# How Do Quantum Operators Affect Position and Momentum Expectations?

• noospace
In summary: However, you may try to have a look directly at "the bible of quantum optics": Mandel & Wolf "Optical coherence and quantum optics" where they give many results of some expectation values (however, you must check carefully their formulas against typing errors).
noospace
I'm trying to evaluate the expectation of position and momentum of

$\exp\left(\xi (\hat{a}^2 - \hat{a}^\dag^2)/2\right) e^{-|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n\rangle}$

where $\hat{a},\hat{a}^\dag$ are the anihilation/creation operators respectively.

Recall $\hat{x} = \sqrt{\hbar/(2m\omega)}(\hat{a}^\dag + \hat{a})$.

The calculation of $\langle \hat{x} \rangle$ and $\langle \hat{p} \rangle$ are easy if one uses the Hadamard lemma

$e^X Y e^{-X} = Y+ [X,Y] + (1/2)[X,[X,Y]] + \cdots$.

I'm running into touble in the evaluation of $\langle\hat{x}^2\rangle,\langle\hat{p}^2\rangle$. In particular in the calculation of

$\exp\left(\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)(\hat{a}^\dag + \hat{a})^2\exp\left(-\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)$.

I was able to show that

$\exp\left(\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)(\hat{a}^\dag + \hat{a})^2\exp\left(-\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right) = (\hat{a}^\dag + \hat{a})^2 + \xi([\hat{a}^2,\hat{a}^\dag^2]+[\hat{a}^\dag\hat{a},\hat{a}^\dag^2-\hat{a}^2]) + \frac{\xi}{2!}[(\hat{a}^\dag+\hat{a})^2,[\hat{a}^2,\hat{a}^\dag^2]+[\hat{a}^\dag\hat{a},\hat{a}^\dag^2-\hat{a}^2]]+\cdots$

but I can't figure out how to simplify this.

Last edited:
You may try to expand $(a^\dagger +a)^2=1+a^\dagger a + (a^\dagger)^2 + a^2$, then $a^\dagger a$ is the photon number operator...

Few years ago I did similar calculations and I was using the book of J.S.Peng and G.X. Li "introduction to modern quantum optics" where they give many results of some expectation values (however, you must check carefully their formulas against typing errors)
Also W.H. Louisell "Quantum statistical properties of radiation" may contain similar calculations.

but you may try to have a look directly at "the bible of quantum optics": Mandel & Wolf "Optical coherence and quantum optics"

noospace said:
I'm trying to evaluate the expectation of position and momentum of

$\exp\left(\xi (\hat{a}^2 - \hat{a}^\dag^2)/2\right) e^{-|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n\rangle}$

where $\hat{a},\hat{a}^\dag$ are the anihilation/creation operators respectively.

Recall $\hat{x} = \sqrt{\hbar/(2m\omega)}(\hat{a}^\dag + \hat{a})$.

The calculation of $\langle \hat{x} \rangle$ and $\langle \hat{p} \rangle$ are easy
oh?
if one uses the Hadamard lemma

$e^X Y e^{-X} = Y+ [X,Y] + (1/2)[X,[X,Y]] + \cdots$.

I'm running into touble in the evaluation of $\langle\hat{x}^2\rangle,\langle\hat{p}^2\rangle$.
it is also easy, if one knows how to do it.

## 1. What is a squeezed gaussian expectation?

A squeezed gaussian expectation is a mathematical concept that describes the expected value of a function in the form of a gaussian distribution that has been squeezed or compressed along one or more axes. This can be visualized as a bell-shaped curve that has been stretched or flattened in certain areas.

## 2. How is a squeezed gaussian expectation calculated?

A squeezed gaussian expectation is calculated by taking the integral of the function multiplied by the gaussian distribution and dividing it by the integral of the gaussian distribution alone. This can be done numerically or analytically depending on the complexity of the function.

## 3. What is the significance of a squeezed gaussian expectation in science?

Squeezed gaussian expectations are commonly used in statistical analysis and modeling in various fields of science, such as physics, engineering, and finance. They provide a way to analyze data and make predictions based on the expected value of a variable that follows a gaussian distribution.

## 4. How does a squeezed gaussian expectation differ from a regular gaussian expectation?

A regular gaussian expectation calculates the expected value of a function using a gaussian distribution with equal standard deviations along all axes. A squeezed gaussian expectation allows for different standard deviations along different axes, providing a more accurate representation of the underlying data.

## 5. Can a squeezed gaussian expectation be used for non-gaussian distributions?

Yes, a squeezed gaussian expectation can be applied to any function or distribution, not just gaussian ones. However, it is most commonly used with gaussian distributions due to their prevalence in scientific data.

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