How Does Time Evolution Affect Operator Expectation Values in Quantum Mechanics?

[mahblah]

Homework Statement

i know that
$H |n> = \varepsilon_n |n>$
$H |i> = \varepsilon_i |i>$

and i want to estimate
$< n | e^{i H t / \hbar} V(t) e^{-i H t / \hbar} |i>$

Homework Equations

$<\psi | \varphi> = \int \psi^*(q) \varphi(q) dq$

The Attempt at a Solution

i don't understand well how the operator interacts with the wave function... I'm pretty sure that the right solution is

$< n | e^{i H t / \hbar} V(t) e^{-i H t / \hbar} |i> = \int{ \left(n e^{- i H t / \hbar}\right)^* V e^{-i \varepsilon_i t / \hbar}} i = \int n^* e^{i \varepsilon_n t / \hbar} V e^{-i \varepsilon_f t / \hbar} i= <n | V(t) | i> e^{i \frac{\varepsilon_n - \varepsilon_i}{\hbar}t}$

but I'm not sure about "why is this the right way to do this" ... for example i don't know why $H |n> = \varepsilon_n |n>$ implies $e^{iHt / \hbar} |n> = e^{i \varepsilon_n t / \hbar } |n>$... or why this way to do is wrong:

$< n | e^{i H t / \hbar} V(t) e^{-i H t / \hbar} |i> = \int \left(n e^{i H t / \hbar}\right)^* e^{i H t / \hbar} V(t) i = \int n^* V(t) i = <n|V(t)|i>$

(i didn't put the "dq" in the integral because i don't know in which variable i should integrate)
if someone can explain me why this is the right way to do this expression i'll be happy.

 

[mathfeel]

for example i don't know why $H |n> = \varepsilon_n |n>$ implies $e^{iHt / \hbar} |n> = e^{i \varepsilon_n t / \hbar } |n>$

Expand the exponential into Taylor series, then operate term-by-term, then recombine.

... or why this way to do is wrong:

$< n | e^{i H t / \hbar} V(t) e^{-i H t / \hbar} |i> = \int \left(n e^{i H t / \hbar}\right)^* e^{i H t / \hbar} V(t) i = \int n^* V(t) i = <n|V(t)|i>$

If $V(t)$ is just a c-number, you can just pull it out, which then makes the problem trivial. So I am guessing that $V(t)$ is an operator, in which case, you cannot commute it with the exponential on the right.

 

[mahblah]

...thanks mathfeel!

 

[vela]

(I didn't put the "dq" in the integral because i don't know in which variable I should integrate.)
If someone can explain me why this is the right way to do this expression, I'll be happy.

When you calculate an amplitude, the integral comes from inserting a complete set$$\int dq\,\lvert q \rangle\langle q \rvert = \mathbf{1}$$ to get
\begin{align*}
\langle \phi \vert \psi \rangle &= \langle \phi \vert \left(\int dq\,\lvert q \rangle\langle q \rvert\right) \vert \psi \rangle \\
&= \int dq\,\langle \phi \vert q \rangle\langle q | \psi \rangle
\end{align*}
Then using the fact that $\langle\phi\vert q\rangle = \phi^*(q)$ and $\langle q\vert\psi\rangle = \psi(q)$, you obtain$$\langle \phi \vert \psi \rangle = \int dq\,\phi^*(q) \psi(q)$$
Note that the bra and ket are related to but are not exactly the same thing as the wave functions. The wave function is the representation of ket relative to some basis.

In this problem, you never need to introduce a complete set and get an integral. Just work with the kets. There's no need to use the wave functions.

 

[paranormal]

I can provide some insights into your questions about the operator in quantum mechanics and its interaction with the wave function.

Firstly, the operator in quantum mechanics represents a physical observable, such as position, momentum, or energy. In your case, the operator H represents the energy of a system. The equation H|n> = ε_n|n> means that when the operator H acts on the state |n>, it returns the energy eigenvalue ε_n times the state |n>. This is a fundamental property of quantum mechanics and is known as the eigenvalue equation.

Now, let's consider the expression <n|e^{iHt/\hbar}V(t)e^{-iHt/\hbar}|i>. This is known as the time-evolution operator, and it tells us how the state |i> evolves into the state |n> over time t. The time-evolution operator is given by the formula e^{-iHt/\hbar}, where H is the Hamiltonian operator (in your case, it is just H) and t is the time. This operator is crucial in understanding how quantum systems evolve over time.

To answer your question about why e^{iHt/\hbar}|n> = e^{iε_nt/\hbar}|n>, we can use the Taylor series expansion of the exponential function. The Taylor series expansion of e^{iHt/\hbar} is given by 1 + (iHt/\hbar) + (iHt/\hbar)^2/2! + ... = 1 + (iε_nt/\hbar) + (iε_nt/\hbar)^2/2! + ... = e^{iε_nt/\hbar}. So, we can see that when the operator H acts on the state |n>, it returns the energy eigenvalue ε_n times the state |n>, which is why e^{iHt/\hbar}|n> = e^{iε_nt/\hbar}|n>.

Now, coming to your question about why the expression <n|e^{iHt/\hbar}V(t)e^{-iHt/\hbar}|i> is the right way to calculate the expectation value of the operator V at time t between the states |n> and |i>, we need to use the definition of the expectation value. The expectation value of an operator A between two states |ψ