# Spectral measure of a unitary operator

The following is an Exercise of Conway's operator theory:

1-Show that if $A$ is hermitian operator, then $U=\text{exp}({iA})$ is unitary.

2- Show that every unitary can be so written.

3-Find the spectral measure of the unitary in terms of that of the hermitian operator.

proof: 1-The first claim is clear.

2- I think the second is false, based on Theorem 2.1.12 of Murphy's operator theory:

And also The following from Takesaki's Operator theory:

3- There is a unique spectral measure $E$ correspondence to $*-$ homomorphism $C(\sigma(A)) \to B(H)$ and $U =\int e^{i\lambda} dE(\lambda)$.

Is it correct? Thanks.

The second claim is indeed true. Take $H = -i\; \log(U)$, where $\log$ is any branch of the natural logarithm that is a bounded Borel function on the unit circle.

• Not in general, if $\sigma(U) = \Bbb T$, then there is not any self-adjoint operator $A$ that $U=exp(iA)$.
– niki
Commented Jan 14, 2015 at 21:46
• What's wrong with the one I specified? Take e.g. the principal branch of log. Commented Jan 14, 2015 at 21:51
• The theorem from Murphy is for a C*-algebra, and the one from Takesaki is for $C(\Gamma)$. In both of those cases you are restricted to continuous functions. But in this case you don't need $H$ to be a continuous function of $U$: any self-adjoint operator is OK. Commented Jan 14, 2015 at 22:58
• @T.A.E. That's exactly it. The Borel functional calculus for a bounded normal operator $U$ gives you something in the von Neumann algebra generated by $U$, not necessarily in the C* algebra. Commented Jan 15, 2015 at 1:31
• @niki One defines $\log: \mathbb T \to \mathbb C$ as follows $\log(e^{i\theta})=i\theta$, where $-\pi<\theta\leq \pi$. This is not a continuous function but it is a Borel function on the circle so one can apply it a unitary operator $U$. Commented Jan 15, 2015 at 8:23

I think that the third claim is correct with the spectral representation

$$A=\int_{\sigma(A)} \lambda \ \mathrm{d}E(\lambda)$$

• Also I mean it.
– niki
Commented Jan 14, 2015 at 21:47
• If it is not obvious, I would encourage you to write $A=\int_{\sigma(A)} \lambda \ \mathrm{d}E(\lambda)$. If $A$ is a unitary operator (thus $\sigma(A) \subset \mathbb{T}$) then by using the pullback measure argument (and by changing variables, see also en.wikipedia.org/wiki/Pushforward_measure) we could write $$A=\int_{\sigma(A)} \lambda \ \mathrm{d}E(\lambda) = \int_{[0, 2\pi)}e^{i\lambda} \ \mathrm{d}E^{\prime}(\lambda),$$ where $E^{\prime}$ is a spectral measure defined on a Borel sigma algebra over $[0, 2\pi)$ and was obtained from $E$ by the pull-back argument. Commented Jan 15, 2015 at 10:38
• @mgn, thanks for the suggestion. In this case $A$ is a hermitian operator and $U=\text{exp}({iA})$ is a unitary operator. Commented Jan 15, 2015 at 11:08