In an operators algebra $\mathcal{A}$ one can consider a self-adjoint (i.e. real) operator $H$ and note that $$U=e^{iH}$$ exists and is unitary. A mathematical question will be whether any unitary operator $U$ is of this form. For there even exist examples where $X,Y$ are self-adjoint and $XY\neq YX$ and $$ e^{iX}e^{iY}\neq e^{i(X+Y)}. $$ I would like to know what information can be deduced for $U$ by knowing that there exists a logarithm $$H=\frac{1}{i}\log U,$$ and what are concrete applications in QM for this.

  • $\begingroup$ Can you give an example where $e^{iX} e^{i Y} \neq e^{i(X+Y)}$? I'm curious. $\endgroup$ – infinitezero Dec 3 '18 at 17:11
  • $\begingroup$ @infinitezero $e^{i\alpha L_z}e^{i\beta L_y}$? $\endgroup$ – ZeroTheHero Dec 3 '18 at 18:05
  • $\begingroup$ @infinitezero Entire quantum theory is based on that simple non-equality. Operators need not necessarily commute. See en.wikipedia.org/wiki/… $\endgroup$ – Avantgarde Dec 3 '18 at 18:23
  • $\begingroup$ Eh yes, of course, I just potatoed there I guess. $\endgroup$ – infinitezero Dec 4 '18 at 15:26

The Stone's theorem proves the following. Consider a group of unitary operators $(U(t))_{t\in\mathbb{R}}$ acting on a Hilbert space $\mathscr{H}$ (i.e. satisfying $U(t+s)=U(t)U(s)$, in more mathematical terms this is a unitary representation of the abelian group $\mathbb{R}$ on $\mathscr{H}$). If in addition such group is strongly continuous, namely is such that for all $\psi\in\mathscr{H}$ $$\lim_{t\to 0} \, \lVert U(t)\psi-\psi\rVert_{\mathscr{H}}=0\; ,$$ then there exists a self-adjoint operator $H$ defined on $D(H)\subseteq\mathscr{H}$ that generates the dynamics, i.e. such that for all $\psi\in D(H)$ $$\lim_{t\to 0}\lVert \frac{1}{t}(U(t)-1)\psi+iH\psi\rVert_{\mathscr{H}}=0\; ,$$ and for all $\phi\in \mathscr{H}$, $U(t)\phi=e^{-itH}\phi$ where the right hand side is defined by the spectral theorem. Also by the spectral theorem, it is in this case "justified" to write $H=i\ln U(1)$.

The above theorem is the one commonly used in quantum mechanics, since it relates the quantum Hamiltonian (the generator $H$) to the unitary dynamics it generates (the group $U(t)$). There are ways to take the "logarithm" of a single unitary operator (e.g. by means of a Cayley transform), however this is not very relevant in physics since the important objects are unitary representations of symmetry groups rather than unitary operators per se.


Time evolution in quantum mechanics* is represented by the action of a unitary operator $U=e^{iHt}$, where $H$ is the Hamiltonian of the system in question. We usually characterize (non-relativistic**) quantum systems by their Hamiltonian; in principle, one could determine*** the Hamiltonian of a system given a time evolution operator $U$ by taking $\frac{1}{it}\log U$. In practice, going in this direction is not usually practical from an experimental perspective, which is why this is not often mentioned.

*In quantum mechanics, the question of which objects evolve in time is a matter of interpretation. In some cases it is easier to think of the wavefunctions as evolving in time (the "Schrodinger picture"), in other cases it is simpler to think of the operators as evolving in time (the "Heisenberg picture"), and in still other cases it is a mix of both that is simplest (the "interaction picture").

**The Hamiltonian is not Lorentz-invariant, which is why you don't often see it in relativistic quantum mechanics/QFT. The Lagrangian, on the other hand, is.

***The log is not necessarily unique, so the Hamiltonian can only be "determined" up to the equivalent of a choice of branch cut.


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