2
$\begingroup$

In class we've been learning about the Schrödinger equation and its solutions. We mentioned that one of the simplest solutions for a free particle are simply plane waves $\psi(x,t)=e^{i(\omega t-k x)}$, but then it was said that those solutions are invalid since they are not normalisable, and that every functions for which a fourier transform exists is a valid solution.

My questions:

  1. Is a plane wave a valid solution to the Schrödinger equation or not, since it is trivial to show that it does solve the S. Equation?

  2. Are only functions for which a fourier transform exists solution or is that not a necessary condition?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

A plane wave is obviously a solution to the Schrödinger equation, but it cannot be normalised and hence cannot represent a particle.

The condition for $\psi(\boldsymbol{x},t)$ to be normalisable, i.e. $ \int \mathrm{d}\boldsymbol{x}\,|\psi(\boldsymbol{x},t)|^2 < \infty, $ is a sufficient condition for the existence of the Fourier transform. However, it is not necessary. For example, the $\delta$ function is not normalisable but does possess a Fourier transform (in fact every tempered distribution does), contradicting what you were told. Thus arguments based on the existence of the Fourier transform should be avoided.

The Fourier transform of any normalisable wave function $\psi(\boldsymbol{x},t)$ ist just a decomposition of $\psi(\boldsymbol{x},t)$ into a spectrum of plane waves. Thus, while each of these waves cannot be normalised, their combination can. Certain problems become much simpler when expressed in the wave domain, i.e. when Fourier transforming the Schrödinger equation (and the wave function). This is why plane waves are important wave functions to consider.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Just one more thing : If I get an initial wave function let's say $\psi(x,0) $, I then find the amplitude function $A(k)$$=\int^{\infty}_{-\infty}\psi(x,0)e^{ik x}dx$, then $\psi(x,t)=\int_{-\infty}^{\infty}A(k)e^{ikx}\space e^{i\omega t}dk$ ,where $k= \omega/c$. So I can "decompose" a "initial condition" wave function in terms of a continous spectrum of plane waves and then "sum" them all together to get the time evolution of the function, since the time evolution of each plane wave is known. Is that correct ? $\endgroup$
    – Luka8281
    Oct 6, 2016 at 18:27
  • $\begingroup$ I made an error with $k=\omega /c$ just forget that part... omega is corresponding to k, thats what I meant to say $\endgroup$
    – Luka8281
    Oct 6, 2016 at 18:48
  • $\begingroup$ Every tempered distribution has a Fourier transform, so also plane waves, deltas, locally integrable functions etc... $\endgroup$
    – yuggib
    Oct 7, 2016 at 13:58
  • $\begingroup$ @Luka8281 I think what you're doing is simply Fourier-transforming the Schrödinger equation, solving it in Fouier space, and transforming back. $\endgroup$
    – Walter
    Oct 13, 2016 at 15:43
  • $\begingroup$ @yuggib You're correct. I have amended my answer. $\endgroup$
    – Walter
    Oct 13, 2016 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.