# Schwinger representation of operators for n-particle 2-mode symmetric states

A bosonic (i.e. permutation-symmetric) state of $n$ particles in $2$ modes can be written as a homogenous polynomial in the creation operators, that is $$\left(c_0 \hat{a}^{\dagger n} + c_1 \hat{a}^{\dagger (n-1)} \hat{b}^{\dagger} +c_2 \hat{a}^{\dagger (n-2)} \hat{b}^{\dagger2} + \ldots+ c_n \hat{b}^{\dagger n}\right)|\Omega\rangle,$$ where $\hat{a}$ and $\hat{b}$ are the annihilation operators, $c_i$ are complex coefficients and $|\Omega\rangle$ is the vacuum state.

Alternatively, one can express the same state as a state in the fully permutation symmetric subspace of $n$ qubits (equivalently - as a state of the maximal total angular momentum, that is, $n/2$).

The question is the following - for a general symmetric operator $$\sum_{perm} (\mathbb{I})^{\otimes (n-n_x-n_y-n_z)} \otimes (\sigma^x)^{\otimes n_x} \otimes (\sigma^y)^{\otimes n_y} \otimes (\sigma^z)^{\otimes n_z},$$ what is its equivalent in the terms of creation and annihilation operators?

Partial solution:

For the simplest cases (i.e. $(n_x,n_y,n_z)\in\{(1,0,0),(0,1,0),(0,0,1)\}$ we get the following: $$\sum_{i=1}^n \sigma^x_i \cong \hat{a}^\dagger \hat{b} + \hat{b}^\dagger \hat{a}$$ $$\sum_{i=1}^n \sigma^y_i \cong -i\hat{a}^\dagger \hat{b} + i \hat{b}^\dagger \hat{a}$$ $$\sum_{i=1}^n \sigma^z_i \cong \hat{a}^\dagger \hat{a} - \hat{b}^\dagger \hat{b}$$ (AFAIR it is called the Schwinger representation). It can be checked directly on Dicke states, i.e. (${n \choose k}^{-1/2}\hat{a}^{\dagger (n-k)} \hat{b}^{\dagger k}|\Omega\rangle$).

For general case it seems that we get $$: \left( \hat{a}^\dagger \hat{b} + \hat{b}^\dagger \hat{a} \right)^{n_x} \left( - i \hat{a}^\dagger \hat{b} + i \hat{b}^\dagger \hat{a} \right)^{n_y} \left( \hat{a}^\dagger \hat{a} - \hat{b}^\dagger \hat{b}\right)^{n_z} :,$$ where :expr: stands for the normal ordering, i.e. putting the creation operators on the left and the annihilation - on the right. However, it's neither checked (besides correlators for 1-2 particles) nor proven.

Of course one can construct the operators recursively, e.g. $$\sum_{i\neq j}^n \sigma^x_i \otimes \sigma^y_j = \left(\sum_{i=1}^n \sigma^x_i\right)\left( \sum_{i=1}^n \sigma^y_i\right) - i \sum_{i=1}^n \sigma^z_i \\ \cong \left( \hat{a}^\dagger \hat{b} + \hat{b}^\dagger \hat{a} \right)\left( -i\hat{a}^\dagger \hat{b} + i \hat{b}^\dagger \hat{a} \right) - i \left( \hat{a}^\dagger \hat{a} - \hat{b}^\dagger \hat{b} \right),$$ but the question is on a general closed-form result.

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## 1 Answer

Previous answer completely rewritten

It seems to me that your hypothesis is true, up to a constant correction: $$\sum_{\textstyle{\pi: \{1,2,\ldots,n\} \to \{\mathbb{I}, \sigma_x, \sigma_y, \sigma_z\} \atop \forall i \in \{x,y,z\}:\ \mathrm{card}(\pi^{-1}(\sigma_i))=n_i}} \!\!\bigotimes_{i=1}^n\ \ \pi(i) \cong \frac1{n_x! n_y! n_z!} \mathopen{:} \left( a^\dagger b + b^\dagger a \right)^{n_x} \left( - i a^\dagger b + i b^\dagger a \right)^{n_y} \left( a^\dagger a - b^\dagger b\right)^{n_z} \mathclose{:},$$ and that this can be proven using multivariate induction.

For generic $n_x, n_y, n_z$, let $$[\![n_x, n_y, n_z]\!]$$ symbolically denote the operator on the left-hand side, in the qubit picture.

The induction step may be based upon the observation $$[\![1,0,0]\!][\![n_x, n_y, n_z]\!] = (n_x+1)[\![n_x+1, n_y, n_z]\!] - i(n_y+1)[\![n_x, n_y+1, n_z-1]\!] + i(n_z+1)[\![n_x, n_y-1, n_z+1]\!] + (n-n_x-n_y-n_z+1)[\![n_x-1, n_y, n_z]\!].$$ (It's a matter of simple combinatorics to find the multipliers.) There are perfectly analogous relations for multiplying by $[\![0,1,0]\!]$ or $[\![0,0,1]\!]$ from the left, too. Defining the double square brackets to be zero when either of the components is negative, this set of relations holds universally and the sufficient set of anchor points is the Wigner representation of $[\![1,0,0]\!]$, $[\![0,1,0]\!]$, $[\![0,0,1]\!]$, where we prove the equivalence to the Fock picture formulas directly.

Now one would just prove either of the relations (arguing by symmetry) to hold for the right-hand side, using re-ordering of the products after a multiplication by $$[\![1,0,0]\!] \cong a^\dagger b + ab^\dagger$$ and $$n \cong a^\dagger a + b^\dagger b$$ from the left. It requires a certain amount of work but should be doable. (I tried but ran into some kind of a numerical error.)

With a future edit on my mind, I will try to finish the proof of the induction step to see if I am right about the additional factor.

Assuming the relation is true, I doubt that there is any more "closed form" than the expansion $$\sum_{j=0}^{n_x} \sum_{k=0}^{n_y} \sum_{l=0}^{n_z} \frac{(-1)^{k+n_z-l} i^{n_y}}{j!k!l!(n_x-j)!(n_y-k)!(n_z-l)!} (a^\dagger)^{j+k+l} a^{n_x+n_y-j-k+l} (b^\dagger)^{n_x+n_y+n_z-j-k-l} b^{j+k+n_z-l}$$ as this does not seem to have any factorisation except for the one employing the normal reordering brackets.

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Yeah, I generally feel it. However, the quest is to actually proof it (or provide a counterexample). – Piotr Migdal Sep 19 '12 at 15:29
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