Overlap integrals and eigenstates problem

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SUMMARY

The discussion focuses on the calculation of probabilities in quantum mechanics using overlap integrals and eigenstates. Specifically, it explains that the probability of finding a particle in a specific position is determined by the square of the wave function, represented as <Ψ|Ψ>. To find the probability of a wave function being in a specific energy eigenstate, the overlap integral is utilized, and its square is taken to ensure the result is a real probability. The necessity of squaring the overlap integral is emphasized as it transforms the complex probability amplitude into a real probability value.

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randybryan
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To find the probability of a particle being at position x we use

<\Psi|\Psi> where the complex conjugate ensures that the answer is real. This means that we're looking at the square of the wave function to determine the probability of finding the particle.

Now to determine the probability of a wave function being in eigenstate n, we use the overlap integral of the wave function and normalised eigenstate (for general harmonic oscillator potential V(x) = 1/2m\omega^{2}x^{2}) over all x. Now I don't understand why we take the square of this answer to find the probability? I know we're not looking for the probability of a position, but an energy eigenstate. I just can't follow where it comes from.
Apologies for not writing out the equation, but the Latex is screwing up for me
 
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We DO take the square of the overlap to find the probability, in either case. The overlap is called the probability amplitude, which in general is complex, whereas a probability must be real.

In your first case, Ψ(x) (or <x|Ψ>) is the probability amplitude and Ψ*(x)Ψ(x) or <Ψ|x><x|Ψ> is the probability of finding the particle at point x. Your second case is exactly the same, except it's <n|Ψ> instead of <x|Ψ>.
 
Thanks! :)
 

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