Vaccuum expectations for a string of creation and annihilation operators

Question

I want to determine the vaccum expectation value of a string of creation and annihilation operators. They have a very specific form:

$$\langle \prod_{i=1}^n \hat{a}(k_i) \, \, \hat{N}_1 \prod_{j=1}^n \hat{a}^\dagger(k_j) \rangle_0,$$

which is the number operator $\hat{N}_1 = \hat{a}^\dagger(k_1) \hat{a}(k_1)$ sandwitched on the left by $n$ annhilation operators and on the right by $n$ creation operators. The subscript $0$ reminds us that it is the vacuum expectation that we are interested in. However, for any $n$ which can be arbitrarily large, there exist related expectation values that I would like to evaluate. For example, I also have

$$\langle \prod_{i=1}^n \hat{a}(k_i) \, \, \hat{a}^\dagger(k_1) \hat{N}_2 \prod_{j=2}^n \hat{a}^\dagger(k_j) \rangle_0,$$

where the number operator is positioned between $\hat{a}^\dagger(k_1)$ and $\hat{a}^\dagger(k_2)$. Note the index $j$ on the second product. This continues for all $\hat{N}_m,$ $m \in [1, n]$, i.e.

$$\langle \prod_{i=1}^n \hat{a}(k_i) \, \, \prod_{j=1}^m \hat{a}^\dagger(k_j) \hat{N}_m \prod_{l=m}^n \hat{a}^\dagger(k_l) \rangle_0.$$

Now the operators obeys the usual commutation relations, but since the arguments $k_i$ are continuous, we have Dirac delta functions:

$$\left[\hat{a}(k_i), \hat{a}^\dagger(k_j)\right] = \delta(k_i - k_j).$$

The whole expectation is in a multidimensional integral $\int dk_1 \cdots dk_n$, and so the re-expression of the expectation values in terms of Dirac delta functions will greatly simplify the problem.

My question is how can I obtain a general expression for the expectation values for arbitrary $n$? I have tried normally ordering the operator string up to $n=3$, but this becomes labourious and I haven't managed to find a general expression yet.

Can the Husimi Q-function be used to do this?

EDIT IN RESPONSE TO ANSWER BY ZEROTHEHERO

A way to solve such VEVs is to use the so-called XD-representation. However, I have been unable to calculate the above VEVs using the method. Any help with this would be appreciated!

Answer 1

The Husimi Q-function is of no use here. What you need is Wick calculus. There is slick trick which is to map $$ \hat a^\dagger (k_i)\to \xi_i\qquad \hat a(k_j) \to \frac{\partial }{\partial \xi_j}\, , $$ where $\xi_i$ is a formal (dummy) variable.
This preserves commutation relations and allows you to convert the string of creation and destruction operators into a string of products of $\xi_j$ and $\partial /\partial \xi_i$, which makes the evaluation easier to track down.
Under this map, $\vert 0\rangle$ maps to the number $1$, $\hat (a^\dagger(k_i))^n\vert 0\rangle $ maps to $\xi^n_i$ etc. The expectation value will be the sum of terms having no powers in any $\xi_i$'s, i.e. the constant terms in the resulting polynomial once you push the derivatives to the right.

---

Here's some more.

In this "formalism", we have $\vert 0\rangle\mapsto 1$ since the action of the destruction operator $\hat a(k_j)\vert 0\rangle =0$ is reproduced by $\frac{\partial}{\partial \xi_j} 1=0$ for all dummies $\xi_j$.

Once you have the correspondence $\vert 0\rangle\mapsto 1$, it is easy to get (for instance) $$ \hat a^\dagger (k_1)\left(\hat a^\dagger (k_2)\right)\mapsto \xi_1\xi_2^2 $$ since the action of $\hat a^\dagger(k_j)$ on a ket is replaced by multiplication of polynomials by $\xi_j$. Hence, taking $n=3$ (for instance) in your original problem, we have $$ \Pi_{j=1}^3\hat a^\dagger(k_j)\vert 0\rangle \mapsto \xi_1\xi_2\xi_3\, . $$ We can now use $$ \hat N_1\Pi_{j=1}^3\hat a^\dagger(k_j)\vert 0=\hat a^\dagger(k_1)\hat a(k_1)\mapsto \xi_1\frac{\partial}{\partial \xi_1}\left(\xi_1\xi_2\xi_3\right)=1\times \xi_1\xi_2\xi_3 $$ and finally $$ \Pi_{i=1}^3 \hat a(k_i)\mapsto \frac{\partial}{\partial\xi_1}\frac{\partial}{\partial\xi_2}\frac{\partial}{\partial\xi_3} $$ on $\xi_1\frac{\partial}{\partial \xi_1}\left(\xi_1\xi_2\xi_3\right)$ to complete the calculation. The result here is $1$ by inspection (and in general) because $\Pi_{i=1}^3 \hat a(k_i)\vert 0\rangle$ creates one excitation in each of the modes labelled by $i$, $\hat N_1$ just counts the number of excitation in mode 1 (which is $1$), and the last $\Pi_{i=1}^3 \hat a(k_i)$ simply removes one excitation from each mode, bringing them back to the ground state.

The trick as stated is therefore not much of a simplification for the specific case you asked, but in general would be much more useful if your initial state were some generic polynomial function $f(\hat a^\dagger_1,\hat a^\dagger_2,\ldots,\hat a^\dagger_n)\vert 0\rangle$. In this case, the state would just become a polynomial $f(\xi_1,\xi_2,\ldots,\xi_n)$ in the variables $\xi_1,\ldots,\xi_n$. Taking $\partial/\partial \xi_j$ on these polynomials amounts to the action of $\hat a(k_j)$, while multiplying the polynomial by $\xi_j$ is the same as using $\hat a^\dagger(k_j)$ on the state.