- #1
Diracobama2181
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- 2
- TL;DR Summary
- I am struggling a bit with the following problem, primarily with the momentum expectation value.
An electric field E(t) (such that E(t) → 0 fast enough as t → −∞)
is incident on a charged (q) harmonic oscillator (ω) in the x direction,
which gives rise to an added ”potential energy” V (x, t) = −qxE(t).
This whole problem is one-dimensional.
(a) Using first-order time dependent perturbation theory, write down
the amplitude cn(t) for finding the system in excited state n at time t
if the system starts in n = 0 at t → −∞.
(b) Find the expectation of the momentum of the oscillator as a function of time.Now part a) seemed straightforward enough.
For that, I got $$C_n=(i/\hbar)\int_{−∞}^{T}(<n|(-qxE(t))|0>)e^{\frac{(-in)(\omega)}{\hbar}}dt$$
Using the position operator $$x=\frac{\hbar}{2m\omega}^{1/2}(a+a^{'})$$,
I obtain $$C_n=(i/\hbar)\int_{−∞}^{T}(-qE(t))(<n|1>)\frac{\hbar}{2m\omega}^{1/2}e^{\frac{(-in)(\omega)}{\hbar}}dt$$
which leaves only $$C_1=(i/\hbar)\int_{−∞}^{T}(-qE(t))\frac{\hbar}{2m\omega}^{1/2}e^{\frac{(-i)(\omega)}{\hbar}}dt$$
Part b is where my issue lies.
I know I can represent the particle state as $$|\psi>=\sum_{n=0}^{\infty}c_n(t)e^{\frac{-iE_nt}{\hbar}}|n> $$
I also know $$p=-(i)\frac{m\hbar\omega}{2}^{1/2}(a-a^{'})$$
When I apply the operator on psi, I get
$$p|\psi>=-(i)\frac{m\hbar\omega}{2}^{1/2}(c_0(t)0-|1>)e^{\frac{-iE_0t}{\hbar}}+...+c_n(t)(\sqrt(n)|n-1>-\sqrt(n+1)|n+1>)e^{\frac{-iE_nt}{\hbar}}$$
When I attempt the expectation value, I find
$$<\psi|P|\psi>=-(i)\frac{m\hbar\omega}{2}^{1/2}(C_0(t)^{*}C_1(t)e^{\frac{i(E_0-E_1)t}{\hbar}}-C_1(t)^{*}C_10t)e^{\frac{i(E_1-E_0)t}{\hbar}}+...C_{n-1}(t)^{*}C_n(t)e^{\frac{i(E_{n-1}-E_n)t}{\hbar}}-C_{n}(t)^{*}C_{n-1}(t)e^{\frac{i(E_{n}-E_{n-1})t}{\hbar}}$$
I suppose I just want to know whether my reasoning has been fine thus far, and if so, how could I go about reducing this equation?
is incident on a charged (q) harmonic oscillator (ω) in the x direction,
which gives rise to an added ”potential energy” V (x, t) = −qxE(t).
This whole problem is one-dimensional.
(a) Using first-order time dependent perturbation theory, write down
the amplitude cn(t) for finding the system in excited state n at time t
if the system starts in n = 0 at t → −∞.
(b) Find the expectation of the momentum of the oscillator as a function of time.Now part a) seemed straightforward enough.
For that, I got $$C_n=(i/\hbar)\int_{−∞}^{T}(<n|(-qxE(t))|0>)e^{\frac{(-in)(\omega)}{\hbar}}dt$$
Using the position operator $$x=\frac{\hbar}{2m\omega}^{1/2}(a+a^{'})$$,
I obtain $$C_n=(i/\hbar)\int_{−∞}^{T}(-qE(t))(<n|1>)\frac{\hbar}{2m\omega}^{1/2}e^{\frac{(-in)(\omega)}{\hbar}}dt$$
which leaves only $$C_1=(i/\hbar)\int_{−∞}^{T}(-qE(t))\frac{\hbar}{2m\omega}^{1/2}e^{\frac{(-i)(\omega)}{\hbar}}dt$$
Part b is where my issue lies.
I know I can represent the particle state as $$|\psi>=\sum_{n=0}^{\infty}c_n(t)e^{\frac{-iE_nt}{\hbar}}|n> $$
I also know $$p=-(i)\frac{m\hbar\omega}{2}^{1/2}(a-a^{'})$$
When I apply the operator on psi, I get
$$p|\psi>=-(i)\frac{m\hbar\omega}{2}^{1/2}(c_0(t)0-|1>)e^{\frac{-iE_0t}{\hbar}}+...+c_n(t)(\sqrt(n)|n-1>-\sqrt(n+1)|n+1>)e^{\frac{-iE_nt}{\hbar}}$$
When I attempt the expectation value, I find
$$<\psi|P|\psi>=-(i)\frac{m\hbar\omega}{2}^{1/2}(C_0(t)^{*}C_1(t)e^{\frac{i(E_0-E_1)t}{\hbar}}-C_1(t)^{*}C_10t)e^{\frac{i(E_1-E_0)t}{\hbar}}+...C_{n-1}(t)^{*}C_n(t)e^{\frac{i(E_{n-1}-E_n)t}{\hbar}}-C_{n}(t)^{*}C_{n-1}(t)e^{\frac{i(E_{n}-E_{n-1})t}{\hbar}}$$
I suppose I just want to know whether my reasoning has been fine thus far, and if so, how could I go about reducing this equation?