1. Using the continuity equation, show that the Schrödinger...

1. Using the continuity equation, show that the Schrödinger equation with a vector potential Hamiltonian, of the form $i\frac{\partial \psi}{\partial t} = \left( \frac{(\hat{P} - e\hat{A})^2}{2m} + \hat{V} \right) \psi$ produces the following current, in the Coulomb gauge: $\mathcal{J} = \frac{e}{2m} (\psi^* \hat{p} \psi - \psi \hat{p} \psi^*) - \frac{e^2}{m} \psi^* \hat{A} \psi$

Transcript

00:01 All right, so we have a probability density that we're defining a si star, psi, and then a probability current density, j, which has a little bit more of a complicated form.

00:12 So this is like negative i, e over 2m times, and i think it is a si star times the gradient of psi minus si times the gradient of psi times the gradient of psi star.

00:30 And we want to show that these satisfy continuity equation, which is that d -row -d -t plus the gradient, or sorry, the divergence of j is equal to zero.

00:42 And so let's prove this.

00:44 Let's look at the shorting, let's look at the left -hand term right here.

00:47 So d -row -d -t.

00:49 So the shorting equation tells us like h -bar, or sorry, h, acting on psi, is i, h -bar, d -sy, d -sy, d -t.

00:58 Which we will need in a moment.

01:02 So if we compute d row dt, what we're going to get is d.

01:08 Sigh star d t times si plus si star d.

01:17 And if we look at the shortyager equation, you can see that for each of these, so let's compute this on the left -hand side.

01:25 D -si star d -d -t is going to be well if we look at her let's write out the conjugate equation first that'll make things a little bit easier so the hamiltonian um is her mission meaning it's the same as it's complex conjugate with the trace so we don't need to worry about how that changes but when we take the conjugate of si we have to take the conjugate of this side as well so it's negative i h bar d si star d t okay so with that in mind the time derivative of si with respect to t is going to be i acting on si star over h bar and this is multiplied by si i'm going to bring it out on the left hand side just for a moment and then if we do the same thing over here well this is a little bit easier is basically just a minus so minus si star i h bar h acting on si divided by h bar so we're going to leave that or actually we're not going to leave it alone but let's let's plug in the substitutions for h acting on si considering that h is negative h squared over 2m times the laplocyan operator plus the potential energy acting on si so what we've got here is if we substitute this into the previous equation we have i times si over h bar is equal to h acting on psi star and now the hamiltonian once again is permission so i don't have to consider how this operator changes under conjugation it's exactly the same operator so if we do this we'll get this term and then we're subtracting from this term something that looks like this i h bar or wait sorry put an extraneous h bar in there i hope you'll forgive me so it's times this put the side here times negative h squared over 2m comes the locloscian plus v acting on si and then this is divided by h bar okay um so if we keep going and in fact one thing we should do is si is on well we can we can cancel the potential terms one thing that's going to know because the potential terms hopefully it's clear um they have no imaginary parts or anything like that.

04:10 So what we'll have is like i we'll have a term that looks like this.

04:13 I, si -star times si -times v, or v acting on si or v acting on si -star...

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