- #1

KDPhysics

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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.