Relationship between commutators and observables

[playoff]

Homework Statement

Suppose A^ and B^ are linear quantum operators representing two observables A and B of a physical system. What must be true of the commutator [A^,B^] so that the system can have definite values of A and B simultaneously?

Homework Equations

I will use the bra-ket notation for the inner product (sorry for lack of latex)

The Attempt at a Solution

So I assumed that for observables A and B to have a definite value, <psi*|A^|psi> and <psi*|B^|psi> have to be normalizable. Call a and b normalization constants of <psi*|A^|psi> and <psi*|B^|psi> respectively, then: a<psi*|A^|psi> = b<psi*|B^|psi> = 1.

Here I made my sloppy assumption that the equation above implies that A^ and B^ are proportional (A^=(b/a)*A^), which leads to that the commutator must be zero.

I made this assumption as both A^ and B^ are being "operated" by the same inner product, so the bolded equation can be reduced to aA^=bB^. Is this an assumption I can make?

Thanks in advance.

 

[fzero]

The equation that you wrote down doesn't lead to any relation between ##\hat{A}## and ##\hat{B}##, since, as long as the expectation values are nonzero, we can always find ##a## and ##b## as the reciprocal of the corresponding expectation values. The values of ##a## and ##b## are unrelated.

The key here is to understand what "definite value" means in this context. For a system to have a definite value of an observable, it must be in an eigenstate of the observable. The problem wants you to figure out what this means when you have two observables

 

[playoff]

@fzero: Yes, I had a totally wrong understanding of a definite value. It makes sense that an operator has a definite value if the normalized wavefunction is its eigenfunction, thus the inner product becomes the eigenvalue.