Sakurai on the time-evolution operator

Question

I have three questions about Sakurai's discussion of the time-evolution operator.

1. First question: In equation 2.12, Sakurai requires the composition property of the time-evolution operator: $$U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$$ Why is this required?

2. Second question: In equation 2.15, Sakurai asserts the requirements of the time-evolution operator are satisfied by $$U(t_0+dt,t_0)=1-i\Omega dt$$ with $\Omega$ a Hermitian operator. Where does this come from?

3. Third question: Why is time evolution represented by an operator at all when, as Sakurai points out, time is not an observable like position or momentum? Observables are represented by operators.

Answer 1

1. The map $t\mapsto \psi(t)$ is essentially a continuous trajectory through the Hilbert space $\mathscr H$ which tracks the state vector of the system as it evolves with time. The operator $U(t_2,t_1)$ maps the state vector at time $t_1$ along the curve to the state vector at time $t_2$. Similarly, the operator $U(t_3,t_2)$ maps the state vector at time $t_2$ further along the curve to the state vector at time $t_3$. I think it is reasonable to expect that moving along the curve from $t_1$ to $t_2$ and then from $t_2$ to $t_3$ would yield the same result as moving along the curve from $t_1$ to $t_3$ all at once, hence the composition property $$U(t_3,t_2) U(t_2,t_1) = U(t_3,t_1)$$
2. I don't understand the question. If you consider Sakurai's assertion and then check the composition property and the unitarity property, then you see that (to first order in $\mathrm dt$) they are satisfied. If you want proof that this is the only form the operator may take for infinitesimal $\mathrm dt$, then you may be interested in Stone's theorem.
3. Observables are represented by operators, but that doesn't mean that all operators must represent observables. The idea is that time evolution maps state vectors of our system into other state vectors, and so at minimum corresponds to a well-defined map from $\mathscr H\rightarrow \mathscr H$. The linearity of the time evolution map is motivated by the superposition principle - if one state vector $\psi$ evolves into $\psi'$ and a second state vector $\phi$ evolves into $\phi'$, then the superposition $c_1 \psi + c_2\phi$ should evolve into $c_1\psi' + c_2\phi'$. Such a thing is not strictly guaranteed of course - nothing in physics ever is - but it is a reasonable guess, and the resulting formalism seems to work rather well.

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In response to your comment on the OP:

(1) He simply points out that the composition property seems reasonable. I guess I’m looking for something more rigorous. (2) Yes, he does show that it works. I want to know how we motivate this without knowing where we are going. (3) I’m still troubled by this. It would help if I understood why it must be an operator.

(1) and (3) are answered by noting that we're doing physics, not mathematics. At the end of the day, both the composition property and the fact that time evolution can be described by a linear operator are postulates. At best, a postulate can be well-motivated by experience and appeal to some higher principle, but fundamentally we have to make a guess, develop the resulting framework, and then see how well it predicts the results of experiments. In that sense, the power to make accurate predictions is the ultimate ex post facto justification of anything we do, and there is never any deeper justification than that.

As for (2), I pointed out Stone's theorem in my original answer, but as a side note - if we don't know where we're going, what's probably going to happen is that we blunder around in the darkness until we get our bearings. Once we've hacked our way through the metaphorical jungle of our ignorance and found a theory which works, we can construct a cleaner and more logical path to it.

If a person walking that path were to say "I want to know how I would have found this precise route if the path hadn't been laid down before me," then the only honest answer would be that it would take a long time and a lot of fumbling around. There's value in understanding the historical route, but by and large in your classes and texts you're being given the nice, clean results of decades of someone else's tedious and confused labor.

Answer 2

Third question: Why is time evolution represented by an operator at all when, as Sakurai points out, time is not an observable like position or momentum? Observables are represented by operators.

In addition to J. Murray's answer, I'd like to add an answer only to this specific question, because I suspect the OP might miss a crucial point. In quantum mechanics observables are represented by a special class of operators, that of self-adjoint (or Hermitian if you wish) operators. This only implies that operators which are not self adjoint cannot represent observables, not that operators (whether self-adjoint or not) cannot represent other properties.

In particular, since time evolution in quantum mechanics is deterministic, that is, since the state at time $t_0$ uniquely determines the state at time $t$, one can intuitively expect to be able to represent this mapping of the states from $t_0$ to $t$ with an operator (which is, in fact, a mapping of the Hilbert space into itself). This intuition is then confirmed by the mathematics, which also shows that the time evolution is unitary, and not self-adjoint.