Simple example of a not strongly continuous operator on a Hilbert space?

101 Views Asked by At

Let $\cal H$ be a Hilbert space. Let $U(t)$ with $t\in \mathbf R$ be a one-parameter family of linear operators on $\cal H$. Strong continuity for $U(t)$ is defined as the condition that $$ \lim_{t\to t_0} U(t)\phi = U(t_0) \phi $$ for any $\phi\in \cal H$.

My question is: naively, this looks like a trivial condition, so could someone provide an elementary example of a one-parameter family of linear operators that does not satisfy this condition?

Note: here the notion of convergence is defined using the norm induced by the hermitian scalar product of $\cal H$.

Context: this is a condition used in Quantum Mechanics to define so-called "Evolution Operators" (with extra conditions though). Such a definition is provided for instance (with explicit mention of the above property) in the textbook "Quantum Mechanics, A New Introduction", by Konichi and Paffuti.

1

There are 1 best solutions below

0
On BEST ANSWER

If $A: H\to H$ is a continuous linear operator and $ g :\mathbb R\to\mathbb R$ is a discontinuous function at $t_0\in\mathbb R$, then the family $U(t)=g(t)A$ does not satisfy the condition: $$\|g(t)A\varphi-g(t_0)A\varphi\|=|g(t)-g(t_0)|\cdot \|A\varphi\|$$ does not converge to $0$.