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[PhD Qualifier] Commutation relation

  1. Jul 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Two quantum mechanical operators obey the following commutation relation.
    Given this commutation relation which of the following are true or false? Justify your answers.
    a) The two observables are simultaneously diagonalizable.
    b) The two satisfy a Heisenberg uncertainty relation that has the form
    c) They are spin operators.
    d) The Hamiltonian [tex]\hat{H}=\hat{A}^2+\hat{B}^2[/tex] could describe a harmonic oscillator system.

    2. Relevant equations

    3. The attempt at a solution
    a) False - simultaneously diagonalizable ==> simultaneously observable. Since they don't commute, they aren't simultaneously observable and can't be simultaneously diagonalizable.
    b) No idea how to get there from here
    c) They could be, but don't have to be. They satisfy a commutation relation for angular momentum (as do spin operators), but other angular momentum operators also satisfy the relation.
    d) True. Dropping a factor of [tex]\hbar[/tex], A and B could be position and momentum operators (respectively), which gives a Hamiltonian of [tex]H=X^2+P^2[/tex].

    I could use a hint or two on b, as well as someone verifying my logic for a, c, and d. Thanks!
  2. jcsd
  3. Jul 26, 2008 #2
    For b., and any other heisenburg type relation, you need to invoke the schwartz inequality.
  4. Jul 26, 2008 #3
    I suppose I can answer (b) using my observation from (d).

    Let [tex]A=\frac{X}{\hbar}, B=P[/tex]. Heisenberg's relation is [tex]\Delta x\Delta p \ge \frac{\hbar}{2}[/tex], so [tex]\Delta a\Delta b\ge\frac{1}{2}[/tex]. Square it to get "true".

    Can someone verify these answers for me?
  5. Jul 26, 2008 #4
    I will tell you that b is true, but that's a tremendously ad hoc way to determine it.

    If you have Griffith's "Quantum Mechanics" he performs the full derivation the uncertainty principle for two arbitrary operators.
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