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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    $\begingroup$ 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. $\endgroup$ – Nikolas Jan 28 '12 at 0:35
  • $\begingroup$ $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ 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. $\endgroup$ – Jonas Meyer Jan 28 '12 at 0:37

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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