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Quantum mechanics, wavefunction problem

  1. Oct 25, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider the wavefunction [itex]\Psi (x,t)=c_1 \psi _1 (x)e^{-\frac{iE_1t}{\hbar}}+c_2 \psi _2 (x)e^{-\frac{iE_2t}{\hbar}}[/itex] where [itex]\psi _1 (x)[/itex] and [itex]\psi _2 (x)[/itex] are normalized and orthogonal. Knowing [itex]\Psi (x,0)[/itex], find the values of [itex]c_1[/itex] and [itex]c_2[/itex].


    2. Relevant equations

    [itex]C^2 \int _{-\infty}^{\infty} |\Psi (x,t)|^2 dx=1[/itex]. I also know that the product of psi 1 by psi 2 is worth 0 (they are orthogonal) so this simplifies the expression to integrate.
    But I'm still left with the integration of both lower case psi functions that I don't know how to handle.
    3. The attempt at a solution
    I'm thinking on how to use the fact that I know Psi at t=0 but so far I'm out of ideas.
     
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  3. Oct 26, 2011 #2

    ehild

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    [itex]\psi _i (x)[/itex]-s are orthonormal, what does it mean on the integrals of [itex]\psi _i ^* (x) \psi _j (x)[/itex]?

    ehild
     
  4. Nov 22, 2011 #3

    fluidistic

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    Sorry for being almost 1 month late. That they are worth 1?

    Edit: Is what I've done right?:
    [itex]\int _{-\infty}^{\infty} \Psi _1 (x) \Psi (x,0)dx=c_1 \int _{-\infty}^{\infty} \Psi _1 ^2 (x)dx=c_1[/itex].
    In the same fashion I get [itex]c_2=\int _{-\infty}^{\infty} \Psi _2 (x) \Psi (x,0)dx[/itex].
    But I didn't use any complex conjugate... I must be missing something.
     
    Last edited: Nov 22, 2011
  5. Nov 22, 2011 #4

    vela

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    That's almost correct. It should be
    [tex]c_i = \int \psi_i^*(x)\Psi(x,0)\,dx[/tex]When you calculate the inner product, you use the complex conjugate of the first function in the integrand.
     
  6. Nov 23, 2011 #5

    fluidistic

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    Let's see if I understand:
    [itex]\int _{- \infty}^{\infty} \Psi _1 ^* (x) \Psi (x,0)dx=c_1 \int _{- \infty}^{\infty} \Psi _1 ^* (x) \Psi _1 (x)dx=c_1 \int _{-\infty} ^{\infty} |\Psi _1 (x)|^2 dx=c_1[/itex].
    Am I right if I say that since [itex]\Psi _1(x)[/itex] and [itex]\Psi _2 (x)[/itex] are orthogonal, so is [itex]\Psi _1^* (x)[/itex] with [itex]\Psi _2 (x)[/itex]? Because I'm using this fact!
    Thanks for your help guys.
     
  7. Nov 23, 2011 #6

    vela

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    Yup. When you say the two are orthogonal, you mean that
    [tex]\int \psi_1^*(x)\psi_2(x)\,dx = 0.[/tex]It does not mean that
    [tex]\int \psi_1(x)\psi_2(x)\,dx = 0.[/tex]The complex conjugation is a vital part of taking the inner product.
     
  8. Nov 23, 2011 #7

    fluidistic

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    Oh, I understand. Wow, thank so you much. My understanding was so poor before I asked help for this problem... now things are getting clearer.
     
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