Suppose to have a chain (of size $L$) with bosons, and $\hat{a}_i^\dagger$,$\hat{a}_i$ are the associated creation and annihilation operators at site $i$. A Fock state can be written as: \begin{equation} | n_1 \dots n_L \rangle = \prod_{i} \frac{1}{\sqrt{n_i!}} \left( \hat{a}_i^\dagger \right)^{n_i} |\rangle \end{equation} where $|\rangle$ is the empty state. Now we define a new set of bosons: \begin{align} \hat{b}_{k} &= \frac{1}{\sqrt{L}} \sum_{j} e^{-ikj} \hat{a}_j & \hat{b}_{k}^\dagger &= \frac{1}{\sqrt{L}} \sum_{j} e^{ikj} \hat{a}_j^\dagger \end{align} where $k$ are such that some boundary condition is fulfilled. Now a Fock state can be written as: \begin{equation} | \dots \tilde{n}_k \dots \rangle = \prod_{k} \frac{1}{\sqrt{\tilde{n}_k!}} \left( \hat{b}_k^\dagger \right)^{\tilde{n}_k} |\rangle \end{equation} The question is: is there a simple formula to express or compute the scalar product $\langle \dots \tilde{n}_k \dots | n_1 \dots n_L \rangle$?
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2$\begingroup$ Do you mean the scalar product $<original | new>$ or $<new | new>$? Introduction of some different notation for the new bosons would help, eg. $|n_1' \ldots n_k' \ldots >$. $\endgroup$– au700Commented Nov 30, 2012 at 10:30
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$\begingroup$ You are right, I edited the question and changed the notation. Anyway, the question is about Fock states in different bases otherwise the answer would be trivial. And of course the two states have the same number of bosons: $\sum_k \tilde{n}_k = \sum_j n_j$. $\endgroup$– HariCommented Nov 30, 2012 at 13:32
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$\begingroup$ You can easily make $< new | original > = \sum c_{\ldots} < original | original >$ by substituting the definition of new creation / anihilation operators in $< new |$. $\endgroup$– au700Commented Nov 30, 2012 at 13:50
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$\begingroup$ If I use your recipe I get: \begin{align} \langle \dots \tilde{n}_k\dots |n_1 \dots n_L \rangle & = \langle | \prod_k \left( \hat{b}_k \right)^{\tilde{n}_k } |n_1 \dots n_L \rangle \\ & = \langle | \prod_k \left( \frac{1}{\sqrt{L}} \sum_j e^{-ikj} \hat{a}_j \right)^{\tilde{n}_k } |n_1 \dots n_L \rangle \end{align} even if now it is quite straightforward which are the terms that survive (the ones that annihilate $|n_1 \dots n_L \rangle$ ) I'm not able to find a "clean" and useful equation ... :( $\endgroup$– HariCommented Nov 30, 2012 at 14:15
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2 Answers
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Define the operators $\hat a(f)=\sum f_j\hat a_j$ and $ |f_1,...,f_n\rangle:=\hat a(f_1)...\hat a(f_n)|vac\rangle$. Then $\langle g_1,...,g_m|f_1,...,f_n\rangle$ vanishes for $m\ne n$ and is a sum of the products $\langle g_1|f_{j_1}\rangle...\langle g_n|f_{j_n}\rangle$ for all possible permutations $(j_1,...,j_n)$ of $1,...,n$. Note that the $\langle g|f\rangle$ are easy to compute.
The formula you requested is a special case of this.
answered Nov 30, 2012 at 14:20
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$\begingroup$ Ok, thank you very much (which is actually the answer I would got expanding the equation which is in the comment I wrote just above). I have to deal with the computation of all the possible permutations...I will let the computer do it for me...:) $\endgroup$– HariCommented Nov 30, 2012 at 14:28
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I realise the question has already been answered, but just a suggestion to the topic starter: finite sums of the form $(\sum_i x_i)^N$ are expanded using objects called multinomial coefficients. They should make your life a little easier.
