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Consider a system of two superconductors (with Hamiltonian $H_{SC1}$ and $H_{SC2}$) weakly tunnel coupled via some intermediate system, i.e. an insulator, a ferromagnet, a quantum dot, etc. (described by a tunneling Hamiltonian $H_{T}$). The full system is described by the Hamiltonian $H=H_{SC1}+H_{SC2}+H_{T}$. Let the superconducting phase difference between the two superconductors be given by $\Delta\phi$. We want to study the system in the low temperature regime so that $k_{B}T$ is much smaller than the eigenenergies of the system.

The usual definition of the Josephson current in this case is given by $$ J^{(1)}_{jos}=\frac{2e}{\hbar}\frac{\partial E_{0}}{\partial\Delta\phi.} $$ with $E_{0}$ being the ground state energy of the system.

Sometimes (for example "Many-Body Quantum Theory in Condensed Matter Physics: An Introduction" by Henrik Bruus,Karsten Flensberg Eq.18.84) people use $$ J^{(2)}_{jos}=\frac{2e}{\hbar}\langle GS|\frac{\partial H}{\partial\Delta\phi}|GS\rangle. $$ with $|GS\rangle$ being a ground state of the system. This seems to be equivalent to the first definition of the Josephson current when the ground state is independent of the phase difference $\Delta\phi$, i.e. $J^{(1)}_{jos}=J^{(2)}_{jos}$. Once the ground state is dependent on the superconducting phase difference the first formula gives a different result, namely $ J^{(1)}_{jos}=\frac{2e}{\hbar}(\langle GS|\frac{\partial H}{\partial\Delta\phi}|GS\rangle+ \langle \frac{\partial GS}{\partial\Delta\phi}|H|GS\rangle + \langle GS|H| \frac{\partial GS}{\partial\Delta\phi}\rangle ) $.

I am somewhat confused now since I am not sure which is the correct formula to use. Personally I expect the Josephson current to be the expectation value of some quantum mechanical current operator.

I would be very happy for some clarification on which definition is the correct one.

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  • $\begingroup$ "...some quantum mechanical current operator...". Precisely. Probability density flux, with coefficients to get current density $$j=\frac{i e \hbar}{2m}(\psi \nabla \psi^{*} - \psi^{*} \nabla \psi )$$ Yet another formula for you to compare with the first two :) $\endgroup$ – LLlAMnYP Sep 18 '15 at 14:37
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Let $|\psi\rangle$ be an eigenstate of $H$ with eigenvalue $E$. $\lambda$ is an external parameter that $H$ depends on. In your case it is $\Delta\phi$. Then

$$ \dfrac{\partial}{\partial\lambda} \langle \psi |H|\psi\rangle=\langle\psi|\dfrac{\partial H}{\partial \lambda}|\psi\rangle+(\dfrac{\partial}{\partial\lambda} \langle\psi|)H|\psi\rangle+\langle \psi|H(\dfrac{\partial}{\partial\lambda} |\psi\rangle) =\langle\psi|\dfrac{\partial H}{\partial \lambda}|\psi\rangle + E[(\dfrac{\partial}{\partial\lambda} \langle\psi|)|\psi\rangle+\langle \psi|(\dfrac{\partial}{\partial\lambda} |\psi\rangle)] =\langle\psi|\dfrac{\partial H}{\partial \lambda}|\psi\rangle+\dfrac{\partial}{\partial\lambda}\langle\psi|\psi\rangle =\langle\psi|\dfrac{\partial H}{\partial \lambda}|\psi\rangle $$ Here in the second equality we use $H|\psi\rangle=E|\psi\rangle$ (also $\langle\psi|H=E\langle\psi|$ because $H$ is Hermitian), and the last equality needs $\langle\psi|\psi\rangle=1$, i.e. $|\psi\rangle$ is normalized. This is called the Hellmann-Feynman theorem.

So the two definitions of the current are the same as long as $|GS\rangle$ is normalized.

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  • $\begingroup$ Is it conceivable that a physical system exists in which $J_{jos}(\Delta\phi=\phi_{0})=0$ for some fixed $\phi_{0}$, e.g. $\phi_{0}=0$, but $\left.\frac{\partial H}{\partial\Delta\phi}\right|_{\Delta\phi=\phi_{0}}\neq0$ on the operator level? $\endgroup$ – IdeA Sep 19 '15 at 7:55

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