0
$\begingroup$

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} ?

$\endgroup$
1
  • $\begingroup$ 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 '21 at 16:03
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.