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If I take the anti-commutator of two positive operators $A,B$ on a Hilbert space, $AB+BA$ is again guaranteed to be Hermitian, but is it also necessarily positive?

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Actually, I just found a counter-example using 2-dimensional complex matrices, so please ignore this question. So, the answer is NO, in general the anti-commutator need not be positive. –  Nikolas Jan 28 '12 at 0:35
    
$A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $B=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ gives a counterexample. You could post an answer to your own question if you want to. –  Jonas Meyer Jan 28 '12 at 0:37
    
Thanks Jonas, I'll do that next time. –  Nikolas Feb 11 '12 at 21:21

1 Answer 1

Since the "next time" never came for the OP, I post the counterexample given by Jonas Meyer.

Let $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}1&1\\1&1\end{pmatrix}$, then $AB+BA=\begin{pmatrix}2&1\\1&0\end{pmatrix}$ has negative determinant.

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