Unitarity of the time-evolution operator

Question

In Sakurai equation 2.15, the infinitesimal time-evolution operator takes the form $$\mathscr{U}\left(t_0+d t, t_0\right)=1-i \Omega d t $$

where $\Omega$ is a Hermitian operator. Below is a subsequent excerpt:

With $(2.15)$ the infinitesimal time-displacement operator satisfies the composition property $$ \mathscr{U}\left(t_0+d t_1+d t_2, t_0\right)=\mathscr{U}\left(t_0+d t_1+d t_2, t_0+d t_1\right) \mathscr{U}\left(t_0+d t_1, t_0\right) ; $$ it differs from the identity operator by a term of order $d t$. The unitarity property can also be checked as follows: $$ \mathscr{U}^{\dagger}\left(t_0+d t, t_0\right) \mathscr{U}\left(t_0+d t, t_0\right)=\left(1+i \Omega^{\dagger} d t\right)(1-i \Omega d t) \simeq 1 $$ to the extent that terms of order $(d t)^2$ or higher can be ignored.

It concerns me greatly that our formulation of the time-evolution operator appears to assume second-order terms are negligible. Does this mean the time-evolution operator is approximate and not exact? If so, are there situations in which the time-evolution operator is a poor approximation, and how do we deal with such situations?

Answer 1

In fact, the infinitesimal version implies that $U$ is unitary. Put differently, we don't need to discard second (and higher) order terms at all, but they will vanish if we take the limit $h\to 0$, where $h:= \mathrm dt$. Indeed, we (at least formally) can do the following:

Observe that

$$U(t+h,t_0) = U(t+h,t)\,U(t,t_0) = U(t,t_0)-ih\,\Omega \, U(t,t_0) + \mathcal O(h^2)\tag{1}$$

and, by taking the adjoint of the above equation,

$$ U^\dagger(t+h,t_0) = U^\dagger(t,t_0) \, U^\dagger(t+h,t)=U^\dagger(t,t_0) + ih\, U^\dagger(t,t_0)\,\Omega + \mathcal O(h^2) \tag{2} \quad .$$

Multiplying these two equations yields

$$U^\dagger(t+h,t_0)\, U(t+h,t_0) - U^\dagger(t,t_0)\,U(t,t_0) = \mathcal O(h^2) \tag{3} \quad .$$

Alternatively, we can obtain $(3)$ as follows: In the last equation of the quote, first relabel $t_0 \rightarrow t$ ($t_0$ is arbitrary here) and then multiply from the left with $U^\dagger(t,t_0)$ and from the right with $U(t,t_0)$.

Introducing $f(t):=U^\dagger (t,t_0)\, U(t,t_0)$, dividing by $h$ and then taking the limit $h\to 0$ in $(3)$ yields $f^\prime(t) = 0$ with $f(t_0)=1$. The unique solution of this initial value problem is $f(t)=1$ for all $t$, i.e. $U^\dagger(t,t_0)\, U(t,t_0)=1$ for all $t,t_0$.

Note that we have to show that $U(t,t_0)\, U^\dagger(t,t_0)=1$, too, but the argument here is the same. We have further used that $\Omega=\Omega^\dagger$; in general, $\Omega$ can be time-dependent.

Answer 2

This is not anything special about the time evolution operator. It's simply a way of expressing calculus using infinitesimals. A similar example would be that if $y=x^2$, then $dy/dx=[(x+dx)^2-x^2]/dx=2x+dx$, where we throw away the second term at the end. People used to habitually express calculus this way for hundreds of years, using procedures like this that were known to work. Around 1900, when the definition of the limit was formalized, people started to get uneasy about whether these styles of reasoning were OK. It turns out they were, as proved in the field of non-standard analysis.

Answer 3

For a time-independent Hamiltonian you can use $$ \hat{U}(t,t_0) = \exp(-i \hat{H} (t-t_0)/\hbar) $$ which is correct for any finite $t$. The state at time $t$ is given by $$ |\psi(t)\rangle = \hat{U}(t,t_0) |\psi(t_0)\rangle $$ You can check that Hermitian $\hat{H}$ here implies unitary $\hat{U}$. Slightly more generally, if $\hat{H}$ at any given time commutes with $\hat{H}$ at other times, then you can use $$ \hat{U}(t,t_0) = \exp\left(-i \int_{t_0}^t \hat{H} dt/\hbar\right). $$ However, the general case, where $\hat{H}$ at one time does not commute with $\hat{H}$ at another time, is a good deal more complicated. This is where the consideration of infinitesimal durations of time may be useful. In this case if each $\hat{U}(t+dt,t)$ is unitary to 1st order in $dt$, then so is the combination $\hat{U}(t+2dt,t+dt)\hat{U}(t+dt,t)$, so we can build up a propagator (that is, time evolution operator) over finite time intervals by taking a product of propagators over small time intervals and taking a limit.

Answer 4

I agree with you that Sakurai's argument is too casual. In fact, it's missing a key assumption: that $\Omega$ is Hermitian at all times, not just at time $t_0$. Here's a more formal version.

Since $t_0$ stays fixed throughout the argument, let's use the shorthand $U(\tau) = \mathscr{U}(t_0 + \tau, t_0)$. The sentence $$\mathscr{U}\left(t_0+dt, t_0\right)=1-i\Omega\,dt$$ is an informal way of saying that $U'(0) = i\Omega$. We can take this as a definition of $\Omega$. More generally, let's define $\Omega(\tau)$ as the operator for which $U'(\tau) = i\Omega(\tau) U(\tau)$. This definition is guaranteed to make sense as long as $U(\tau)$ is invertible. For simplicity, let's just assume that $U(\tau)$ is invertible at all times, although we could actually prove it as part of our argument.

We want to show that $U(\tau)$ is unitary at all times, given that $\Omega(\tau)$ is Hermitian at all times and $U(0) = 1$. In other words, we want to show that $\langle U(\tau)\,\psi, U(\tau)\,\phi \rangle = \langle \psi, \phi \rangle$ at all times. To check whether this is true, let's see how $\langle U(\tau)\,\psi, U(\tau)\,\phi \rangle$ changes over time—hoping, of course, that it won't change at all.

Because the inner product $\langle\;\;,\;\;\rangle$ is conjugate-linear in the first argument, linear in the second argument, and continuous, it follows the product rule, which tells us that $$\begin{align*} \langle U(\tau)\,\psi, U(\tau)\,\phi \rangle' & = \langle U'(\tau)\,\psi, U(\tau)\,\phi \rangle + \langle U(\tau)\,\psi, U'(\tau)\,\phi \rangle \\ & = \langle i\Omega(\tau)U(\tau)\,\psi, U(\tau)\,\phi \rangle + \langle U(\tau)\,\psi, i\Omega(\tau)U(\tau)\,\phi \rangle. \end{align*}$$ Using those linearity properties again, we see that $$\begin{align*} \langle U(\tau)\,\psi, U(\tau)\,\phi \rangle' & = -i\langle \Omega(\tau)U(\tau)\,\psi, U(\tau)\,\phi \rangle + i\langle U(\tau)\,\psi, \Omega(\tau)U(\tau)\,\phi \rangle \\ & = -i\langle \Omega(\tau)U(\tau)\,\psi, U(\tau)\,\phi \rangle + i\langle \Omega(\tau)^\dagger U(\tau)\,\psi, U(\tau)\,\phi \rangle. \end{align*}$$ Since we're assuming that $\Omega(\tau)$ is Hermitian, we can substitute $\Omega(\tau)$ for $\Omega(\tau)^\dagger$ and conclude that $$\begin{align*} \langle U(\tau)\,\psi, U(\tau)\,\phi \rangle' & = -i\langle \Omega(\tau)U(\tau)\,\psi, U(\tau)\,\phi \rangle + i\langle \Omega(\tau) U(\tau)\,\psi, U(\tau)\,\phi \rangle \\ & = 0. \end{align*}$$ We now see, as we'd hoped, that $\langle U(\tau)\,\psi, U(\tau)\,\phi \rangle$ never changes. Since $U(0) = 1$, we know $\langle U(0)\,\psi, U(0)\,\phi \rangle = \langle \psi, \phi \rangle$, so it follows that $\langle U(\tau)\,\psi, U(\tau)\,\phi \rangle = \langle \psi, \phi \rangle$ at all times. In other words, $U(\tau)$ is unitary at all times.