# Expectation value on coherent states of $(\hat{a}+\hat{a}^\dagger)^n$

I need to evaluate the following expectation value $$\langle \alpha \vert (\hat{a}+\hat{a}^\dagger)^n \vert \alpha\rangle$$ The formulation is very easy, but I can not tackle the problem. Any hint?

• Try to look at a simple case (for instance $n=2$), observe what happens, then you should be able to generalize Mar 13, 2021 at 18:57
• @Emmy i tried but it is easy for small n, but for large and arbitrary n i was not able to find a closed formula Mar 13, 2021 at 18:59
• iirc this exact question has been asked on this site at least once before, and the answers have the solutions you're looking for Mar 13, 2021 at 20:49
• @Eletie I can’t actually find the question you are referring to Mar 13, 2021 at 20:55
• @ArtemAlexandrov i m not sure since coherent states are not orthogonal Mar 13, 2021 at 20:56

A general parameter $$n$$ involves hypergeometric functions.

For that expression, we make one observation that in natural unit, the position operator $$\hat x$$ can be written as follows $$\hat x = \frac{1}{{\sqrt 2 }}\left( {\hat a + {{\hat a}^\dagger }} \right) \tag{1}\label{posop}.$$

We also need the position-representation of a general coherent state $$|\alpha\rangle$$

$$\alpha(x) = \left\langle {x} \mathrel{\left | {~ \alpha } \right. } \right\rangle = {\pi ^{ - 1/4}}\exp \left[ { - \frac{{{{\left( {x - \sqrt 2 {\alpha _1}} \right)}^2}}}{2} + ix\sqrt 2 {\alpha _2}} \right],$$ with the complex number $$\alpha = \alpha_1+i\alpha_2$$.

Now we have all the ingredients for the solution.

The quantity we are after reads $$\mathcal{I} = \left\langle \alpha \right|{\left( {\hat a + {{\hat a}^\dagger }} \right)^n}\left| \alpha \right\rangle = {2^{n/2}}\left\langle \alpha \right|{{\hat x}^n}\left| \alpha \right\rangle \tag{2}\label{ans},$$ where we have used Eq. \eqref{posop} for the position operator.

Now we insert an identity inside Eq. \eqref{ans} as follows \begin{align} \mathcal{I} &= \int {dx \cdot } \left\langle \alpha \right|{\left( {\hat a + {{\hat a}^\dagger }} \right)^n}\left| x \right\rangle \left\langle {x} \mathrel{\left | {\vphantom {x \alpha }} \right.} {\alpha } \right\rangle , \\ & = {2^{n/2}} \cdot { } \int {dx \cdot } \left\langle \alpha \right|{{\hat x}^n}\left| x \right\rangle \alpha \left( x \right), \\ & = {2^{n/2}} \cdot { } \int {dx \cdot } {x^n}\left\langle {\alpha } \mathrel{\left | \right. } {x} \right\rangle \alpha \left( x \right), \\ & = {2^{n/2}} \cdot { } \int {dx \cdot } {x^n}\cdot{\alpha ^*}\left( x \right)\alpha \left( x \right), \\ & = {2^{\frac{n}{2} - 1}}\left[ {\left( {{{\left( { - 1} \right)}^n} + 1} \right)\Gamma \left( {\frac{{n + 1}}{2}} \right){}_1{F_1}\left( { - \frac{n}{2};\frac{1}{2}; - 2{\alpha _1}^2} \right) - \sqrt 2 {\alpha _1}\left( {{{\left( { - 1} \right)}^n} - 1} \right)n\Gamma \left( {\frac{n}{2}} \right){}_1{F_1}\left( { - \frac{n}{2} + \frac{1}{2};\frac{3}{2}; - 2{\alpha _1}^2} \right)} \right], \end{align}

where $$_1{F_1}$$ is the hypergeometric function.

For the first few values of $$n$$, the expression read as follows:

Some properties of the above table, for the final expression of $$\mathcal{I}$$ in Eq. \eqref{ans}, are listed here.

• It is always real, since the operator $$({\hat a + {{\hat a}^\dagger }})^n$$ is Hermitian for any $$n$$.
• The final expression does not depend on the imaginary part of $$\alpha$$, since the position and momentum coordinates are decoupled in an idealized coherent state. An interesting thing would be to couple these two degrees of freedom (position and momentum), e.g., via spin-orbit interaction, and look at the moments of the respective operators.
• For $$\alpha_1=0$$, i.e., a coherent state at the origin of the phase space, the expression is tractable. It becomes zero for odd values of $$n$$, and for even values of $$n$$ it reads $$\mathcal{I} = (n-1)!!$$.
• That sound very interesting. Thank you. Do you think we can generalize this formula for an arbitrary rotated power of the position operator, namely $\langle \alpha \vert \hat{x}_{\psi}^n \vert \alpha \rangle$, where $\hat{x}_\psi = \hat{a}e^{-i\psi} + \hat{a}^\dagger e^{i\psi}$? Mar 14, 2021 at 8:55
• Interesting direction. Thank you! As far I understand, a rotation of the position is accompanied by an appropriate rotation of the coherent state as well. From the back-of-the-envelop calculation, I do not see any change in the final expression. Please let me know if you find something different. Mar 14, 2021 at 11:18
• Actually, I tried to go back to the previous case by make a change of the creation and annihilation operators, namely $\hat{a} = e^{-i\psi} \hat{b}$. In this way, the coherent state in terms of the new creation operator is slightly different, that is $\vert \alpha\rangle_a = \vert \alpha e^{-i\psi} \rangle_b$, where the subscript stands for the pair of creation to which the coherent states belong. In this way, the expectation value is the same but we should make to following substitution: $\alpha_1 \to \alpha_1\cos(\psi) +\alpha_2\sin(\psi)$. Do you agree? Mar 16, 2021 at 12:06
• Exactly! Now I realize that, this is spin-orbit interaction, since the momentum $\alpha_2$ is now at play. One observation (maybe not be entirely correct): This rotated 'position operator' actually measures the 'sizes (I mean moments)' of the coherent state along different angular directions, centred around its own origin; and this leads to the spin-orbit interaction, for the expectation value of $\hat{x}_\psi^n$. Thank you raskolnikov! I understood my own answer better. If you want, please go ahead and edit my answer to add the part on rotated position operator and the related discussion. Mar 16, 2021 at 18:42

The coherent state $$|\alpha\rangle = \exp(\alpha \hat{a}^\dagger)|0\rangle$$ as a function of $$\alpha$$ satisfies the following equation, $$(\hat{a}+\hat{a}^\dagger)|\alpha\rangle = \left (\alpha+\frac{\partial}{\partial\alpha}\right )|\alpha\rangle .$$ Hence, the following relation is valid, $$\langle\beta|(\hat{a}+\hat{a}^\dagger)^n|\alpha\rangle = \left (\alpha+\frac{\partial}{\partial\alpha}\right)^n\exp(\beta^*\alpha)$$ for any $$\beta$$ and $$\alpha$$.

Now, considering $$\alpha^*$$ independent of $$\alpha$$, for the expectation value, we obtain $$\exp(-\alpha^*\alpha)\left(\alpha+\frac{\partial}{\partial\alpha}\right )^n\exp(\alpha^*\alpha) = \left (\alpha+\alpha^*+\frac{\partial}{\partial\alpha}\right )^n\ 1.$$ In mathematics & probability, the Hermite polynomials are defined by the following relation, $$H_n(x) = (-1)^n \exp(x^2/2)\frac{d^n}{dx^n}\exp(-x^2/2) = \left (x-\frac{d}{dx}\right )^n\ 1.$$ Thus, it is easy to see that the answer to the question is $$i^n H_n\left(\frac{\alpha+\alpha^*}i\right).$$

This answer coincides with that given by Mehedi Hasan for $$n = 0,\dots,8$$.

• I hope you didn't mind my copyediting, for readability, your sparse, truly elegant answer. Mar 14, 2021 at 20:48
• @CosmasZachos, I don't mind. Thank you!
– Gec
Mar 14, 2021 at 20:54
• I suspect you might well be the only person on the planet interested in this one. Mar 14, 2021 at 21:02
• @CosmasZachos, You are right, I am interested. I was not aware of that discussion at the moment when I answered this question. I think the accepted answer in the topic You've mentioned might be wrong.
– Gec
Mar 15, 2021 at 11:56
• @CosmasZachos, there is no need to get to the bottom of that answer. In my opinion, the very first formula there is wrong because it is valid only for normally ordered operators.
– Gec
Mar 15, 2021 at 13:11

There is an identity, associated with the Baker-Campbell-Hausdorff-formula, written as $$\exp(\hat{X})\hat{Y}\exp(-\hat{X})=\hat{Y}+[\hat{X},\hat{Y}] +\frac{1}{2!}[\hat{X},[\hat{X},\hat{Y}]]+...$$ It will allow you to solve the problem if you express the coherent states in terms of displacement operators operating on the vacuum.