In section 18.7 of Bruus & Flensberg the authors provide a microscopic derivation of the Josephson effect.

The hamiltonian on both sides of the tunnelling junction is just the typical BCS hamiltonian, on one side (with fermion operators $c$) \begin{equation} H_c = \sum_{k,\sigma} \epsilon_k c_{k,\sigma}^\dagger c_{k,\sigma} - \sum_{k}(\Delta e^{-i\phi_c}c^\dagger_{k,\uparrow}c^\dagger_{-k,\downarrow}+\Delta e^{i\phi_c} c_{-k,\downarrow}c_{k,\uparrow}) \end{equation} and on the other side (with fermion operators $f$) \begin{equation} H_f = \sum_{k,\sigma} \epsilon_k f_{k,\sigma}^\dagger f_{k,\sigma} - \sum_{k}(\Delta e^{-i\phi_c}f^\dagger_{k,\uparrow}f^\dagger_{-k,\downarrow}+\Delta e^{i\phi_c} f_{-k,\downarrow}f_{k,\uparrow}) \end{equation} Here we let the gap parameter for both superconductors have the same magnitude $\Delta$ but different phases $\phi_c$ and $\phi_f$. We then introduce a tunnelling Hamiltonian coupling the two superconductors \begin{equation} H_t = \sum_{k,p,\sigma} (tc^\dagger_{k\sigma} f_{p,\sigma}+t^* f^\dagger_{p\sigma} c_{k,\sigma}) \end{equation}

We can deal with the phases by introducing a gauge transformation $c\rightarrow e^{-i\phi_c/2} c$ and $f \rightarrow e^{-i\phi_f/2} f$ so that the tunnelling coefficients acquire a phase $t \rightarrow e^{-i\phi/2} t$ with $\phi=\phi_c-\phi_f$. Then we see that the Josephson current is \begin{equation} I_J = \langle I \rangle = -2e \langle \dot{N} \rangle = -2e \bigg\langle \frac{\partial H_t}{\partial \phi}\bigg\rangle \end{equation} where we used the Heisenberg equations of motion $\dot{N} = -i[N,H]$ and the fact that $N=-i\frac{\partial}{\partial \phi}$. The current can be calculated perturbatively using the Dyson series \begin{equation} \exp(-\beta H) = \exp(-\beta H_0)\bigg[1-\int_0^\beta d\tau \ \hat{H}_t(\tau)\bigg]+o((H_t)^2) \end{equation} where $\hat{H}(\tau)$ is in the interaction picture. Then \begin{equation} I_J = -2e\bigg[\bigg\langle \frac{\partial H_t}{\partial \phi}\bigg\rangle_0-\frac{1}{2}\bigg\langle \int_0^\beta d\tau \ \hat{H}_t(\tau) H_t\bigg\rangle_0+\ ...\bigg] \end{equation} where $\langle \rangle_0$ denotes thermal averaging with respect to $H_0=H_c+H_f$. The first order contribution is \begin{equation} \bigg\langle \frac{\partial H_t}{\partial \phi}\bigg\rangle_0 = \frac{\partial}{\partial \phi}\text{Tr}(e^{-\beta H_0} H_t) = \frac{\partial}{\partial \phi} \sum_m e^{-\beta E^0_m} \langle m|H_t|m\rangle \end{equation} which I presume (but I am really not too sure about this) vanishes since $H_t$ moves an odd number of fermions from one superconductor to the other, so its action on any eigenstate $|m\rangle$ of the BCS hamiltonian will produce a state orthogonal to it. The second order contribution on the other hand is given as \begin{equation} \frac{\partial}{\partial \phi} \int_0^\beta d\tau \ (t^2 e^{i\phi} \mathcal{F}_{\downarrow \uparrow}(\textbf{k},\tau) \mathcal{F}^*_{\downarrow \uparrow}(\textbf{p},-\tau)+c.c.) \end{equation} where we defined $\mathcal{F}(\textbf{k},\tau)=-\langle \mathcal{T} c^\dagger_{k,\downarrow}(\tau)c^\dagger_{-k,\uparrow}(0)\rangle$. However I have a hard time making sense of this equation. I find that \begin{equation} \frac{1}{2}\frac{\partial}{\partial \phi} \int_0^\beta d\tau \ \bigg \langle \sum_{k,p,\sigma}\sum_{k',p',\sigma'} (tc^\dagger_{k\sigma} f_{p,\sigma}+t^* f^\dagger_{p\sigma} c_{k,\sigma})(tc^\dagger_{k'\sigma'} f_{p',\sigma'}+t^* f^\dagger_{p'\sigma'} c_{k',\sigma'})\bigg \rangle \end{equation} but I'm not sure how this simplifies to the result by Bruus & Flensberg.


1 Answer 1


First, in the last equation of your problem, the $t$ there should be $te^{i\phi}$, then it's clear why the result only contains the anomalous Green's function $\mathcal{F}(\textbf{k},\tau)$: terms like $\left\langle (tc^\dagger_{k\sigma} f_{p,\sigma}t^* f^\dagger_{p'\sigma'} c_{k',\sigma'})\right \rangle$ have no $\phi$ dependence so you can drop them. What's left is the \begin{equation} \frac{1}{2}\frac{\partial}{\partial \phi} \int_0^\beta d\tau t^2\ \sum_{k,p,\sigma}\sum_{k',p',\sigma'} e^{i\phi}\left\langle c^\dagger_{k\sigma} f_{p,\sigma}c^\dagger_{k'\sigma'} f_{p',\sigma'}\right\rangle+e^{-i\phi}\left\langle f^\dagger_{p\sigma} c_{k,\sigma} f^\dagger_{p'\sigma'} c_{k',\sigma'}\right\rangle \end{equation} by wick theorem and counting the spin index properly, and finally Fourier tranforming back to the real space you could get the result.

  • $\begingroup$ Aaah, thanks that was crystal clear. $\endgroup$ Apr 7 at 6:51
  • $\begingroup$ Just one doubt, why does "wick theorem and counting the spin index properly, and finally Fourier tranforming back to the real space" impose $k=-k'$, $p=-p'$ and $\sigma=-\sigma'$? $\endgroup$ Apr 7 at 6:56
  • $\begingroup$ $k--k',p=-k$ comes from the conservation of momentum, and $\sigma=-\sigma'$ comes from you choose singlet pairing in Hamiltonian $c_{\uparrow}^\dagger c^\dagger_{\downarrow}$,Which means total spin is conserved, Besides, the original Hamiltonian is exactly solvable in Nambu representation, yields a ground state annihilated by bogoliubov quasi particle annihilation operator. You could directly check them. $\endgroup$ Apr 7 at 15:11
  • $\begingroup$ The BCS hamiltonian is exactly solvable for the ground state but I've never seen an expression for the excited states which are required to evaluate the thermal expectation value right? Sorry for my daftness but why does conservation of momentum occur? Is it because the BCS hamiltonian conserves momentum? In that case wouldn't the tunnelling hamiltonian break the momentum conservation? $\endgroup$ Apr 11 at 11:47
  • $\begingroup$ For example in a normal metal I am led to believe that one would get $k=k'$, $p=p'$ and $\sigma=\sigma'$, so what makes the superconducting case different? $\endgroup$ Apr 12 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.