Microscopic derivation of Josephson effect

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In summary, the conversation discusses the microscopic derivation of the Josephson effect, which involves a tunnelling Hamiltonian coupling two superconductors. The Josephson current is calculated perturbatively using the Dyson series, and the first order contribution vanishes due to the "odd number theorem" in superconductivity. The correct expression for the second order contribution is also provided.
  • #1
KDPhysics
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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.
 
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  • #2


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!
 

Related to Microscopic derivation of Josephson effect

1. What is the Josephson effect?

The Josephson effect is a phenomenon in quantum physics where a supercurrent (a current without resistance) can flow between two superconducting materials separated by a thin insulating barrier, known as a Josephson junction.

2. What is the microscopic derivation of the Josephson effect?

The microscopic derivation of the Josephson effect involves using quantum mechanical equations to describe the behavior of electrons in a superconducting material. By considering the wave functions of the electrons and the properties of the material, scientists can mathematically derive the equations that govern the Josephson effect.

3. What are the key equations in the microscopic derivation of the Josephson effect?

The key equations in the microscopic derivation of the Josephson effect include the Schrödinger equation, which describes the wave function of a particle, and the BCS theory, which describes the behavior of electrons in a superconducting material. These equations are used to derive the Josephson equations, which govern the behavior of the supercurrent in a Josephson junction.

4. How does the Josephson effect relate to superconductivity?

The Josephson effect is closely related to superconductivity, as it can only occur between two superconducting materials. Superconductivity is a state of matter where electrons can flow without resistance, and the Josephson effect is a manifestation of this behavior.

5. What are the practical applications of the Josephson effect?

The Josephson effect has a variety of practical applications, including in superconducting electronics, quantum computing, and metrology (the science of measurement). It has also been used in the development of the SQUID (superconducting quantum interference device), which is a highly sensitive magnetometer used in a range of fields such as medical imaging and geology.

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