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$$ (\;\int_{-\infty}^{\infty} \left |\Psi_n(q)\right |^2dq=1 \; \forall n \;)\;\;\&\;\; (\;\Psi(q) =\;\sum a_n\Psi_n(q)\;) \implies\; \sum \left |a_n\right |^2= \int_{-\infty}^{\infty} \left |\Psi(q)\right |^2dq $$ $$ n\in \mathbb{N},\; \Psi_n(q) \in \mathbb{C},\; \Psi(q) \in \mathbb{C},\; a_n \in \mathbb{C}, \; q \in \mathbb{R} $$
The book attempts to justify this statement with some vague arguments. I am looking for a rigourous proof of this statement. This appears to be a standard result. So, if I have missed any necessary conditions before the implies sign, please fill them in.

Also, in the book, q is treated as a general configuration space, instead of a one dimentional space as I have considered here and the integrals are done over the entire configuration space. If possible, also give proof for this general case.

Prajwal Samal
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1 Answers1

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Start from

$$ \psi(q) = \sum_n a_n \psi_n(q) \tag{1} $$

take the complex conjugate

$$ \psi^*(q) = \sum_m a_m^* \psi_m(q)^* \tag{2} $$

Multiply them

$$ \psi(q)\psi^*(q) = |\psi(q)|^2 = \sum_{mn} a_n a_m^* \psi_n(q)\psi_m^*(q)\tag{3} $$

Now, if $\{\psi_n\}_n$ is an orthogonal set, then when you integrate (3) you get

$$ \int {\rm d}q~ |\psi(q)|^2 = \sum_{mn} a_n a_m^* \underbrace{\int{\rm d}q~\psi_n(q)\psi_m^*(q)}_{\delta_{mn}} = \sum_{mn} a_n a_m^*\delta_{mn} = \sum_n a_n a_n^* = \sum_n|a_n|^2 \tag{4} $$

caverac
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  • Are you sure it cant be done without the orthogonality assumption? May with a weaker assumption. Btw, in the book, this statement has been justified first and then, using this, orthogonality has been proved. I dont know whether the book is wrong or right or has toned down the maths or what. – Prajwal Samal May 09 '18 at 10:44
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    @PrajwalSamal Orthogonality is not such a strong requirement, specially in QM: Schrodinger's equation is a Sturm-Liouville problem, which naturally leads to orthogonal eigenstates – caverac May 09 '18 at 10:46