10
$\begingroup$

If one were to look at Schroedinger's equation for two interacting electrons in a one dimensional infinite square well, it would something like this:

$$-\frac{\hbar^2}{2m}\partial^2_{x_1}\psi(x_1,x_2)-\frac{\hbar^2}{2m}\partial^2_{x_2}\psi(x_1,x_2)+\frac{e^2}{4\pi\epsilon_0\lvert x_1-x_2\rvert}\psi(x_1,x_2)=E\psi(x_1,x_2)$$

where $(x_1,x_2)$ (representing the coordinates of electron one and electron two respectively) is constrained to $[-L/2,L/2]\times[-L/2,L/2]$ where $\psi$ has to be zero on the boundary.

Without the repulsion energy, one can separate the variables in the equation and solve quickly. However this equation is not separable. Can one potentially solve this by changing the domain of this problem over to a square in the complex plane where $x_1$ becomes $x$ and $x_2$ becomes $iy$ and $z=x+iy$? I think this might change the equation over to:

$$-\frac{\hbar^2}{2m}\frac{d^2}{dz^2}\psi(z)+\frac{e^2}{4\pi\epsilon_0}\frac{z}{\mid z\mid^2}\psi(z)=E\psi(z)$$

One could solve this on the real line on interval $[-L/2,L/2]$ with $\psi$ vanishing on the boundary yielding an equation:

$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)+\frac{e^2}{4\pi\epsilon_0}\frac{1}{\mid x \mid}\psi(x)=E\psi(x)$$

which has one variable. Once a solution is obtained, then one could extend this function analytically over the complex plane.

However, I see several problems with this approach:

  1. Since $\psi$ must be a differentiable function, and hence holomorphic, $\nabla^2$ of this function would be zero.

  2. The analytic extension of this function might not vanish on the boundaries in the complex plane.

Is there a way to remedy this and solve this equation?

$\endgroup$
  • $\begingroup$ Are you sure about $d^2/dz^2 = d^2/dx^2 + d^2/dy^2$? I would have written $\frac{d}{dz} \frac{d}{d\bar z}$ instead. Then I don't see why $\psi$ should be holomorphic, and the ansatz might not be too helpful... I don't have other ideas right now, though. $\endgroup$ – Noiralef Dec 19 '15 at 3:03
  • $\begingroup$ If you had a parabolic confinement, you could work out separate equations for the center of mass motion and the relative motion. Would this approach make sense here too? $\endgroup$ – flaudemus Feb 20 at 17:51
1
$\begingroup$

Might have a solution but not by the complex variable method proposed.

The original equation has $1/|x_1-x_2|$ multiplying $e^2\psi/4\pi\epsilon_0$ but this is converted into $z/|z|^2$. In the first case, this factor is infinite whenever $x_1=x_2$ but after it's converted to complex, it is infinite only when $z=0$ which happens only when both $x_1$ and $x_2$ are zero.

The correct conversion is $1/|x_1-x_2| = 1/|\textrm{Re}(z)-\textrm{Im}(z)|$ which isn't nearly as nice as $z/|z|^2$. But even if $z/|z|^2$ were the correct conversion that function isn't holomorphic (or regular) anywhere so it doesn't give a useful complex function in terms of reducing the number of variables from 2 real to 1 complex.

And even if we did succeed in rewriting the equation in terms of holomorphic complex functions the boundary does not become the end points of the line $[-L/2,L/2]$ but instead it is still a box in the complex plane.

$\endgroup$

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.