# Operators satisfying abstract commutation relation; then finding an eigenvalue.

1. Feb 7, 2012

### arp777

So, my problem statement is:

Suppose that two operators P and Q satisfy the commutation relation [Q,P] = Q .
Suppose that ψ is an eigenfunction of the operator P with eigenvalue p. Show that Qψ is also an eigenfunction of P, and find its eigenvalue.

This shouldn't be too difficult, but I'm wrestling with this one. I can't seem to figure out the significance of the commutation relation in the prompt, or whether or not it's saying anything particularly relevant to finding the eigenvalue of P..

Any help is greatly appreciated.

2. Feb 7, 2012

QP-PQ=Q

Pψ=pψ

Now PQψ=?

3. Feb 7, 2012

### arp777

Following my knowns, I'm tempted to say that:

PQ = QP - Q ... based on the commutation relation.

Which would mean that PQψ = QPψ - Qψ

Showing you what I worked out before posting might help you understand where my confusion's coming from. I started with:

relation: [Q,P] = Q

which means, QPψ - PQψ = Q(pψ) - [Q(pψ) - Qψ] = Qψ

So, how do I show that Qψ is also an eigenfunction? What needs to be true?

Thanks again.

4. Feb 7, 2012

### Staff: Mentor

I hope you aren't slipping a homework problem in here

What you're trying to show is that: PQψ = κQψ; that is, Qψ is an eigenfunction of P with eigenvalue κ.

You got as far as PQψ = QPψ - Qψ; from there if you substitute the known value of Pψ you'll end up with your answer pretty quickly.

5. Feb 7, 2012

### arp777

Hahah, well, it is a homework problem! But I get better help on here than I do in the study center without being given the answer in either place. (Believe me, I'm IN the study center right now). Kinda sad, huh!? Thanks for the help, ya'll.