Orthogonality question

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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).
 

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  • #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.
 
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  • #3
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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.
I guess then that my question is how do you numerically evaluate that improper integral?
 
  • #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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