Can Qpsi be an eigenfunction of P given the commutation relation [P,Q]=P?

This will give you the desired relationship between the eigenvalues of QY and P.In summary, the operators P and Q satisfy the commutation relation [P,Q]=Q. If psi is an eigenfunction of P with eigenvalue p, then Qpsi is also an eigenfunction of P with eigenvalue q, where q=p+1. This can be shown by using the commutation relation and the fact that p is just a number.
  • #1
lilsalsa74
3
0

Homework Statement


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.


Homework Equations





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.
 
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  • #2


lilsalsa74 said:
I thought [P,Q]=PQ-QP=0 if the two operators commute.
The question doesn't say that P & Q commute does it?
 
  • #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


lilsalsa74 said:
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?
 
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  • #5


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


lilsalsa74 said:
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 ...)
 
  • #7


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


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


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


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.
 

FAQ: Can Qpsi be an eigenfunction of P given the commutation relation [P,Q]=P?

Can you explain the concept of an eigenfunction?

An eigenfunction is a special type of function that, when operated on by a linear operator, returns a scalar multiple of itself. This scalar multiple is known as the eigenvalue. In other words, an eigenfunction is a function that is unchanged (up to a scalar multiple) when operated on by a specific operator.

What is a commutation relation?

A commutation relation is a mathematical relationship between two operators, A and B, that describes how they behave when they are applied to the same function. In other words, it describes how the order in which the operators are applied affects the final result. Operators that commute with each other will give the same result regardless of the order they are applied in, while operators that do not commute may give different results depending on the order.

How do you determine if a show operator is an eigenfunction of another operator?

To determine if a show operator is an eigenfunction of another operator, you need to check if the show operator and the other operator commute with each other. If they do, then the show operator can be an eigenfunction of the other operator. This means that the show operator will return a scalar multiple of itself when operated on by the other operator.

What is the significance of a show operator being an eigenfunction of another operator?

If a show operator is an eigenfunction of another operator, it means that the show operator and the other operator commute with each other. This is significant because it allows us to simplify mathematical calculations and make predictions about the behavior of the system. It also allows us to find the eigenvalues and eigenfunctions of the system, which can provide important information about its properties.

Can an operator have multiple eigenfunctions?

Yes, an operator can have multiple eigenfunctions. This is because an operator can have multiple scalar multiples (eigenvalues) for a given eigenfunction. In other words, there can be multiple functions that are unchanged (up to a scalar multiple) when operated on by the same operator. These different eigenfunctions can also have different eigenvalues, which can provide additional information about the system.

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