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Qm problem

  1. Feb 1, 2005 #1
    Hi, I have a problem.

    I want to show that

    [tex] \frac{d}{dt} \int_{-\infty}^{\infty} \psi_1^{*}\psi_2 dx = 0 [/tex]

    for any two (normalizable) solutions to the Schrödinger equation. I have tried rearranging the Schrödinger equation to yield expressions for [tex] \psi_1^{*} [/tex] and [tex] \psi_2 [/tex] like this:

    [tex] \psi_1^{*} = \frac1{V(x)}\left( ih\frac{\partial \psi_1^{*}}{\partial t} + \frac{h^2}{2m}\frac{\partial^2 \psi_1^{*}}{\partial x^2} \right) [/tex]

    [tex] \psi_2 = \frac1{V(x)}\left( ih\frac{\partial \psi_2}{\partial t} + \frac{h^2}{2m}\frac{\partial^2 \psi_2}{\partial x^2} \right) [/tex]

    It gets me nowhere...
  2. jcsd
  3. Feb 1, 2005 #2

    Tom Mattson

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    You've got 2 quantum states Ψ1 and Ψ2, which are superpositions of basis states {φkexp(iEkt/hbar)}.



    Take the product of Ψ1* and Ψ2 and integrate. If j=k, then the exponential terms cancel and the term is manifestly time independent. If j !=k, then the basis states are orthogonal and the space integral vanishes.
  4. Feb 1, 2005 #3

    Tom Mattson

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    That should be:

    [tex] \psi_1^{*} = \frac1{V(x)}\left(- ih\frac{\partial \psi_1^{*}}{\partial t} + \frac{h^2}{2m}\frac{\partial^2 \psi_1^{*}}{\partial x^2} \right) [/tex]

    You forgot the negative sign when you took the complex conjuage of the energy term.
  5. Feb 1, 2005 #4
    Thanks, the problem is, though, that this assignment is in the first chapter, where the author hasn't introduced stationary states and he hasn't mentioned the fact that every wave function can be written as a superposition like that. There must be some other way...
  6. Feb 1, 2005 #5


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    Here's the most rigurous and elegant approach:
    Schroedinger's equation:
    [tex] \frac{d|\psi (t)\rangle}{dt} =\frac{1}{i\hbar}\hat{H}(t)|\psi (t)\rangle [/tex] (1)


    Show that for two arbitary solutions of equation (1) we have:
    [tex] \frac{d}{dt}\langle \psi_{1}(t)|\psi_{2}(t)\rangle = 0 [/tex] (2)


    (2) is equivalent to:
    [tex] \langle \psi_{1}(t)|\psi_{2}(t)\rangle=\langle \psi_{1}(t_{0})|\psi_{2}(t_{0})\rangle [/tex] (3)

    Using the formalism of UNITARY TIME EVOLUTION OPERATORS:
    [tex] \langle \psi_{1}(t)|=\langle \psi_{1}(t_{0})| \hat{U}^{\dagger}(t,t_{0}) [/tex] (4)

    [tex] |\psi_{2}(t)\rangle =\hat{U}(t,t_{0})|\psi_{2}(t_{0})\rangle [/tex] (5)

    Making the scalar product and taking into account the UNITARY character of the time evolution operator,the conclusion is immediate...

    Last edited: Feb 1, 2005
  7. Feb 1, 2005 #6


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    It's quite elaborate, but you can do it from the Schrodinger equation directly by bringing the derivative inside of the integral.

    [tex] \frac{d}{dt} \int_{-\infty}^{\infty} \psi_1^{*}\psi_2 dx = \int_{-\infty}^{\infty} \frac{\partial}{\partial t}(\psi_1^{*}\psi_2) dx=\int_{-\infty}^{\infty} \left(\frac{\partial \psi_1^*}{\partial t}\psi_2 +\psi_1^*\frac{\partial \psi_2}{\partial t}\right)dx[/tex]

    Then invoke the Schrodinger equation:
    [tex]\frac{\partial \psi}{\partial t}=\frac{i\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2}-\frac{i}{\hbar}V\psi[/tex]
    (Note that you have the complex conjugate of this equation for [itex]\psi_1^*[/itex].)

    Plug it in the equation, use the fact that V is real. Then rewrite the integrand as a derivative of x, use the fundamental theorem of calculus and that any normalized wavefunction goes to 0 as x goes to infinity.

    BTW: It is actually not true that square integrability of a function f implies that f(x) goes to zero when x goes to infinity. You can even make continuous functions with this property, although they do not look like acceptable physical solutions for a wavefunction.
  8. Feb 1, 2005 #7
    There is no assumption here, this expression can easily be proved...The only thing that you need to know is the fact that taking the first derivative with respect to time yields the energy. Yes, i know, even this fact, we have already proven...Don't get to carried away with postulate number seven.


    I also favour the approach of Galileo, it is more logic and straightforeward (ofcourse that is only my opinion)
  9. Feb 1, 2005 #8


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    It is a premise...In the context of the problem it has the value of a postulate.The fact that it is not really a postulate is irrelevant for this specific problem...Not really a fan of LOGICS,are u?? :wink:

    :rofl: That's the best joke i ever heard in years... :tongue2:

  10. Feb 2, 2005 #9
    I found out by using Galileo's approach, which was the only one that could be carried out using the knowledge in the first chapter. Thanks.
  11. Feb 2, 2005 #10


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    You're welcome. :smile:

    Note that, if we take [itex]\Psi_1=\Psi_2 = \Psi[/itex] then you've just proved the following nontrivial fact:

    [tex]\frac{d}{dt}\int \limits_{-\infty}^{+\infty}|\Psi|^2dx =0[/tex]

    That is, if a wavefunction is normalized at one instant in time, it will stay normalized for all future time.
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