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Homework Help: Violation of the laws of quantum physics?

  1. Nov 14, 2008 #1
    If I want to calculate <x> of an electron that is in a state described by [tex]\psi (x,t)=\frac{1}{\sqrt{2}}[\psi_{0}(x,t) + \psi_{1}(x,t)]][/tex] where [tex]\psi_{0} [/tex] and [tex]\psi_{1}[/tex] are the two lowest energy states of an electron in a one dimensional box, can I then simply calculate <x> for [tex]\frac{1}{\sqrt{2}}\psi_{0}[/tex] and [tex]\frac{1}{\sqrt{2}}\psi_{1}[/tex] alone and add them?

    Is it even possible for an electron to have this energy? Doesn't it violate the laws of quantum physics? I find it strange that this was an exam problem in quantum physics at my university recently.
    Last edited: Nov 14, 2008
  2. jcsd
  3. Nov 14, 2008 #2
    You know, what <x> means?
    It is [tex]\langle x\rangle=\frac12\int_{-\infty}^{\infty}dx x (\psi_0^2+\psi_1^2+2\psi_0\psi_1)[/tex].
    Indeed, [tex]\int_{-\infty}^{\infty}dx 2\psi_0\psi_1=0[/tex]
    but [tex]\int_{-\infty}^{\infty}dx 2x\psi_0\psi_1\neq0[/tex].
    So you can't calculate <x> for $\psi_0$ and $\psi_1$ separately.
    Sure, it's a legal state in quantum physics.

    http://www.shareapic.net/content.php?id=12668640&owner=dabi [Broken]
    Last edited by a moderator: May 3, 2017
  4. Nov 14, 2008 #3
    The energy must still be either E0 or E1, right?
  5. Nov 14, 2008 #4
    If you measure the energy, you will find either E0 or E1. And that is the special property of quantum mechanics, that the probability is for either case 50%.
  6. Nov 14, 2008 #5
    But it's not possible to measure E with certainty, right?

    And the [tex]\frac{1}{\sqrt{2}}[/tex] is there just to normalise the probability function?
  7. Nov 14, 2008 #6
    Given your other post about <p>, first read your textbook!
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