I am reading a introduction to quantum mechanics right now. There is a part about the time evolution operator:

\begin{align*} i\hbar \partial_t \,\psi(\vec r, t) = \hat H(t)\, \psi(\vec r,t) \end{align*} is the time dependent Schrödinger-equation. If we assume that for each $\psi(\vec r, t_0)$ there is a unique solution $\psi(\vec r, t)$, then we can define an operator

$$U(t,t_0): \mathcal H \to \mathcal H,\, \psi(\vec r, t_0) \mapsto \psi(\vec r, t)$$

This operator is linear, since the Schrödinger equation is linear and it is unitary, since $\partial_t \langle\psi(\vec r, t)| \psi(\vec r, t)\rangle = 0$. I am totally happy with that. I can also accept, that $U(t,t_0) = e^{-i(t-t_0)\hat H/ \hbar}$, if $\hat H$ is time independent, where $e^{-i(t-t_0)\hat H/ \hbar}$ is defined over how it acts on the eigenvectors of $\hat H$.

But I have no idea, what the next sentence in my book means, and there is no good explanation. Is says there:

The differential equation, together with the initial condition ($U(t_0,t_0) = Id$) is equivalent to the integral equation: \begin{align*} U(t,t_0) = 1 - \frac{i}{\hbar } \int_{t_0}^t ds\, \hat H(s) U(s,t_0) \end{align*}

So my problem is basically, I don't understand this at all :/. How can I integrate operators, what does that even mean? Are there any good examples, where this integral makes sense? This is probably a really stupid question, but I am happy if someone could spare two minutes to help me.

  • $\begingroup$ The lhs and the rhs of your last equation are, formally, allowed to act on wavefunctions, by means of $U(t,t_0) \psi(t_0) = \psi(t_0) - \frac{i}{\hbar} \int_{t_0}^t ds \hat{H}(s)U(s,t_0) \psi(t_0)$. Now the rhs is just the usual integral. $\endgroup$
    – Creo
    Oct 15, 2018 at 20:28

1 Answer 1


This is a very good question, but a mathematical one. The expression you quoted from the book is the part-"summation" of the Dyson expansion of the unitary evolution operator. To quote from Reed and Simon, theorem X.69 (Vol. II, p. 282)

Let $t\mapsto H(t)$ a strongly continuous map of $\mathbb{R}$ into the bounded self-adjoint operators on a Hilbert space $\mathcal{H}$. Then there is a unitary propagator on $\mathcal{H}$ so that, for all $\psi\in\mathcal{H}$, $$\phi_s (t) = U(t,s) \psi $$ satisfies $$ \frac{d}{dt} \phi_s (t) = -i H(t) \phi_s (t) \ ; \ \phi_s (s) = \psi$$

The proof starts by explicitely exhibiting the unitary propagator as

$$ U(t,s) \phi = 1 +\sum_{n=1}^{\infty} (-i)^n \int_{s}^{t} \int_{s}^{t_1} ... \int_{s}^{t_n} H(t_1)... H(t_n) \phi \ dt_n ... \ dt_1 $$

What the book did is just ~resum~ the infinite expression to the right of $ H(t_1) $ into another U (the minus vs. plus sign after the unit vector comes from the different convention for evolution). Now we have no longer an integral of a product of operators, but of Hilbert space-valued functions. This is just an iteration of a Bochner-type integral.

  • 1
    $\begingroup$ Thanks a lot.I didn't know this book exists. It's really cool. I want to make sure, that I understand it right... The proof is roughly: 1.) Show that the series is absolutely convergent with respect to the operator norm for each t, thus U(t,s) exists, since operator spaces are complete. 2.) check all of its properties. The fact that U(t,s) satisfies the differential equations, comes from the fact, that I am allowed to differentiate series term by term, if I am inside the convergence radius. $\endgroup$
    – Nico
    Oct 16, 2018 at 7:32
  • $\begingroup$ @N.Beck That is correct. $\endgroup$
    – DanielC
    Oct 16, 2018 at 16:20

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