$e^{itH}$ notation

Question

recently I saw the notation $e^{itH}$, and just wondering how should I interpret it?

In my understanding, $u(t,x) = e^{itH} u_0$ is, for example, a solution to Schrodinger-type equation $i\partial_tu = -H u$ with the initial data $u_0$. In case $H = \Delta$, the solution to Schrodinger equation is known to involve the Schrodinger kernel in the integrand. In such case, does $e^{itH}$ is a short-hand notation for the operator involving the Schrodinger kernel?

Or should I interpret $e^{itH}$ as the Taylor series with $H^k$ terms involved? In this case, does the (operator) series converge once applied to the element in the domain of $H$?

Also, I would be very glad to get a reference to read more on this type of operators. Thank you very much!

Modulo convergence issues those interpretations are equivalent. The convergence issues are fine as long as $H$ is bounded but of course for physically interesting examples that won't be the case. — Qiaochu Yuan, Aug 06 '20 at 21:06
This isn't my area of expertise but this Wikipedia page and its references might be a good place to start: https://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups — Qiaochu Yuan, Aug 06 '20 at 21:19

score 3 · Answer 1 · answered Aug 06 '20 at 20:59

The underlying Hilbert space has an orthonormal basis consisting of eigenfunctions $u_\lambda$ of $H$ with eigenvalue $\lambda$. Then $e^{itH}u_\lambda:=e^{it\lambda}u_\lambda$. Forcing $e^{itH}$ to be linear and continuous then determines $e^{itH}u$ in general.

score 2 · Answer 2 · answered Aug 06 '20 at 20:55

2

From functional calculus, your operator is computed by a Taylor series as you say: exp(itH)=1+itH+(itH)^2/2!+(itH)^3/3!...

Further reading: https://en.wikipedia.org/wiki/Holomorphic_functional_calculus#Need_for_a_general_functional_calculus

answered Aug 06 '20 at 20:55

EEatWork

99

2

This leads to convergence problems, as the OP guesses, since $H$ is unbounded. – Bananach Aug 06 '20 at 21:11

Disintegrating By Parts · Accepted Answer · 2020-08-21T02:31:47.630

Let $X$ be a real or complex Banach space, and let $\mathcal{L}(X)$ denote the bounded linear operators on $X$. A $C^0$ semigroup on $X$ is a function $$ T : [0,\infty)\rightarrow\mathcal{L}(X) $$ such that \begin{align} (i) & \;\;\; T(0) = I, \\ (ii) & \;\;\; T(t)T(t')=T(t+t'), \\ (iii) & \;\;\; \lim_{t\downarrow 0}T(t)x = x,\;\; \forall x \in X. \end{align} For any such operator, let $\mathcal{D}(A)$ denote the set of all $\in X$ for which the following limit exists $$ \lim_{h\downarrow 0} \frac{1}{h}(T(h)x-x), $$ and let $Ax$ denote this limit. Then $H : \mathcal{D}(A)\subseteq X\rightarrow X$ is a densely-defined linear operator, and we write $T(t)=e^{tA}$ to summarize these properties.

If $T(t)$ is unitary for all $t > 0$, then $A=iH$, where $H$ is self-adjoint. Then it is customary to write $T(t) = e^{itH}$. This is typical of the time-invariant Schrodinger equation, for example. For such an operator, there is the Borel functional calculus, where $f(H)=$ is defined for any Borel measurable function on $\mathbb{R}$. Using this, $e^{itH}=f(H)$ where $f(s)=e^{its}$.

$e^{itH}$ notation

3 Answers3