How does one solve the Schroedinger equation for a 2D, time-dependent harmonic potential?

This is the Schroedinger equation with a particular 2D harmonic potential:

$$\begin{multline}i\hbar\frac{\partial}{\partial t}\Psi(x_1,x_2,t) = \\ \Biggl[-\frac{\hbar^2}{2m}\nabla^2 + \frac{1}{2} m\omega^2\Biggl(\biggl(x_1 - \frac{x_0(t)}{2}\biggr)^2 + \biggl(x_2 + \frac{x_0(t)}{2}\biggr)^2\Biggr)\Biggr]\Psi(x_1,x_2,t)\end{multline}$$

Can anyone please tell me what the upside down triangle means? I know its the second derivative, but since the problem I have is of variables $x_1$, $x_2$, and $t$, do I take the second derivative of time? Or does that upside down triangle only does it for $x_1$ and $x_2$?

$x_0$ is the separation distance between $x_1$ and $x_2$ and its a function of time because the two double HO potential wells are moving apart from each other.

Also, how would I go about doing separation of variables here? Would I move all the time pieces to one side and all the x pieces to the other side?

Won't I have to do separation of variables twice? Where the second time is when I have to separate $x_1$ and $x_2$?

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Tip: you can use LaTeX markup for math. – Emilio Pisanty Sep 13 '12 at 16:31
I think this would be more appropriate on Mathematics - any thoughts from anyone else? – David Z Sep 13 '12 at 17:50
I think it's fine here. We'd like undergrads with this kinds of questions on the Schrödinger equation to come here, right? – Emilio Pisanty Sep 14 '12 at 9:08

1 Answer

The symbol $\nabla^2$ usually means the sum of second derivatives if you are in a cartesian-like system. Thus for your case $$\nabla^2=\frac{\partial^2}{\partial x^2_1}+\frac{\partial^2}{\partial x^2_2}.$$ To do separation of variables, you usually postulate separate dependences on each of the variables, $$\Psi(x_1,x_2,t)=T(t)\psi_1(x_1)\psi_2(x_2).$$ However, the time-dependence of the well centres will make matters rather more difficult, and as it stands all you can hope to do is a separation of the type $$\Psi(x_1,x_2,t)=\psi_1(x_1,t)\psi_2(x_2,t).$$

I don't quite understand, though, your comment that

x0 is the separation distance between x1 and x2 and its a function of time because the two double HO potential wells are moving apart from each other.

If $x_0=x_2-x_1$, then it is time-independent. If it's the well centres that move apart from each other, then all you can hope for is the above.

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X0(t) = (v2 - v1)*t, so that means x0 = x2 - x1. So its time-independent and I can get it in the form, T(t)*psi(x1)*psi(x2)? I don't know how to write equation in Latex here. :( – QEntanglement Sep 13 '12 at 17:00
The real question is whether $x_0$ is a number or a coordinate. What are $v_1$ and $v_2$? Since the $x_i$ are coordinates, it is not usually the case that $\dot{x}_i=v_i$. – Emilio Pisanty Sep 14 '12 at 9:11
I consulted a physics professor and he told me that x0 can actually be treated as a constant because the the HO potential wells are moving apart from each other, while holding the two particles in the centers of their wells, very slowly, adiabatically. I'm thinking this helps a lot. – QEntanglement Sep 16 '12 at 20:17