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I have some basic doubts about Stone's theorem. 1) Can we apply Stone's theorem to conclude that given any Unitary operator U, we can find a self adjoint operator A such that U = exp(i A). That is, is any unitary operator part of a one parameter group? 2) Is there some version of Stone's theorem for real Hilbert spaces like in finite dimensions?

user157106
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    Over the reals, you'd want to use not the orthogonal group (the obvious analog of the unitary group) but rather the special orthogonal group. The point is that the set of possible determinants, the set of scalars of absolute value 1, is connected in the complex case but not in the real case. – Andreas Blass Jul 17 '14 at 15:45

2 Answers2

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Stone's theorem works in both senses. Given a one-parameter unitary group of operators $U(t)$ on a Hilbert space, there is one self-adjoint operator $H$, called its generator, such that $i\partial_t U(t)=HU(t)$. Conversely, given a self-adjoint operator we can define a unitary group of operators generated by it.

Stone's theorem obviously works also on real Hilbert spaces.

yuggib
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    Seperability isn't needed, Stone's theorem holds true also for non-seperable Hilbert spaces. An the OP asks for the following: Given $U$, find a unitary group $U(\cdot)$ such that $U(1) = U$. – martini Jul 17 '14 at 07:47
  • @martini noted and edited. – yuggib Jul 17 '14 at 07:51
  • So, can any U be written as a member of some one parameter unitary group? – user157106 Jul 17 '14 at 07:54
  • No, I don't think it is possible. And surely not in an unique way: e.g. the identity operator is an unitary operator member of infinitely many unitary groups. – yuggib Jul 17 '14 at 08:01
  • I never asked for uniqueness. Can you give me an example of U which does not belong to a 1 parameter unitary group? – user157106 Jul 17 '14 at 08:50
  • There are unitary two-parameter groups that formally can be written as $U(t,s)=\exp(-i\int_s^tH(\tau)d\tau)$, where ${H(t)}{t\in\mathbb{R}}$ is a family of self-adjoint operators. The $U(t,s)$ exist under suitable conditions on ${H(t)}{t\in\mathbb{R}}$, are unitary and satisfy a two-parameter group condition but cannot be written as the exponential of a self-adjoint operator. I'm sure there are other simpler examples that does not come to my mind ;-). – yuggib Jul 17 '14 at 09:05
  • Seems beyond me, but still intuitively, it seems that $\int H(\tau)d\tau$ should be self adjoint too. About my second doubt, it clearly isnt true for real Hilbert spaces since reflection is unitary(for example reflection about x axis in 2 dimensions) and cannot be written as the exp(iA) for some real matrix A. – user157106 Jul 17 '14 at 10:28
  • As I underlined $\int H(\tau)d\tau$ is just a formal statement in this case, to give an intuition. Even if you can define $U(t,s)$, it will not be generated by a self-adjoint operator in the Stone's theorem sense. – yuggib Jul 17 '14 at 10:32
  • So, the answer is in negative to my first question(it is definitely an affirmative in finite dimensional case) though I would love to hear a more lucid/simple explanation. – user157106 Jul 17 '14 at 10:39
  • "Stone's theorem obviously works also on real Hilbert spaces." How? Stone's original paper started off assuming complex Hilbert spaces. Am I missing anything here? – Argyll Jun 06 '15 at 04:28
  • @Argyll The proof, stated in modern terminology, works also for real Hilbert spaces. If you want to explicitly avoid complex quantities you may consider the unitary group $e^{-tA}$, $t\in\mathbb{R}$, where $A=-A^*$ is the skew-selfadjoint generator. – yuggib Jun 07 '15 at 16:13
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    @yuggib: Would you be able to point me to some references please? – Argyll Jun 10 '15 at 00:23
  • @Argyll you may look for example at the book by Pazy on semigroups of operators, or any book on mathematical methods of quantum mechanics (Reed-Simon vol.1, Yosida's functional analysis, Kato's perturbation of linear operators) they all should contain proofs of the Stone's theorem (none however treat explicitly the case of real Hilbert spaces, but the proof by Pazy for example is straightforward to adapt). – yuggib Jun 10 '15 at 05:17
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As mentioned in the previous post, you won't have uniqueness of the unitary group. But I think you can find one using the spectral theorem for $U$. To find $U$, write $$ U = \int_{T} \lambda dE(\lambda), $$ where $T$ is the unit circle in the complex plane, and $E$ is the Borel spectral measure for $U$. Define $$ U(t) = \int_{T} e^{it\arg(\lambda)} dE(\lambda), $$ where, for example, you take $\arg(\lambda) \in [0,2\pi)$ such that $e^{i\arg(\lambda)}=\lambda$ on $T$. The generator $A$ of this semigroup is the bounded selfadjoint operator $A=\int_{T}\arg(\lambda)dE(\lambda)$. Notice that $\sigma(A)\subseteq [0,2\pi]$ and $2\pi \notin\sigma_{p}(A)$, which may make such $A$ unique. (?)

Disintegrating By Parts
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