Thermal Stability of Superconductors

In the mainstream viewpar ; sch ; tin , the thermal properties of superconductors are discussed within the framework of the phenomenological equation by Ginzburg and Landaugin (GL) and the BCS theorybcs . However, since this work is aimed at accounting for the stability of the superconducting state with respect to the normal one, we shall develope an alternative approach, based on thermodynamicslan , the properties of the Fermi gasash and recent resultssz5 ; sz4 , claimed to be valid for all superconductors, including low and high $T_{c}$ materials.

The outline is as follows : the specific heat of the superconducting phase is calculated in section $2$, which enables us to assess its binding energy and thereby to confirm and refine a necessary condition, established previously for the existence of persistent currentssz4 ; section $3$ is concerned with the inter-electron coupling, mediated by the electron-phonon and hyperfine interactions; new experiments, dedicated at validating this analysis, are discussed in section $4$ and the results are summarised in the conclusion.

As in our previous worksz4 ; sz5 ; sz1 ; sz2 ; sz3 ; sz7 , the present analysis will proceed within the framework of the two-fluid model, for which the conduction electrons comprise bound and independent electrons, in respective temperature dependent concentration $c_{s}(T),c_{n}(T)$. They are organized, respectively, as a many bound electronsz5 (MBE) state, characterised by its chemical potential $\mu(c_{s})$, and a Fermi gasash of Fermi energy $E_{F}(T,c_{n})$. The Helmholz free energy of independent electrons per unit volume $F_{n}$ and $E_{F}$ on the one hand, and the eigenenergy per unit volume $\mathcal{E}_{s}(c_{s})$ of bound electrons and $\mu$ on the other hand, are relatedash ; lan , respectively, by $E_{F}=\frac{\partial F_{n}}{\partial c_{n}}$ and $\mu=\frac{\partial\mathcal{E}_{s}}{\partial c_{s}}$. At last, according to Gibbs and Duhem’s lawlan , the two-fluid model fulfilssz4 at thermal equilibrium

$$E_{F}(T,c_{n}(T))=\mu(c_{s}(T)),\quad c_{0}=c_{s}(T)+c_{n}(T),$$

(1)

with $c_{0}$ being the total concentration of conduction electrons. Solutions of Eq.(1) are given for $T=0,T_{c}$ in Fig.1. Besides, Eq.(1) has been shownsz5 ; sz4 to read for $T=T_{c}$ (see $B$ in Fig.1)

$$E_{F}(T_{c},c_{0})=\mu(c_{s}=0)=\varepsilon_{c}/2\quad,$$

(2)

with $\varepsilon_{c}$ being the energy of a bound electron pairsz5 . Note that Eq.(2) is consistent with the superconducting transition, occuring at $T_{c}$, being of second orderlan , whereas it has been shownsz5 to be of first order at $T<T_{c}$, if the sample is flown through by a finite current. Then the binding energy of the superconducting state $E_{b}(T<T_{c})$ has been workedsz5 ; gor out as

$$E_{b}(T)=\int_{T}^{T_{c}}\left(C_{s}(u)-C_{n}(u)\right)du\quad,$$

(3)

with $C_{s}(T),C_{n}(T)=\frac{\left(\pi k_{B}\right)^{2}}{3}\rho(E_{F})T,$ being the electronic specific heat of a superconductor, flown through by a vanishing currentsz5 and that of a degenerate Fermi gasash ($k_{B},\rho(E_{F})$ stand for Boltzmann’s constant and the one-electron density of states at the Fermi energy). Due to Eq.(3), a stable superconducting phase $\Leftrightarrow E_{b}>0$ requires $C_{s}(T)>C_{n}(T)$, which is confirmed experimentallypar ; ash , namely $C_{s}(T_{c})\approx 3C_{n}(T_{c})$.

The bound and independent electrons contribute, respectively,

$$\begin{array}[]{l}\mathcal{E}_{s}(c_{s})=\int_{0}^{c_{s}}\mu(u)du\\ \mathcal{E}_{n}(T,c_{n})=\int_{\epsilon_{b}}^{\epsilon_{u}}\epsilon\rho(\epsilon)f(\epsilon-E_{F},T)d\epsilon\end{array}\quad,$$

to the total electronic energy $\mathcal{E}=\mathcal{E}_{n}+\mathcal{E}_{s}$. The symbols $\epsilon,f(\epsilon-E_{F},T)$ refer to the one-electron energy and Fermi-Dirac distribution, while $\epsilon_{b},\epsilon_{u}$ designate the lower and upper limits of the conduction band. Then, thanks to Eq.(1) ($\Rightarrow dc_{n}+dc_{s}=dE_{F}-d\mu=0$), $C_{s}(T)=\frac{d\mathcal{E}}{dT}$ is inferred to read

$$C_{s}=\frac{\partial\mathcal{E}_{n}}{\partial T}-E_{F}\frac{\partial c_{n}}{\partial T}+\frac{dE_{F}}{dT}\left(\frac{\partial\mathcal{E}_{n}}{\partial E_{F}}-E_{F}\frac{\partial c_{n}}{\partial E_{F}}\right)\quad,$$

(4)

with $c_{n}=c_{n}(T),c_{s}=c_{s}(T),E_{F}=E_{F}\left(T,c_{n}(T)\right)$. Because the independent electrons make up a degenerate Fermi gas ($\Rightarrow T<<T_{F}=E_{F}/k_{B}$), the following expressions can be obtained owing to the Sommerfeld expansionash up to $T^{2}$

$$\begin{array}[]{c}\frac{\partial\mathcal{E}_{n}}{\partial E_{F}}=E_{F}\rho+\left(2\rho^{\prime}+E_{F}\rho^{\prime\prime}\right)\frac{\left(\pi k_{B}T\right)^{2}}{6}\\ \frac{\partial\mathcal{E}_{n}}{\partial T}=\left(\rho+E_{F}\rho^{\prime}\right)\frac{\left(\pi k_{B}\right)^{2}}{3}T\\ \frac{\partial c_{n}}{\partial E_{F}}=\rho+\rho^{\prime\prime}\frac{\left(\pi k_{B}T\right)^{2}}{6}\quad,\quad\frac{\partial c_{n}}{\partial T}=\rho^{\prime}\frac{\left(\pi k_{B}\right)^{2}}{3}T\end{array}\quad,$$

(5)

with $\rho=\rho(E_{F}),\rho^{\prime}=\frac{d\rho}{dE_{F}}(E_{F}),\rho^{\prime\prime}=\frac{d^{2}\rho}{dE_{F}^{2}}(E_{F})$. Then Eq.4 is finally recast into

$$C_{s}(T)=\frac{\left(\pi k_{B}\right)^{2}}{3}\rho T\left(1+\frac{dE_{F}}{dT}\frac{\rho^{\prime}}{\rho}T\right)\quad.$$

(6)

Applying Eq.6 at $T=T_{c}$ yields

$$C_{s}(T_{c})=C_{n}(T_{c})\left(1+\frac{dE_{F}}{dT}(T_{c}^{-})\frac{\rho^{\prime}}{\rho}T_{c}\right)\quad.$$

(7)

Hence it is in order to work out the expressions of $\frac{dE_{F}}{dT}(T>T_{c})$ and $\frac{dE_{F}}{dT}(T\leq T_{c})$.

Due to $c_{n}(T>T_{c})=c_{0}$, $\frac{dE_{F}}{dT}$ is deducedash to read

$$\frac{dE_{F}}{dT}(T>T_{c})=-\frac{\frac{\partial c_{n}}{\partial T}}{\frac{\partial c_{n}}{\partial E_{F}}}=-\frac{\left(\pi k_{B}\right)^{2}}{3}\frac{\rho^{\prime}}{\rho}T\quad,$$

(8)

which is integrated with respect to $T$ to yield

$$E_{F}(T=0,c_{0})-E_{F}(T,c_{0})=\frac{\left(\pi k_{B}\right)^{2}}{6}\frac{\rho^{\prime}}{\rho}T^{2}\quad.$$

(9)

Then consistency with Fig.1 requires $\rho^{\prime}(E_{F})>0$ so that $C$ goes toward $B$ for $T\searrow T_{c}$. Assuming $\rho(\epsilon)=\rho_{f}(\epsilon)\propto\sqrt{\epsilon}\Rightarrow\rho_{f}^{\prime}(\epsilon)>0,\forall\epsilon$, with $\rho_{f}(\epsilon)$ being the density of states of three-dimensional free electrons, leads to

$$1-\frac{E_{F}(T>T_{c})}{E_{F}(0,c_{0})}=\frac{\pi^{2}}{12}\left(\frac{T}{T_{F}}\right)^{2}\quad.$$

A numerical application with a typical value $T_{F}=3\times 10^{4}K$ yields $T_{F}(300K)-T_{F}(0)\approx 3K<<T_{F}\Rightarrow\left|\frac{dT_{F}}{dT}\left(T>T_{c}\right)\right|<<1$.

Taking advantage of Eq.1, the expression of $\frac{dE_{F}}{dT}(T\leq T_{c})$ is obtained to read

$$\left.\begin{array}[]{l}dc_{n}=\frac{\partial c_{n}}{\partial E_{F}}dE_{F}+\frac{\partial c_{n}}{\partial T}dT\\ dc_{s}=\frac{\partial c_{s}}{\partial\mu}d\mu=-dc_{n}\end{array}\right\}\Rightarrow\frac{dE_{F}}{dT}=-\frac{\frac{\partial c_{n}}{\partial T}}{\beta(T)}\quad,$$

(10)

with $\beta(T)=\frac{\partial c_{n}}{\partial E_{F}}+\frac{\partial c_{s}}{\partial\mu}$. The Sommerfeld expansion (see Eq.(5)) leads to

$$\alpha=\frac{dE_{F}}{dT}(T_{c}^{-})\frac{\rho^{\prime}}{\rho}T_{c}=-\frac{\left(\pi k_{B}\rho^{\prime}T_{c}\right)^{2}}{3\rho\beta(T_{c})}\quad.$$

(11)

Thus, looking back at Eq.7, it is realized that the observedpar ; ash relation $C_{s}(T_{c})\approx 3C_{n}(T_{c})$ requires $\alpha>0\Rightarrow\beta(T_{c})<0$, which had been already identifiedsz4 as a necessary condition for the superconducting state to be at thermal equilibrium. At last, $\alpha$ reads in case of $\rho=\rho_{f}$

$$\alpha=\frac{\pi^{2}}{12}\left(\frac{T}{T_{F}}\right)^{2}\rho\frac{\partial E_{F}}{\partial c_{n}}\left(1+\frac{\partial E_{F}}{\partial c_{n}}\frac{\partial c_{s}}{\partial\mu}\right)^{-1}\quad.$$

Due to $\frac{T}{T_{F}}<<1$ and $\rho\frac{\partial E_{F}}{\partial c_{n}}\approx 1$, getting $\alpha\approx 2$ requires $\beta(T_{c})\approx 0\Rightarrow\frac{\partial E_{F}}{\partial c_{n}}+\frac{\partial\mu}{\partial c_{s}}\approx 0$, so that the stability criterion of the superconducting state reads finally

$$\frac{\partial E_{F}}{\partial c_{n}}(T_{c},c_{0})=-\frac{\partial\mu}{\partial c_{s}}(0),\quad\rho^{\prime}(E_{F}(T_{c},c_{0}))>0\quad.$$

(12)

Because of $\frac{\partial E_{F}}{\partial c_{n}}(T_{c},c_{0})\approx\frac{1}{\rho}>0$, Eq.(12) is seen to be consistent with $\frac{\partial\mu}{\partial c_{s}}(c_{s})<0$, established previously as a prerequisite for persistent currentssz4 and the Josephson effectsz6 . At last, note that there is $\frac{dT_{F}}{dT}(T\leq T_{c})>>1$ but inversely $0<-\frac{dT_{F}}{dT}(T>T_{c})<<1$.

In order to grasp the significance of the constraint expressed by Eq.(12), let us elaborate the case for which Eq.(12) is not fulfilled ($\Rightarrow C_{s}(T<T_{c})<C_{n}(T)$). Accordingly the hatched area in Fig.1 is equal to the difference in free energy at $T=0$ between the superconducting phase and the normal one, and thence also equal to $E_{b}(0)>0$ because the entropy of the normal state vanisheslan at $T=0$. However applying Eq.(3) with $C_{s}(T<T_{c})<C_{n}(T)$ yields $E_{b}(0)<0$, which contradicts the above opposite conclusion $E_{b}(0)>0$, and thereby entails that the MBE state, associated with $A$ in Fig.1, is not observable at thermal equilibrium in case of unfulfilled Eq.(12), even though it is definitely a MBE eigenstateja1 ; ja2 ; ja3 of the Hubbard Hamiltonian, accounting for the motion of correlated electrons, and its energy is indeed lower than that of the Fermi gas $\mathcal{E}_{n}(T=0,c_{0})$.

Since energy and free energy are equallan at $T=0$, $E_{b}(0)$ reads

$$E_{b}(0)=\int_{0}^{c_{s}(0)}\left(E_{F}(0,c_{0}-c_{s})-\mu(c_{s})\right)dc_{s}\quad.$$

In order to work out an upper bound for $E_{b}(0)$, $E_{F}(T,c_{0}-c_{s})-\mu(c_{s})$ will be approximated by its Taylor expansion at first order with respect to $c_{s}-c_{s}(T)$, which yields

$$E_{F}(T,c_{0}-c_{s})-\mu(c_{s})\approx\frac{m}{e^{2}}\gamma\left(c_{s}(T)-c_{s}\right)\quad,$$

(13)

with $\gamma=\frac{\partial E_{F}}{\partial c_{n}}(c_{n}(T))+\frac{\partial\mu}{\partial c_{s}}(c_{s}(T))$. Since it has been shownsz5 that $E_{F}(T,c_{0}-c_{s})-\mu(c_{s})<10^{-5}eV$, Eq.(13) turns out to be very accurate. Likewise, due to $c_{s}(T)\geq c_{s}\geq 0$ (see Fig.1) and $\gamma>0$ being a necessary conditionsz4 for $A$ in Fig.1 to correspond to a stable equilibrium, Eq.(13) entails

$$E_{F}(T,c_{0}-c_{s})-\mu(c_{s})\leq E_{F}(T,c_{0})-\mu(0)\quad.$$

Then by taking advantage of Eqs.(2,9), we get

$$E_{F}(T,c_{0})-\mu(0)\leq E_{F}(0,c_{0})-E_{F}(T_{c},c_{0})=\frac{\left(\pi k_{B}T_{c}\right)^{2}}{6}\frac{\rho^{\prime}}{\rho},$$

with $\rho^{\prime}>0$ as required by Eq.(12). At last, assuming $\rho(\epsilon)=\rho_{f}(\epsilon)$, the searched upper bound per electron is obtained to read

$$\frac{E_{b}(0)}{c_{0}E_{F}(T_{c},c_{0})}\leq\frac{\pi^{2}}{12}\left(\frac{T_{c}}{T_{F}}\right)^{2}\quad.$$

Applying this formula to $Al$ ($T_{c}=1.2K,T_{F}\approx 3\times 10^{4}K$) gives $\frac{E_{b}(0)}{c_{0}E_{F}(T_{c},c_{0})}<10^{-8}$. Moreover, that latter result had enabled us to realizesz2 that the formula $E_{b}(0)=\mu_{0}H_{c}(0)^{2}/2$, albeit ubiquitous in textbookspar ; sch ; tin ($H_{c}(T\leq T_{c}),\mu_{0}$ refer to the critical magnetic field and the magnetic permeability of vacuum, respectively), underestimates $E_{b}(0)$ by ten orders of magnitude.

Since fulfilling Eq.(12) is tantamount to $\beta(T_{c})=0$, which entails $\frac{dE_{F}}{dT}(T\rightarrow T_{c}^{-})\rightarrow\infty$ and thence $C_{s}(T\rightarrow T_{c}^{-})\rightarrow\infty$, it must be checked that $\mathcal{E}(T)=\int_{0}^{T}C_{s}(u)du$ remains still finite for $T\rightarrow T_{c}^{-}$. To that end, let us work out the Taylor expansion of $\mu(c_{s}),E_{F}(T,c_{n})$ up to the second order around $T=0,c_{s}=0$

$$\begin{array}[]{l}\mu(c_{s})=\mu(0)+\frac{\partial\mu}{\partial c_{s}}(0)c_{s}+\frac{\partial^{2}\mu}{\partial c_{s}^{2}}(0)\frac{c_{s}^{2}}{2}\\ E_{F}(T,c_{n})=E_{F}(T_{c},c_{0})-\frac{c_{s}}{\rho}-\frac{\rho^{\prime}}{\rho^{3}}\frac{c_{s}^{2}}{2}\\ \quad\quad\quad\quad\quad+\frac{\left(\pi k_{B}\right)^{2}}{6}\frac{\rho^{\prime}}{\rho}\left(T_{c}^{2}-T^{2}\right)\end{array}\quad,$$

for which we have used $c_{n}=c_{0}-c_{s},c_{s}=c_{s}(T),\frac{\partial E_{F}}{\partial c_{n}}=\frac{1}{\rho}\Rightarrow\frac{\partial^{2}E_{F}}{\partial c_{n}^{2}}=-\frac{\rho^{\prime}}{\rho^{3}}$. Then taking advantage of Eqs.(1,2) ($\Rightarrow E_{F}(T,c_{n})=\mu(c_{s}),E_{F}(T_{c},c_{0})=\mu(0)$) and Eq.(12) ($\Rightarrow\beta(T_{c})=\frac{\partial\mu}{\partial c_{s}}(0)+\frac{1}{\rho}=0$) results into

$$c_{s}(T\rightarrow T_{c}^{-})=\pi k_{B}\sqrt{\frac{\rho^{\prime}\left(T_{c}^{2}-T^{2}\right)}{3\left(\rho\frac{\partial^{2}\mu}{\partial c_{s}^{2}}(0)+\frac{\rho^{\prime}}{\rho^{2}}\right)}}\propto\sqrt{T_{c}-T}\quad.$$

It should be noticed that the GL equation predictstin rather $c_{s}(T\rightarrow T_{c}^{-})\propto T_{c}-T$.

Likewise, let us calculate similarly the Taylor expansion of $\beta(T)\propto\frac{\partial E_{F}}{\partial c_{n}}+\frac{\partial\mu}{\partial c_{s}}$ up to the first order around $T=T_{c},c_{s}=0$

$$\begin{array}[]{c}\beta(T\rightarrow T_{c}^{-})\propto\left(\frac{\partial^{2}\mu}{\partial c_{s}^{2}}(0)-\frac{\partial^{2}E_{F}}{\partial c_{n}^{2}}(T_{c},c_{0})\right)c_{s}\Rightarrow\\ \beta(T\rightarrow T_{c}^{-})\propto\sqrt{T_{c}-T}\Rightarrow\mathcal{E}(T\rightarrow T_{c}^{-})\propto\sqrt{T_{c}-T}\end{array}\quad,$$

whence $\mathcal{E}(T\rightarrow T_{c}^{-})$ is concluded to remain indeed finite.

At last, we shall work out the expression of $j_{M}(T\rightarrow T_{c}^{-})$, the maximum current density $j_{s}$, conveyed by bound electrons which was shownsz5 to read

$$j_{M}=ec_{m}(T)\sqrt{\frac{2}{m}\left(E_{F}\left(T,c_{0}-c_{m}(T)\right)-\mu(c_{m}(T))\right)}\quad,$$

with $e,m$ standing for the charge and effective mass of the electron, while $c_{m}(T)=\frac{2}{3}c_{s}(T)$ designates the corresponding value of $c_{s}$, i.e. $j_{s}(c_{m})=j_{M}$. Hence $j_{M}$ readssz5 finally

$$\begin{array}[]{c}j_{M}(T)=\frac{er}{\sqrt{m}}\left(\frac{2}{3}c_{s}(T)\right)^{1.5}\\ r=\sqrt{\frac{\partial E_{F}}{\partial c_{n}}(c_{n}(T))+\frac{\partial\mu}{\partial c_{s}}(c_{s}(T))}\end{array}\quad.$$

It ensues from $\beta(T_{c})=0$ that the leading term of the Taylor expansion of $r$ around $T=T_{c},c_{s}=0$ reads

$$\begin{array}[]{c}r\left(T\rightarrow T_{c}^{-}\right)\propto\sqrt{c_{s}(T)}\Rightarrow r\propto\left(T_{c}-T\right)^{\frac{1}{4}}\Rightarrow\\ j_{M}\left(T\rightarrow T_{c}^{-}\right)\propto T_{c}-T\end{array}\quad,$$

which is to be compared with the maximum persistent current densitysz5 $j_{c}\left(T\rightarrow T_{*}^{-}\right)\propto\sqrt{T_{*}-T}$ with $T_{*}<T_{c}$.

Substituting, in a superconducting material, an atomic species of mass $M$ by an isotope, is well-knownpar ; sch ; tin to alter $T_{c}$. This isotope effect was ascribed to the electron-phonon coupling, on the basis of the observed relation $T_{c}\sqrt{M}=constant$. The ensueing theoretical treatmentpar ; sch ; tin capitalisedfro on Froehlich’s perturbationlan2 calculation of the self-energy of an independent electron induced by the electron-phonon coupling. However since the BCS picturebcs has subsequently ascertained the paramount role of inter-electron coupling, we shall rather focus hereafter on the effective phonon-mediated interaction between two electrons.

Thus let us consider independent electrons of spin $\sigma=\pm 1/2$, moving in a three-dimensional crystal, containing $N$ sites. The dispersion of the one-electron band reads $\epsilon(k)$ with $\epsilon(k),k$ being the electron, spin-independent ($\Rightarrow\epsilon(-k)=\epsilon(k)$) energy and a vector of the Brillouin zone, respectively. Their motion is governed, in momentum space, by the Hamiltonian $H_{d}$

$$H_{d}=\sum_{k,\sigma}\epsilon(k)c^{+}_{k,\sigma}c_{k,\sigma}\quad,$$

with the sum over $k$ to be carried out over the whole Brillouin zone. Then the $c^{+}_{k,\sigma},c_{k,\sigma}$’s are Fermi-like creation and annihilation operatorssch on the Bloch state $\left|k,\sigma\right\rangle$

$$\left|k,\sigma\right\rangle=c^{+}_{k,\sigma}\left|0\right\rangle\quad,\quad\left|0\right\rangle=c_{k,\sigma}\left|k,\sigma\right\rangle\quad,$$

with $\left|0\right\rangle$ being the no electron state. Let us introduce now the electron-phononpar ; sch ; tin ; fro coupling $h_{e-\phi}$

$$h_{e-\phi}=\frac{g_{q}}{\sqrt{N}}\sum_{k,k^{\prime},\sigma}c^{+}_{k,\sigma}c_{k^{\prime},\sigma}\left(a^{+}_{q}+a_{-q}\right)\quad,$$

with $q=k^{\prime}-k$ and $g_{q}\propto\left(\omega_{q}M\right)^{-1/2}$ being the coupling constant characterising the electron-phonon interaction. Likewise, $\omega_{q}$ is the phonon frequency, while the $a^{+}_{q},a_{q}$’s are Bose-like creation and annihilation operatorssch on the $n_{q}\in\mathcal{N}$ phonon state $\left|n_{q}\right\rangle$

$$a^{+}_{q}\left|n_{q}\right\rangle=\sqrt{n_{q}+1}\left|n_{q}+1\right\rangle\quad,\quad a_{q}\left|n_{q}\right\rangle=\sqrt{n_{q}}\left|n_{q}-1\right\rangle.$$

Because of $\left\langle k\left|h_{e-\phi}\right|k^{\prime}\right\rangle=0,\forall k,k^{\prime}$ with $\left.|k\right\rangle=c^{+}_{k,+}c^{+}_{-k,-}\left.|0\right\rangle,\left.|k^{\prime}\right\rangle=c^{+}_{k^{\prime},+}c^{+}_{-k^{\prime},-}\left.|0\right\rangle$, we shall deal with $h_{e-\phi}$ as a perturbation with respect to $H_{d}$, in order to reckon $\left\langle k\left|k^{\prime}_{2}\right.\right\rangle$ with $\left|k^{\prime}_{2}\right\rangle$ denoting $\left|k^{\prime}\right\rangle$ perturbed at second orderlan2 . Accordingly, we first introduce the unperturbed electron-phonon eigenstates

$$\left|\widetilde{k}\right\rangle=\left|k\right\rangle\otimes\frac{\left|n_{q}\right\rangle+\left|n_{-q}\right\rangle}{\sqrt{2}},\quad\left|\widetilde{k^{\prime}}\right\rangle=\left|k^{\prime}\right\rangle\otimes\frac{\left|n_{q}\right\rangle+\left|n_{-q}\right\rangle}{\sqrt{2}}\quad,$$

with $n_{q}=n_{-q}=n$. Their respective energies read $E(k)=2\epsilon(k)+n\hbar\omega_{q}$, $E(k^{\prime})=2\epsilon(k^{\prime})+n\hbar\omega_{q}$. Then we reckon $\left|\widetilde{k^{\prime}_{2}}\right\rangle$ and further project it onto $\left|\widetilde{k}\right\rangle$, which yields

$$\begin{array}[]{l}\left\langle\widetilde{k}\left|\widetilde{k^{\prime}_{2}}\right.\right\rangle=\frac{g^{2}_{q}}{2N}\left(\left\langle\widetilde{k}\left|h_{e-\phi}\right|\varphi_{+}\right\rangle\left\langle\varphi_{+}\left|h_{e-\phi}\right|\widetilde{k^{\prime}}\right\rangle\right.\\ \quad\quad\quad\quad\quad\quad\left.+\left\langle\widetilde{k}\left|h_{e-\phi}\right|\varphi_{-}\right\rangle\left\langle\varphi_{-}\left|h_{e-\phi}\right|\widetilde{k^{\prime}}\right\rangle\right)\\ \varphi_{+}=c^{+}_{k,+}c^{+}_{-k^{\prime},-}\left|0\right\rangle\otimes\left(\frac{\sqrt{n+1}}{D_{+}}\left|n_{q}+1\right\rangle+\frac{\sqrt{n}}{D_{-}}\left|n_{-q}-1\right\rangle\right)\\ \varphi_{-}=c^{+}_{k^{\prime},+}c^{+}_{-k,-}\left|0\right\rangle\otimes\left(\frac{\sqrt{n+1}}{D_{+}}\left|n_{-q}+1\right\rangle+\frac{\sqrt{n}}{D_{-}}\left|n_{q}-1\right\rangle\right)\end{array},$$

with $D_{\pm}=\epsilon_{k}-\epsilon_{k^{\prime}}\pm\hbar\omega_{q}$. The searched $x_{k,k^{\prime}}=N\left\langle k\left|k^{\prime}_{2}\right.\right\rangle$ is then inferred to read

$$x_{k,k^{\prime}}=\frac{\left(2n(T)+1\right)g^{2}_{q}}{\left(\left(\epsilon_{k}-\epsilon_{k^{\prime}}\right)^{2}-\left(\hbar\omega_{q}\right)^{2}\right)}\quad,$$

with $n(T)=\left(e^{\frac{\hbar\omega_{q}}{k_{B}T}}-1\right)^{-1}$ being the thermal average of $n_{\pm q}$. Moreover it can be checked that $x_{k,k^{\prime}}=x_{k^{\prime},k}$. Thus, for $q$ not close to the Brillouin zone center (the most likely occurence), there is $x_{k,k^{\prime}}>0$, whereas $x_{k,k^{\prime}}<0$ can be found only for $q\approx 0$. Likewise, though the hereabove expression is redolent of one derived by Froehlichfro , their respective significances are unrelated, since Froehlich interpreted the self-energy of one electron and one phonon bound together in terms of virtual transitions between various electron-phonon states, whereas $x_{k,k^{\prime}}$ refers to the dot product of two-electron-states.

Projecting the hermitian BCS Hamiltonianbcs ; ja1 ; ja2 ; ja3 $H$ onto the basis $\left\{\left|k_{2}\right\rangle,\left|k^{\prime}_{2}\right\rangle\right\}$ yields

$$\begin{array}[]{l}H_{k_{2},k_{2}}=2\left(\epsilon_{k}+\frac{x_{k,k^{\prime}}U}{N^{2}}+\frac{x^{2}_{k,k^{\prime}}}{N^{2}}\epsilon_{k^{\prime}}\right)\\ H_{k_{2},k^{\prime}_{2}}=\frac{U}{N}\left(1+\frac{x^{2}_{k,k^{\prime}}}{N^{2}}\right)+2\frac{x_{k,k^{\prime}}}{N}\left(\epsilon_{k}+\epsilon_{k^{\prime}}\right)\\ H_{k^{\prime}_{2},k^{\prime}_{2}}=2\left(\epsilon_{k^{\prime}}+\frac{x_{k,k^{\prime}}U}{N^{2}}+\frac{x^{2}_{k,k^{\prime}}}{N^{2}}\epsilon_{k}\right)\end{array}\quad,$$

whence it can be concluded within the thermodynamic limit ($N\rightarrow\infty$) that the diagonal matrix elements $H_{k,k}$ remains unaltered by the electron-phonon coupling, whereas $U$ is slightly renormalised to $U+2x_{k,k^{\prime}}\left(\epsilon_{k}+\epsilon_{k^{\prime}}\right)$. Anyhow, since, as noted above, $x_{k,k^{\prime}}>0$ is the most likely case, it is hard to figure out how the phonon-mediated isotope effect could lessen $U$, as concluded by Froehlichfro .

Because, in some materials, the observed isotope effect does not comply with $T_{c}\sqrt{M}=constant$, it has been ascribed tentativelysab to the hyperfineabr interaction, coupling the nuclear and electron spin, provided the electron wave-function includes some $s$-like character. We shall derive the corresponding $x_{k,k^{\prime}}$, by proceeding similarly as above for the electron-phonon one and keeping the same notations.

The Hamiltonian reads for nuclear spins $=1/2$ in momentum space

$$H_{h}=\frac{A}{\sqrt{N}}\sum_{k,k^{\prime}}c^{+}_{k,+}c_{-k^{\prime},-}I_{q}^{-}+c^{+}_{-k,-}c_{k^{\prime},+}I_{q}^{+}\quad,$$

with $A$ being the hyperfine constant, $\pm$ referring to the two spin directions and $q=k+k^{\prime}$. Likewise, the $I^{\pm}=\frac{\sigma_{x}\pm i\sigma_{y}}{2}$’s, with $\sigma_{x},\sigma_{y}$ being Pauli’s matricesabr characterising the nuclear spin, operate on nuclear spin states $\left|\pm\right\rangle$. Note that the term $\propto\sigma_{z}$ has been dropped because it turned out to contribute nothing to $x_{k,k^{\prime}}$. The unperturbed eigenstates read

$$\left|\widetilde{k}\right\rangle=\left|k\right\rangle\otimes\frac{\left|+\right\rangle_{q}+\left|-\right\rangle_{q}}{\sqrt{2}},\quad\left|\widetilde{k^{\prime}}\right\rangle=\left|k^{\prime}\right\rangle\otimes\frac{\left|+\right\rangle_{q}+\left|-\right\rangle_{q}}{\sqrt{2}}\quad.$$

Their respective energies are $E(k)=2\epsilon(k)$, $E(k^{\prime})=2\epsilon(k^{\prime})$. Then $x_{k,k^{\prime}},\left\langle k\left|k^{\prime}_{2}\right.\right\rangle$ read in this case

$$\begin{array}[]{l}x_{k,k^{\prime}}=-\frac{A^{2}}{4\left(\epsilon_{k^{\prime}}-\epsilon_{k}\right)^{2}}\\ \left\langle k\left|k^{\prime}_{2}\right.\right\rangle=\frac{x_{k,k^{\prime}}}{N}\left\langle\widetilde{k}\left|h_{h}\right|\varphi\right\rangle\left\langle\varphi\left|h_{h}\right|\widetilde{k^{\prime}}\right\rangle\\ \varphi=c^{+}_{k,+}c^{+}_{k^{\prime},+}\left|0\right\rangle\otimes\left|-\right\rangle_{q}+c^{+}_{-k,-}c^{+}_{-k^{\prime},-}\left|0\right\rangle\otimes\left|+\right\rangle_{q}\end{array}.$$

Except for having the opposite sign, $x_{k,k^{\prime}}$ has the same properties as in case of the electron-phonon coupling, which causes $U$ to be renormalised to a slightly lesser value.

Three experiments, enabling one to assess the validity of this analysis, will be discussed below. The first one addresses the determination of $\frac{\partial\mu}{\partial c_{s}}$, which plays a key role for the existence of persistent currentssz4 and the stability of the superconducting phase (see Eq.(12)). As shown elsewheresz5 , the partial pressure $p(T\leq T_{c})$, exerted by the conduction electrons, and their associated compressibility coefficientlan3 $\chi(T)$ read

$$\begin{array}[]{l}p=c_{n}E_{F}(c_{n})-F_{n}(c_{n})+c_{s}\mu(c_{s})-\mathcal{E}_{s}(c_{s})\Rightarrow\\ \chi=-\frac{dV}{Vdp}=\left(c_{n}^{2}\frac{\partial E_{F}}{\partial c_{n}}+c_{s}^{2}\frac{\partial\mu}{\partial c_{s}}\right)^{-1}\end{array}\quad,$$

(14)

with $c_{n}=c_{n}(T),c_{s}=c_{s}(T)$ and $V$ being the sample volume. For $T\rightarrow T_{c}$, there is $c_{s}\rightarrow 0$, so that it might be impossible to measure the contribution of bound electrons $\propto c_{s}^{2}\frac{\partial\mu}{\partial c_{s}}(0)$ to $\chi$ in Eq.(14). Such a hurdle might be dodged by making the kind of differential measurement to be described now. A square-wave current $I(t+t_{p})=I(t),\forall t$, such that $I\left(t\in\left[-\frac{t_{p}}{2},0\right]\right)=0,I\left(t\in\left[0,\frac{t_{p}}{2}\right]\right)=I_{c}$ ($I_{c}$ stands for the critical current), is flown through the sample, so that the sample switches periodically from superconducting to normal. Then using a lock-in detection procedure for the $\chi$ measurement might enable one to discriminate $c_{s}^{2}\frac{\partial\mu}{\partial c_{s}}$ against $c_{n}^{2}\frac{\partial E_{F}}{\partial c_{n}}$, despite $c_{s}\left(T\rightarrow T_{c}\right)<<c_{n}\approx c_{0}$ and thence to check the validity of Eq.(12).

The validity of Eq.(1) can be assessed by shining $UV$ light of variable frequency $\omega$ onto the sample and measuring the electron work functionash $w(T\leq T_{c})$ by observing two distinct photoemission thresholds $w_{1}=\hbar\omega_{1}=E_{F}(T),w_{2}=\hbar\omega_{2}=2\mu(T)$, associated respectively with single electron and electron pair excitation. Observing $\omega_{2}=2\omega_{1}$ would validate Eq.(1). Besides, if $c_{s}(T)$ is known from skin-depth measurementssz1 , $\mu(c_{s})$ could be charted. Note also that, if such an experiment were to be carried out in a material, exhibiting a superconducting gap $E_{g}$, a large decrease of $E_{F}$ from $E_{F}(T_{c})$ down to $E_{F}(0)=\mu(c_{0})=\epsilon_{b}-E_{g}$ should be expected ($\epsilon_{b}$ designates the bottom of the conduction band).

For $T>10K$, the electron specific heat is overwhelmedash by the lattice contribution $C_{\phi}$, so that there are no accurate experimental datalor for $C_{s}(T)$. Such a difficulty might be overcome by using again the differential technique, described above. A constant heat power $W$ is fed into a thermally insulated sample, while its time-dependent temperature $T(t)$ is monitored. Thus $T(t)$ can be obtained owing to

$$\begin{array}[]{l}W=\left(C_{\phi}(T)+C_{s}(T)\right)\dot{T}\left(t\in\left[-\frac{t_{p}}{2},0\right]\right)\\ W=\left(C_{\phi}(T)+C_{n}(T)\right)\dot{T}\left(t\in\left[0,\frac{t_{p}}{2}\right]\right)\end{array}\quad,$$

with $\dot{T}=\frac{dT}{dt}$. Feeding again the square-wave current, mentioned above, into the sample, while using the same lock-in detection technique, could enable one to extract $C_{s}(T)-C_{n}(T)$ from the measured signal $\dot{T}(t)$, despite $C_{\phi}>>C_{s},C_{n}$. Notesz5 that $C_{\phi},C_{n}$, unlike $C_{s}$, do not depend on the current $I$ and $C_{n}$ can always be measured at low $T$ and then extrapolatedash up to $T_{c}$ thanks to $C_{n}(T)=\frac{\left(\pi k_{B}\right)^{2}}{3}\rho(E_{F})T$.