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If $\hat O$ is an operator and $c\in \mathbb C$

Then can I safely say that $[\exp(c\hat O)]^\dagger=\exp(c^*\hat O^\dagger)$?

My reasoning is by taking the transpose term by term in the power series definition of $\exp$ but I don't know if that is valid or not!

Thanks!

George

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Yes ... you just have to verify, that $O \mapsto O^\dagger$ is continuous in the operator topology you are using. –  martini Oct 1 '12 at 8:47
    
@martini: Thank you! I am wondering suppose it is given that $O \mapsto O^\dagger$ is continuous, can the transpose of all power series be defined by its term by term transpose? –  George Oct 1 '12 at 8:50
    
@martini: Btw, I am not too sure what it means for $O \mapsto O^\dagger$ to be continuous in my operator topology... Would you mind briefly explaining or provide a link to some (relatively simple) references? (I haven't done operator topology.) –  George Oct 1 '12 at 8:52
    
I'll write an answer, but you have to explain me one thing: How do you want the powerseries $\exp(O) = \sum_k \frac 1{k!}O^k$ to be defined? –  martini Oct 1 '12 at 8:56
    
@martini: Thanks! I am not sure what you mean by how to define the power series...? –  George Oct 1 '12 at 8:58

1 Answer 1

up vote 2 down vote accepted

So I suppose that $\hat O\colon X \to X$ is a continous operator on some complex Hilbert space $X$. Then one usually defines \[ \exp(\hat O) := \sum_{k=0}^\infty \frac 1{k!} \hat O^k = \lim_{K \to \infty} \sum_{k=0}^K \frac 1{k!} \hat O^k \] where the limit is meant with respect to the operator norm \[ \|T\| = \sup_{\|x\| = 1} \|Tx\| \] The above series converges in this norm as we have \[ \left\|\sum_{k=L+1}^K \frac 1{k!}\hat O^k\right\| \le \sum_{k=L+1}^K \frac 1{k!}\|\hat O\|^k \to 0, \quad K,L \to \infty\] So $\exp(\hat O)$ is a well defined operator $X \to X$.

Concernig it's transpose, if we try to compute it formally we have \begin{align*} \exp(\hat O)^\dagger &= \left(\sum_{k=0}^\infty \frac 1{k!} \hat O^k \right)^\dagger \\&= \left(\lim_{K\to \infty} \sum_{k=0}^K \frac 1{k!} \hat O^k \right)^\dagger \\&\stackrel?= \lim_{K\to \infty} \left(\sum_{k=0}^K \frac 1{k!} \hat O^k \right)^\dagger \\&= \lim_{K\to \infty} \sum_{k=0}^K \frac 1{k!} {\hat O^\dagger}^k \\&= \exp(\hat O^\dagger) \end{align*} The only step which doesn't follow from the linearity and antimultiplicativity of $(-)^\dagger$ is marked with $\stackrel?=$, we want to know if $(-)^\dagger$ exchanges with limits with respect to the operator topology, that is, if $(-)^\dagger\colon L(X) \to L(X)$ is continuous (where $L(X)$ denotes the set of linear continous functions $X \to X$). As taking the transpose is linear, it suffices to prove it is bounded. We have - denoting the scalar product of $X$ by $(-,-)$ - for $T \in L(X)$: \begin{align*} \|T^\dagger\| &= \sup_{\|x\| = 1} \|T^\dagger x\| \\&= \sup_{\|x\| = 1} \sup_{\|y\| = 1} \left|(T^\dagger x, y)\right| \\&= \sup_{\|x\| = 1} \sup_{\|y\| = 1} \left|(x, Ty)\right| \\&= \sup_{\|y\| = 1} \sup_{\|x\| = 1} \left|(Ty, x)\right| \\&= \sup_{\|y\| = 1} \|Ty\| \\&= \|T\|, \end{align*} so $T \mapsto T^\dagger$ is bounded (even an isometry) and hence, $\stackrel?=$ is verified, we do indeed have $\exp(\hat O)^\dagger = \exp(\hat O^\dagger)$.

The continuity of $(-)^\dagger$ implies of course, that for every convergent series of operators we can apply the transpose termwise.

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Thank you, martini, for the very detailed answer! :) –  George Oct 1 '12 at 9:30

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