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Showing posts from September, 2024

Unbounded generators of unitary groups

Stone's theorem tells us when a one-parameter unitary group has a self-adjoint generator. If $U(t)$ is a group --- i.e., it satisfies $U(t_1 + t_2) = U(t_1) U(t_2)$ --- then we can write $U(t) = e^{i H t}$ for some unbounded $H$ if and only if we have $\lim_{t \to t_0} U(t) |\psi \rangle = U(t_0) |\psi\rangle$ for every $|\psi\rangle$ in Hilbert space. This is the condition that $U(t)$ is strongly continuous . What if we have a unitary group with multiple parameters? Within any one-parameter subgroup we should be able to find a generator. But is there a connection between the generators of different subgroups? Concretely, imagine we have a two-parameter subgroup that forms a representation of $\mathbb{R}^2.$ I.e., assume we have $$U(t_1 + t_2, s_1 + s_2) = U(t_1, s_1) U(t_2, s_2).$$ Assume further that the map from $\mathbb{R}^2$ to unitary operators is strongly continuous. Then for any one-parameter subgroup we have a generator, in particular, we have $$U(t,0) = e^{i H_1 t}$$ and...

Pick functions and operator monotones

Any time you can order mathematical objects, it is productive to ask what operations preserve the ordering. For example, real numbers have a natural ordering, and we have $x \geq y \Rightarrow x^k \geq y^k$ for any odd natural number $k$. If we further impose the assumption $y \geq 0,$ then order preservation holds for $k$ any positive real number. Self-adjoint operators on a Hilbert space have a natural (partial) order as well. We write $A \geq 0$ for a self-adjoint operator $A$ if we have $$\langle \psi | A | \psi \rangle \geq 0$$ for every vector $|\psi\rangle,$ and we write $A \geq B$ for self-adjoint operators $A$ and $B$ if we have $(A - B) \geq 0.$ Curiously, many operations that are monotonic for real numbers are not monotonic for matrices. For example, the matrices $$P = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$ and $$Q = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$ are both self-adjoint and positive, so we have $P+Q \geq P \geq 0$, but a str...