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?


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!

  • $\begingroup$ @AccidentalFourierTransform, Wicks theorem is the standard approach and searches for all combinatorics (i.e. contractions) in the boson string. This is effective if the boson string is not very large but becomes increasingly cumbersome for large $n$. Is there an alternative method? $\endgroup$
    – Sid
    Commented Mar 16, 2016 at 17:42

1 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.

  • $\begingroup$ After seeing this [article] (arxiv.org/abs/0704.3116), I am convinced of the method you give in your answer. However, I haven't still been able to derive the VEV shown in the question. Could you give details? Also, I'm not sure why the vacuum ket $\vert 0\rangle$ maps to 1 and why $\vert n\rangle$ maps to $\sqrt{n!}\xi^n$. $\endgroup$
    – Sid
    Commented Jan 25, 2017 at 11:49
  • $\begingroup$ sure... will find time to expand my answer in the next 48hrs. $\endgroup$ Commented Jan 25, 2017 at 12:18
  • $\begingroup$ @Sid : should be enough now. Be well. $\endgroup$ Commented Jan 25, 2017 at 13:57

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