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Normalisation of Schrodinger Eq.

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose you assume that you have normalised a wave function at t = 0. How do you know that it will stay normalised as time goes on? Show explicitly that the Schrodinger equation has the property that it preserves normalistion over time.

    2. Relevant equations

    From my notes I have deduced:

    [tex]\frac{d}{dt} \int_{-\infty}^{+\infty} \left| \Psi\left(x,t\right)\right|^2 dx = \int_{-\infty}^{+\infty} \frac{\partial}{\partial t} \left| \Psi\left(x,t\right)\right|^2 dx = 0[/tex]

    However I have failed to come up with a solution.

    EDIT: Should I rearrange the Schrodinger equation for d/dt of psi then solve that way? I'm confused.
     
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2

    Hurkyl

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    You know how to compute derivatives, right?

    (bra-ket notation might be easier, though.... no sense in using integrals when you can just do algebra!)
     
  4. Feb 22, 2009 #3

    malawi_glenn

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    The schrödinger equation

    [tex]i\hbar \frac{\partial}{\partial t} \Psi (x,t) = E\Psi (x,t)[/tex]

    suggest that you can write the time-evolution of a state according to:

    [tex] \Psi(x,t) = \psi(x) e^{-i{E t\, \hbar}}[/tex]

    Now, what will this give you?

    (this is the way I was taught how to show that [itex]
    \int_{-\infty}^{+\infty} \left| \Psi\left(x,t\right)\right|^2 dx = [/itex] is constant over time. )
     
  5. Feb 22, 2009 #4
    So I get: [tex]\frac{\partial}{\partial t} \left|\Psi\right|^2 = \Psi \frac{\partial \Psi^*}{\partial t} + \Psi^* \frac{\partial \Psi}{\partial t}[/tex]
    What should I use for my wave function? [tex]\Psi\left(x,t\right) = Ae^{i(kx-wt)/\hbar}[/tex] ?
     
  6. Feb 22, 2009 #5

    malawi_glenn

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  7. Feb 22, 2009 #6

    Hurkyl

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    What's wrong with just calling it [itex]\Psi[/itex], and working with it symbolically?

    Although... you could prove the theorem first for stationary states, as malawi_glenn is implicitly suggesting....


    Don't you know something about that? ....
     
  8. Feb 22, 2009 #7

    malawi_glenn

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    [tex]
    \frac{\partial \Psi}{\partial t}
    [/tex]

    hint: The Schrödinger equation.

    And for [tex]
    \frac{\partial \Psi^*}{\partial t}
    [/tex]

    The complex conjugate of the schrödinger equation.
     
  9. Feb 22, 2009 #8
    How's this...

    [tex]E \Psi(x,t) = i\hbar \frac{\partial}{\partial t} \Psi(x,t), \Rightarrow E \Psi^*(x,t) = -i\hbar \frac{\partial}{\partial t} \Psi^*(x,t)[/tex]

    Giving:

    [tex]\frac{\partial \Psi}{\partial t} = -\frac{iE}{\hbar} \Psi(x,t)[/tex] and [tex]\frac{\partial \Psi^*}{\partial t} = \frac{iE}{\hbar} \Psi^*(x,t)[/tex]

    So:

    [tex]\frac{\partial}{\partial t} |\Psi|^2 = |\Psi|^2 (\frac{iE}{\hbar} - \frac{iE}{\hbar}) = 0[/tex]

    Is this making sense?
     
  10. Feb 22, 2009 #9

    malawi_glenn

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    yes, very good! Recall that this is possible since E is real.
     
  11. Feb 22, 2009 #10
    So the integral of zero is zero thus proving the initial statement. Yay!

    Cheers for the help guys, all seems so easy now!
     
  12. Feb 22, 2009 #11

    Hurkyl

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    But only in the case of stationary states! You still have to do some more work for an arbitrary solution to the Schrödinger equation.

    Fortunately, it's not much work, since the general case features the unitary operator H rather than the real number E. Fortunately, unitary operators share enough of the properties of real numbers that the proof works out pretty much unchanged.



    Incidentally, I think the main lesson here is when you don't know what you're supposed to do, instead try what you know how to do. You know how to take derivatives, you know to use the Schrödinger equation to compute time derivatives of quantum states -- you just had to try and string them along together. A great many problems can be solved this way: just do what you can and eventually you get to a point where you can see how to connect what you can do to what you're trying to do, and then you've done it!

    (Special cases can help too -- e.g. trying stationary states first because it's simpler)
     
  13. Feb 23, 2009 #12

    malawi_glenn

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    Is H unitary? I thought one use the fact that since [tex]H^{\dagger}=H[/tex], i.e hermitian, the energy eigenvaules E are real.

    But Hurkyl has correct that using that E is real still only proves the stationary case.

    What the real proof is about is that [tex]H^{\dagger}=H[/tex], the Schrödinger equation is
    [tex]
    H\Psi(x,t) = i\hbar \frac{\partial}{\partial t} \Psi(x,t)
    [/tex]

    And the complex conjugated will use [tex]H^{\dagger}[/tex]

    I agree on Hurkyl's advice, don't think TOO much, use what you know and see if it works.
     
    Last edited: Feb 23, 2009
  14. Feb 23, 2009 #13

    Hurkyl

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    Bleh, that's what I meant. Thanks for the correction!
     
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