Check Answers to Quantum Minima Qs: In-State Vector Operator

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The discussion revolves around the interpretation of state vectors and operators in quantum mechanics. When an operator, such as momentum (p), is placed between a bra and a ket, it transforms the state vector, resulting in a new ket, denoted as phi. The matrix element obtained from this operation, phi(x), is a complex number representing the wavefunction of the state but does not correspond to the eigenvalue of the operator p. The participants clarify that phi(x) is a generic wavefunction, emphasizing its distinction from spectral values of the operator. This highlights the nuanced understanding of wavefunctions and operators in quantum mechanics.
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I'm trying to "sure"up my Quantum Mechanics and found an sheet of mostly conceptual review questions. A sort of "Quantum minima".

I'd like to check my answers to a couple of these.

Firstly:

In terms of state vector the wavefunction is the expansion coefficient (probability amplitude) of a state i.e

\psi = \langle x \left | \psi \right\rangle

What happens if you put an operator (e.g. p) between the bra and ket as if you were calculating an expectation value? i.e.

\left\langle x \left | p \right | \psi \right \rangle

My guess is that the action of the operator on the state vector \left|\psi\right\rangle will collapse the vector to a single basis vector of p \left| p \right\rangle the expression above would reduce to an element of a basis transformation matrix \left\langle x | p \right\rangle

Is that correct...at least in part?
 
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Not quite. The action of p on the ket psi leads to a ket phi, generally different. Then the matrix element with the bra x leads to a complex number, namely phi(x).
 
Thanks for the reply.

What would the physical interpretation of the complex number phi(x) be? The wavefunction of the state with eigenvalue of p?
 
No, it's a generic wavefunction. No relation to spectral values of p.
 
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