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

by lilsalsa74
Tags: commutation, eigenfunction, operator, relation
 P: 3 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.
Emeritus
Sci Advisor
PF Gold
P: 9,789
 Quote by lilsalsa74 I thought [P,Q]=PQ-QP=0 if the two operators commute.
The question doesn't say that P & Q commute does it?
 P: 3 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?
Emeritus
Sci Advisor
PF Gold
P: 9,789

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

 Quote by 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?
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?
 P: 3 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?
Sci Advisor
P: 1,185
 Quote by 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?
Yep. (Except that, in your first post, you say [P,Q]=P, not Q; you switched to Q in a later post ...)
 P: 28 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.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,534 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.
 P: 28 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.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,534 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.

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