$\displaystyle U_{B_{1g}}$ $\displaystyle=u-v+3w,$ $\displaystyle u_{B_{1g}}$
$\displaystyle=u+v+w,$ $\displaystyle U_{B_{2g}}$ $\displaystyle=u-v-w,$
$\displaystyle u_{B_{2g}}$ $\displaystyle=u+v-w,$ (125)

repeated in Eq. (56) of the main text. Combining Eqs. (105), (116), and (124)
then gives the variational free energy (54) of the main text, which we repeat
here for convenience:

$\displaystyle f_{v}$
$\displaystyle=\\!\frac{T}{2V}\\!\\!\sum_{k}\\!\mathrm{tr}\log\left(\mathcal{G}_{k}^{-1}\right)+2\left[r_{0}\\!-\\!R_{0}\\!+\\!U_{A_{1g}}\Pi^{A_{1g}}\right]\\!\Pi^{A_{1g}}$
$\displaystyle\quad-\\!2\\!\\!\sum_{n\in\mathbb{G}_{\mathbb{R}}}\left[\Phi^{n}\\!-\\!U_{n}\Pi^{n}\right]\Pi^{n}-\\!\\!\\!\sum_{n\in\mathbb{G}_{\mathbb{C}}}\\!\\!\\!\left[\left(\phi^{n}\\!-\\!u_{n}\pi^{n}\right)\bar{\pi}^{n}\\!+\mathrm{c.c.}\right].$
(126)

Now, let us briefly prove the relation (57) of the main text,

$\displaystyle\frac{\partial f_{v}}{\partial X_{i}}\Big{|}_{\Pi^{n},\pi^{n}}$
$\displaystyle=0\,,$ (127)

stating that the partial derivative of the variational free energy (126) with
respect to any of its variational parameters
$X_{i}\in\left\\{R_{0},\Phi^{n},\phi^{n},\bar{\phi}^{n}\right\\}$ vanishes if
$\Pi^{n}$ and $\pi^{n}$ are kept constant. The relation (127) can directly be
read off using the derivative

$\frac{\partial}{\partial
X_{i}}\frac{T}{2V}\\!\\!\sum_{k}\\!\mathrm{tr}\log\left(\mathcal{G}_{k}^{-1}\right)=\begin{cases}2\Pi^{A_{1g}}&,X_{i}=R_{0}\\\
2\Pi^{n}&,X_{i}\in\left\\{\Phi^{n}\right\\}\\\
\bar{\pi}^{n}&,X_{i}\in\left\\{\phi^{n}\right\\}\\\
\pi^{n}&,X_{i}\in\left\\{\bar{\phi}^{n}\right\\}\end{cases}\,.$ (128)

Therefore, we obtain Eq. (127), and the minimization of the variational free
energy follows the steps explained in Sec. V.2.

We finish this Appendix by demonstrating that the expectation value of a
bilinear reduces to its expectation value with respect to the trial action
within the variational approach, i.e. we derive Eq. (66) of the main text. The
expectation values of the (uniform) bilinears are given by:

$\displaystyle\left\langle\Psi_{q=0}^{n}\right\rangle$
$\displaystyle=\\!\sum_{k}\left\langle\\!\hat{\boldsymbol{\Delta}}_{k}^{\dagger}M^{n}\hat{\boldsymbol{\Delta}}_{k}\\!\right\rangle,$
$\displaystyle\left\langle\psi_{q=0}^{n}\right\rangle$
$\displaystyle=\\!\sum_{k}\left\langle\\!\hat{\boldsymbol{\Delta}}_{k}^{\dagger}m^{n}\hat{\boldsymbol{\Delta}}_{k}\\!\right\rangle.$
(129)

To proceed, we introduce the external conjugate fields $Y_{0}^{n}$ and
$(y_{0}^{n},\bar{y}_{0}^{n})$ that couple linearly to the bilinear
combinations via:

$\displaystyle\mathcal{S}_{Y}$
$\displaystyle=-\sum_{n\in\mathbb{G}_{\mathbb{R}}}\Psi^{n}Y_{0}^{n}-\sum_{n\in\mathbb{G}_{\mathbb{C}}}\left(\psi^{n}\bar{y}_{0}^{n}+\bar{\psi}^{n}y_{0}^{n}\right)\,.$
(130)

The new partition function

$\displaystyle\mathcal{Z}^{Y}$
$\displaystyle=\int\mathcal{D}(\boldsymbol{\Delta},\bar{\boldsymbol{\Delta}})\,e^{-\mathcal{S}-\mathcal{S}_{Y}},$
(131)

allows for a direct computation of the above expectation values

$\displaystyle\left\langle\Psi_{q=0}^{n}\right\rangle$
$\displaystyle=\frac{\partial\log\mathcal{Z}^{Y}}{\partial
Y_{0}^{n}}\Big{|}_{Y_{0}^{n}=y_{0}^{n}=0},$
$\displaystyle\left\langle\psi_{q=0}^{n}\right\rangle$
$\displaystyle=\frac{\partial\log\mathcal{Z}^{Y}}{\partial\bar{y}_{0}^{n}}\Big{|}_{Y_{0}^{n}=y_{0}^{n}=0},$
(132)

for $n\in\mathbb{G}_{\mathbb{R}}$ and $n\in\mathbb{G}_{\mathbb{C}}$,
respectively. We rewrite the new partition function (131) to systematically
correctly embed it into the framework of the variational approach, cf. Eq.
(49),

$\displaystyle\mathcal{Z}^{Y}$ $\displaystyle=\mathcal{Z}_{0}^{Y}\langle
e^{-\left(\mathcal{S}-\mathcal{S}_{0}\right)}\rangle_{0}^{Y},$ (133)

where $\langle\mathcal{O}\rangle_{0}^{Y}$ denotes the expectation value with
respect to $\mathcal{S}_{0}+\mathcal{S}_{Y}$,

$\displaystyle\langle\mathcal{O}\rangle_{0}^{Y}$
$\displaystyle\equiv\big{[}\mathcal{Z}_{0}^{Y}\big{]}^{-1}\,\int\mathcal{D}(\boldsymbol{\Delta},\bar{\boldsymbol{\Delta}})\,\,\mathcal{O}\,\,e^{-\mathcal{S}_{0}-\mathcal{S}_{Y}},$
(134)

and
$\mathcal{Z}_{0}^{Y}\equiv\int\mathcal{D}(\boldsymbol{\Delta},\bar{\boldsymbol{\Delta}})\,e^{-\mathcal{S}_{0}-\mathcal{S}_{Y}}$
is the associated partition function. Now, we employ the convexity inequality
(48) on the expression (133), to derive the corresponding variational free
energy in the presence of the external field,

$\displaystyle f_{v}^{Y}$
$\displaystyle=-\frac{T}{V}\log\mathcal{Z}_{0}^{Y}+\frac{T}{V}\left\langle\mathcal{S}-\mathcal{S}_{0}\right\rangle_{0}^{Y}.$
(135)

The derivative of the free energy (135) with respect to the conjugate field
gives the bilinear expectation values (132). To compute these derivatives, we
note that $\mathcal{S}_{0}+\mathcal{S}_{Y}$ only depends on the variable
combinations $\tilde{\Phi}^{n}=\Phi^{n}-\frac{T}{V}Y_{0}^{n}$ and
$\tilde{\phi}^{n}=\phi^{n}-2\frac{T}{V}y_{0}^{n}$. Correspondingly, we define
$\tilde{f}_{v}^{Y}=f_{v}^{Y}-\frac{T}{V}\left\langle\mathcal{S}_{Y}\right\rangle_{0}^{Y}$
such that $\tilde{f}_{v}^{Y}$ also only depends on $\tilde{\Phi}^{n}$ and
$\tilde{\phi}^{n}$. Then, derivatives of the type (132) vanish by
construction, for example:

$\displaystyle\frac{\partial\tilde{f}_{v}^{Y}}{\partial
Y_{0}^{n}}\Big{|}_{Y_{0}^{n}=y_{0}^{n}=0}$
$\displaystyle=\frac{\partial\tilde{f}_{v}^{Y}}{\partial\tilde{\Phi}^{n}}\frac{\partial\tilde{\Phi}^{n}}{\partial
Y_{0}^{n}}\Big{|}_{Y_{0}^{n}=y_{0}^{n}=0}=-\frac{T}{V}\frac{\partial
f_{v}}{\partial\Phi^{n}}=0.$

Thus, rewriting the free energy (135) as
$f_{v}^{Y}=\tilde{f}_{v}^{Y}+\frac{T}{V}\left\langle\mathcal{S}_{Y}\right\rangle_{0}^{Y}$
is convenient as only its second term contributes to the expectation values
(132). With $\mathcal{S}_{Y}$ being already linear in
$(Y_{0}^{n},y_{0}^{n},\bar{y}_{0}^{n})$, see Eq. (130), the derivatives are
straightforwardly evaluated:

$\displaystyle\left\langle\Psi_{q=0}^{n}\right\rangle$
$\displaystyle=\left\langle\Psi_{q=0}^{n}\right\rangle_{0}=2\Pi^{n},$
$\displaystyle\left\langle\psi_{q=0}^{n}\right\rangle$
$\displaystyle=\left\langle\psi_{q=0}^{n}\right\rangle_{0}=2\pi^{n}.$ (136)

Here, we inserted the previously derived expressions (114). As expected,
within the Gaussian variational approach, the expectation values of the
bilinears reduce to their trial expectation values.

##  Appendix C One-component superconductor in the variational approach

As discussed in the main text, the straightforward extension of the
variational ansatz to include the possibility of long-range superconducting
order yields unreasonable results. The issue is not particular to vestigial
phases or to multi-component superconductivity, as it emerges already in the
simpler case of a one-component superconductor, indicating that this is likely
a drawback of the variational ansatz itself. In this Appendix, we show that
the variational ansatz applied to a one-component superconductor gives a
first-order superconducting transition.

The action describing this system is given by

$\displaystyle\mathcal{S}$
$\displaystyle=\frac{V}{T}\sum_{k}\bar{\Delta}_{k}\left(r_{0}+f_{\boldsymbol{k}}^{A_{1g}}\right)\Delta_{k}+\int_{x}u\,|\Delta|^{4},$
(137)

with the gradient term the same as before. For a one-component superconductor,
the Nambu basis $\hat{\boldsymbol{\Delta}}_{k}=(\Delta_{k},\bar{\Delta}_{-k})$
has only two components. As an additional variational parameter, we introduce
the non-zero expectation value of the superconducting order parameter,
$\delta=|\delta|e^{\mathsf{i}\varphi}$, which in the Nambu basis becomes
$\hat{\boldsymbol{\delta}}=(\delta,\bar{\delta})$. The modified variational
trial action is given by:

$\mathcal{S}_{0}=\frac{1}{2}\frac{V}{T}\sum_{k}\left(\hat{\boldsymbol{\Delta}}_{k}^{\dagger}-\hat{\boldsymbol{\delta}}^{\dagger}\delta_{k,0}\right)\,\mathcal{G}_{k}^{-1}\,\left(\hat{\boldsymbol{\Delta}}_{k}-\hat{\boldsymbol{\delta}}\delta_{k,0}\right).$
(138)

Since there are only two bilinears allowed in this case,
$\Phi^{A_{1g}}=R_{0}-r_{0}$ and $\phi^{A_{1g}}$, the Green’s function
simplifies to

$\displaystyle\mathcal{G}_{k}^{-1}$
$\displaystyle=2\left(R_{0}+f_{\boldsymbol{k}}^{A_{1g}}\right)M^{A_{1g}}+\left[\phi^{A_{1g}}(m^{A_{1g}})^{\dagger}+\mathrm{H.c.}\right],$
(139)

with the $2\times 2$ matrices $M^{A_{1g}}=\frac{1}{2}\sigma^{0}$ and
$m^{A_{1g}}=\sigma^{-}$. For convenience, we parameterize the inverted matrix
by

$\displaystyle\mathcal{G}_{k}$
$\displaystyle=2G_{k}^{A_{1g}}M^{A_{1g}}+\left[g_{k}^{A_{1g}}(m^{A_{1g}})^{\dagger}+\mathrm{H.c.}\right],$
(140)

and introduce
$D^{A_{1g}}\equiv\hat{\boldsymbol{\delta}}^{\dagger}M^{A_{1g}}\hat{\boldsymbol{\delta}}=|\delta|^{2}$
and
$d^{A_{1g}}\equiv\hat{\boldsymbol{\delta}}^{\dagger}m^{A_{1g}}\hat{\boldsymbol{\delta}}=\delta^{2}$.
To obtain the variational free energy, we follow the same steps as detailed in
App. B. The only two technical differences are the fact that now the
superconducting order parameter $\boldsymbol{\delta}$ is non-zero, and that we
want to derive an expression for the superconducting susceptibility $\chi$. To
accomplish the latter, we introduce an external conjugate field
$\hat{\boldsymbol{h}}_{k}=(h_{k},\bar{h}_{-k})$ that couples to the uniform
superconducting order parameter in the action via:

$\displaystyle\mathcal{S}_{h}$
$\displaystyle=-\hat{\boldsymbol{\Delta}}_{0}^{\dagger}\hat{\boldsymbol{h}}_{0}\,.$
(141)

Then, the partition function subjected to the external field,

$\displaystyle\mathcal{Z}[\boldsymbol{h}]$
$\displaystyle=\int\mathcal{D}(\boldsymbol{\Delta},\bar{\boldsymbol{\Delta}})\,e^{-\mathcal{S}-\mathcal{S}_{h}},$
(142)

allows for a direct evaluation of the superconducting susceptibility:

$\displaystyle\chi_{ij}$
$\displaystyle=\frac{\partial^{2}\log\mathcal{Z}[\boldsymbol{h}]}{\partial\hat{h}_{i,0}^{*}\partial\hat{h}_{j,0}}\Bigg{|}_{\boldsymbol{h}=0}=\left\langle\hat{\Delta}_{i,0}\hat{\Delta}_{j,0}^{*}\right\rangle-\Big{\langle}\hat{\Delta}_{i,0}\Big{\rangle}\left\langle\hat{\Delta}_{j,0}^{*}\right\rangle.$
(143)

The incorporation of the new partition function (142) into the variational
framework follows the same logic outlined in the previous section, around Eq.
(133). Correspondingly, the variational free energy—in the presence of the
external field $\boldsymbol{h}$—becomes

$\displaystyle f_{v}[\boldsymbol{h}]$
$\displaystyle=-\frac{T}{V}\log\mathcal{Z}_{0}^{h}+\frac{T}{V}\left\langle\mathcal{S}-\mathcal{S}_{0}\right\rangle_{0}^{h},$
(144)

with expectation values defined as

$\displaystyle\langle\mathcal{O}\rangle_{0}^{h}$
$\displaystyle\equiv\big{[}\mathcal{Z}_{0}^{h}\big{]}^{-1}\,\int\mathcal{D}(\boldsymbol{\Delta},\bar{\boldsymbol{\Delta}})\,\,\mathcal{O}\,\,e^{-\mathcal{S}_{0}-\mathcal{S}_{h}},$
(145)

and the associated partition function
$\mathcal{Z}_{0}^{h}\equiv\int\mathcal{D}(\boldsymbol{\Delta},\bar{\boldsymbol{\Delta}})\,e^{-\mathcal{S}_{0}-\mathcal{S}_{h}}$.

The calculation of the free energy density (144) is tedious but
straightforward. For convenience, we show the intermediate results

$\displaystyle\mathcal{Z}_{0}^{h}$
$\displaystyle=\exp\left(\frac{T}{2V}\hat{\boldsymbol{h}}_{0}^{\dagger}\mathcal{G}_{0}\hat{\boldsymbol{h}}_{0}+\hat{\boldsymbol{h}}_{0}^{\dagger}\hat{\boldsymbol{\delta}}\right)\prod_{k}\Big{[}\\!\det\left(VT\mathcal{G}_{k}^{-1}\right)\\!\Big{]}^{-\frac{1}{2}}\\!,$
$\displaystyle\frac{\mathcal{Z}_{0}^{h}[\boldsymbol{j}]}{\mathcal{Z}_{0}^{h}}$
$\displaystyle=\exp\left(\frac{T}{2V}\sum_{k}\hat{\boldsymbol{j}}_{k}^{T}\mathcal{G}_{k}\hat{\boldsymbol{j}}_{k}^{*}+\hat{\boldsymbol{j}}_{k}^{T}\Big{[}\hat{\boldsymbol{\delta}}+\frac{T}{V}\mathcal{G}_{0}\hat{\boldsymbol{h}}_{0}\Big{]}\right),$
(146)

where we introduced an auxiliary field $\boldsymbol{j}_{k}$ similar as in Eq.
(107) that allows for the direct computation of expectation values. For
example, one obtains

$\displaystyle\langle\hat{\boldsymbol{\Delta}}_{k}\rangle_{0}^{h}$
$\displaystyle=\frac{\delta}{\delta\hat{\boldsymbol{j}}_{k}}\left.\frac{\mathcal{Z}_{0}^{h}[\boldsymbol{j}]}{\mathcal{Z}_{0}^{h}}\right|_{\boldsymbol{j}=0}=\delta_{k,0}\left(\hat{\boldsymbol{\delta}}+\frac{T}{V}\mathcal{G}_{0}\hat{\boldsymbol{h}}_{0}\right),$
(147)
$\displaystyle\langle\hat{\Delta}_{i,k}^{*}\hat{\Delta}_{j,k^{\prime}}\rangle_{0}^{h}$
$\displaystyle=\frac{\delta}{\delta\hat{j}_{i,k}^{*}}\frac{\delta}{\delta\hat{j}_{j,k^{\prime}}}\left.\frac{\mathcal{Z}_{0}^{h}[\boldsymbol{j}]}{\mathcal{Z}_{0}^{h}}\right|_{\boldsymbol{j}=0}$
$\displaystyle=\frac{T}{V}\delta_{k,k^{\prime}}\mathcal{G}_{k}^{ji}+\delta_{k,0}\delta_{k^{\prime},0}\hat{\delta}_{h,i}^{*}\hat{\delta}_{h,j},$
(148)

with
$\hat{\boldsymbol{\delta}}_{h}=\hat{\boldsymbol{\delta}}+\frac{T}{V}\mathcal{G}_{0}\hat{\boldsymbol{h}}_{0}$,
and similarly for higher order terms. Eventually, one arrives at the
expression for the variational free energy (144). For convenience, we separate
the field-dependent and field-independent parts,
$f_{v}[\boldsymbol{h}]=f_{v}^{0}+\delta f_{v}^{h}$ with
$f_{v}^{0}=f_{v}[\boldsymbol{0}]$, and $\delta
f_{v}^{h}=f_{v}[\boldsymbol{h}]-f_{v}[\boldsymbol{0}]$. The first part is
given by

$\displaystyle f_{v}^{0}$
$\displaystyle=\frac{1}{2}\frac{T}{V}\sum_{k}\mathrm{tr}\log\left(\mathcal{G}_{k}^{-1}\right)+\left[r_{0}-R_{0}+2u\Pi^{A_{1g}}\right]\Pi^{A_{1g}}$
$\displaystyle-\frac{1}{2}\left[\left(\phi^{A_{1g}}-u\pi^{A_{1g}}\right)\bar{\pi}^{A_{1g}}+\mathrm{c.c.}\right]+\frac{1}{2}\hat{\boldsymbol{\delta}}^{\dagger}\Big{(}\mathcal{K}_{0}^{-1}+\frac{1}{6}\mathcal{D}_{0}^{-1}\Big{)}\hat{\boldsymbol{\delta}},$
(149)

with the integrals:

$\displaystyle\Pi^{A_{1g}}$
$\displaystyle=\frac{T}{V}\sum_{k}\mathrm{tr}\left[\mathcal{G}_{k}M^{A_{1g}}\right]=\frac{T}{V}\sum_{k}G_{k}^{A_{1g}},$
(150) $\displaystyle\pi^{A_{1g}}$
$\displaystyle=\frac{T}{V}\sum_{k}\mathrm{tr}\left[\mathcal{G}_{k}m^{A_{1g}}\right]=\frac{T}{V}\sum_{k}g_{k}^{A_{1g}},$
(151)

and the $2\times 2$ matrices:

$\displaystyle\mathcal{K}_{0}^{-1}$
$\displaystyle=\left(r_{0}+4u\Pi^{A_{1g}}\right)2M^{A_{1g}}+2u\left[\bar{\pi}^{A_{1g}}m^{A_{1g}}+\mathrm{H.c.}\right],$
(152) $\displaystyle\mathcal{D}_{0}^{-1}$
$\displaystyle=8uD^{A_{1g}}M^{A_{1g}}+2u\left[\bar{d}^{A_{1g}}m^{A_{1g}}+\mathrm{H.c.}\right].$
(153)

The field-dependent part becomes

$\displaystyle\delta f_{v}^{h}$
$\displaystyle=-\frac{T^{2}}{V^{2}}\hat{\boldsymbol{h}}_{0}^{\dagger}\Big{[}\mathcal{G}_{0}-\frac{1}{2}\mathcal{G}_{0}\left(\mathcal{K}_{0}^{-1}+4uD^{A_{1g}}M^{A_{1g}}\right)\mathcal{G}_{0}\Big{]}\hat{\boldsymbol{h}}_{0}$
$\displaystyle+\frac{T^{2}}{V^{2}}u\left(\hat{\boldsymbol{h}}_{0}^{\dagger}\mathcal{G}_{0}\hat{\boldsymbol{\delta}}\right)^{2}-\frac{T}{V}\hat{\boldsymbol{h}}_{0}^{\dagger}\Big{[}\text{1}-\mathcal{G}_{0}\Big{(}\mathcal{K}_{0}^{-1}+\frac{1}{3}\mathcal{D}_{0}^{-1}\Big{)}\Big{]}\hat{\boldsymbol{\delta}}.$
(154)

Here, terms of the order $\mathcal{O}\big{(}\boldsymbol{h}^{3}\big{)}$ are
neglected as they do not contribute to the susceptibility (143). In between,
we used the identity
$\hat{\boldsymbol{\delta}}^{\dagger}\mathcal{D}_{0}^{-1}\hat{\boldsymbol{\delta}}=12u\left(D^{A_{1g}}\right)^{2}$.

Let us now minimize the free energy (149) with respect to the variational
parameters. The corresponding derivatives have the same structure as Eq. (58).
The saddle-point equations for the variables
$X_{i}\in\left\\{R_{0},\,\phi^{A_{1g}},\bar{\phi}^{A_{1g}}\right\\}$ becomes:

$\frac{df_{v}^{0}}{dX_{i}}=V_{A_{1g}}\frac{\partial\Pi^{A_{1g}}}{\partial
X_{i}}+v_{A_{1g}}\frac{\partial\pi^{A_{1g}}}{\partial
X_{i}}+\\!\bar{v}_{A_{1g}}\frac{\partial\bar{\pi}^{A_{1g}}}{\partial
X_{i}}=0,$ (155)

with:

$\displaystyle V_{A_{1g}}$ $\displaystyle=\frac{\partial
f_{v}^{0}}{\partial\Pi^{A_{1g}}}=r_{0}-R_{0}+4u\left(\Pi^{A_{1g}}+D^{A_{1g}}\right),$
(156) $\displaystyle v_{A_{1g}}$ $\displaystyle=\frac{\partial
f_{v}^{0}}{\partial\pi^{A_{1g}}}=-\frac{1}{2}\bar{\phi}^{A_{1g}}+u\left(\bar{\pi}^{A_{1g}}+\bar{d}^{A_{1g}}\right).$
(157)

The saddle-point equation related to the variational parameter $\delta$ is
given by:

$\displaystyle\frac{df_{v}}{d\hat{\boldsymbol{\delta}}^{*}}$
$\displaystyle=\left[\mathcal{K}_{0}^{-1}+\frac{1}{3}\mathcal{D}_{0}^{-1}\right]\,\hat{\boldsymbol{\delta}}=0.$
(158)

Above $T_{c}$, we can set $\delta=0$, which automatically solves the saddle-
point equation (158). Moreover, since the matrix encapsulated by Eq. (155) is
non-singular, the saddle-point equations above $T_{c}$ give
$V_{A_{1g}}=v_{A_{1g}}=0$. Interestingly, $v_{A_{1g}}=0$ can only be solved by
$\phi^{A_{1g}}=0$, since $u>0$. Thus, a single-component superconductor does
not support a vestigial charge-$4e$ order.

Below $T_{c}$, while the matrix encoded in Eq. (155) remains non-singular,
there is an additional constraint imposed by Goldstone’s theorem, which
reduces the number of variational parameters from 3 to 2 by imposing a
condition $R_{0}=R_{0}\left(\Phi^{A_{1g}},\phi^{A_{1g}}\right)$ . To see that,
we need to first calculate the superconducting susceptibility. From the field-
dependent free energy density (154), we can directly compute both the
superconducting expectation value, as well as the superconducting
susceptibility (143). One obtains

$\displaystyle\langle\hat{\boldsymbol{\Delta}}_{0}\rangle$
$\displaystyle=\Big{[}1-\mathcal{G}_{0}\Big{(}\mathcal{K}_{0}^{-1}+\frac{1}{3}\mathcal{D}_{0}^{-1}\Big{)}\Big{]}\,\hat{\boldsymbol{\delta}}=\hat{\boldsymbol{\delta}}\,,$
(159) $\displaystyle\frac{V}{T}\chi$
$\displaystyle=\mathcal{G}_{0}-\mathcal{G}_{0}\tilde{\mathcal{D}}_{0}^{-1}\mathcal{G}_{0}\,,$
(160)

where, in the first line, we inserted the saddle-point equation (158) and, in
the second line, we defined the matrix

$\displaystyle\tilde{\mathcal{D}}_{0}^{-1}$
$\displaystyle=-\mathcal{G}_{0}^{-1}+\mathcal{K}_{0}^{-1}+\mathcal{D}_{0}^{-1}$
$\displaystyle=V_{A_{1g}}2M^{A_{1g}}+2\left[v_{A_{1g}}m^{A_{1g}}+\mathrm{H.c.}\right].$
(161)

Note that, above the superconducting transition temperature, where the saddle-
point equation gives $V_{A_{1g}}=v_{A_{1g}}=0$, we have
$\tilde{\mathcal{D}}_{0}^{-1}=0$, thus recovering the expected result
$\frac{V}{T}\chi=\mathcal{G}_{0}$, where the susceptibility only diverges if
$\det\mathcal{G}_{0}^{-1}=0$. For later convenience, we “rotate” the
susceptibility to the basis of real and imaginary components of $\Delta_{k}$.
This is accomplished by performing a unitary transformation

$\displaystyle\chi_{b}$ $\displaystyle=U_{b}^{\dagger}\,\chi\,U_{b},$ with
$\displaystyle U_{b}$
$\displaystyle=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&\mathsf{i}\\\
1&-\mathsf{i}\end{array}\right).$ (164)

Inside the superconducting phase, Goldstone’s theorem requires the transverse
component of the susceptibility (160) to be divergent at all temperatures,
which translates into one eigenvalue of $\mathcal{G}_{0}^{-1}$ being
constrained to be zero. Diagonalizing the inverse Green’s function matrix
gives

$\displaystyle U^{\dagger}\mathcal{G}_{0}^{-1}U$
$\displaystyle=\left(\begin{array}[]{cc}\Lambda_{-}&0\\\
0&\Lambda_{+}\end{array}\right),$ $\displaystyle U$
$\displaystyle=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&e^{\mathsf{i}\varphi_{A}}\\\
-e^{-\mathsf{i}\varphi_{A}}&1\end{array}\right),$ (169)

where $\Lambda_{\pm}\equiv R_{0}\pm\left|\phi^{A_{1g}}\right|$ are the
eigenvalues and the phase $\varphi_{A}$ is defined according to
$\phi^{A_{1g}}=|\phi^{A_{1g}}|e^{\mathsf{i}\varphi_{A}}$. Since $R_{0}\geq 0$,
only the eigenvalue $\Lambda_{-}$ can vanish, leading to the constraint
$R_{0}=|\phi^{A_{1g}}|$ inside the superconducting state. As a result, $R_{0}$
and $\phi^{A_{1g}}$ are no longer independent, which must be taken into
account when minimizing the variational free energy. The saddle-point equation
is then given by:

$\displaystyle\frac{df_{v}^{0}}{d\phi^{A_{1g}}}$ $\displaystyle=\frac{\partial
f_{v}^{0}}{\partial\phi^{A_{1g}}}\Big{|}_{R_{0}}+\frac{\partial
f_{v}^{0}}{\partial R_{0}}\frac{\partial R_{0}}{\partial\phi^{A_{1g}}}=0,$
(170)

with a similar expression for $\bar{\phi}^{A_{1g}}$. Using the fact that
$\partial R_{0}/\partial\phi^{A_{1g}}=\frac{1}{2}e^{-\mathsf{i}\varphi_{A}}$,
we can recast the saddle point equation as:

$\displaystyle e^{\mathsf{i}\varphi_{A}}\frac{df_{v}^{0}}{d\phi^{A_{1g}}}\pm
e^{-\mathsf{i}\varphi_{A}}\frac{df_{v}^{0}}{d\bar{\phi}^{A_{1g}}}$
$\displaystyle=-\frac{T}{2V}\mathrm{tr}\left[\tilde{\mathcal{D}}_{0}^{-1}Z^{\pm}\right]=0,$
(171)

where we introduced

$\displaystyle Z^{\pm}$
$\displaystyle=\sum_{k}\mathcal{G}_{k}\Big{[}2M^{A_{1g}}\frac{1\pm
1}{2}\\!+e^{\mathsf{i}\varphi_{A}}(m^{A_{1g}})^{\dagger}\\!\pm
e^{-\mathsf{i}\varphi_{A}}m^{A_{1g}}\Big{]}\mathcal{G}_{k}.$ (172)

Using the Green’s function parametrization (140) and the property
$g_{k}^{A_{1g}}=-\big{|}g_{k}^{A_{1g}}\big{|}e^{\mathsf{i}\varphi_{A}}$, the
expressions above simplify to:

$\displaystyle Z^{+}$
$\displaystyle=\\!\left[2M^{A_{1g}}\\!+\\!e^{\mathsf{i}\varphi_{A}}(m^{A_{1g}})^{\dagger}\\!+\\!e^{-\mathsf{i}\varphi_{A}}m^{A_{1g}}\right]\\!\sum_{k}\\!\\!\Big{(}\\!G_{k}^{A_{1g}}\\!\\!-\\!|g_{k}^{A_{1g}}|\Big{)}^{\\!2},$
(173) $\displaystyle Z^{-}$
$\displaystyle=\\!\left[e^{\mathsf{i}\varphi_{A}}(m^{A_{1g}})^{\dagger}\\!-\\!e^{-\mathsf{i}\varphi_{A}}m^{A_{1g}}\right]\\!\sum_{k}\\!\Big{[}(G_{k}^{A_{1g}})^{2}\\!-\\!|g_{k}^{A_{1g}}|^{2}\Big{]}.$
(174)

Substituting them back in Eq. (171) gives:

$\displaystyle 0$
$\displaystyle=V_{A_{1g}}+v_{A_{1g}}e^{\mathsf{i}\varphi_{A}}+\bar{v}_{A_{1g}}e^{-\mathsf{i}\varphi_{A}},$
(175) $\displaystyle 0$
$\displaystyle=v_{A_{1g}}e^{\mathsf{i}\varphi_{A}}-\bar{v}_{A_{1g}}e^{-\mathsf{i}\varphi_{A}},$
(176)

which, together with Eq. (158), defines the saddle-point equations below
$T_{c}$.

Before proceeding, let us compute explicitly the superconducting
susceptibility $\chi$. First, we note that the unitary matrix $U$ in Eq. (169)
also diagonalizes the matrix $\tilde{\mathcal{D}}_{0}^{-1}$ in Eq. (161),

$\displaystyle U^{\dagger}\tilde{\mathcal{D}}_{0}^{-1}U$
$\displaystyle=\mathrm{diag}\left(\tilde{V},0\right),$ (177)

where we used Eqs. (175)-(176) and defined

$\displaystyle\tilde{V}$
$\displaystyle=V_{A_{1g}}-v_{A_{1g}}e^{\mathsf{i}\varphi_{A}}-\bar{v}_{A_{1g}}e^{-\mathsf{i}\varphi_{A}}.$
(178)

Using Eq. (177), the application of the same unitary matrix $U$ (169)
simplifies the superconducting susceptibility given in Eq. (160),

$\displaystyle\frac{V}{T}\chi$
$\displaystyle=U\,\mathrm{\,diag}\Big{(}\frac{\Lambda_{-}-\tilde{V}}{\Lambda_{-}^{2}},\frac{1}{\Lambda_{+}}\Big{)}\,\,U^{\dagger}.$

Upon rotating it to the basis of real and imaginary components of the gap,
i.e. upon computing $\chi_{b}$ in Eq. (164), we find:

$\displaystyle\frac{V}{T}\chi_{b}$
$\displaystyle=\left(\\!\\!\frac{\Lambda_{-}\\!\\!-\\!\tilde{V}}{2\Lambda_{-}^{2}}\\!+\\!\frac{1}{2\Lambda_{+}}\\!\right)\\!\sigma^{0}\\!-\\!\left(\\!\\!\frac{\Lambda_{-}\\!\\!-\\!\tilde{V}}{2\Lambda_{-}^{2}}\\!-\\!\frac{1}{2\Lambda_{+}}\\!\right)\\!\left(\\!\\!\begin{array}[]{cc}\cos\varphi_{A}&\\!\\!\sin\varphi_{A}\\\
\sin\varphi_{A}&\\!\\!\text{-}\cos\varphi_{A}\end{array}\\!\\!\right)\\!.$

We now choose to condense the superconducting order parameter along the real
axis, i.e. we choose a real $\delta$. Then, according to Goldstone’s theorem,
the susceptibility associated with the real component (longitudinal
susceptibility) has to be finite, whereas the susceptibility associated with
the imaginary component (transverse susceptibility) has to be divergent at all
temperatures below $T_{c}$. Of course, the Goldstone mode will end up gapped
via the Anderson-Higgs mechanism, which is not included in our model since
there is no coupling to the electromagnetic fields in the action. Because
$\Lambda_{-}=0$ inside the superconducting state, a vanishing transverse
susceptibility can only be achieved by setting the charge-$4e$ order-parameter
phase $\varphi_{A}=0$. In this case, we find

$\displaystyle\frac{V}{T}\chi_{b}\left(\varphi_{A}=0\right)$
$\displaystyle=\mathrm{diag}\Big{(}\frac{1}{\Lambda_{+}},\frac{\Lambda_{-}-\tilde{V}}{\Lambda_{-}^{2}}\Big{)}.$
(179)

Note that inclusion of the charge-$4e$ parameter $\phi^{A_{1g}}$ is essential
to ensure that the longitudinal and transverse susceptibilities are different.

Moving on to the superconducting saddle-point equation (158) for the case of a
real $\delta$, we note that the unitary matrix $U$ (169) also diagonalizes the
matrix $\mathcal{D}_{0}^{-1}$ (153), yielding
$U^{\dagger}\mathcal{D}_{0}^{-1}U=2u\,\delta^{2}\left(2\sigma^{0}-\sigma^{z}\right)$.
Since
$\mathcal{K}_{0}^{-1}=\tilde{\mathcal{D}}_{0}^{-1}+\mathcal{G}_{0}^{-1}-\mathcal{D}_{0}^{-1}$,
and using the fact that $U$ diagonalizes all three matrices
$\tilde{\mathcal{D}}_{0}^{-1}$, $\mathcal{G}_{0}^{-1}$, and
$\mathcal{D}_{0}^{-1}$, the saddle-point equation (158) simplifies to:

$\displaystyle\frac{df_{v}}{d\hat{\boldsymbol{\delta}}^{*}}=0$
$\displaystyle=\Big{[}\mathcal{G}_{0}^{-1}+\tilde{\mathcal{D}}_{0}^{-1}-\frac{2}{3}\mathcal{D}_{0}^{-1}\Big{]}\hat{\boldsymbol{\delta}}$
$\displaystyle 0$
$\displaystyle=\mathrm{diag}\Big{(}\Lambda_{-}+\tilde{V}-\frac{4}{3}u\,\delta^{2},\Lambda_{+}-4u\delta^{2}\Big{)}\,\,\hat{\boldsymbol{\delta}}_{d},$
(180)

where
$\hat{\boldsymbol{\delta}}_{d}=U^{\dagger}\hat{\boldsymbol{\delta}}=\sqrt{2}\delta\,(0,1)^{T}.$
Thus, a non-trivial solution exists only when the second diagonal matrix
element of Eq. (180) vanishes. Imposing this condition, we find:

$\delta^{2}=\frac{\Lambda_{+}}{4u}=\frac{R_{0}}{2u}\,,$ (181)

where, in the last step, we used the facts that $\Lambda_{+}\equiv
R_{0}+\left|\phi^{A_{1g}}\right|$ and $R_{0}=\left|\phi^{A_{1g}}\right|$. The
parameter $R_{0}$ is given by the remaining two saddle-point equations (175)
and (176). Since $\varphi_{A}=0$, $\phi^{A_{1g}}$ is real, and thus the
saddle-point equation (176) is automatically satisfied. As for the first
saddle-point equation (175), we obtain:

$\displaystyle r_{0}$
$\displaystyle=-R_{0}-2u\left(2\Pi^{A_{1g}}-\big{|}\pi^{A_{1g}}\big{|}\right).$
(182)

where

$\displaystyle\Pi^{A_{1g}}\left(R_{0}\right)$
$\displaystyle=\frac{T}{V}\\!\\!\sum_{k}\\!\\!\frac{R_{0}+f_{\boldsymbol{k}}^{A_{1g}}}{\big{(}2R_{0}\\!+\\!f_{\boldsymbol{k}}^{A_{1g}}\big{)}\\!f_{\boldsymbol{k}}^{A_{1g}}}\,,$
(183) $\displaystyle\big{|}\pi^{A_{1g}}\left(R_{0}\right)\big{|}$
$\displaystyle=\frac{T}{V}\\!\\!\sum_{k}\\!\\!\frac{R_{0}}{\big{(}2R_{0}\\!+\\!f_{\boldsymbol{k}}^{A_{1g}}\big{)}\\!f_{\boldsymbol{k}}^{A_{1g}}}\,.$
(184)

Equation (182) is an implicit equation for $R_{0}$ as a function of the
temperature $r_{0}$. To proceed, we note that, according to Eq. (182), a
second-order superconducting transition is signaled by the vanishing of
$R_{0}$. It is therefore convenient to define the reference temperature
$r_{0}^{c}$ by setting $R_{0}=0$ in Eq. (182), yielding
$r_{0}^{c}=-4u\,\Pi^{A_{1g}}\left(R_{0}=0\right)$, where
$\Pi^{A_{1g}}\left(R_{0}=0\right)=\frac{T}{V}\sum_{k}\big{(}f_{\boldsymbol{k}}^{A_{1g}}\big{)}^{-1}$.
Note that the last integral is infrared-divergent in $d=2$, reflecting Mermin-
Wagner theorem. Introducing $\delta r_{0}=r_{0}-r_{0}^{c}$ , we rewrite Eq.
(182) as

$\displaystyle\delta r_{0}$
$\displaystyle=-R_{0}-2u\left[2\Pi^{A_{1g}}-2\Pi^{A_{1g}}\left(R_{0}=0\right)-\big{|}\pi^{A_{1g}}\big{|}\right],$
(185)

where it holds $\delta r_{0}\left(R_{0}=0\right)=0$, by construction. In order
for the superconducting transition to be second-order, it must hold that
$\frac{\partial R_{0}}{\partial\delta r_{0}}\big{|}_{\delta r_{0}=0}<0$, which
implies $\frac{\partial\delta r_{0}}{\partial R_{0}}\big{|}_{R_{0}=0}<0$. A
straightforward calculation gives

$\displaystyle\frac{\partial\delta r_{0}}{\partial R_{0}}\Big{|}_{R_{0}=0}$
$\displaystyle=-1+6u\frac{T}{V}\sum_{k}\frac{1}{\big{(}f_{\boldsymbol{k}}^{A_{1g}}\big{)}^{2}},$
(186)

which contains an infrared divergent integral for $d\leq 4$. This means that
for $d\leq 4$, the derivative (186) is always positive and, thus, the
superconducting transition is not second-order, but first-order for this
variational ansatz.

##  Appendix D Two-component superconductor in the variational approach

In this Appendix, we show how the formulas presented in App. C are modified in
the case of a two-component superconductor. We demonstrated in App. C that the
Gaussian variational ansatz does not correctly capture the superconducting
transition, as it always leads to a first-order transition. Nonetheless, for
the sake of completeness, here we summarize the description of the two-
component superconductor below $T_{c}$.

The derivation follows the same steps outlined in App. B, and in particular,
App. C. Extending the trial action (96) as done in Eq. (138), we include the
superconducting variational parameter
$\hat{\boldsymbol{\delta}}=(\boldsymbol{\delta},\bar{\boldsymbol{\delta}})^{T}$
with $\boldsymbol{\delta}=(\delta_{1},\delta_{2})$. For later convenience, we
introduce the symmetry-classified bilinear combinations:

$\displaystyle D^{n}$
$\displaystyle=\hat{\boldsymbol{\delta}}^{\dagger}\,M^{n}\,\hat{\boldsymbol{\delta}},$
$\displaystyle d^{n}$
$\displaystyle=\hat{\boldsymbol{\delta}}^{\dagger}\,m^{n}\,\hat{\boldsymbol{\delta}},$
(187)

for $n\in\mathbb{G}_{\mathbb{R}}^{0}$ and $n\in\mathbb{G}_{\mathbb{C}}$,
respectively. Because the superconducting order parameter is now non-zero, the
variational free-energy density (126), acquires an additional contribution
$f_{\delta}$, i.e. $f_{v}\rightarrow f_{v}+f_{\delta}$. In close analogy to
the one-component superconducting case [cf. Eq. (149)], this extra
contribution evaluates to

$\displaystyle f_{\delta}$
$\displaystyle=\\!\frac{1}{2}\hat{\boldsymbol{\delta}}^{\dagger}\big{[}\mathcal{K}_{0}^{-1}+\frac{1}{6}\mathcal{D}_{0}^{-1}\big{]}\hat{\boldsymbol{\delta}},$
(188)

where we introduced the $4\times 4$ matrices

$\displaystyle\mathcal{D}_{0}^{-1}$
$\displaystyle=\\!\\!\\!\sum_{n\in\mathbb{G}_{\mathbb{R}}^{0}}\\!\\!\\!2U_{n}D^{n}M^{n}+\sum_{n\in\mathbb{G}_{\mathbb{C}}}u_{n}\left[\bar{d}^{n}m^{n}+d^{n}(m^{n})^{\dagger}\right],$
(189) $\displaystyle\mathcal{K}_{0}^{-1}$
$\displaystyle=\left(r_{0}+2U_{A_{1g}}\Pi^{A_{1g}}\right)2M^{A_{1g}}+\sum_{n\in\mathbb{G}_{\mathbb{R}}}4U_{n}\Pi^{n}M^{n}$
$\displaystyle\quad+\sum_{n\in\mathbb{G}_{\mathbb{C}}}2u_{n}\left[\bar{\pi}^{n}m^{n}+\pi^{n}(m^{n})^{\dagger}\right].$
(190)

Note the relationship:

$\displaystyle
u\big{(}D^{A_{1g}}\big{)}^{\\!2}\\!+\\!v\big{(}D^{A_{2g}}\big{)}^{\\!2}\\!+\\!w\big{(}D^{B_{1g}}\big{)}^{\\!2}$
$\displaystyle=\frac{1}{12}\hat{\boldsymbol{\delta}}^{\dagger}\mathcal{D}_{0}^{-1}\hat{\boldsymbol{\delta}}.$
(191)

Due to Eq. (188), the partial derivatives in the free energy minimization (58)
acquire an explicit $\boldsymbol{\delta}$ dependence:

$\displaystyle\\!\\!V_{A_{1g}}$ $\displaystyle=\frac{\partial
f_{v}}{\partial\Pi^{A_{1g}}}=2\left(r_{0}\\!-\\!R_{0}\right)\\!+\\!2U_{A_{1g}}\\!\left(2\Pi^{A_{1g}}\\!+\\!D^{A_{1g}}\right),\\!\\!\\!\\!$
(192) $\displaystyle\\!\\!V_{n}$ $\displaystyle=\frac{\partial
f_{v}}{\partial\Pi^{n}}=-2\Phi^{n}+2U_{n}\left(2\Pi^{n}+D^{n}\right),$ (193)
$\displaystyle\\!\\!v_{n}$ $\displaystyle=\frac{\partial
f_{v}}{\partial\pi^{n}}=-\bar{\phi}^{n}+u_{n}\left(2\bar{\pi}^{n}+\bar{d}^{n}\right),$
(194)

for $n\in\mathbb{G}_{\mathbb{R}}$ and $n\in\mathbb{G}_{\mathbb{C}}$,
respectively. Variation of the free energy with respect to
$\boldsymbol{\delta}$ yields:

$\displaystyle\frac{df_{v}}{d\hat{\boldsymbol{\delta}}^{*}}$
$\displaystyle=\big{(}\mathcal{K}_{0}^{-1}+\frac{1}{3}\mathcal{D}_{0}^{-1}\big{)}\hat{\boldsymbol{\delta}}=0.$
(195)

For the superconducting susceptibility, after a somewhat tedious derivation
similar to that in App. C, we find

$\displaystyle\frac{V}{T}\chi$
$\displaystyle=\mathcal{G}_{0}-\mathcal{G}_{0}\tilde{\mathcal{D}}_{0}^{-1}\mathcal{G}_{0},$
(196)

where

$\displaystyle\tilde{\mathcal{D}}_{0}^{-1}$
$\displaystyle=-\mathcal{G}_{0}^{-1}+\mathcal{K}_{0}^{-1}+\mathcal{D}_{0}^{-1}$
(197)
$\displaystyle=\sum_{n\in\mathbb{G}_{\mathbb{R}}^{0}}V_{n}M^{n}+\sum_{n\in\mathbb{G}_{\mathbb{C}}}\left(\bar{v}_{n}(m^{n})^{\dagger}+v_{n}m^{n}\right),$
(198)

and with $V_{n}$, $v_{n}$ given in Eqs. (192)-(194).

As explained in App. C, to properly describe the SC state, one has to ensure
that the transverse susceptibility remains divergent below $T_{c}$.
Consequently, one has to ensure that one eigenvalue of $\mathcal{G}_{0}^{-1}$
always vanishes below $T_{c}$, which leads to an additional constraint
$R_{0}=R_{0}\left(\Phi^{n},\phi^{n}\right)$. Being subjected to this
constraint, the free energy $f_{v}$ can be minimized. For a given
superconducting ground state, saddle-point equations similar to Eqs.
(175)-(176) are obtained.

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