Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rewriting a wave function superposition

  1. Aug 16, 2007 #1


    User Avatar


    I'm looking at this wave function:

    [tex]\psi(x,t) = \frac{4}{5}{\psi}_{1} + \frac{3}{5}{\psi}_{2}[/tex]

    The functions involved here are the typical eigenfunctions for the ground state and first excited level in an infinitely-deep 1-D square well.

    [tex]A = 4/5.\sqrt{2/a}[/tex]
    [tex]B = 3/5.\sqrt{2/a}[/tex]
    [tex]K = \pi/a[/tex]

    I might have rewritten this as

    [tex]\psi(x,t) = Asin(Kx).exp(-\frac{iE_1.t}{\hbar}) + Bsin(2Kx).exp(-\frac{iE_2.t}{\hbar})[/tex]

    However, the text restates this as

    [tex]\psi(x,t) = Asin(Kx) + Bsin(2Kx).exp(-\frac{i\Delta.t}{\hbar})[/tex]

    [tex]\Delta = E_{2} - E_{1}[/tex]

    Can someone tell me how the time element has been attached to only one of the eigenfunctions like that? I expect it's obvious, but I'm just not seeing it at the moment! (It's evidently been done this way to make the expectation value calculation that follows simpler).

    Many thanks!
  2. jcsd
  3. Aug 16, 2007 #2
    You can always multiply a wave function with an arbitrary phase factor without changing the state physically. So there the author changes the wave function with

    \psi(t,x)\mapsto e^{iE_1 t/\hbar}\psi(t,x)

    and it remains physically same. It could have been clearer to use a different symbol for the wave function with an additional phase factor. For example

    \tilde{\psi}(t,x) = e^{iE_1 t/\hbar}\psi(t,x)
  4. Aug 16, 2007 #3


    User Avatar

    Good. Thank you. Mathematically, that was the only way I could think of doing it, but the author didn't bother to use a different symbol for the shifted function, or to explain what he was doing physically.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook