Expectation value of operators and squeezing in the even cat state

[eigenpsi]

Homework Statement

I need to derive the equations 7.126 and 7.127(image posted) from Gerry/Knight's quantum optics book. The equations are the expectation values of the square of the quadrature operators for the even cat state.

Relevant Equations

<X1^2>=0.25 + ( (alpha)^2 )/(1+exp(-2alpha^2))

<X2^2>=0.25 - (exp(-2alpha^2)*(alpha)^2)/(1+exp(-2alpha^2))

I started and successfully showed that the expectation of X_1 and X_2 are zero. However the expectation value of X1^2 and X2^2 which I am getting is <X1^2> = 0.25 + \alpha^2 and <X2^2> = 0.25.

How do I derive the given equations?

 

[member 553083]

To derive the given equations, we can use the definition of expectation value. The expectation value of a random variable X is given by E[X] = ∑xP(x), where P(x) is the probability of the random variable X taking the value x. For X_1 and X_2, since they are independent random variables with uniform distribution over the interval [-α, α], we have: E[X_1] = E[X_2] = 0.Now to calculate the expectation values of X_1^2 and X_2^2, we use the definition again. We have: E[X_1^2] = ∑x^2P(x) = ∫x^2P(x)dx = ∫_{-α}^{α}x^2dx/2α = (x^3/3)|_{-α}^{α} = (α^3- (-α)^3)/3α = (2α^3)/3α = 2α^2/3 Similarly, E[X_2^2] = ∫_{-α}^{α}x^2dx/2α = (x^3/3)|_{-α}^{α} = (α^3- (-α)^3)/3α = (2α^3)/3α = 2α^2/3 Therefore, we have E[X_1^2] = 0.25 + α^2 and E[X_2^2] = 0.25.