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Harmonic oscillators and commutators.

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data

    If we have a harmonic oscillator with creation and annhilation operators [itex]a_{-} a_{+} [/itex], respectively. The commutation relation is well known:

    [tex] [a_{+},a_{-}] = I [/tex]

    However, if we have two independent oscillators with operators [itex]a'_{-} a'_{+} [/itex]

    As the operators are the same function would they commute if mixed.

    i..e is [tex] [a_{+},a'_{-}] = I [/tex]

    Still valid?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 23, 2011 #2

    vela

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    When you say two operators A and B commute, it means AB=BA so that [A,B]=AB-BA=0.

    As to your question about the commutation relation, the answer is no. The creation and annihilation operators for the one oscillator commute with the operators for the other oscillator.
     
    Last edited: Oct 23, 2011
  4. Oct 23, 2011 #3
    So [itex][a_{+},a'_{-}] = 0 [/itex]

    Is there any reason for this beside the operators for one oscillator have no effect on the states of another?

    Would assume that any other mix of these operators would commute?

    e.g [a'_{-}a_{+},a'_{+},a_{+}] = 0
     
  5. Oct 23, 2011 #4

    vela

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    The position and momentum operators, x1 and p1, for one oscillator commute with the position and momentum operators, x2 and p2 for the second oscillator:
    \begin{align*}
    [x_i, x_j] &= 0 \\
    [p_i, p_j] &= 0 \\
    [x_i, p_j] &= i\hbar\delta_{ij}
    \end{align*}It follows from these that a primed and unprimed operator commute.
     
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