Microscopic derivation of Josephson effect

[KDPhysics]

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 $\dot{N} = [N,H]$ 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 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.

 

[eljose79]

Hello,

Thank you for your detailed explanation of the Josephson effect. As a fellow scientist, I would like to provide some clarification on the equations you have presented.

Firstly, the reason why the first order contribution vanishes is because the operator $H_t$ changes the number of fermions in the system, and thus it cannot have a non-zero expectation value in the ground state. This is known as the "odd number theorem" in superconductivity.

Secondly, the second order contribution you have written is not quite correct. The correct expression is
$$
\frac{1}{2}\frac{\partial}{\partial \phi} \int_0^\beta d\tau \ \bigg \langle \sum_{k,p,\sigma}\sum_{k',p',\sigma'} (tc^\dagger_{k\sigma} f_{p,\sigma}+t^* f^\dagger_{p\sigma} c_{k,\sigma})(tc^\dagger_{k'\sigma'} f_{p',\sigma'}+t^* f^\dagger_{p'\sigma'} c_{k',\sigma'})\bigg \rangle_0
$$
where the thermal averaging is with respect to the Hamiltonian $H_0$. This is because the operator $H_t$ is already in the interaction picture, so there is no need to include the exponential factor in the thermal averaging.

I hope this clears up any confusion. Keep up the good work!