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The Hamiltonian for Josephson Junctions

Physics Asked on November 6, 2021

In the Feynman Lectures on physics, Feynman says the amplitudes across a Josephson Junction should be related by the following, where the subscript denotes the side of the junction that the amplitude or potential belongs to:
$ihbar frac{partial psi_1}{partial t} = U_1psi_1 + Kpsi_2 tag{1}$
$ihbar frac{partial psi_2}{partial t} = U_2psi_2 + Kpsi_1 tag{2}$
Why is this the correct Hamiltonian? Certainly having a term proportional to the potential energy is standard, but why is there a term proportional to the amplitude on the other side? Is this because we expect some of the amplitude to tunnel through, and hence be proportional the amplitude on the other side? While I understand that our Hamiltonian can be any Hermitian operator, I just don’t see the full justification for choosing this one.

2 Answers

In the physics of nanostructures one often uses something like: $$hat{H} = -E_J coshat{varphi},$$ where $hat{varphi}=hat{varphi}_1 - hat{varphi}_2$ is the superconductor phase difference, related to the number of particles via the commutation relation $$[hat{varphi}_j, hat{N}_j] = -i.$$

Answered by Roger Vadim on November 6, 2021

In the same spirit as the tight-binding approach for atomic orbitals in a lattice, you assume that the spatial wavefunctions (let's call them $phi_{1,2}(mathbf{r})$ here) on either site of the barrier are approximately orthogonal,

$$ int phi_1^* phi_2 , dmathbf{r} simeq 0,$$

whereas the barrier allows particle to tunnel through and lifts this orthogonality

$$int phi_1^* V(mathbf{r}) phi_2 = K.$$

Integrating out the spatial coordinates from the Schrödinger equation,

$$i hbar dot{Psi} = [hat{H}_0 + V(mathbf{r})] Psi,$$

where the total wavefunction of the system is factorized as $Psi = psi_1 phi_1(mathbf{r}) + psi_1 phi_2(mathbf{r})$ and $psi_{1,2}$ are the probability coefficients (or particle coefficients if one considers superconductors or condensates). You then have,

$$ i hbar begin{pmatrix} dot{psi}_1 \ dot{psi}_2 end{pmatrix} = begin{pmatrix} U_1 & K \ K & U_2 end{pmatrix} begin{pmatrix} psi_1 \ psi_2 end{pmatrix}.$$

Here, $U_{1,2}$ are the bulk energies (or un-perturbed energies) of the wavefunction at each side.

Answered by Naloo on November 6, 2021

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