I am working on the perturbation theory problem with the infinite potential well. However, I am not sure how to write a wavefunction for different than the infinite potential well from $0$ to $a$ and in my problem the well is from $0$ to $2a$.

Is it necessary to solve the potential well all over again? Maybe it is possible just to apply the transformation as $a\rightarrow2a$, for the initial wavefunction

$$\psi(x)=\sqrt\frac{2}{a}\sin\left(\frac{n \pi x}{a}\right)$$ ?

  • $\begingroup$ Yes, you need to solve a problem all over again, and perturbation theory is a way of solving it (admittedly, not the only one). $\endgroup$ – Ivan Neretin Apr 18 '20 at 11:16
  • $\begingroup$ @IvanNeretin, yes, but the first-order energy corrections $E^{(1)}=\int \psi^{(0)}_n V(x) \psi^{(0)}_n$ where $\psi^{(0)}$ is the unperturbed Hamiltonian wave function. In this case, I am not sure how to recalculate the unperturbed state wavefunction. $\endgroup$ – aerospace Apr 18 '20 at 11:23
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    $\begingroup$ The unperturbed function is the one before the perturbation; you already have it, I suppose. Wait! Now I got it: you have the unperturbed function for a different well. Yes, then just apply... I wouldn't even call that a transformation. You see the equation, where $a$ is any number; you plug any number in place of $a$, and it just works. Oh, and it is not just numbers that you can plug. $\endgroup$ – Ivan Neretin Apr 18 '20 at 11:30
  • $\begingroup$ @IvanNeretin, yes, I was talking about that! Thank you very much for the help! $\endgroup$ – aerospace Apr 18 '20 at 11:42

Well, yes; the original length $a$ is just a symbol and you can replace it with whatever other quantity you wish. The relevant wavefunctions are thus just

$$\psi_n = \sqrt{\frac{1}{a}}\sin\left(\frac{n\pi x}{2a}\right)$$

You can verify that these wavefunctions are still normalised correctly by explicit integration.


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