Unitary Time Evolution Operator

Question

I am a physics undergraduate reading through section 2.1 of Sakurai's Modern Quantum Mechanics (3ed). Note that I am dealing with a time independent Hamiltonian.

I have been having a hard time parsing the notation of the section since the same symbols seem to be used for different mathematical manipulations. In particular,

Q: How do I write the action of the unitary time-evolution operator $\mathcal{U}(t_0, t)$ on a state ket $\alpha$ precisely? To my understanding, the content of the section does not make sense if you treat $\mathcal{U}(t_0, t)$ as a function of $t$$^{[1]}$. Instead, it seems one should treat the operator like $$ \mathcal{U}(t_0, t') |_t \cdot | \alpha \rangle.$$

In other words, that when one acts with the unitary time-evolution operator, one is implicitly evaluating the operator at a particular value $t$ and then acting on the state ket.

Is this accurate?

[1] Suppose $A$ is an observable such that $[A, H] = 0$, i.e., $A$ and the Hamiltonian are compatible observables. Suppose there is another observable $B$ which does not necessarily commute with $A$ nor $H$. Sakurai writes

$$\langle B \rangle = [\sum_{a'}c^*_{a'}\langle a'|\exp(\frac{iE_{a'}t}{\hbar})]\cdot B \cdot [\sum_{a''}c_{a''}\exp(\frac{-iE_{a'}t}{\hbar})|a''\rangle]$$ $$ = \sum_{a'}\sum_{a''}c^*_{a'}c_{a''}\langle a'|B|a''\rangle] \exp(\frac{iE_{a'}t}{\hbar})\exp(\frac{-iE_{a'}t}{\hbar}).$$

This seems to imply to treat the exponentials (the unitary time-evolution operators) as constants. However, this brings up another confusion since the end result $\langle B \rangle$ is said to depend on time. Thus, I interpret $\langle B \rangle$ to be a function of time. But, how can a function of time emerge if we are treating the $t$ found in the unitary time-evolution operator as a constant?

Answer 1

When one acts with the unitary time-evolution operator, one is implicitly evaluating the operator at a particular value t and then acting on the state ket.

Is this accurate?

Yes, this is accurate, except for the word "implicitly". However, there is no magic here and nothing is hidden. Maybe you should think carefully about what you mean by "treating $\mathcal U(t_0, t)$ as a function of $t$" and what the difference would be whether you "treat it as a function of $t$" or not. I think you will find that there is no difference.

For each value of $t$, the expression $\mathcal U(t_0, t)$ yields an operator on the Hilbert space. The expression therefore represents a function $\mathcal U(t_0, \cdot): \mathbb R \to \mathcal B(\mathcal H)$ (where I use $\mathcal B(\mathcal H)$ to denote the space of operators on the Hilbert space $\mathcal H$). Similarly, the expression $\mathcal U(t_0, t) |\alpha\rangle$ evaluates to some state in the Hilbert space, by first evaluating $\mathcal U(t_0, t)$ and then acting with that operator on $|\alpha\rangle$. In short, you put in a $t$ and you get out a Hilbert space vector, that is the very definition of a function $\mathbb R \to \mathcal H$.

It is correct, and important to understand, that time plays a different role here than, for example, spatial coordinates. Let's consider the Hilbert space of a free particle, $\mathcal H = L^2(\mathbb R^3)$. At each time, the system state is described by a different vector $|\psi_t\rangle \in \mathcal H$, but each vector $|\psi_t\rangle$ contains the complete spatial dependence of the wave function as $\psi_t(\vec x) = \langle \vec x | \psi_t \rangle$. From the point of view of the vectors in the Hilbert space, $t$ is just a parameter and an expression like your $\exp(\frac{-\mathrm iE_{a''}t}{\hbar}) |a''\rangle$ really just means multiplying the vector $|a''\rangle$ by the number $\exp(\frac{-\mathrm iE_{a''}t}{\hbar})$.

Answer 2

I am learning Quantum Mechanics from Griffiths, and to my understanding, you are right. To your confusion in [1], this is not exactly implying the exponential that depends on $t$ is a constant. It's saying that you can treat the operator as a number. Based on the definition of the time evolution operator, it's clear that it ends up being some complex number that depends on time. When your operator is just a number, to apply it you just multiply the function by that number, so you are free to move it to the back. But a number is different from a constant, as a number can be dependent on t, but a constant can't. So $<B>$ is dependent on time.