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If $L$ is an Hermitian matrix associated with an apparatus acting on state $\Psi$, how does the state vector $\Psi$ collapse or change according to matrix arithmetic?

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  • $\begingroup$ Your language is really unclear... $\endgroup$ – childofsaturn Dec 26 '15 at 15:21
  • $\begingroup$ How much quantum mechanics have you studied? What about linear algebra? $\endgroup$ – Kyle Kanos Dec 26 '15 at 16:29
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    $\begingroup$ It's really not clear what you are asking. The question itself lacks a proper verb ("how do we [verb!] $\Psi$ collapses"), and you should be aware that there are interpretations of QM without collapse. $\endgroup$ – ACuriousMind Dec 26 '15 at 19:24
  • $\begingroup$ Wave function collapse is a separate postulate. The collapse occurs when a projector acts on a state vector. These are all vectors, there are no matrices involved. $\endgroup$ – Sidd Dec 30 '15 at 21:24
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The expectation value for an operator $\hat L$ acting on a wave $\Psi$ is:

$ \langle \Psi \vert \hat L \vert \Psi \rangle =\\ \int_\Omega [\Psi(\mathbf r)]^\dagger \hat L \Psi(\mathbf r) dV $

(Is that what you asked?) (Seems like a density matrix is also used for computing the expected value in more complicated quantum mechanics)

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  • $\begingroup$ no I didn't ask for that, anyways I think got the answer to my question, just see my answer. $\endgroup$ – Manish Kumar Singh Dec 26 '15 at 18:34
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Actually while studying, I had an issue as to when should $\Psi$ collapse and when should it just modify. I now have my answers . Its basically that we say $\Psi$ collapse only if we exclude apparatus outside the quantum system, otherwise when includes, the combined system is in a bell state which upon inspection of apparatus gives the the final state- what we think of a the state $\Psi$ has collapsed to. So in a nutshell there is no dis ambiguity on whether to collapse or just modify.

P.S question was asked by me and now i guess I have an answer.

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