# Microscopic derivation of the Josephson effect

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

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

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 $$$$\frac{1}{2}\frac{\partial}{\partial \phi} \int_0^\beta d\tau t^2\ \sum_{k,p,\sigma}\sum_{k',p',\sigma'} e^{i\phi}\left\langle c^\dagger_{k\sigma} f_{p,\sigma}c^\dagger_{k'\sigma'} f_{p',\sigma'}\right\rangle+e^{-i\phi}\left\langle f^\dagger_{p\sigma} c_{k,\sigma} f^\dagger_{p'\sigma'} c_{k',\sigma'}\right\rangle$$$$ by wick theorem and counting the spin index properly, and finally Fourier tranforming back to the real space you could get the result.
• 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 $\sigma=-\sigma'$? Apr 7 at 6:56
• $k--k',p=-k$ 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. Apr 7 at 15:11
• 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? Apr 12 at 16:39