1

I'm struggling to write the Fourier expansion of the wave-function associated to a particle confined in an infinite potential well. Suppose that $V(x) = \left\{\begin{array}{c c} \infty & \text{ if } x\notin(0,a)\\ 0 & \text{ if } x\in(0,a)\\ \end{array}\right.\,.$

Then, the solution to the time-independent Schrödinger equation is:

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

In this case, the momentum is a discrete variable, so I would use a discrete Fourier expansion, that is, I use the following formulas for any function $f(x)$

$f(x) = \sum\limits_{m=-\infty}^{\infty}f_m\,e^{i\frac{2m\pi}{a}x}\;\;;\;\;f_m = \frac{1}{a}\int\limits_0^af(x)\,e^{-i\frac{2m\pi}{a}x}\,dx$.

If I use the formula for$f_m$ to obtain the expression in the momentum's domain of $\psi_n(x)$ and using $\sin(x) = \frac{1}{2i}\left(e^{ix}-e^{-ix}\right)$ I got to the following equation:

$f_m = \frac{1}{2ia}\int\limits_0^a\left[e^{i\frac{x\pi}{a}(n-2m)}-e^{-i\frac{x\pi}{a}(n+2m)}\right]\,dx$

which does not seem to have a simple orthogonality relation, because $n = \pm 2m$ does the trick to the zero-argument for the exponential, but if $n\neq \pm 2m$ the condition for the integrals to be zero would the that $\frac{n}{2}\pm m\in\mathbb{Z}$ which forces me to discard all the non-pair $n$'s!

If I think of a Fourier transformation with the same "argument-type" as the wave-function, that is, instead of $\frac{2\pi m}{a}x$ using $\frac{\pi m}{a}x$ so that the expansion is:

$f(x) = \sum\limits_{m=-\infty}^{\infty}f_m\,e^{i\frac{m\pi}{a}x}$,

I got cornered much in the same way. Does anyone have an idea what I'm doing wrong?

Condereal
  • 55
  • In your expression for $f_m$, I guess "$2 l$" should be $2 m$", right? – Clara Diaz Sanchez Mar 08 '20 at 11:44
  • yes, totally right, I'll edit the post! Thank you! – Condereal Mar 08 '20 at 11:57
  • Related: What's the deal with momentum in the infinite square well? – Emilio Pisanty Mar 08 '20 at 12:13
  • What is your question? Also your expression for the potential is not confining, is that a typo? – lcv Mar 08 '20 at 12:22
  • omg, I typed it fast and of course made I mistake. It's supposed to be a confining potential well with value 0 in the interval (0,a) and $\infty$ everywhere else. – Condereal Mar 08 '20 at 12:33
  • Nevertheless, I've found that if I use a base given by $\sin(\frac{n\pi}{a}x)$ instead of the exponential functions, I arrive to the correct result. Then, the expansion I should use is just $f(x) = \sum_{m=1}^{\infty}c_n\sin\left(\frac{n\pi}{a}x\right)$ – Condereal Mar 08 '20 at 12:36
  • No problem. But you didn't say what's the question – lcv Mar 08 '20 at 12:36
  • It seems that you are trying to compute the Fourier series of the eigenfunctions (in particular for $n$ odd). But why would you want to do that? – lcv Mar 08 '20 at 12:38
  • @lcv I expected that the result of the integrals in $f_m$ had a Kroenecker's delta form! – Condereal Mar 08 '20 at 12:39
  • Why did you expect this? – lcv Mar 08 '20 at 12:40
  • @lcv because I want, as you said, to compute the Fouries series of the eigenfunctions. I want to do this because I'm interested particularly in writing the retarded Green-function of this problem in the momentum-domain. – Condereal Mar 08 '20 at 12:43
  • You don't need to compute the Fourier series of the eigenfunctions to do that. You only need the eigenfunctions. Note that momentum is not conserved but you have a quasi-momentum whose values are $k_n = n \pi/a$. Any meaningful definition of momentum domain, is the set of such $k_n$'s. – lcv Mar 08 '20 at 12:57

0 Answers0