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

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.

Hootenanny

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

lilsalsa74

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?

Hootenanny

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?

lilsalsa74

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?

Avodyne

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

Kvm90

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.

vela

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.

Kvm90

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.

vela

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.