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


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?


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – levitopher
    Commented Apr 3, 2013 at 23:10
  • 1
    $\begingroup$ 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.. $\endgroup$
    – nervxxx
    Commented Apr 3, 2013 at 23:16

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