1
$\begingroup$

I am solving the Schrödinger equation for a particle in a hybrid system. Specifically I have to solve the following differential equation

$$ - \frac{\hbar^2}{2}\frac{d}{dx}\left(\frac{1}{m^{*}(x)}\frac{d\psi}{dx}\right) -\frac{\hbar^2}{2m^{*}(x)} \frac{d^2\psi}{dy^2} + V(x)\psi = E\psi \ \ \ \tag{1} $$

I would like to know: Is this problem separable? I.e. can the solutions be written as $\psi(x,y)=\Phi(x)\chi(y)$?

To investigate this question, I plugged the separable solution into (1) and found after some rearrangement:

$$ \frac{1}{\chi(y)}\frac{d^2\chi(y)}{dy^2} + \frac{m^{*}(x)}{\Phi(x)} \cdot \frac{d}{dx}\left(\frac{1}{m^{*}(x)}\frac{d\Phi}{dx}\right) + 2m^{*}(x) \cdot \frac{V(x)-E}{\hbar^2} = 0 $$

Now, the first term depends only on y while the two other terms depend only on x. Therefore one can separate the above into two differential equations, a simple one for $\chi(y)$ and a nasty one for $\Phi(x)$. Does this prove that the solutions of (1) are separable? And say they are, how would I go about calculating the solutions to the differential equation for $\Phi(x)$. The spatial dependence of the mass $m^{*}(x)$ is rather annoying.

$\endgroup$

1 Answer 1

-1
$\begingroup$

Yes. Remember what role mathematics plays in models of the physical world. Once we have modeled the physical scenario in terms of a mathematical model, we "forget" the physical world and simply solve the mathematical problem presented by the model. Once we have a solution, we can test it against the physical scenario with the aid of experiments to see if the solution is valid.

In this particular case, the solution that you would get as a separable product of functions would be a valid solution of the mathematical problem. However, you would get a whole set of such solutions. The physical scenario that the model describes may be obtained as a superposition of these solutions.

As for solving this particular set of equations, there are various approaches. Much of it would depend on the details of $m^*(x)$ and $V(x)$. So I don't think I can give a generic answer to this part of the question, without more information.

$\endgroup$

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.