Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
That's it, thanks. –  Ramashalanka Feb 26 '11 at 5:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.