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Working out uncertainty in superimposed coherent states using the density operator!

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a quantum oscillator in a mixed state described by the density operator [itex]\rho = \frac{1}{2}( |\alpha><\alpha| + |-\alpha><-\alpha| ) [/itex]. Calculate [itex] \Delta (\hat{X}^2)_1 [/itex] and [itex] \Delta (\hat{X}^2)_2 [/itex] in this case.

    Where X1 and X2 are the dimensionless position and momentum operators:
    [tex] \hat{X}_1 = \frac{1}{2}(a + a^{\dagger}) [/tex]
    and
    [tex] \hat{X}_2 = \frac{1}{2i}(a - a^{\dagger}) [/tex]

    Also, the state [itex] |\alpha> [/itex] (and -alpha) are coherent states, and so are eigenfunctions of the creation/annihilation operators.



    2. Relevant equations
    Well, I know how to calculate the uncertainty in a measurement, given the expectation values of X1, X2, X1^2 and X2^2. I also know that [itex] < \hat{A} > = Tr[ \hat{\rho} \hat{A} ] [/itex].

    Since the states in the density operator are coherent, we know that:
    [tex] <\alpha|\beta> = e^{ -|\alpha^2|/2 - |\beta^2|/2 + \beta^{\ast}\alpha} [/tex]


    3. The attempt at a solution
    Based on what I know, to get the uncertainty in X1 and X2, I need 2 calculate the expectation values for X1, X1^2 and X2, X2^2, which means I need to expand everthing out into the 'a' and 'a-dagger' relations... right? Taking X1 as an example:

    [tex]<X_1> = \frac{1}{4} ( \sum{m} <m|\alpha><\alpha| a + a^{\dagger} |-\alpha><-\alpha|X_1|m> [/tex].

    I was wondering if this is the way to do this question, because thats alot of work to expand that out and yet this is the easier expression! I then have 2 expand out X1^2 in terms of a and a\dagger's, and work all that out, so before I really get down to doing pages of maths, I just wanna check that this is the right way to do this question.
     
  2. jcsd
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