
(16)

where
$\mathcal{F}(\varepsilon,\varepsilon^{\prime},p_{z},p_{z}^{\prime},s,s^{\prime},\ell,\ell^{\prime},k_{\perp},\theta)=\mathcal{F}$
is some scalar function that does not depend on $\varphi_{k}$ (see Eqs.(15),
(87), (86), (88)). Note that the conservation of the energy and the
longitudinal momentum provided by the corresponding delta functions
automatically leads to the conservation of the transverse momentum:

$\delta(\omega+\varepsilon^{\prime}-\varepsilon)=\delta\left(\sqrt{k_{\perp}^{2}+(p_{z}-p_{z}^{\prime})^{2}}+\varepsilon^{\prime}-\varepsilon\right)=\frac{\varepsilon-\varepsilon^{\prime}}{\kappa}\delta(k_{\perp}-\kappa),$
(17)

where

$\kappa=\sqrt{(\varepsilon-\varepsilon^{\prime})^{2}-(p_{z}-p_{z}^{\prime})^{2}}\geq
0$ (18)

is the transverse momentum of the photon. The photon spectra here coincide
with those in the standard quantum synchrotron radiation theory, described,
for instance, in [10, 11, 15].

Let us now return to the derivation of the evolved state of the emitted
photon. The integral over momenta in the Eq. (3) can be taken in cylindrical
coordinates over $k_{z}$ and $k_{\perp}$ with the aid of Eq. (17), resulting
in

$\ket{\gamma}_{ev}=(\varepsilon-\varepsilon^{\prime})\mathcal{F}\sum_{\lambda=\pm
1}\int_{0}^{2\pi}d\varphi_{k}\ket{\bm{k},\lambda}e^{i(\ell-\ell^{\prime})\varphi_{k}}.$
(19)

In this expression, the following equalities are understood:

$\theta=\arctan\left(\frac{\kappa}{p_{z}-p_{z}^{\prime}}\right),$ (20)
$\bm{k}=(\kappa\cos\varphi_{k},\kappa\sin\varphi_{k},p_{z}-p_{z}^{\prime}),$
(21)

such that in the integrand of Eq. (19) only the exponential
$e^{i(\ell-\ell^{\prime})\varphi_{k}}$ and the state $\ket{\bm{k},\lambda}$
depend on $\varphi_{k}$.

Note that the expression (19) is essentially an expansion of the state
$\ket{\gamma}_{ev}$ in terms of the plane wave states $\ket{\bm{k},\lambda}$
with the same energy $\omega=\varepsilon-\varepsilon^{\prime}$, longitudinal
momentum $k_{z}=p_{z}-p_{z}^{\prime}$, and a transverse momentum $\kappa$.
This means that the emitted photon possesses a definite energy, definite
longitudinal and transverse momenta, but not an azimuthal component of the
momentum, which hints that $\ket{\gamma}_{ev}$ is a so-called Bessel beam
propagating along the field axis on average [8]. This can be confirmed by the
explicit calculation of the photon TAM projection. One can easily check that
due to the chosen expansion (10) of the polarization vector
$\bm{e}_{\lambda}(\bm{k})$ (see discussion on this choice in Ref.[9]) it does
not contribute to the TAM of the photon:

$\hat{j}_{z}^{(\gamma)}\bm{e}_{\lambda}(\bm{k})=0,$ (22)

where
$\hat{\bm{j}}^{(\gamma)}=\hat{\bm{l}}^{(\gamma)}+\hat{\bm{s}}^{(\gamma)}$ is
the TAM operator for a photon. Thus, the operator $\hat{j}^{(\gamma)}_{z}$
acts on the vector potential of the evolved state (4) simply as the
differentiation operator $\hat{l}^{(\gamma)}_{z}=-i\partial_{\varphi_{k}}$.
Therefore, we finally find that in addition to the energy, the transverse
momentum and the longitudinal one the emitted photon also has a definite TAM
projection:

$\hat{j}_{z}^{(\gamma)}\bm{A}^{(ev)}(\bm{k})=(\ell-\ell^{\prime})\bm{A}^{(ev)}(\bm{k}).$
(23)

This equation illustrates that the emitted photon generally represents a
twisted state rather than a plane wave. The value of $\ell-\ell^{\prime}$ can,
however, be zero in case of a transition with the change of the radial quantum
number $s$ only. Transitions without change of both $\ell$ and $s$ are also
possible, but only for the fixed value of the longitudinal momentum
$p_{z}^{\prime}$. As we are interested in the probability integrated over
$p_{z}^{\prime}$ in the following section, such transitions are not considered
here. The conservation of the TAM projection during the photon emission is of
no surprise due to the azimuthal symmetry of the problem and the electron
states (12) and (13) being the eigenstates of $\hat{j}_{z}$. Importantly, the
TAM of the above Bessel beam cannot be unambiguously split into the orbital
part and the spin one. In this problem, this ambiguity is closely related to
the same non-uniqueness of separation of the electron TAM into the OAM and the
spin, which is why the operator of the latter is also ambiguously defined (see
Appendix A).

We would like to stress that the Bessel beam naturally emerged in our
derivations as the state of the electromagnetic field evolving directly from
the process. In contrast, in Refs. [12, 13, 14] the four-vector potential of a
Bessel photon was explicitly substituted into the S-matrix amplitude (5),
meaning that the photon was supposed to be detected as a Bessel state with the
definite parameters. An advantage of the current approach is that it allows
one to derive all the parameters (the transverse momentum, the TAM, and so
forth) of the Bessel beam as they evolve from the process itself instead of
making an ansatz.

The reader may well ask why we have obtained the Bessel beam, which is the
simplest and somewhat idealized model of a twisted photon. Indeed, the fact
that the Bessel beam naturally arises in this problem may seem surprising,
because it is known to be non-integrable in the transverse plane. Since both
the initial and the final states of the electron are transversely localized
wave packets, one may expect the emitted photon to be localized as well,
representing a normalizable Laguerre-Gaussian-like beam, or at least a
weighted superposition of the Bessel states [8]. Clearly, this is not true, as
well as for a scalar charged particle in magnetic field [15]. This is because
we consider the initial and final electron states possessing definite
longitudinal momenta, $p_{z}$ and $p_{z}^{\prime}$ respectively, which means
that they are delocalized along the z-axis. This results in a delta function
$\delta(k_{z}+p_{z}^{\prime}-p_{z})$ in Eq. (16) and allows us to use Eq.
(17), which gives rise to the definite transverse momentum of the emitted
photon - a hallmark of a non-normalizable Bessel beam.

Let us analyze what happens if the initial electron represents a
longitudinally localized packet with a mean momentum value $\left\langle
p_{z}\right\rangle$. For simplicity, we take a Gaussian distribution only over
$p_{z}$ and consider the final electron with a definite $p_{z}^{\prime}$. An
example of the “modified” initial electron state can be

$\displaystyle\Psi^{\prime}_{i}(x)=\bigintss_{-\infty}^{\infty}dp_{z}\frac{1}{\sqrt{2\pi\sigma_{p}^{2}}}\,e^{-\frac{(p_{z}-\left\langle
p_{z}\right\rangle)^{2}}{2\sigma_{p}^{2}}}N_{i}\begin{pmatrix}(m+\varepsilon)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
0\\\ p_{z}\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
-ieH\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon+i\ell\varphi+ip_{z}z}.$
(24)

In this expression $\varepsilon$ depends on $p_{z}$ making the state
nonstationary. The integration over $p_{z}$ in Eq. (24) is successively
transferred to the S-matrix element and to the evolved state. Thus, the
modified evolved state simply becomes

$\ket{\gamma}_{ev}^{\prime}=\frac{1}{\sqrt{2\pi\sigma_{p}^{2}}}\int_{-\infty}^{\infty}dp_{z}e^{-\frac{(p_{z}-\left\langle
p_{z}\right\rangle)^{2}}{2\sigma_{p}^{2}}}\ket{\gamma}_{ev}.$ (25)

Note that here $\ket{\gamma}_{ev}$ given by the Eq. (19) is a function of
$p_{z}$. Since $\kappa$ included in the expression for $\ket{\gamma}_{ev}$
depends on $p_{z}$, $\ket{\gamma}_{ev}^{\prime}$ represents an intricate
superposition of the Bessel beams with some distribution over both the
longitudinal and transverse momenta. Although it is impossible to explicitly
factor out the Gaussian distribution over the $\kappa(p_{z})$ in this
expression, it is tempting to say that Eq.(25) is a Bessel-Gauss-like state,
which is normalizable [8]. Thus, the “unphysical” Bessel beam would not emerge
in our derivation if we considered the nonstationary and longitudinally
localized states of the electron. This underlines the importance of
nonstationary states for the correct description of the electron in an
external field, which has already been noted in [32] and applied to the
dynamics of the electron r.m.s. radius in a solenoid.

##  5 Emission probability and intensity

Having derived and analyzed the state $\ket{\gamma}_{ev}$ in which the photon
is emitted, we now turn to the radiation probability. Recalling the expression
(3) for the evolved state of the photon we can find the probability of
transition between the electron states with some fixed quantum numbers as

$\displaystyle
W^{(1)}_{s^{\prime},\ell^{\prime}}(p_{z}^{\prime})=\leftindex_{ev}{\innerproduct{\gamma}{\gamma}}_{ev}=\sum_{\lambda,\lambda^{\prime}=\pm
1}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{d^{3}k^{\prime}}{(2\pi)^{3}}\bra{\bm{k}^{\prime},\lambda^{\prime}}\ket{\bm{k},\lambda}S_{fi}^{(1)}(\bm{k},\lambda)S_{fi}^{(1)*}(\bm{k}^{\prime},\lambda^{\prime})=$
(26) $\displaystyle=\sum_{\lambda=\pm
1}\int\frac{d^{3}k}{(2\pi)^{3}}\left|S_{fi}^{(1)}\right|^{2},$ (27)

where we have used the orthogonality relation for plane waves.

The total radiation probability is then given by the summation over all
possible quantum numbers of the final electron state
($s^{\prime},\ell^{\prime}$) and the integration over the final longitudinal
momentum $p_{z}^{\prime}$:

$W^{(1)}=\sum_{s^{\prime},\ell^{\prime}}\int
W^{(1)}_{s^{\prime},\ell^{\prime}}(p_{z}^{\prime})\frac{dp_{z}^{\prime}}{2\pi}L,$
(28)

where $L$ is the normalization length. However, since we are interested in the
emission of the photons with definite values of the TAM, we will not sum over
the discrete quantum numbers and we only integrate Eq. (26) over
$p_{z}^{\prime}$ introducing the following probability:

$W_{s^{\prime},\ell^{\prime}}^{(1)}\equiv\int
L\frac{dp_{z}^{\prime}}{2\pi}\;W^{(1)}_{s^{\prime},\ell^{\prime}}(p_{z}^{\prime})=\sum_{\lambda=\pm
1}\int\frac{dp_{z}^{\prime}}{2\pi}L\frac{d^{3}k}{(2\pi)^{3}}\left|S_{fi}^{(1)}\right|^{2}.$
(29)

When squaring the delta functions in the matrix element, we use the rule

$(\delta(\omega+\varepsilon^{\prime}-\varepsilon))^{2}(\delta(p_{z}-p_{z}^{\prime}-k_{z}))^{2}\rightarrow\frac{T}{2\pi}\delta(\omega+\varepsilon^{\prime}-\varepsilon)\frac{L}{2\pi}\delta(p_{z}-p_{z}^{\prime}-k_{z}),$
(30)

and for the integration over $d^{3}k$ in cylindrical coordinates we use Eq.
(17). The corresponding emission probability per unit time is found as

$\dot{W}^{(1)}_{s^{\prime},\ell^{\prime}}=\frac{W^{(1)}_{s^{\prime},\ell^{\prime}}}{T}=\frac{1}{(2\pi)^{4}}L\int
dp_{z}^{\prime}\int k_{\perp}dk_{\perp}d\varphi_{k}dk_{z}\sum_{\lambda=\pm
1}\left|S_{fi}^{(1)}\right|^{2}.$ (31)

For instance, for the “up-up” (no spin-flip) transition substituting the
amplitude (15) into the Eq. (31) yields

$\dot{W}^{(1)}_{s^{\prime},\ell^{\prime}}=\pi L^{2}e^{2}N_{i}^{2}\int
dp_{z}^{\prime}N_{f}^{2}\times\\\ \sum_{\lambda=\pm
1}\Bigg{|}d_{0\lambda}^{(1)}(\theta)\rho_{H}^{\ell+\ell^{\prime}+1}\left[m(p_{z}+p_{z}^{\prime})+p_{z}^{\prime}\varepsilon+p_{z}\varepsilon^{\prime}\right]F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}-1/2}(y)+\\\
4\sqrt{2}\rho_{H}^{\ell+\ell^{\prime}}\left[(m+\varepsilon^{\prime})d_{-1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell+1/2,\ell^{\prime}-1/2}(y)-(m+\varepsilon)d_{1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}+1/2}(y)\right]\Bigg{|}^{2},$
(32)

where $N_{f}$, $\varepsilon^{\prime}$, $\sin\theta=\kappa/\omega$, and
$\kappa$ depend on $p_{z}^{\prime}$.

Figure 1: The emission probability (32) (left) and the corresponding intensity
(33) (right) for $H=H_{c}$, $p_{z}=10^{-3}mc$, and no spin-flip. For the solid
lines $s=s^{\prime}=20$, the dashed lines correspond to the twisted photons
with a simultaneous change of the radial quantum number $s\to s^{\prime}\neq
s$, the dash dotted lines correspond to the untwisted photons with the TAM
$j_{z}=\ell-\ell^{\prime}=0$. Figure 2: The same as in Fig 1, but for
$H=0.1H_{c}$ and $s=5$. Figure 3: The emission probability (32) (left) and the
corresponding intensity (33) (right) for $H=0.01H_{c}$, $p_{z}=10^{-3}mc$, and
no spin-flip. For the solid lines $s=s^{\prime}=5$, the dashed lines
correspond to the twisted photons with a simultaneous change of the radial
quantum number $s\to s^{\prime}\neq s$. The magenta dash-dotted line
corresponds to the transitions with increase of the electron OAM, so that the
photon TAM is $\ell-\ell^{\prime}=-1$. Other dash dotted lines correspond to
the untwisted photons with $j_{z}=\ell-\ell^{\prime}=0$.

Correspondingly, the radiation intensity is obtained by multiplying the
integrand by the photon energy $\omega=\varepsilon-\varepsilon^{\prime}$,
which also depends on $p_{z}^{\prime}$:

$I_{s^{\prime},\ell^{\prime}}^{(1)}\equiv\int
L\frac{dp_{z}^{\prime}}{2\pi}\;\omega\dot{W}^{(1)}_{s^{\prime},\ell^{\prime}}(p_{z}^{\prime}).$
(33)

The integration over $p_{z}^{\prime}$ in equations (32) and (33) is to be
carried out numerically over the region allowed by the condition $\kappa\geq
0$ (see Eq. (18) and Ref.[15] for more detail):

$p_{z}^{\prime}\in\left[p_{z}-\frac{p_{\perp}^{2}-(p_{\perp}^{\prime})^{2}}{2(\varepsilon-
p_{z})},p_{z}+\frac{p_{\perp}^{2}-(p_{\perp}^{\prime})^{2}}{2(\varepsilon+p_{z})}\right].$
(34)

The radiation probability $\dot{W}^{(1)}_{s^{\prime},\ell^{\prime}}$ for three
other types of possible transitions can be obtained similarly from the
amplitudes (86) - (88). For the sake of brevity we do not present the
corresponding expressions here.

Due to dependence of the radiated photon energy on $\ell$ and $\ell^{\prime}$,
the emission probability and the intensity depend slightly differently on
$\ell^{\prime}$. For “up-up” transitions this dependence is shown in Fig. 1,
Fig. 2 and Fig. 3. Electron is supposed to be nonrelativistic with
$p_{z}=10^{-3}m$ and $H=H_{c}$, $H=10^{-1}H_{c}$, and $H=10^{-2}H_{c}$,
respectively. Note that the results in Figs. 1, 2 and 5 are presented for the
same parameters as in Ref. [15] for a spinless particle. It is important to
note that the absolute values of probability and intensity agree with the
predictions of the scalar QED within at least 15% while qualitatively the
dependencies are absolutely the same. While the probability of the transitions
$\ell\to\ell^{\prime}=\ell-1$ always dominates, their intensity becomes
slightly less than that of $\ell\to\ell^{\prime}=\ell-2$ starting from
$\ell^{\prime}\geq 20$ at $H\sim 0.1H_{c}$. Moreover, in subcritical magnetic
field $H=0.1H_{c}$ even the intensities of the transitions
$\ell\to\ell^{\prime}=\ell-3,\ell-4$ also exceed that for
$\ell\to\ell^{\prime}=\ell-1$ starting from $\ell^{\prime}\geq 50$ (see Fig.
2). As can be seen from the Figs. 1-3, the transitions with radiation of
“untwisted” photons are generally suppressed compared to those with
$\ell\to\ell^{\prime}\neq\ell$. The transitions with the increase of the
electron TAM $\ell\to\ell^{\prime}>\ell$ are even more damped, as demonstrated
in Fig. 3.

Figure 4: The ratio of four possible types of transition probabilities to the
probability of emission by a scalar charge derived in [15]. $H=10^{-3}H_{c}$
(left) and $H=H_{c}$ (right); $\ell-\ell^{\prime}=3$, $s=s^{\prime}=20$ for
all transitions.

Let us now evaluate the probabilities of three other types of the transitions
– “up-down”, “down-up” and “down-down” – and compare these results with those
obtained in [15], where the same process is considered for a scalar charged
particle. Since the initial and final states of a spinless particle in
magnetic field are characterized by the integer values of TAM ($\ell$ and
$\ell^{\prime}$, respectively), it seems sensible to compare a “scalar”
transition $\ell\to\ell^{\prime}$ with the “spin” transitions
$\ell+1/2\to\ell^{\prime}+1/2$ (shown in Fig. 4) or
$\ell-1/2\to\ell^{\prime}-1/2$. The latter comparison is not presented in the
figures as it does not qualitatively differ from the former. Thus, we consider
the ratio
$\dot{W}^{(1)}_{s^{\prime},\ell^{\prime}}/\dot{\mathcal{W}}^{(1)}_{s^{\prime},\ell^{\prime}+1/2}$,
where $\dot{\mathcal{W}}^{(1)}_{s^{\prime},\ell^{\prime}}$ is the radiation
probability per unit time for a scalar charge. Two regimes are presented in
figure 4: $H=10^{-3}H_{c}$ (left) and $H=H_{c}$ (right). It can be seen that
generally the probabilities of “up-up” and “down-down” transitions are both of
the same order and are in good agreement with the predictions of the scalar
QED. On the contrary, the probabilities of “up-down” and “down-up” transitions
are significantly suppressed as compared to the transitions without the spin
flip. The “up-down” probabilities, in turn, exceed those for the inverse spin
flip by many orders, which is a manifestation of a well-know effect of self-
polarization predicted by Sokolov and Ternov [10] and lately investigated for
the case of weakly excited electron states in ultrastrong magnetic fields
[12]. Due to the spin-flip transitions being considerably damped, the
probabilities of “up-up” transitions shown in Figs. 1 \- 3 can be viewed as
summed over the spin states of the final electron with a good accuracy.

Figure 5: The dependence of the emission probability (left) and the intensity
(right) on the electron momentum $p_{z}$ for $H=0.1H_{c}$, $s=s^{\prime}=20$.
The transition $20\frac{1}{2}\to 19\frac{1}{2}$ means $\ell=20\frac{1}{2}$,
$\ell^{\prime}=19\frac{1}{2}$, $s=s^{\prime}=20$; those with
$\ell:20\frac{1}{2}\to 20\frac{1}{2}$ correspond to the untwisted photons with
$j_{z}=0$. The green line overlaps with the pink dashed one on the left; the
cyan line on the left overlaps with the blue one on the right. The magenta
dash-dotted line corresponds to increase of the electron OAM during the
emission (so that the photon TAM is $\ell-\ell^{\prime}=-1$).

Fig. 5 illustrates the radiation probability per second and the intensity for
the definite discrete quantum numbers of the final electron as a function of
the initial momentum $p_{z}$ along the magnetic field. Similarly to Figs. 1 \-
3 we consider here only “up-up” transitions. Both probability and intensity
turn out to be symmetric functions of $p_{z}$ with a maximum at $p_{z}=0$
although the bispinors (LABEL:Psi_Plus) and (58) depend on the sign of
$p_{z}$. Whereas the momentum-dependent probabilities for the untwisted
photons emission are not necessarily the lowest ones, the intensities are
almost independent on the momentum $p_{z}$ and demonstrate the domination of
the twisted radiation with the TAM $\ell-\ell^{\prime}\sim 1,2$ and unchanged
$s$. In scalar QED, however, the radiation intensity was shown to be
completely independent on $p_{z}$ [15].

##  6 Discussion and conclusion

The dynamics of electrons in magnetic fields and the synchrotron radiation
have been studied for a very long time both in classical and quantum
formalisms. However, the transfer of the quantized angular momentum from
electrons to the electromagnetic field has only recently been discovered and
analyzed within classical [1, 2, 3, 4] and quantum [33, 15] frameworks. Here,
we have ascertained that the definite TAM projection is a natural property of
the photon state evolving from the process itself rather than a consequence of
the choice of the detector.

We have found that the vast majority of the emitted photons are indeed
twisted, representing the Bessel beams with the TAM projection
$j_{z}=\ell-\ell^{\prime}$, although a small part of them are emitted due to
the transitions without a change of the angular momentum. The radiation of
such non-twisted photons is not predicted by classical electrodynamics. For
critical fields, $H\lesssim H_{c}$, the energies of the twisted photons can
belong to the hard X-ray and $\gamma$-range [15].

We have also argued that the reason why the transversely localized electron
packet emits a delocalized Bessel photon is that here we do not account for
the longitudinal localization of the electron. Considering the longitudinally
localized packets would lead to the radiation of fully localized, yet
diffractively spreading photon beams. Spatial localization of the electron
Landau states in all three dimensions can be especially important for proper
description of the transitions in metals and solids exposed to magnetic
fields, as well as in Penning traps, because the resultant twisted photon
packets will also be localized. Say, in the critical magnetic fields their
transverse coherence length can be as small as of the order of the electron
Compton wavelength [15]. Depending on the spatial coherence, these twisted
photons can interact differently with other electrons, ions, phonons, cosmic
plasma, and so forth.

Although the spin states of the electron in magnetic field can be defined
ambiguously, the emitted photons turn out to be twisted regardless of the
electron polarization description. The choice of the concrete spin operator
corresponds to the specific separation of the TAM operator, the latter being
uniquely defined, into its orbital and spin parts. However, it is the TAM
projection of the photon that is observable in this problem and its value is
determined only by the TAM of the initial and final electron states, which do
not depend on the choice of the electron spin operator. The differential
probability and intensity of transitions, nevertheless, depend on the electron
polarization revealing the self-polarization effect.

In conclusion, we note that the angular momentum of the emitted photons does
not usually manifest itself at the terrestrial experiments (e.g. at storage
rings) as it is quantized along the field, while the photons are being
detected at the angles close to the orbital plane where the wave front looks
almost flat. The helical structure of the wave front can be noticed only when
observing the radiation at angles close to the magnetic field axis, especially
in the critical and subcritical fields, $H\lesssim H_{c}$, typical for
magnetars. The importance of twisted photons in astrophysics lies primarily in
their ability to interact with the medium differently from the plane waves,
e.g., to excite the Landau electrons to higher angular momentum states or
induce stimulated emission, which can have implications for studying stellar
nucleosynthesis [13]. Finally, the dynamics of the spreading photon packets
emitted by the localized electron states – say, during the propagation through
the interstellar medium – may also be an interesting problem because the cross
section of elastic scattering of twisted photons on charged particles usually
has a maximum at much larger angles than it is for the plane-wave photons.

We are thankful to S. Baturin, A. Chaikovskaia, D. Glazov, D. Grosman, G.
Sizykh, A. Volotka, and, especially, to A. Di Piazza for useful discussions
and criticism. The studies in Sec. II are supported by the Government of the
Russian Federation through the ITMO Fellowship and Professorship Program. The
studies in Sec. III are supported by the Russian Science Foundation (Project
No. 21-42-04412; https://rscf.ru/en/project/21-42-04412/). The work on the
evolved state in Sec.IV (by D. Karlovets) was supported by the Foundation for
the Advancement of Theoretical Physics and Mathematics “BASIS”. The studies in
Sec.V are supported by the Russian Science Foundation (Project No.
23-62-10026; https://rscf.ru/en/project/23-62-10026/).

##  Appendix A Appendix: Electron states in magnetic field

###  A.1 Solution of the Dirac equation

In this section, we describe a simple method of obtaining a solution of Dirac
equation in a constant homogeneous magnetic field utilizing an approach of
[26]. Slightly different realization of the procedure described below was
lately proposed in [34].

First, let us define the Klein-Gordon operator:

$\hat{K}\equiv(\hat{p}-e\mathcal{A})^{2}-m^{2}=(i\partial^{\mu}-e\mathcal{A}^{\mu})^{2}-m^{2}=-\square-2ie\mathcal{A}_{\mu}\partial^{\mu}+e^{2}\mathcal{A}^{2}-m^{2}.$
(35)

The Klein-Gordon equation then simply reads as

$\hat{K}\Phi(x)=0.$ (36)

We consider constant magnetic field $\bm{H}=\\{0,0,H\\}$, which can be
described by a vector-potential in two different gauges – the Landau gauge and
the symmetric one – resulting in the Hermite-Gaussian beams and in the axially
symmetric Laguerre-Gaussian packets, respectively. We choose the four-vector
potential in the later symmetric gauge:

$\displaystyle\mathcal{A}^{\mu}=\\{0,\bm{A}\\};$ (37)
$\displaystyle\bm{A}=\frac{H\rho}{2}\bm{e}_{\varphi}=\frac{H}{2}\\{-y,x,0\\}.$

Then the Klein-Gordon in cylindrical coordinates takes the following form:

$\left[-\partial_{t}^{2}+\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial}{\partial\rho}\right)+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\varphi^{2}}+\partial_{z}^{2}-ieH\partial_{\varphi}-e^{2}\frac{H^{2}\rho^{2}}{4}-m^{2}\right]\Phi(x)=0.$
(38)

By taking the following ansatz for the solution

$\Phi(x)\equiv\Phi(\rho)e^{-i\varepsilon t+i\ell\varphi+ip_{z}z}$ (39)

we obtain the equation for the transverse part of the wave function:

$\left[\partial_{\rho}^{2}+\frac{1}{\rho}\partial_{\rho}-\frac{\ell^{2}}{\rho^{2}}+e\ell
H-\left(\frac{eH\rho}{2}\right)^{2}+\varepsilon^{2}-p_{z}^{2}-m^{2}\right]\Phi(\rho)=0.$
(40)

Following [10], the solution of (6) can be written as

$\Phi(\rho)=\mathcal{N}\left(\frac{2\rho^{2}}{\rho_{H}^{2}}\right)^{\ell/2}L_{s}^{\ell}(2\rho^{2}/\rho_{H}^{2})e^{-\rho^{2}/\rho_{H}^{2}},$
(41)

where $\mathcal{N}$ is a normalization constant,
$\rho_{H}=\sqrt{\frac{4}{\absolutevalue{e}H}}$, $L_{s}^{\ell}$ are the
associated Laguerre polynomials and the energy dispersion is

$\varepsilon^{2}_{s,\ell}-m^{2}-p_{z}^{2}\equiv
p_{\perp}^{2}=2\absolutevalue{e}H(s+\ell+1/2).$ (42)

Let us define the transverse function without the prefactor as

$\Phi_{s,\ell}(\rho)\equiv\rho^{\ell}L_{s}^{\ell}(2\rho^{2}/\rho_{H}^{2})e^{-\rho^{2}/\rho_{H}^{2}}.$
(43)

Second, we consider the Dirac equation

$\left[\gamma^{\mu}(\hat{p}_{\mu}-eA_{\mu})-m\right]\Psi(x)=0,$ (44)

where $\gamma^{\mu}$ are the gamma matrices in the standard representation.
Instead of solving the first-order equation (44) directly, one can first
transform it into a second-order equation by applying the “projection”
operator $\gamma^{\mu}(\hat{p}_{\mu}-eA_{\mu})+m$ [26]:

$\left[(\hat{p}-eA)^{2}-m^{2}+e\bm{\Sigma}\cdot\bm{H}-ie\bm{\alpha}\cdot\bm{E}\right]\Phi(x)=0.$
(45)

Here matrices $\bm{\alpha}$ and $\bm{\Sigma}$ are defined as follows:

$\bm{\alpha}=\begin{pmatrix}0&\bm{\sigma}\\\ \bm{\sigma}&0\\\
\end{pmatrix},\;\bm{\Sigma}=\begin{pmatrix}\bm{\sigma}&0\\\ 0&\bm{\sigma}\\\
\end{pmatrix},$ (46)

where $\bm{\sigma}$ are the Pauli matrices. The third and the fourth terms
here arise from the products of gamma-matrices and the electromagnetic tensor.
The solution of the first-order equation $\Psi(x)$ is then obtained by
applying the projection operator to $\Phi(x)$:

$\Psi(x)=\left[\gamma^{\mu}(\hat{p}_{\mu}-eA_{\mu})+m\right]\Phi(x)$ (47)

Substituting the expression for the magnetic field to (45) we get

$\left[(\hat{p}-eA)^{2}-m^{2}+eH\begin{pmatrix}1&0&0&0\\\ 0&-1&0&0\\\
0&0&1&0\\\ 0&0&0&-1\end{pmatrix}\right]\Phi(x)=0.$ (48)

This is a system of four Klein-Gordon-like equations

$\left[\hat{K}\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{pmatrix}+eH\begin{pmatrix}1&0&0&0\\\ 0&-1&0&0\\\ 0&0&1&0\\\
0&0&0&-1\end{pmatrix}\right]\Phi(x)=0,$ (49)

that leads to two independent equations

$\begin{cases}(\hat{K}+eH)\Phi_{+}^{KG}=0\\\
(\hat{K}-eH)\Phi_{-}^{KG}=0.\end{cases}$ (50)

Then the solution of Eq. (49) is

$\Phi(x)\equiv\begin{pmatrix}c_{1}\Phi_{+}^{KG}\\\ c_{2}\Phi_{-}^{KG}\\\
c_{3}\Phi_{+}^{KG}\\\ c_{4}\Phi_{-}^{KG}\end{pmatrix}.$ (51)

where $c_{i}$ are the arbitrary complex-valued constants. A naive choice of
$\Phi_{\pm}^{KG}$ could be

$\Phi_{\pm}^{KG}=\Phi_{s,l}(\rho)e^{-it\varepsilon_{s,l\mp
1/2}+il\varphi+ip_{z}z},$ (52)

However, such a solution is not physically relevant because $\Phi_{+}^{KG}$
and $\Phi_{-}^{KG}$ should correspond to the same energy. To satisfy this
condition, one should choose $\Phi_{\pm}^{KG}$ in another form:

$\Phi_{\pm}^{KG}=\Phi_{s,\ell\mp
1/2}(\rho)e^{-it\varepsilon_{s,\ell}+i(\ell\mp 1/2)\varphi+ip_{z}z}$ (53)

(here $\ell$ is a half-integer).

First, we consider the following two “auxillary” solutions of (49):

$\Phi_{s,\ell}^{\uparrow}(x)=N^{\uparrow}\Phi_{s,\ell-1/2}(\rho)e^{-it\varepsilon_{s,\ell}+i(\ell-1/2)\varphi+ip_{z}z}\begin{pmatrix}1\\\
0\\\ 0\\\ 0\end{pmatrix}$ (54)

and

$\Phi_{s,\ell}^{\downarrow}(x)=N^{\downarrow}\Phi_{s,\ell+1/2}(\rho)e^{-it\varepsilon_{s,\ell}+i(\ell+1/2)\varphi+ip_{z}z}\begin{pmatrix}0\\\
1\\\ 0\\\ 0\end{pmatrix}$ (55)

(the notation $\uparrow$ and $\downarrow$ will become clear later on). These
are clearly the eigenfunctions of the operators
$\frac{1}{2}\widehat{\Sigma}_{z}$, $\hat{l}_{z}=-i\partial_{\varphi}$ and
$\hat{j}_{z}=\frac{1}{2}\widehat{\Sigma}_{z}+\hat{l}_{z}$, but not yet the
solutions of the Dirac equation (44). By acting with the projection operator
on Eq. (54) and using the relations

$\displaystyle
xL_{s}^{\ell}(x)=(s+\ell)L_{s}^{\ell-1}(x)-(s+1)L_{s+1}^{\ell-1}(x);$ (56)
$\displaystyle L_{s}^{\ell}(x)=L_{s}^{\ell+1}(x)-L_{s-1}^{\ell+1}(x);$
$\displaystyle\frac{d}{dx}L_{s}^{\ell}(x)=-L_{s-1}^{\ell+1}(x),$

we obtain the following solution of the Dirac equation:

$\displaystyle\Psi_{s,\ell}^{\uparrow}(x)=N^{\uparrow}\begin{pmatrix}(m+\varepsilon)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
0\\\ p_{z}\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
-ieH\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon_{s,\ell}+i\ell\varphi+ip_{z}z};$
(57)
$\displaystyle\varepsilon_{s,l}=\sqrt{m^{2}+p_{z}^{2}+p_{\perp}^{2}},\>p_{\perp}^{2}=\frac{8}{\rho_{H}^{2}}(s+\ell+1/2)=2m^{2}\frac{H}{H_{c}}(s+\ell+1/2).$

One can similarly consider second auxiliary solution: after applying the
projector to Eq. (55) we find

$\Psi_{s,\ell}^{\downarrow}(x)=N^{\downarrow}\begin{pmatrix}0\\\
(m+\varepsilon)\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\\\
-2i(\ell+s+1/2)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
-p_{z}\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon_{s,\ell}+i\ell\varphi+ip_{z}z}.$
(58)

Notably, the energy $\varepsilon_{s,\ell}$ here is the same as that in Eq.
(LABEL:Psi_Plus): it is determined by the TAM projection $\ell$ rather than by
the spin or the orbital momentum separately. These states are exactly the
solutions obtained in [10] with another method and coincide with those
described in [34] when the TAM projection $\ell$ is positive. We employ these
states for all the calculations in the main text of the paper. Let us
normalize the states as follows:

$\int
d^{3}x\>j^{0}(x)=\int_{0}^{2\pi}d\varphi\int_{-L/2}^{L/2}dz\int_{0}^{\infty}\rho
d\rho\Psi_{s,\ell}^{\uparrow}(x)^{\dagger}\Psi_{s,\ell}^{\uparrow}(x)=1.$ (59)

Note that according to Eq. (7.414) in [35]

$\int_{0}^{\infty}\Phi_{s,\ell}^{2}(\rho)\rho
d\rho=\frac{\rho_{H}^{2\ell+2}}{2^{\ell+2}}\frac{(s+\ell)!}{s!}.$ (60)

Then the normalization constants are

$N^{\uparrow}=\sqrt{\frac{2^{\ell-1/2}s!}{L\pi(s+\ell-1/2)!}}\frac{\rho_{H}^{-\ell-1/2}}{\sqrt{\varepsilon(\varepsilon+m)}},$
(61)
$N^{\downarrow}=\sqrt{\frac{2^{\ell+1/2}s!}{L\pi(s+\ell+1/2)!}}\frac{\rho_{H}^{-\ell-3/2}}{\sqrt{\varepsilon(\varepsilon+m)}}.$
(62)

One can easily check that the states $\Psi_{s,\ell}^{\uparrow}$ and
$\Psi_{s^{\prime},\ell^{\prime}}^{\downarrow}$ are orthogonal in the sense of
the integral

$\int_{0}^{\infty}\rho
d\rho\Psi_{s,\ell}^{\uparrow}(x)^{\dagger}\Psi_{s^{\prime},\ell^{\prime}}^{\downarrow}(x)=0$
(63)

even for $s=s^{\prime}$ and $\ell=\ell^{\prime}$. The states of the same type
with the different quantum numbers $s$, $\ell$ and $p_{z}$ are also
orthogonal:

$\int_{0}^{2\pi}d\varphi\int_{-\infty}^{\infty}dz\int_{0}^{\infty}\rho
d\rho\Psi_{s,\ell}^{\uparrow,\downarrow}(x)^{\dagger}\Psi_{s^{\prime},\ell^{\prime}}^{\uparrow,\downarrow}(x)\propto\delta_{ss^{\prime}}\delta_{\ell\ell^{\prime}}\delta(p_{z}-p_{z}^{\prime}).$
(64)

Importantly, $\Psi_{s,\ell}^{\uparrow,\downarrow}$ are the eigenfunctions of
z-projection of the total angular momentum (TAM) operator

$\displaystyle\hat{j}_{z}\Psi_{s,\ell}^{\uparrow,\downarrow}=\left(\hat{\ell}_{z}+\frac{1}{2}\widehat{\Sigma}_{z}\right)\Psi_{s,\ell}^{\uparrow,\downarrow}=\ell\Psi_{s,\ell}^{\uparrow,\downarrow};$
(65) $\displaystyle\widehat{\Sigma}_{z}=\begin{pmatrix}\sigma_{z}&0\\\
0&\sigma_{z}\\\ \end{pmatrix},$

but clearly not the eigenfunctions of $\hat{\ell}_{z}$ and
$\widehat{\Sigma}_{z}$ separately.

###  A.2 Spin properties of the electron

In a longitudinal magnetic field the spin projection onto the field is not
conserved in the sense that the Pauli spin operator
$\hat{S}=\frac{1}{2}\hat{\bm{\Sigma}}$ does not commute with the Dirac
Hamiltonian. Many attempts have been made to introduce other spin operators
that commute with the Hamiltonian and adequately describe the polarization of
the particle [10, 29, 31]. Since the TAM operator
$\hat{\bm{j}}=\hat{\bm{r}}\times\hat{\bm{p}}+\frac{1}{2}\hat{\bm{\Sigma}}$ is
well-defined for the Dirac equation and undisputed, different definitions of
the spin operator imply different position operators, and hence different
orbital angular momentum operators [29, 30]. Thus, any specific choice of the
spin operator is equivalent to the splitting of the TAM into spin and orbital
parts, which is ambiguous.

As the investigation of the polarization phenomena, such as well-studied
Sokolov-Ternov effect [10, 12], is not the purpose of this work, we do not
follow the approach mentioned above. Alternatively, we claim that two types of
states obtained in the previous section are sufficient to describe the states
of the relativistic electron in magnetic field and derive the evolved state of
the emitted photon because it is the total angular momentum projection $j_{z}$
of the electron that is observable in this problem and not $S_{z}$ alone.

Let us, however, evaluate the mean values of the spin operator $\hat{S}_{z}$
for the above states $\Psi_{s,\ell}^{\uparrow}$ and
$\Psi_{s,\ell}^{\downarrow}$:

$S_{z}^{\uparrow,\downarrow}\equiv\frac{1}{2}\int
d^{3}\bm{r}\>\left(\Psi_{s,\ell}^{\uparrow,\downarrow}{}\right)^{\dagger}\hat{\Sigma}_{z}\Psi_{s,\ell}^{\uparrow,\downarrow}=\pm\frac{1}{2}\left(1-\frac{p_{\perp}^{2}}{\varepsilon(\varepsilon+m)}\right).$
(66)

Importantly, when the transverse momentum $p_{\perp}$ becomes non-
relativistic, $p_{\perp}\ll m$, in the limit $H\ll H_{c}$, the spin
projections tend to the values $\pm 1/2$. For this reason, we refer to the
state $\Psi_{s,\ell}^{\uparrow}$ as “spin-up” and to
$\Psi_{s,\ell}^{\downarrow}$ as “spin-down”.

###  A.3 Alternative choice of the solutions

The bispinors (54) and (55) are simple and intuitive, yet certainly not the
only possible solutions of the Dirac equation (44). One can generally choose
arbitrary constants $c_{i}$ in Eq. (51) (see the discussion in [10]) but this
arbitrariness does not affect the main property of the emitted photon: it
represents a twisted Bessel beam with the TAM projection transferred from the
initial electron, regardless of the values of $c_{i}$. The constants in Eq.
(51), however, govern the spin properties of the states and the transition
probabilities. Let us now consider another pair of “auxillary” solutions of
(49) as, for instance,

$\tilde{\Phi}_{s,\ell}^{\uparrow}(x)=\tilde{N}^{\uparrow}\Phi_{s,\ell-1/2}(\rho)e^{-it\varepsilon_{s,\ell}+i(\ell-1/2)\varphi+ip_{z}z}\begin{pmatrix}0\\\
0\\\ 1\\\ 0\end{pmatrix};$ (67)
$\tilde{\Phi}_{s,\ell}^{\downarrow}(x)=\tilde{N}^{\downarrow}\Phi_{s,\ell+1/2}(\rho)e^{-it\varepsilon_{s,\ell}+i(\ell+1/2)\varphi+ip_{z}z}\begin{pmatrix}0\\\
0\\\ 0\\\ 1\end{pmatrix}.$ (68)

The corresponding “projected” solutions of the Dirac equation are

$\tilde{\Psi}_{s,\ell}^{\uparrow}(x)=\tilde{N}^{\uparrow}\begin{pmatrix}-p_{z}\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
ieH\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\\\
(m-\varepsilon)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
0\end{pmatrix}e^{-it\varepsilon_{s,\ell}+i\ell\varphi+ip_{z}z};$ (69)
$\tilde{\Psi}_{s,\ell}^{\downarrow}(x)=\tilde{N}^{\downarrow}\begin{pmatrix}2i(\ell+s+1/2)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
p_{z}\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\\\ 0\\\
(m-\varepsilon)\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon_{s,\ell}+i\ell\varphi+ip_{z}z}.$
(70)

The mean value of the spin projection is (compare with Eq.(66))

$\tilde{S}_{z}^{\uparrow,\downarrow}=\frac{1}{2}\int
d^{3}\bm{r}\>\left(\tilde{\Psi}_{s,\ell}^{\uparrow,\downarrow}\right)^{\dagger}\hat{\Sigma}_{z}\tilde{\Psi}_{s,\ell}^{\uparrow,\downarrow}=\pm\frac{1}{2}\left(1-\frac{p_{\perp}^{2}}{\varepsilon(\varepsilon-m)}\right).$
(71)

For these states the orthogonality relations (63) and (64) are also satisfied
and still
$\hat{j}_{z}\tilde{\Psi}_{s,\ell}^{\uparrow,\downarrow}=\ell\tilde{\Psi}_{s,\ell}^{\uparrow,\downarrow}$.
Although the probabilities and intensities shown in Fig. 6 differ from those
in Fig. 4 — in the left figure the ratio of “up-down” transition probability
to the “down-up” one is two orders of magnitude lower — the self-polarization
effect is present and qualitatively the picture is the same. Another
comparison of the emission probabilities calculated for different polarization
states is presented in Fig. 7, demonstrating their qualitative agreement. This
fact alongside with the orthogonality does not allow us to say that any of the
two sets of states – either
$\\{\Psi_{s,\ell}^{\uparrow},\Psi_{s,\ell}^{\downarrow}\\}$ or
$\\{\tilde{\Psi}_{s,\ell}^{\uparrow},\tilde{\Psi}_{s,\ell}^{\downarrow}\\}$ –
is more preferable for describing an electron in the magnetic field without
discussing the spin detector properties. We stress, however, that these spin
subtleties only affect the differential probability but not the total one and
not the evolved state of the emitted photon, which is defined by the
difference of the electron total angular momenta, $\ell-\ell^{\prime}$.

Figure 6: The ratio of four possible types of transition probabilities to the
probability of emission by a scalar charge for the alternative choice of
“spin-up” and “spin-down” states given by Eqs. (69) and (70). $H=10^{-3}H_{c}$
(left) and $H=H_{c}$ (right); $\ell-\ell^{\prime}=3$ for all transitions. The
same as in Fig. 4, but for the alternative choice of “spin-up” and “spin-down”
states given by Eqs. (69) and (70). Figure 7:  Comparison of the emission
probability (left) and the corresponding intensity (right) for “up-up”
transitions calculated for the “spin-up” states (54) (solid lines) and for the
alternative states (67) (dashed lines). Parameters: $H=0.01H_{c}$,
$p_{z}=10^{-3}mc$ and $s=s^{\prime}=20$ unless stated otherwise.

Let us also discuss the auxiliary solutions which represent the sums of
$\Phi_{s,\ell}^{\uparrow,\downarrow}$ and
$\tilde{\Phi}_{s,\ell}^{\uparrow,\downarrow}$:

$\phi_{s,\ell}^{\uparrow}(x)=n^{\uparrow}\Phi_{s,\ell-1/2}(\rho)e^{-it\varepsilon_{s,\ell}+i(\ell-1/2)\varphi+ip_{z}z}\begin{pmatrix}1\\\
0\\\ 1\\\ 0\end{pmatrix};$ (72)
$\phi_{s,\ell}^{\downarrow}(x)=n^{\downarrow}\Phi_{s,\ell+1/2}(\rho)e^{-it\varepsilon_{s,\ell}+i(\ell+1/2)\varphi+ip_{z}z}\begin{pmatrix}0\\\
1\\\ 0\\\ 1\end{pmatrix}.$ (73)

The corresponding “projected” solutions of the Dirac equation are

$\psi_{s,\ell}^{\uparrow}(x)=n^{\uparrow}\begin{pmatrix}(m-p_{z}+\varepsilon)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
ieH\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\\\
(m+p_{z}-\varepsilon)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
-ieH\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon_{s,\ell}+i\ell\varphi+ip_{z}z};$
(74)
$\psi_{s,\ell}^{\downarrow}(x)=n^{\downarrow}\begin{pmatrix}2i(\ell+s+1/2)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
(m+p_{z}+\varepsilon)\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\\\
-2i(\ell+s+1/2)\Phi_{s,\ell-1/2}(\rho)e^{-i\varphi/2}\\\
(m-p_{z}-\varepsilon)\Phi_{s,\ell+1/2}(\rho)e^{i\varphi/2}\end{pmatrix}e^{-it\varepsilon_{s,\ell}+i\ell\varphi+ip_{z}z}.$
(75)

These states, however, have certain shortcomings. First of all,
$\psi_{s,\ell}^{\uparrow}$ and $\psi_{s,\ell}^{\downarrow}$ are not
orthogonal:

$\int_{0}^{\infty}\rho
d\rho\>\psi_{s,\ell}^{\uparrow}(x)^{\dagger}\psi_{s,\ell}^{\downarrow}(x)\neq
0,$ (76)

which does not allow to speak about two distinct polarization states, yet the
orthogonality with respect to $s$, $\ell$ and $p_{z}$ (See Eq. (64)) still
holds. Second, the dependence of the mean values of $\hat{s}_{z}$ on the sign
of $p_{z}$ arises:

$s_{z}^{\uparrow,\downarrow}=\frac{1}{2}\int
d^{3}\bm{r}\>\left(\psi_{s,\ell}^{\uparrow,\downarrow}\right)^{\dagger}\hat{\Sigma}_{z}\psi_{s,\ell}^{\uparrow,\downarrow}=\pm\frac{1}{2}\left(1-\frac{p_{\perp}^{2}}{\varepsilon(\varepsilon\mp
p_{z})}\right).$ (77)

If $p_{z}\gg\varepsilon$ or $p_{z}\ll\varepsilon$, Eq. (77) enables the
effective spin flip, meaning that $s_{z}^{\uparrow}\to-\frac{1}{2}$ and
$s_{z}^{\downarrow}\to\frac{1}{2}$, respectively, which means that the sign of
$s_{z}$ is frame dependent and, thus, $s_{z}$ itself is not a good quantum
number.

Figure 8: The same as in Figs. 4 and 6, but for the “spin-up” state (74) and
the “spin-down” state (75): ratio of four possible types of the transition
probabilities to the probability of emission by a scalar charge.
$H=10^{-3}H_{c}$ (left) and $H=H_{c}$ (right); $\ell-\ell^{\prime}=3$,
$s=s^{\prime}=20$ for all transitions. In contrast to Figs. 4 and 6, the scale
here is not logarithmic.

Finally, although the mean spin projections (77) tend to $\pm 1/2$ for the
weak field, these states do not predict the self-polarization effect even when
$H\ll H_{c}$. Fig. 8 demonstrates that the probabilities of “up-down” and
“down-up” transitions are indistinguishable both in critical and subcritical
magnetic field. Moreover, in a critical field the transition probabilities of
all four types are close to those for a “scalar electron”, which confirms that
any effects associated with the spin alone are not adequately described with
these states. It is also noteworthy that according to Figs. 4, 6 and 8, the
probabilities of no spin-flip transitions are of the order of the probability
for a scalar particle regardless of the spin states basis and the field
strength.

##  Appendix B Appendix: Transition currents and amplitudes

The transition current components for a transition between two “spin-up”
states (12) and (13) are

$j_{\uparrow\uparrow}^{0}(x)=N_{i}^{\uparrow}N_{f}^{\uparrow}\rho^{\ell+\ell^{\prime}-1}e^{-2\rho^{2}/\rho_{H}^{2}+i(p_{z}-p_{z}^{\prime})z+i(\ell-\ell^{\prime})\varphi-i(\varepsilon-\varepsilon^{\prime})t}\times\\\
\times\Bigg{[}\left(m(m+\varepsilon+\varepsilon^{\prime})+p_{z}p_{z}^{\prime}+\varepsilon\varepsilon^{\prime}\right)L_{s}^{\ell-1/2}(2\rho^{2}/\rho_{H}^{2})L_{s^{\prime}}^{\ell^{\prime}-1/2}(2\rho^{2}/\rho_{H}^{2})+4\rho^{2}/\rho_{H}^{2}L_{s}^{\ell+1/2}(2\rho^{2}/\rho_{H}^{2})L_{s^{\prime}}^{\ell^{\prime}+1/2}(2\rho^{2}/\rho_{H}^{2})\Bigg{]};$
(78)

$j_{\uparrow\uparrow}^{1}(x)=-ieHN_{i}^{\uparrow}N_{f}^{\uparrow}\rho^{\ell+\ell^{\prime}}e^{-2\rho^{2}/\rho_{H}^{2}+i(p_{z}-p_{z}^{\prime})z+i(\ell-\ell^{\prime})\varphi-i(\varepsilon-\varepsilon^{\prime})t}\times\\\
\times\Bigg{[}e^{i\varphi}\left(m+\varepsilon^{\prime}\right)L_{s}^{\ell+1/2}(2\rho^{2}/\rho_{H}^{2})L_{s^{\prime}}^{\ell^{\prime}-1/2}(2\rho^{2}/\rho_{H}^{2})-e^{-i\varphi}(m+\varepsilon)L_{s}^{\ell-1/2}(2\rho^{2}/\rho_{H}^{2})L_{s^{\prime}}^{\ell^{\prime}+1/2}(2\rho^{2}/\rho_{H}^{2})\Bigg{]};$
(79)

$j_{\uparrow\uparrow}^{2}(x)=-eHN_{i}^{\uparrow}N_{f}^{\uparrow}\rho^{\ell+\ell^{\prime}}e^{-2\rho^{2}/\rho_{H}^{2}+i(p_{z}-p_{z}^{\prime})z+i(\ell-\ell^{\prime})\varphi-i(\varepsilon-\varepsilon^{\prime})t}\times\\\
\times\Bigg{[}e^{i\varphi}\left(m+\varepsilon^{\prime}\right)L_{s}^{\ell+1/2}(2\rho^{2}/\rho_{H}^{2})L_{s^{\prime}}^{\ell^{\prime}-1/2}(2\rho^{2}/\rho_{H}^{2})+e^{-i\varphi}(m+\varepsilon)L_{s}^{\ell-1/2}(2\rho^{2}/\rho_{H}^{2})L_{s^{\prime}}^{\ell^{\prime}+1/2}(2\rho^{2}/\rho_{H}^{2})\Bigg{]};$
(80)

$j_{\uparrow\uparrow}^{3}(x)=2N_{i}^{\uparrow}N_{f}^{\uparrow}\rho^{\ell+\ell^{\prime}-1}e^{-2\rho^{2}/\rho_{H}^{2}+i(p_{z}-p_{z}^{\prime})z+i(\ell-\ell^{\prime})\varphi-i(\varepsilon-\varepsilon^{\prime})t}\times\\\
\times\big{(}m(p_{z}+p_{z}^{\prime})+p_{z}^{\prime}\varepsilon+p_{z}\varepsilon^{\prime}\big{)}L_{s}^{\ell-1/2}(2\rho^{2}/\rho_{H}^{2})L_{s^{\prime}}^{\ell^{\prime}-1/2}(2\rho^{2}/\rho_{H}^{2}).$
(81)

To calculate the S-matrix amplitude we first take the integrals over $t$, $z$
and $\varphi$ in Eq. (5) and leave the radial part unevaluated. The azimuthal
integral can be taken by employing the integral representation of the Bessel
functions

$\int_{0}^{2\pi}\frac{d\varphi}{2\pi}e^{i\ell\varphi+ix\cos\varphi}=i^{\ell}J_{\ell}(x).$
(82)

Thus, we arrive at the following expression

$S_{\uparrow\uparrow}^{(1)}=\int_{0}^{\infty}\tilde{S}_{\uparrow\uparrow}^{(1)}d\rho,$
(83)

where the integrand is

$\tilde{S}_{\uparrow\uparrow}^{(1)}=ei^{-\ell+\ell^{\prime}+1}(2\pi)^{3}\frac{N_{i}^{\uparrow}N_{f}^{\uparrow}}{\sqrt{2\omega
V}}\delta(\omega+\varepsilon^{\prime}-\varepsilon)\delta(k_{z}+p_{z}^{\prime}-p_{z})e^{i(\ell-\ell^{\prime})\varphi_{k}-2\tilde{\rho}^{2}}\times\\\
\times\Bigg{[}\big{(}m(p_{z}+p_{z}^{\prime})+p_{z}^{\prime}\varepsilon+p_{z}\varepsilon^{\prime}\big{)}d_{0\lambda}^{(1)}(\theta)\tilde{\rho}^{\ell+\ell^{\prime}}\rho_{H}^{\ell+\ell^{\prime}}J_{\ell-\ell^{\prime}}(y\tilde{\rho})L_{s}^{\ell-1/2}(2\tilde{\rho}^{2})L_{s^{\prime}}^{\ell^{\prime}-1/2}(2\tilde{\rho}^{2})-\\\
\sqrt{2}eH(m+\varepsilon^{\prime})d_{-1\lambda}^{(1)}(\theta)\tilde{\rho}^{\ell+\ell^{\prime}+1}\rho_{H}^{\ell+\ell^{\prime}+1}J_{\ell-\ell^{\prime}+1}(y\tilde{\rho})L_{s}^{\ell+1/2}(2\tilde{\rho}^{2})L_{s^{\prime}}^{\ell^{\prime}-1/2}(2\tilde{\rho}^{2})+\\\
\sqrt{2}eH(m+\varepsilon)d_{1\lambda}^{(1)}(\theta)\tilde{\rho}^{\ell+\ell^{\prime}+1}\rho_{H}^{\ell+\ell^{\prime}+1}J_{\ell-\ell^{\prime}-1}(y\tilde{\rho})L_{s}^{\ell-1/2}(2\tilde{\rho}^{2})L_{s^{\prime}}^{\ell^{\prime}+1/2}(2\tilde{\rho}^{2})\Bigg{]}.$
(84)

Here we have denoted $\tilde{\rho}\equiv\rho/\rho_{H}$ and $y\equiv
k_{\perp}\rho_{H}$. Then we use the following integral from [35] (there is,
however, a misprint $m\leftrightarrow n$)

$\int\limits_{0}^{\infty}dx\,x^{\ell+\ell^{\prime}+1}\,L_{s}^{\ell}(2x^{2})L_{s^{\prime}}^{\ell^{\prime}}(2x^{2})\,J_{\ell-\ell^{\prime}}(yx)\,e^{-2x^{2}}=\\\
\frac{(s^{\prime}+\ell^{\prime})!}{s!}\frac{1}{2^{3(s-s^{\prime})+2\ell-\ell^{\prime}+2}}\,y^{2(s-s^{\prime})+\ell-\ell^{\prime}}\,L_{s^{\prime}+\ell^{\prime}}^{s-s^{\prime}+\ell-\ell^{\prime}}\left(y^{2}/8\right)L_{s^{\prime}}^{s-s^{\prime}}\left(y^{2}/8\right)\,e^{-y^{2}/8}\equiv
F^{\ell,\ell^{\prime}}_{s,s^{\prime}}(y)$ (85)

to obtain the expression (15).

The S-matrix elements for three other types of transitions are

$S_{\uparrow\downarrow}^{(1)}=ei^{-\ell+\ell^{\prime}}(2\pi)^{3}\frac{N_{i}^{\uparrow}N_{f}^{\downarrow}}{\sqrt{2\omega
V}}\delta(\omega+\varepsilon^{\prime}-\varepsilon)\delta(k_{z}+p_{z}^{\prime}-p_{z})e^{i(\ell-\ell^{\prime})\varphi_{k}}\times\\\
\times\Bigg{[}d_{0\lambda}^{(1)}(\theta)\rho_{H}^{\ell+\ell^{\prime}+1}\left[-\left(m+\varepsilon\right)\left(1+2s^{\prime}+2\ell^{\prime}\right)F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}-1/2}(y)+4(m+\varepsilon^{\prime})F_{s,s^{\prime}}^{\ell+1/2,\ell^{\prime}+1/2}(y)\right]-\\\
-\sqrt{2}\rho_{H}^{\ell+\ell^{\prime}+2}\big{(}m(p_{z}^{\prime}-p_{z})+p_{z}^{\prime}\varepsilon-
p_{z}\varepsilon^{\prime}\big{)}d_{1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}+1/2}(y)\Bigg{]};$
(86)

$S_{\downarrow\uparrow}^{(1)}=ei^{-\ell+\ell^{\prime}}(2\pi)^{3}\frac{N_{i}^{\downarrow}N_{f}^{\uparrow}}{\sqrt{2\omega
V}}\delta(\omega+\varepsilon^{\prime}-\varepsilon)\delta(k_{z}+p_{z}^{\prime}-p_{z})e^{i(\ell-\ell^{\prime})\varphi_{k}}\times\\\
\times\Bigg{[}d_{0\lambda}^{(1)}(\theta)\rho_{H}^{\ell+\ell^{\prime}+1}\left[-\left(m+\varepsilon^{\prime}\right)\left(1+2s+2\ell\right)F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}-1/2}(y)+4(m+\varepsilon)F_{s,s^{\prime}}^{\ell+1/2,\ell^{\prime}+1/2}(y)\right]-\\\
-\sqrt{2}\rho_{H}^{\ell+\ell^{\prime}+2}\big{(}m(p_{z}^{\prime}-p_{z})+\varepsilon
p_{z}^{\prime}-\varepsilon^{\prime}p_{z}\big{)}d_{-1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell+1/2,\ell^{\prime}-1/2}(y)\Bigg{]};$
(87)

$S_{\downarrow\downarrow}^{(1)}=ei^{-\ell+\ell^{\prime}+1}(2\pi)^{3}\frac{N_{i}^{\downarrow}N_{f}^{\downarrow}}{\sqrt{2\omega
V}}\delta(\omega+\varepsilon^{\prime}-\varepsilon)\delta(k_{z}+p_{z}^{\prime}-p_{z})e^{i(\ell-\ell^{\prime})\varphi_{k}}\times\\\
\times\Bigg{[}d_{0\lambda}^{(1)}(\theta)\rho_{H}^{\ell+\ell^{\prime}+3}\big{(}m(p_{z}+p_{z}^{\prime})+p_{z}^{\prime}\varepsilon+p_{z}\varepsilon^{\prime}\big{)}F_{s,s^{\prime}}^{\ell+1/2,\ell^{\prime}+1/2}(y)-\\\
\sqrt{2}\rho_{H}^{\ell+\ell^{\prime}+2}\left[\left(1+2s^{\prime}+2\ell^{\prime}\right)(m+\varepsilon)d_{-1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell+1/2,\ell^{\prime}-1/2}(y)-\left(1+2s+2\ell\right)(m+\varepsilon^{\prime})d_{1\lambda}^{(1)}(\theta)F_{s,s^{\prime}}^{\ell-1/2,\ell^{\prime}+1/2}(y)\right]\Bigg{]}.$
(88)

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