# Show operator can be an eigenfunction of another operator given commutation relation

1. Oct 16, 2008

### lilsalsa74

1. The problem statement, all variables and given/known data
Suppose that two operators P and Q satisfy the commutation relation: [P,Q]=P. Suppose that psi is an eigenfunction of the operator P with eigenvalue p. Show that Qpsi is also an eigenfunction of P, and find its eigenvalue.

2. Relevant equations

3. The attempt at a solution
First off, I know that if psi is an eigenfunction of P it means that P(psi)=p*psi. If Qpsi is also an eigenfunction of P it means that P(Qpsi)=q*Qpsi. p and q would be the eigenvalues. I also know that I have to use the commutation relation to manipulate these two equations. What I don't understand is how [P,Q] can equal Q. I thought [P,Q]=PQ-QP=0 if the two operators commute.

2. Oct 16, 2008

### Hootenanny

Staff Emeritus
Re: Show operator can be an eigenfunction of another operator given commutation relat

The question doesn't say that P & Q commute does it?

3. Oct 16, 2008

### lilsalsa74

Re: Show operator can be an eigenfunction of another operator given commutation relat

Correction: The two operators P and Q satisfy the commutation relation [P,Q]=Q.
It doesn't say that they commute but that they satisy the relation. How else can they satisfy the relation if they don't commute?

4. Oct 16, 2008

### Hootenanny

Staff Emeritus
Re: Show operator can be an eigenfunction of another operator given commutation relat

What condition must two operators satisfy to be said to commute?

(HINT: You said it yourself in your first post)

Edit: Perhaps I'm being a little too cryptic here. My point was merely that to commute P and Q must satisfy [P,Q] = 0, since they don't they do not commute. However, does because they do not commute doesn't mean they cannot satisfy a general commutation relation.

Does that make sense?

Last edited: Oct 16, 2008
5. Oct 16, 2008

### lilsalsa74

Re: Show operator can be an eigenfunction of another operator given commutation relat

So P and Q satisfy the given relation...this means that PQ-QP=Q? Is this the correct expression I should be using to evaluate the eigenvalues?

6. Oct 16, 2008

### Avodyne

Re: Show operator can be an eigenfunction of another operator given commutation relat

Yep. (Except that, in your first post, you say [P,Q]=P, not Q; you switched to Q in a later post ...)

7. Nov 16, 2010

### Kvm90

Re: Show operator can be an eigenfunction of another operator given commutation relat

Hey I'm working on the same problem. Are you saying that Q=0? I don't understand why P and Q 'must' commute to 0.

8. Nov 16, 2010

### vela

Staff Emeritus
Re: Show operator can be an eigenfunction of another operator given commutation relat

They don't. If P and Q commute, that means [P,Q]=0. You're given that [P,Q]=P (or [P,Q]=Q), so P and Q obviously don't commute.

9. Nov 16, 2010

### Kvm90

Re: Show operator can be an eigenfunction of another operator given commutation relat

Ok so here's my thinking:

Let's say Y is Psi--

[P,Q] = PQ - QP = Q
= PQY - QPY = QY plug in (PY=pY)
= PQY - QpY = QY
PQY = QY + QpY

is the eigenvalue of QY then QY + QpY? i'm pretty sure the answer to that question is no, but I don't know where to go from here.

10. Nov 16, 2010

### vela

Staff Emeritus
Re: Show operator can be an eigenfunction of another operator given commutation relat

The eigenvalue p is just a number, so it commutes with Q in the last term. Then you can factor QY out on the RHS of the equation.