Operator in quantum mechanics

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Homework Statement


i know that
[itex] H |n> = \varepsilon_n |n>[/itex]
[itex] H |i> = \varepsilon_i |i>[/itex]

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

Homework Equations



[itex] <\psi | \varphi> = \int \psi^*(q) \varphi(q) dq [/itex]


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

[itex] < 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} [/itex]

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

[itex] < 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> [/itex]

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

  • #2
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for example i dont know why [itex] H |n> = \varepsilon_n |n> [/itex] implies [itex] e^{iHt / \hbar} |n> = e^{i \varepsilon_n t / \hbar } |n> [/itex]
Expand the exponential into Taylor series, then operate term-by-term, then recombine.

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

[itex] < 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> [/itex]
If [itex]V(t)[/itex] is just a c-number, you can just pull it out, which then makes the problem trivial. So I am guessing that [itex]V(t)[/itex] is an operator, in which case, you cannot commute it with the exponential on the right.
 
  • #3
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...thanks mathfeel!
 
  • #4
vela
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(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[tex]\int dq\,\lvert q \rangle\langle q \rvert = \mathbf{1}[/tex] 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 [itex]\langle\phi\vert q\rangle = \phi^*(q)[/itex] and [itex]\langle q\vert\psi\rangle = \psi(q)[/itex], you obtain[tex]\langle \phi \vert \psi \rangle = \int dq\,\phi^*(q) \psi(q)[/tex]
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.
 

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