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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?

I am a physics undergraduate reading through section 2.1 of Sakurai's Modern Quantum Mechanics (3ed).

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?

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?

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Operators vs. Observables Unitary Time Evolution Operator

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