Let us consider a Fock space of continuous modes $a(\omega)$ with $[a(\omega),a^\dagger (\tilde{\omega})]=\delta(\omega-\tilde{\omega})$ where the vaccum state is defined as $a(\omega)|0\rangle = 0$.

Is there a general formula for the inner product of two Fock states \begin{equation} \langle 0 |a(\omega_N)\ldots a(\omega_1)a^\dagger(\nu_1)\ldots a^\dagger(\nu_N)|0\rangle \end{equation} ?

  • $\begingroup$ This is basically the content of Wick's theorem: en.wikipedia.org/wiki/Wick%27s_theorem. Although Wick's theorem usually deals with converting time-ordered modes into normal-ordered modes, the only modification needed in your case is that you only contract an $a$ with an $a^{\dagger}$ if the former is to the left of the latter. The end result is equivalent to the combinatorics described by @ZeroTheHero. $\endgroup$ Jun 10, 2021 at 16:03

1 Answer 1


The process is recursive. Start with \begin{align} a(\omega_1)a^\dagger(\nu_1) &=\delta(\omega_1-\nu_1)\mathbb{1}+a^\dagger(\nu_1)a(\omega_1)\, , \tag{1}\\ a(\omega_1)a^\dagger(\nu_1)a^\dagger(\nu_2) &=\delta(\omega_1-\nu_1)a^\dagger(\nu_2)+ a^\dagger(\nu_1)\left(a(\omega_1)a^\dagger(\nu_2) \right)\, , \tag{2} \end{align} and now use Eq.(1) to unwrap $a(\omega_1)a^\dagger(\nu_2)$ in Eq.(2) and so forth.

When you take the expectation value of this on the ground state, there are simplifications since $a(\omega_k)\vert 0\rangle=0$, so you are left with a sum of products of $\delta$-functions.

\begin{align} N=2:&\langle 0\vert a(\omega_2)a(\omega_1)a^\dagger(\nu_1)a^\dagger(\nu_2)\vert 0\rangle\\ &= \delta \left(\omega _1-\nu _2\right) \delta \left(\omega _2-\nu _1\right)+\delta \left(\omega _1-\nu _1\right) \delta \left(\omega _2-\nu _2\right)\, ,\\ N=3: &\delta \left(\omega _1-\nu _3\right) \delta \left(\omega _2-\nu _2\right) \delta \left(\omega _3-\nu _1\right)+\delta \left(\omega _1-\nu _2\right) \delta \left(\omega _2-\nu _3\right) \delta \left(\omega _3-\nu _1\right)\\ &+\delta \left(\omega _1-\nu _3\right) \delta \left(\omega _2-\nu _1\right) \delta \left(\omega _3-\nu _2\right)+\delta \left(\omega _1-\nu _1\right) \delta \left(\omega _2-\nu _3\right) \delta \left(\omega _3-\nu _2\right)\\ &+\delta \left(\omega _1-\nu _2\right) \delta \left(\omega _2-\nu _1\right) \delta \left(\omega _3-\nu _3\right)+\delta \left(\omega _1-\nu _1\right) \delta \left(\omega _2-\nu _2\right) \delta \left(\omega _3-\nu _3\right) \, , \end{align} and you can see that what you have is the sum of products of the type $\delta(\omega_1-\nu_{\gamma(1)})\delta(\omega_2-\nu_{\gamma(2)})\delta(\omega_3-\nu_{\gamma(3)})$ where the sum is over all permutations $\gamma$ in $S_N$. In other words, for $N=3$, you sum over all permutations $\gamma\in S_3$ of $\gamma(1),\gamma(2),\gamma(3)$ the products $\delta(\omega_1-\nu_{\gamma(1)})\delta(\omega_2-\nu_{\gamma(2)})\delta(\omega_3-\nu_{\gamma(3)})$.

The same result generalizes for any $N$ but of course the permutation group $S_N$ contains $N!$ elements so the sum over permutations gets terribly long.

If you are a Mathematica user, check this question and related answer for a short Mathematica script to deal with normal ordering of the kind you're interested in.


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