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Quantum mechanics inner products

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Given [tex] (\psi_1, \psi_2)=\int dx \psi_1^*(x) \psi_2(x) [/tex], show [tex] (\psi_1, \psi_2)=\int dk \phi_1^*(k) \phi_2(k) [/tex], where [tex] \phi_{1,2}(k)= \int dx \psi_k^*(x) \psi_{1,2}(x) [/tex] and [tex] psi_k(x)=\frac{1}{\sqrt{2 \pi}} e^{ikx} [/tex].

    2. Relevant equations

    [tex] \psi (x)= \int dk \phi(k) \psi_k(x) [/tex]

    [tex] \psi(x)=\int dk \phi(k) \psi_k(x) [/tex]

    3. The attempt at a solution

    [tex] (\psi_1 , \psi_2)= \int dx \left \{ \int dk \phi_1^*(k) \psi_k^*(x) \right \}\left \{ \int dk \phi_2(k) \psi_k(x) \right \} [/tex]

    [tex] =\int dx \left \{ \int dk \phi_1^*(k) \frac{1}{\sqrt {2 \pi}}e^{-ikx} \int dk \phi_2(k) \frac{1}{\sqrt {2 \pi}}e^{ikx} \right \} [/tex]

    [tex] = \frac{1}{2 \pi}\int dx \left \{ \int dk \phi_1^*(k) \int dk \phi_2(k) \right \} [/tex]

    Is this correct so far? How do I proceed from here? It looks like a Fourier Transform with the 1/2pi. And I have two integrals within another one for the dx. Can I separate them some how?
     
  2. jcsd
  3. Oct 30, 2011 #2

    diazona

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    When you substitute in the integrals for [itex]\psi_1^*(x)[/itex] and [itex]\psi_2(x)[/itex], the [itex]k[/itex] in one integral isn't the same variable as the [itex]k[/itex] in the other integral. If you use the same variable for both, you're almost certainly going to confuse yourself. In this case, you can't cancel out the [itex]e^{-ikx}[/itex] factor from one integral with [itex]e^{ikx}[/itex] from the other.
     
  4. Nov 1, 2011 #3
    [tex] =\int dx \left \{ \int dk_1 \phi_1^*(k_1) \frac{1}{\sqrt {2 \pi}}e^{-ik_1x} \int dk_2 \phi_2(k_2) \frac{1}{\sqrt {2 \pi}}e^{ik_2x} \right \} [/tex]

    I have revised my equation above. But now that the exponents don't cancel, how should I proceed with this?
     
  5. Nov 1, 2011 #4

    vela

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    You want to use the fact that [tex]\delta(k_1-k_2) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{i(k_1-k_2)x}\,dx[/tex]
     
  6. Nov 2, 2011 #5
    [tex] (\psi_1, \psi_2)=\delta(k_2-k_1) \int dk_1 \phi_1^*(k_1) \int dk_2 \phi_2(k_2) [/tex]

    Is the delta function equal to 1 in this case? And has the equation given in the problem [tex] (\psi_1, \psi_2)=\int dk \phi_1^*(k) \phi_2(k) [/tex] combined k_1 and k_2?
     
  7. Nov 2, 2011 #6

    vela

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    The delta function depends on k1 and k2. You can't just pull it out front like that.

    By using the delta function, you can perform one of the integrations. I suggest you read about the delta function to see how that works.
     
  8. Nov 2, 2011 #7
    I obtained [tex] (\psi_1, \psi_2)=\int dk_2 \phi_2(k_2) \int dk_1 \phi_1^*(k_1) \delta(k_2-k_1) = \int dk_2 \phi_2(k_2) \phi_1^* (k_2) [/tex] by using [tex] \psi(x)= \int dx' \psi(x') \delta(x-x') [/tex]

    Is it correct? Thanks for the help once again.
     
  9. Nov 2, 2011 #8

    vela

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    Yes, that's correct.
     
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