4
$\begingroup$

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: $i\hbar \frac{\partial \psi_1}{\partial t} = U_1\psi_1 + K\psi_2 \tag{1}$ $i\hbar \frac{\partial \psi_2}{\partial t} = U_2\psi_2 + K\psi_1 \tag{2}$ 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.

$\endgroup$
1
$\begingroup$

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,

$$ \int \phi_1^* \phi_2 \, d\mathbf{r} \simeq 0,$$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.