Expressing Wavefunction psi(x) Using Eigenstates |ai> for State |psi>

[jrevill]

Given a state |psi> and an operator A with (non-degenerate) eigenstates |ai> corresponding to real eigenvalues ai, express the wavefunction psi(x) corresponding to the state |psi> in terms of |psi>

Now this was a question in last year's exam that my lecturer went through, but his answer on the whiteboard was:

psi(r) = <r|psi>

But the question asked for psi(x), not psi(r). Are the two then interchangeable?

 

[dextercioby]

Do you know what a spectral decomposition of a linear selfadjoint operator is ? If so, then write it down and then act with it on |psi>.

Daniel.

 

[jrevill]

Okay, well from Wikipedia: "The spectral decomposition of an operator A which has an orthonormal basis of eigenvectors is obtained by grouping together all vectors corresponding to the same eigenvalue."

So you're referring to a spectral decomposition of the r operator? And the orthonormal basis of eigenvectors would be... x, y and z?

 

[dextercioby]

This is the spectral decomposition of A (self-adjoint)

A=\sum_{i} a_{i}|a_{i}\rangle \langle a_{i}|

, where the sigma symbol means, as usually, the sum over the discrete spectrum and an integral over parameter space for the continuous spectrum.

Since you need \psi (x)= \langle x|\psi\rangle, and A is self-adjoint, therefore its (possibly generalized) eigenvectors span a basis in the (rigged) Hilbert space of the system, so then |a_{i}\rangle form a complete set, all you need to do is find \psi (x) in terms of a_{i}(x).

Daniel.