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Relationship between commutators and observables

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data
    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?

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

    3. 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.
     
  2. jcsd
  3. Jun 3, 2015 #2

    fzero

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    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
     
  4. Jun 4, 2015 #3
    @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.
     
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