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

1. Oct 25, 2009

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

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

Also, the state $|\alpha>$ (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 $< \hat{A} > = Tr[ \hat{\rho} \hat{A} ]$.

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

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:

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

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.