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Assume that we have a Hamiltonian eigenvalue problem with continuous energy eigenvalues $E$. Griffiths says that the inner product of an eigenstate $ψ$ with the total wavefunction $Ψ$ gives the amplitude $c(E)$. I'm curious if this amplitude, $c $, is always the coefficient of the expansion of the total wavefunction in the continuous eigenstate basis - in other words if the total wavefunction $Ψ$, can be always written as the integral of $(c(E)*ψ)dE$.

In the case of a free particle this is true - but I guess that's a result of Fourier transform.

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In general, the basis of eigenvectors of a Hermitian operator is always complete. Because the Hamiltonians are Hermitian, the set of energy eigenstates will always form a complete basis, and therefore you will be able to describe any state as a linear combination of them. Be this linear combination an integral, in the case of a continuous spectrum, or a series for discrete spectra.

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    $\begingroup$ But will the coefficients of this linear combination always be the probability amplitudes in the continuous case? $\endgroup$
    – MTYS
    Dec 24, 2022 at 20:48
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    $\begingroup$ Yes, by definition a probability amplitude is the projection of your total state onto a particular state, and hence it will be the "expansion coefficient" associated to it. $\endgroup$
    – FranDahab
    Dec 24, 2022 at 20:51
  • $\begingroup$ This is clear for me in the discrete state but not in the continuous. Born's rule says that the probability amplitude is (ψ,Ψ). Why would this be the coefficient of the expansion? How do we prove this? If we accept as a definition what you're saying, then we also need to be able to prove that the coefficients in the continuous case are (ψ,Ψ), right? $\endgroup$
    – MTYS
    Dec 24, 2022 at 20:55
  • $\begingroup$ Sure, proving this is a good exercise. It's clearer with a nondegenerate spectrum, where this is a direct consequence of the fact that eigenvectors of Hermitian operators associated to different eigenvalues are orthogonal. Using this, the derivation should be pretty direct. $\endgroup$
    – FranDahab
    Dec 24, 2022 at 21:01
  • $\begingroup$ But that's for the discrete spectrum - trivial. In the continuous spectrum the coefficients are inside the integral. I don't know how to proceed in the continuous case. $\endgroup$
    – MTYS
    Dec 24, 2022 at 21:10

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