# Notational techniques for dealing with creation operators on Fock space

This question is trying to see if anyone has some simple notation (or tricks) for dealing with operators acting on coherent states in a Fock space. I use bosons for concreteness; what I'm interested in might not be applicable to fermions.

If I have a multiparticle state defined by

$$|\phi\rangle=|n_1 n_2 ...n_k\rangle=\frac{(a_1^\dagger)^{n_1}}{\sqrt{n_1 !}}\frac{(a_2^\dagger)^{n_2}}{\sqrt{n_2 !}}...\frac{(a_k^\dagger)^{n_k}}{\sqrt{n_k !}}|0\rangle$$

I can write this compactly as

$$|\phi\rangle=\prod_{i=1}^{k}\frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i !}}|0\rangle.$$

Now, I would like to act on this state with some operator consisting of creation and annihilation operators; this could be quite complicated, like

$$a_ja_ka^\dagger_l.$$

Now, IF the above product was a sum, I could find the answer very easily using commutation relations and delta functions:

\begin{align} a_ja_ka^\dagger_l|\phi\rangle&=a^\dagger_la_j\sum_i\frac{(\delta_{ki}+a^\dagger_{i}a_k)}{\sqrt{n_i!}}|0\rangle\\ &=a^\dagger_la_j\left(\frac{1}{\sqrt{n_k!}}|0\rangle \right) \end{align}

...continue until you get it to normal form, and you are done. Of course, this is a different problem but I am just illustrating how neat this notation is.

You can't just throw around delta functions in the product above, since they would make the entire product vanish. What I want is a clean way to denote "I have passed $a_k$ through all $i\neq k$, used the commutation relations to get $1+a^\dagger_ia_k$, and am now ready to hit this with another creation opertaor $a_l$."

I think it's fairly clear what I am looking for; anyone have any notation or tricks to make calculations using arbitrary operators like the one I have above easier?

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I don't really understand your point about 'if the product was a sum', and 'this is not a Fock space'.

Let's first consider a state with only 1 species of boson. Then you know that \begin{align} a | \phi \rangle = a | n \rangle = \sqrt{n} | n-1 \rangle, \end{align} which follows from the bosonic commutation relations.

So if you have a state with multiple species of bosons, we have \begin{align} a_i | \phi \rangle = a_i | n_1, \cdots n_{i-1}, n_i, n_{i+1}, \cdots \rangle = \sqrt{n_i} | n_1, \cdots n_{i-1}, n_i - 1, n_{i+1}, \cdots \rangle, \end{align} since $a_i$ commutes with any operator $i \neq j$.

You can show this explicitly: \begin{align} a_i | \phi \rangle & = a_i \prod_k \frac{(a_k^\dagger)^{n_k!}}{\sqrt{n_k}} |0\rangle \nonumber \\ & = \left(\prod_{k < i} \frac{(a_k^\dagger)^{n_k}}{\sqrt{n_k!}}\right) (1 + a_i^\dagger a_i) \frac{(a_i^\dagger)^{n_i-1}}{\sqrt{n_i!}} \left( \prod_{k > i} \frac{(a_k^\dagger)^{n_k}} {\sqrt{n_k!}}\right)|0\rangle \nonumber \\ & = \left[ \left(\prod_{k < i} \frac{(a_k^\dagger)^{n_k}}{\sqrt{n_k!}}\right) \frac{(a_i^\dagger)^{n_i-1}}{\sqrt{n_i!}} \left( \prod_{k > i} \frac{(a_k^\dagger)^{n_k}} {\sqrt{n_k!}}\right) \right. \nonumber \\ &~~~ \left. + \left(\prod_{k < i} \frac{(a_k^\dagger)^{n_k}}{\sqrt{n_k!}}\right) (a_i^\dagger a_i)\frac{(a_i^\dagger)^{n_i-1}}{\sqrt{n_i!}} \left( \prod_{k > i} \frac{(a_k^\dagger)^{n_k}} {\sqrt{n_k!}}\right) \right]|0\rangle \nonumber \\ & = \left[ \left(\prod_{k < i} \frac{(a_k^\dagger)^{n_k}}{\sqrt{n_k!}}\right) \frac{(a_i^\dagger)^{n_i-1}}{\sqrt{n_i!}} \left( \prod_{k > i} \frac{(a_k^\dagger)^{n_k}} {\sqrt{n_k!}}\right) \right. \nonumber \\ &~~~ \left. + \left(\prod_{k < i} \frac{(a_k^\dagger)^{n_k}}{\sqrt{n_i!}}\right) (a_i^\dagger)^1(1+a_i^\dagger a_i)\frac{(a_i^\dagger)^{n_i-2}}{\sqrt{n_k!}} \left( \prod_{k > i} \frac{(a_k^\dagger)^{n_k}} {\sqrt{n_k!}}\right) \right]|0\rangle \nonumber \\ & = \cdots \nonumber \\ & = \left(\prod_{k < i} \frac{(a_k^\dagger)^{n_k}}{\sqrt{n_k!}}\right) \sqrt{n_i} \frac{(a_i^\dagger)^{n_i-1}}{\sqrt{(n_i-1)!}} \left( \prod_{k > i} \frac{(a_k^\dagger)^{n_k}} {\sqrt{n_k!}}\right)|0\rangle \nonumber \\ & = \sqrt{n_i}| n_1,\cdots n_{i-1}, n_i - 1, n_{i+1},\cdots \rangle, \end{align} which is precisely 'throwing delta functions' (but $i = k$ so the delta function is $1$) in the product above. Note that for fermions there are extra factors of -1s picked up from anticommuting.

So to answer your question, acting $a_i$ on a multiparticle state simply reduces the occupation number of the $i$-th 'orbital' by 1. And writing the state in terms of occupation numbers is a Fock state.

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Your comment about the Fock space is correct; that is now changed. But you are illustrating my point - that is complicated! I know how to get the answer that way (call it 'brute force'). I want a sneaky way, so that if my operators get too complicated I don't have 5 or 6 pages of "product except i" and so forth. –  levitopher Apr 3 '13 at 23:10
Well the sneaky way is just to know its action $a_i | n_1, \cdots n_{i-1}, n_i, n_{i+1}, \cdots \rangle = \sqrt{n_i} | n_1, \cdots n_{i-1}, n_i - 1, n_{i+1}, \cdots \rangle$? say you act on your state with $a_j a_k a_l^\dagger$. Then it increases the $l$-th orbital by 1, and decreases the $j$ and $k$-th orbitals by 1 each, with some prefactors? you don't actually have to do the commuting if you know a priori what the lowering and raising operators' actions on a Fock state are.. –  nervxxx Apr 3 '13 at 23:16