Baker–Campbell–Hausdorff and Gaussian density operators for fermions

Question

I'm trying to understand a passage from the paper "Gaussian operator bases for correlated fermions", J. F. Corney and P. D. Drummond (https://arxiv.org/abs/quant-ph/0511007), specifically going from the first to second line of Eq. (2.5), which is about Gaussian density operators:

The operators $\hat{b}_i$ are fermionic, and so obey the usual anticommutation relations. The problem is, the term $\hat{b}_i^\dagger\hat{b}_j$ does not necessarilly commute with $\hat{b}_k^\dagger\hat{b}_l$, so I don't see how you can take the summation out of the exponential as a simple product. From the BCH formulas (https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula), I would expect there to be another term that takes into accout the commutators $[X,Y]$, $[X,[X,Y]]$, etc.

Can someone shed some light into why are those terms zero/not necessary in this situation?

Qmechanic · Accepted Answer · 2020-07-20T16:08:27.620

1

This is because "operators" (more precisely: symbols/functions, cf. my Phys.SE answer here) super-commute under the normal-order sign.

edited Jul 20 '20 at 16:08

answered Jul 20 '20 at 15:50

Qmechanic

201,751

I see. Thank you. This normal-ordering is a new concept to me. Since in this case the normal-ordering operatiion is applied to a sum of symbols, I suppose we make use of a linearity of this operation. In other comments of the link you mention I see this linearity property mentioned, but the wikipedia page (https://en.wikipedia.org/wiki/Normal_order#Notation) afirms: "Attempting to apply normal ordering to a sum of operators is not useful as normal ordering is not a linear operation". Now I'm confused, is normal-ordering a linear operation or not? – Lucas Baldo Jul 20 '20 at 17:19
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$\uparrow$ It's linear. – Qmechanic Jul 20 '20 at 17:23

Baker–Campbell–Hausdorff and Gaussian density operators for fermions

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