7
$\begingroup$

This question and the comments and answers it received encouraged me to ask this question, although I know that there will be some people who think that this belongs in the math forum. But I think that this topic is more relevant to mathematical physicists than to pure mathematicians.

Motivation: One of the answers to this question explains that if $f\colon\mathbb C\to\mathbb C$ is suitable function and $A\colon\text{Dom}(A)\subset H\to H$ is a suitable operator, we can define \begin{equation} f(A):=\int_{\mathbb C}f\,\mathrm{d}P_A \end{equation} where $P_A\colon B(\mathbb C)\to B(H)$ is a measure. However, it's much easier to understand the definition in terms of convergent series, e.g. in the case of the exponential or the logarithm. (In statistical physics, $S=k_B\langle\ln\rho\rangle$ is the entropy, when $\rho$ is the density operator.) That's why I'd like to know:

Is it also possible to write $f(A)$ in terms of a converging series when $f$ has a taylor expansion around some point?

In case that the answer is yes, I also wonder if there is a relatively easy way to see how the integral and the series are equivalent. (As far as I know, integrals - even $\int_{\mathbb C}f\,\mathrm{d}P_A$ - can be expressed as limit of some series, so maybe that would be a good starting point).

Examples: The expression \begin{equation} \sum_{n=0}^\infty \frac{1}{n!} A^n \end{equation} makes sense whenever $A$ is an element of a complete normed space and converges to $\mathrm{e}^A=\int_{\mathbb C}\text{exp}\,\mathrm{d}P_A$ when $A$ is suitable operator (source).

It is even known that \begin{equation} \left(\sum_{k=1}^N (-1)^{k+1}\frac{(A-\text{id})^k}{k}\right)_{N\in\mathbb N} \end{equation} converges to $\text{ln}(A)$ under certain circumstances (see here and here), so I was wondering if there is a general rule. That is, if we have \begin{equation} f(x)=\sum_{n=0}^{\infty}a_n(x-b)^n, \end{equation} in a neighborhood of $b$, is \begin{equation} f(A)=\sum_{n=0}^{\infty}a_n(A-b\cdot\text{id})^n? \end{equation}

$\endgroup$
  • $\begingroup$ Does this answer your question? A rigorous definition of the exponential of an operator in QM? $\endgroup$ – jacob1729 Nov 30 '20 at 13:02
  • 2
    $\begingroup$ (In particular, you can define any function on an operator via its spectral decomposition as in ACuriousMind's answer there.) $\endgroup$ – jacob1729 Nov 30 '20 at 13:05
  • $\begingroup$ @jacob1729 I made a huge edit to my post, I hope that it's clearer now. $\endgroup$ – Filippo Nov 30 '20 at 15:06
  • 1
    $\begingroup$ If the operator is normal and bounded you can use the series in its convergence circle. That is due to the fact that the convergence is uniform therein and that has a good interplay with the uniform operator topology... $\endgroup$ – Valter Moretti Nov 30 '20 at 17:36
9
$\begingroup$

If the operator $A$ belongs to $B(H)$ (the space of everywhere defined bounded operator on the Hilbert space $H$) and is normal: $$A^*A=AA^*$$ then it admits a spectral decomposition $$A = \int_{\mathbb{C}} z dP(z) = \int_{\sigma(A)} z dP(z)$$ and, with an obvious notation, $|\sigma(A)| \leq ||A|| <+\infty$.

In this case (and also in the general case where $A$ is unbounded (densely-defined, closed, normal)), $$f(A) := \int_{\sigma(A)} f(z) dP(z)$$ for every Borel measurable function $f: \sigma(A) \to \mathbb{C}$. In this case the answer is relatively easy.

Proposition.

Let $A \in B(H)$ be normal and consider $f: \Omega \to \mathbb{C}$ an analytic function on the open set $\Omega \subset \sigma(A) \subset \mathbb{C}$.

If $z_0 \in \Omega$ and the Taylor expansion of $f$ around $z_0$ $$f(z) = \sum_{n=0}^{+\infty} a_n (z-z_0)^n$$ has convergence radius $R+\epsilon$ for some $\epsilon>0$, and finally $$\sigma(A) \subset C_R(z_0):= \{ z\in \mathbb{C}\:|\: |z-z_0| <R\}$$ then $$f(A) = \sum_{n=0}^\infty a_n (A-z_0)^n$$ where the convergenge of the right-hand side is in the norm of $B(H)$.

PROOF. We start from the inequality $$||\int_{\mathbb{C}} g(z) dP(z)|| \leq ||g||_\infty$$ which is valid if $g$ is Borel-measurable and bounded. This inequality holds true even if $A$ is not bounded.

Exploiting that inequality we have $$\left|\left|\int_{\sigma(A)} \left[\sum_{n=0}^N a_n(\lambda -\lambda_0)^n - f(z)\right] dP(z)\right|\right| \leq \sup_{z\in \sigma(A)}\left|\sum_{n=0}^N a_n(\lambda -\lambda_0)^n - f(z)\right| \to 0$$ for $N\to +\infty$ since the convergence of the Taylor expansion is uniform in every compact in the convergence disk. Notice that $\sigma(A)$ is indeed a compact included in $C_{R+\epsilon}(z_0)$.

Using definition of $g(A)$, we therefore have that $$\int_{\sigma(A)} \sum_{n=0}^N a_n(\lambda -\lambda_0)^n dP(z) \to \int_{\sigma(A)} f(z) dP(z)$$ with respect to the norm of $B(H)$. In other words, if $N\to +\infty$ $$\sum_{n=0}^{+\infty} a_n (A-z_0I)^n = f(A)$$ in that topology. That is the thesis. QED

There are other results which can be proved analogously and which encompass the case of $A$ unbounded (densely defined, closed, and normal). If $\psi$ belongs to the projection space of $\int_{E} 1 dP(z)$, where $E \subset C_{R}(z_0)$ is a bounded Borel set (so that $\psi$ is an analytic vector of $A$), then $$f(A)\psi = \sum_{n=0}^\infty a_n (A-z_0)^n\psi$$ where now the convergence is in the Hilbert space norm.

It is difficult to produce a finer result.


(As references I can quote my books https://doi.org/10.1007/978-3-319-70706-8 and https://doi.org/10.1007/978-3-030-18346-2)

$\endgroup$
  • $\begingroup$ Thank you very much - as always it's an honour to get an elaborated answer from you. I am happy that you also included the formula $f(A)\psi = \sum_{n=0}^\infty a_n (A-z_0)^n\psi$ for an unbounded $A$, as I think that this rather corresponds to the situations one encounters in physics. If I'm not mistaken, elementary results about absolutely convergent series suffice to show that the series converge, but it's good to know that they also converge to the right thing :) $\endgroup$ – Filippo Nov 30 '20 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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