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The operators in this question are in Hilbert space, but I'm going to word it as a discrete spectrum and a finite dimensional (linear algebra) answer is fine by me.

Operators $A$ and $\Delta$ are bounded, self-adjoint and non-commutative, and $B\equiv e^{- A} \Delta e^{ A}$. Some of the eigenvalues of $A$ or $\Delta$ may be negative. I want to prove that the $B \Delta$ is positive semi-definite and that the eigenvalue $0$ has the same multiplicity in $B\Delta$ as it does in $\Delta$ (I'm not sure that this second part is true).

$B$ is a similarity transformation, so the eigenvalues, $\Delta_i$, of $\Delta$ are the same as those of $B$, but the eigenvectors are different, so the eigenvalues of $B\Delta$ are not $\Delta_i^2$. The $e^A$ form (positive eigenvalues) is important, since the product of general similar operators doesn't have a non-negative spectrum in general.

I'm tempted to expand in bases of the eigenvectors $|x\rangle = \sum_i b_i |B_i\rangle = \sum_j d_j |\Delta_j\rangle$, $|B_i\rangle = \sum_j c_{i,j} |\Delta_j\rangle \implies d_j = \sum_i b_ic_{i,j}$, to find

$$ \begin{aligned} \langle x | B\Delta | x \rangle & = \sum_{i,j} b_i^* d_j \langle B_i | B\Delta | \Delta_j \rangle = \sum_{i,j} b_i^* d_j \Delta_i \Delta_j \langle B_i| \Delta_j \rangle =\sum_{i,j,k} b_i^* d_j c_{i,k}^* \Delta_i\Delta_j \delta_{kj} \\ & = \sum_{i,j} b_i^* d_j c_{i,j}^* \Delta_i\Delta_j \end{aligned} $$

which doesn't help much, because I haven't used that $e^A$ is positive. If I expand in the basis of eigenvectors of $A$ and $\Delta$, I get a similar but slightly more complicated expression, and I think there must be a simpler approach.

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up vote 1 down vote accepted

Consider the similarity transformation

$$e^{A/2}B\Delta e^{-A/2}=e^{-A/2}\Delta e^{A/2} e^{A/2} \Delta e^{-A/2}=M^\dagger M$$

with $M:=e^{A/2}\Delta e^{-A/2}$. An operator of the form $M^\dagger M$ is positive semi-definite, since $\langle x \vert M^\dagger M \vert x \rangle=\langle x M \vert M x \rangle \ge 0$.

That the eigenvalue $0$ has the same multiplicity in $B\Delta$ as it does in $\Delta$ follows because the eigenvalues of $M^\dagger M$ are the squared absolute values of the singular values of $M$, the number of non-zero singular values of $M$ is the rank of $M$, the number of non-zero eigenvalues of $\Delta$ is the rank of $\Delta$, and these two ranks are equal since $M$ is obtained from $\Delta$ by a similarity transformation.

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That's it, thanks. – Ramashalanka Feb 26 '11 at 5:21

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