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I Orthogonality question

  1. Mar 11, 2017 #1
    Consider two momentum eigenstates ##\phi_1## and ##\phi_2## representing momenta ##p_1## and ##p_2##. For the sake of easy numbers, ##p_1=1*\hbar## (with ##k=1##) and ##p_2=2*\hbar## (with ##k=2##). Thus, ##\phi_1=e^{ix}## and ##\phi_2=e^{2ix}##. Orthogonality states that
    ##\int \phi_1^*\phi_2dx=\int e^{-ix}e^{2ix}dx=0##
    Why is this?

    I understand how orthogonality would work with dirac deltas (i.e. I know why position eigenfunctions are orthogonal in position space and why momentum eigenfunctions are orthogonal in momentum space, etc.) but I am unclear of how it works with plane waves.

    Also, I am specifically asking why ##\int e^{-ix}e^{2ix}dx=0##, not why orthogonality works in general (I understand its derivation using the definition of hermitian operators and the inner product).
     
  2. jcsd
  3. Mar 11, 2017 #2

    hilbert2

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    You could first try to plot some graphs of real functions ##\sin k_1 x \sin k_2 x## where ##k_2## is much larger than ##k_1## and convince yourself of the fact that there's practically as much "positive" as "negative" surface area between the graph and the x-axis.
     
  4. Mar 11, 2017 #3
    I guess then that my question is how do you numerically evaluate that improper integral?
     
  5. Mar 11, 2017 #4

    hilbert2

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    When talking about rigorous mathematics, the improper integral does not converge, but I guess you can show that if you calculate an integral ##\int_{-\infty}^{\infty} e^{-a|x|}e^{-ix}e^{2ix}dx## (which does converge) and take the limit ##a \rightarrow 0+##, then you get value 0.
     
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