Are the relations $\ln (e^{\widehat A}) = \widehat A$ and $e^{\ln \widehat A} = \widehat A$ true for operators?

Question

For the real numbers, $x\in \mathbb R$, we have the relations $$\ln (e^x) = x = e^{\ln x}.$$ Since operators aren't numbers, these equations don't necessarily hold for operators, that is, if we replace $x$ with an operator $\widehat A$. So my question is: Are these relations valid for operators? And why / why not?

My attempt to far:

As far as I know, operator functions used in quantum mechanics are defined as a power series expansion of that function, using the operator as the argument:

$$e^{\widehat B} \equiv \sum_{n=0}^\infty \frac{{\widehat B}^n}{n!}.$$

For $\ln x$ I found different power series for different domains of $x$, for example the series

$$\ln x = \sum_{n=1}^\infty \frac{2}{2n-1} \left(\frac{x-1}{x+1}\right)^{2n-1},$$

which is valid for $x>0$ . But my attempt to use these power series (and I'm not even sure what it would mean for an operator to be greater than $0$, as required by this log series) goes nowhere useful:

\begin{align*} e^{\ln \widehat A} &= \sum_{n=0}^\infty \frac{(\ln\widehat A)^n}{n!}\\ &= \sum_{n=0}^\infty \frac{1}{n!}\left( \sum_{k=1}^\infty \frac{2}{2k-1} \left(\frac{\widehat A-1}{\widehat A+1}\right)^{2k-1} \right)^n \\ &\stackrel ?= \widehat A \end{align*}

I can't see whether those power series reduce to just $\widehat A$ or not. Any help or clarification is greatly appreciated!

Answer 1

In QM most operators are Hermitean, and thus diagonalizable, $\hat A= U^\dagger D U$, for some unitary U and real diagonal D. So all your series expressions $f(\hat A)=0$ are essentially $U^\dagger f(D) U=0$, that is, a tower of the same relations for each diagonal component of D, each eigenvalue.

If it so happens that all eigenvalues of $\hat A$ are positive definite, your equation holds (if your respective expansion expression held for each one of them), otherwise you must be very careful with logarithms, a tall order: hardly realistic. (Exponentials, however, are normally fine, since they are single-valued functions. It is the logarithms that are dangerous. See WP article linked.)

If you are sure $\hat A$ has no null eigenvectors, but have doubts about the sign of its real eigenvalues, consider $\hat {A} \hat A $ instead, whose eigenvalues are guaranteed to be positive definite.

To get a grip on your procedures, practice with, e.g. $\hat A = \sigma_1$, so $D=\sigma_3$, and $\hat A \hat A = 1\!\!1$. You readily see how your hyperbolic tangent expansion is aggressively ill-defined for $\hat A$ but fine, and trivial, for its square.

Answer 2

In general, the Taylor expansion does not work with operators for many reasons (especially problem with domains), unless the operator is (a) everywhere defined and (b) bounded.

In this case, $e^A$ can be safely defined via Taylor expansion, whereas $\ln A$ also requires $||A-I||< 1$ (exactly as for complex numbers and I am thinking here of the standard expansion of $\ln(1+y)$ for $|y| <1$, your series needs a more difficult analysis because it is not a power series and one cannot automatically extend the popular results from complex numbers to operators). The inversion formulas are valid accordingly.

In case $A: D(A) \to H$ is closed and normal ($D(A)$ is a dense subspace of $H$ and normal means $A^\dagger A = AA^\dagger$, and this is true in particular if $A$ is selfadjoint) then one can exploit the standard functional calculus based on the spectral theorem.

Under the said hypotheses, the spectral decomposition holds $$A = \int_{\sigma(A)} z dP^{(A)}(z)$$ where $\sigma(A) \subset \mathbb{C}$ is the spectrum of $A$.

Here one can define $$e^A := \int_{\sigma(A)} e^z dP^{(A)}(z)$$ and (paying attention to to the fact that $\ln$ is multivalued on $\mathbb{C}$, so that some precautions need if $\sigma(A)$ for instance includes the semiaxis $Re z <0$), $$ \ln A := \int_{\sigma(A)} \ln z dP^{(A)}(z)\:.$$ Regarding domains, it turns out that $$D(f(A)) = \left\{x \in H \:\left|\: \int_{\sigma(A)} |f(z)|^2 d\mu_{x,x}(z)\right.< +\infty\right\}$$ where the complex measure appearing in the right-hand side is defined as
$$\mu_{x,y}(E) = \langle x|P^{(A)}_E y\rangle$$ for every Borel set $E \subset \mathbb{C}$.

It turns out that, if $D(A)=H$ and $||A|| <+\infty$, then the above definition coincides to the Taylor-expansion definition.

Since, when the written composition makes sense, it holds $$\int_{\sigma(A)} f(g(z)) dP^{(A)}(z) = \int_{\sigma(A)} f(z) dP^{(g(A))}(z) $$ you can conclude that $e^{\ln A}= A$ and $\ln e^A =A$ provided the relevant left-hand side is well defined, according to the discussion above.