Suppose you have a bosonic Fock space with a vacuum $|0\rangle$. A particular state is labeled by the parameter $N \in \mathbb{Z}$. You can construct states like $$ | n_{N} \rangle = \frac{ \left( \hat{a}_{N}^{\dagger}\right)^{n_N}}{\sqrt{n_N!}} | 0 \rangle $$ which means that there are $n_N$ particles in the state $N$.

If you were to compute a trace using these states as a basis, how would you do this? Is it something like $$ \mathrm{Tr}\left[ \hat{A} \right] = \prod_{N \in \mathbb{Z}} \sum_{n_{N}} \langle n_{N} | \hat{A} | n_{N} \rangle \ \ \ \ \ ? $$ (where $\hat{A}$ is some operator). Or is it more complicated than this?

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    $\begingroup$ A basis state is specified by $N$ and $n_N$, so you sum over those. You don't multiply, that's not what a trace is and it wouldn't be linear in $A$. $\endgroup$
    – Javier
    Commented Aug 7, 2018 at 2:08
  • $\begingroup$ As @Javier mentions, there is no product of the type you write. However, there is also no reason that all states are of the form $|n_N\rangle$. The product comes into play when you construct states of the form $(a_1^\dagger)^{n_1}\cdot (a_2^\dagger)^{n_2}\cdots |-\rangle$ with your normalization factors. $\endgroup$
    – user178876
    Commented Aug 7, 2018 at 3:09

1 Answer 1


There is no reason only to consider states of the form $(a_N^\dagger)^{n_N}|-\rangle$. Rather you need to consider all states of the form $$ |\vec n\rangle~:=~|n_1,n_2,\dots\rangle ~=~ \prod\limits_i \frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}}|-\rangle\;.$$ And if you sum those, the trace becomes $$ \text{Tr}\,A~=~\sum\limits_{\vec n}\left\langle \vec n| A |\vec n\right\rangle\;,$$ where of course $|\vec n\rangle$ involves a product.

  • $\begingroup$ Can you be more explicit in the way that the product appears in $\mathrm{Tr}[A]$? $\endgroup$ Commented Aug 10, 2018 at 14:35

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