If there are $N$ single-particle states labeled by $1,2,3,\cdots,N$, it is said that the general two-boson state is given by

$$|\Psi\rangle=\sum_{i,j=1}^N \omega_{ij}a_i^\dagger a_j^\dagger |0\rangle$$

where $\omega_{ij}$ is a complex symmetric matrix.

My question is why $\omega_{ij}$ needs to be symmetric.


$$a_i^\dagger a_j^\dagger = a_j^\dagger a_i^\dagger ,$$

only $\omega_{ij}+\omega_{ji}$ needs to be fixed.

It seems that the matrix $\omega_{ij}$ is not unique.

Is it true that the requirement for $\omega_{ij}$ to be symmetric is just for convenience?

  • 5
    $\begingroup$ The short answer is yes. The anti-symmetric part of $\omega$ does not contribute anything to $\lvert \Psi\rangle$, so it's arbitrary. It's just convenient to fix $\omega_{ij} - \omega_{ji} = 0$. $\endgroup$ Jan 25, 2016 at 16:27

1 Answer 1


Every matrix $\omega$ can be written as the sum $\omega = \omega^S + \omega^A$, where $\omega^S = \frac{1}{2}(\omega+\omega^T)$ is symmetric and $\omega^A = \frac{1}{2}(\omega-\omega^T)$ is antisymmetric.

Since $a^\dagger_i$ commutes with all $a^\dagger_j$, the sum $\sum_{i,j}(\omega^A)_{ij}a^\dagger_i a^\dagger_j$ is zero since $(\omega^A)_{ij}$ is anti-symmetric in $i,j$ while $a^\dagger_i a^\dagger_j$ is symmetric in $i,j$. Hence $$ \sum_{i,j}\omega_{ij}a^\dagger_i a^\dagger_j = \sum_{i,j}(\omega^S)_{ij}a^\dagger_i a^\dagger_j$$ and we may take $\omega$ to be symmetric without loss of generality. A symmetric $\omega$ is preferable because you can recover it from the state as $$ \omega_{ij} = \langle 0 \vert a_j a_i\vert \Phi \rangle$$ while you obviously can't recover anti-symmetric parts because they are irrelevant.


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