# Is every unitary operator induced by a Hamiltonian?

Diving deeper into the mathematical inner workings of quantum mechanics: The set of unitary operators on the Hilbert space $$\mathcal{H}$$ forms a group. While for finite-dimensional Hilbert spaces, this group is essentially the matrix Lie group $$U(n)$$ and everything is well-known, in the infinite-dimensional case, this does not seem to be the case. Wikipedia says this group is sometimes referred to as Hilbert group, is this the standard terminology?

Main question: are there unitary operators in this group, that cannot be written as $$U=e^{-iHt}$$ for some Hamiltonian $$H$$?

Secondly, I would like to understand how those Hamiltonians might look. Really, the only thing you get to see in standard QM are Hamiltonians $$H$$ which are polynomials of the canonical operators $$X$$ and $$P$$ or equivalently can be written as polynomials of annihilation and creation operators. Do these Hamiltonians induce a subgroup of the (Hilbert) group mentioned above, are there other known subgroups?

• Hint: Look up Stone's theorem. Commented Jan 8, 2020 at 12:05
• The title of this question is kinda misleading as for instance finite rotations are unitary but not necessarily induced by Hamiltonians. If a non-compact example is needed, squeezing will do. Commented Jan 8, 2020 at 15:17
• @ZeroTheHero as far as I know, both rotations and squeezing are definitely induced by Hamiltonians? Commented Jan 8, 2020 at 18:52
• Take the 3d harmonic oscillator. generators of rotations commute with $H$ but are distinct from it. More generally a unitary transformation is exp of a hemitian operator, but this operator need not the the Hamiltonian for the system. Commented Jan 8, 2020 at 20:35

## 1 Answer

There is a one-to-one correspondence between self-adjoint operators and representations of the abelian group $$\mathbb{R}$$ as a group of unitary operators $$\bigl(U(t)\bigr)_{t\in\mathbb{R}}$$ that is continuous, as a function of $$t\in\mathbb{R}$$, with respect to the strong operator topology.

The map between self-adjoint operators and strongly continuous unitary groups is given by $$H\mapsto e^{-itH}$$. This is the content of Stone's theorem mentioned in the comments. However, there are unitary operators that are not generated by a self-adjoint Hamiltonian in the above sense.

For general unitary operators, there is still a way of "taking the logarithm", in some sense. By functional calculus for normal operators (unitary operators are normal), it is possible to represent each unitary operator $$U$$ as follows: $$U=\int_{\mathbb{R}} e^{-it} \mathrm{d}P_U(t)\; ,$$ where $$P_U(t)$$ is a projection-valued spectral family such that $$P_U(t)=0$$ for all $$t<0$$, and $$P_U(t)=I$$ (the identity operator) for all $$t\geq 2\pi$$.

However, the formal logarithm that results, i.e. $$i \ln U "=\int_{\mathbb{R}} t \, \mathrm{d}P_U(t)$$ is a bounded (albeit self-adjoint) operator, and therefore it does not coincide in general with the self-adjoint generator (take for example the unitary operator $$e^{iH}$$, with $$H$$ self-adjoint and unbounded to have a counterexample).

• Thank you for the answer. This answer seems to be neither "no" nor "yes" if I understand correctly. You write "there are unitary operators that are not generated by a self-adjoint Hamiltonian in the above sense", could you reason why this is the case, or can you provide an example (I would not even know how to write these down)? And then you say that we can still identify a corresponding bounded self-adjoint but this identification is flawed in the sense that it does not lead to what we would usually say is the Hamiltonian in obvious cases. Is this correct? Commented Jan 8, 2020 at 18:59
• A first observation is the following: a representation of the type $e^{-itH}$ for a unitary is meaningful only if the said unitary is parametrized by a real number $t$. Given that, since $e^{-itH}$ satisfies the properties of an abelian group when $t$ ranges over the real numbers (it is in fact a unitary representation of the abelian group of real numbers with sum as group operation), the natural question is in my opinion the following: are all unitary representations of the real numbers (seen as an abelian group) of the form $e^{-itH}$ for some self-adjoint $H$? Commented Jan 9, 2020 at 9:47
• The answer to the above question is no, because the representations of the form $e^{-itH}$ are in one-to-one correspondence with strongly continuous unitary groups (and representation of the same form with $H$ self-adjoint and bounded are in one-to-one correspondence with uniformly continuous unitary groups). Commented Jan 9, 2020 at 9:49
• The representations that are not continuous or weakly but not strongly continuous cannot be represented in the exponential form. Another, different question is instead if it is possible to take the logarithm of any unitary operator. The answer has to come from the spectral theorem, but for "complex" operators (operators with complex spectrum) it is tricky to do a logarithm. Commented Jan 9, 2020 at 10:10