# Expectation value of operators and squeezing in the even cat state

• eigenpsi
In summary, the expectation values of X_1 and X_2 are zero, and to derive the given equations for E[X_1^2] and E[X_2^2], we use the definition of expectation value and the fact that X_1 and X_2 are independent random variables with uniform distribution over the interval [-α, α].
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?

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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.

## 1. What is the expectation value of an operator in the even cat state?

The expectation value of an operator in the even cat state is the average value that would be obtained if the system were measured many times. It is calculated by taking the inner product of the operator with the even cat state and then squaring the result.

## 2. How is the expectation value of an operator related to the uncertainty principle?

The expectation value of an operator is related to the uncertainty principle because it represents the average of all possible measurements of a physical quantity. The uncertainty principle states that the more precisely one quantity is measured, the less precisely the other can be known. Therefore, the expectation value can give us information about the uncertainty of a physical quantity.

## 3. What is squeezing in the even cat state?

Squeezing in the even cat state refers to the phenomenon where the uncertainty in one physical quantity is reduced at the expense of increasing the uncertainty in another. In the even cat state, squeezing can occur between the position and momentum operators, resulting in a reduction in the uncertainty in one of these quantities.

## 4. How is squeezing related to the even cat state?

The even cat state is a special type of quantum state that exhibits squeezing. This means that the even cat state can have a reduced uncertainty in one physical quantity, such as position or momentum, while still satisfying the uncertainty principle by having a larger uncertainty in the other quantity.

## 5. What are the applications of the even cat state and squeezing?

The even cat state and squeezing have various applications in quantum information and quantum computing. They can be used to improve the precision of measurements and to enhance the performance of quantum devices such as quantum sensors and quantum computers. Squeezing in the even cat state can also be used to study and manipulate quantum entanglement, which is crucial for many quantum information protocols.

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