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- 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?