Orthogonality of eigenfunctions with continuous eigenvalues

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 5K views
jazznaz
Messages
23
Reaction score
0

Homework Statement



With knowledge of the orthogonality conditions for eigenfunctions with discrete eigenvalues, determine the orthonormal set for eigenfunctions with continuous eigenvalues. Use the definition of completeness to show that | a(k) |^2 = 1.

2. The attempt at a solution

The first step is:

http://img55.imageshack.us/img55/5229/81115215vg1.jpg
http://g.imageshack.us/img55/81115215vg1.jpg/1/

Since the integral is only equal to 0 when k' = k. (The same condition as the kronecker delta.)

Next:

http://img352.imageshack.us/img352/9519/94213081to0.jpg
http://g.imageshack.us/img352/94213081to0.jpg/1/

After here I get a bit lost, even though I think this is almost the solution. My work differs from the course notes and a QM book I've looked through. They both have an integration with respect to k for the RHS, not x.

Any pointers would be greatly appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
jazznaz said:

Homework Statement



With knowledge of the orthogonality conditions for eigenfunctions with discrete eigenvalues, determine the orthonormal set for eigenfunctions with continuous eigenvalues. Use the definition of completeness to show that | a(k) |^2 = 1.

2. The attempt at a solution

The first step is:

http://img55.imageshack.us/img55/5229/81115215vg1.jpg
http://g.imageshack.us/img55/81115215vg1.jpg/1/

Since the integral is only equal to 0 when k' = k. (The same condition as the kronecker delta.)

no. the right hand side is always equal to zero except when k'=k

Next:

http://img352.imageshack.us/img352/9519/94213081to0.jpg
http://g.imageshack.us/img352/94213081to0.jpg/1/

After here I get a bit lost, even though I think this is almost the solution. My work differs from the course notes and a QM book I've looked through. They both have an integration with respect to k for the RHS, not x.

Any pointers would be greatly appreciated.

The very first line is incorrect. The RHS should not be integrated over x but rather over k.
 
Last edited by a moderator:
Should I integrate the RHS w.r.t. k, multiply both sides by the conjugate of Phi and then integrate both sides over x?
 
jazznaz said:
Should I integrate the RHS w.r.t. k, multiply both sides by the conjugate of Phi and then integrate both sides over x?

First... get the definition of [itex]\psi(x)[/itex] in terms of a(k) and [itex]\phi_k(x)[/itex] correct.

Then square psi(x) (which is in terms of a(k) and phi(k,x) now) and integrate over x using the "first step" you mention in the OP.

Use the fact that psi(x) is normalized to see that a(k) is is normalized.