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Show operator can be an eigenfunction of another operator given commutation relation 
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#1
Oct1608, 06:09 PM

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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]=PQQP=0 if the two operators commute. 


#2
Oct1608, 06:13 PM

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PF Gold
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#3
Oct1608, 06:16 PM

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? 


#4
Oct1608, 06:21 PM

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PF Gold
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Show operator can be an eigenfunction of another operator given commutation relation
(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? 


#5
Oct1608, 09:35 PM

P: 3

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



#6
Oct1608, 11:48 PM

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#7
Nov1610, 09:50 PM

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.



#8
Nov1610, 10:06 PM

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PF Gold
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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
Nov1610, 10:17 PM

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. 


#10
Nov1610, 10:19 PM

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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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