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Hi there. I'm dealing with this problem, which says:

Well, I computed the transition probability using time dependent perturbation theory at first, cause I thought it would be easier than solving exactly.

I have:

##H=H_0+\lambda \hat W\\

E_n^0=\hbar\omega\left (n + \frac{1}{2}\right ) \\

H_0=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2\\

\ W(t) = \left\{ \begin{array}{ll}

qEX & \mbox{if $t \geq 0$};\\

0 & \mbox{if $t < 0$}.\end{array} \right. \ \\

P_{if}=\frac{1}{\hbar^2} \left | \int_0^t e^{i\omega_{fi}t'}W_{fi}dt' \right |^2 \\ ##

Then using some harmonic oscillator algebra I have:

##W_{10}=qE\sqrt {\frac{\hbar}{2m\omega}}\\

\omega_{10}=\omega##

And the probability of transition using time dependent perturbation theory gave me:

##P_{01}=\frac{q^2E^2}{\hbar m \omega^3}\left (1-\cos(\omega t) \right )##

Then I've tried to solve the Schrödinger equation exactly, which is easy to solve by completing the square.

That gives the eigenfunctions:

##\phi_n'(x)=\phi_n(x+\frac{qE}{m\omega^2})##

The effect of the electric field is just a shift in x.

So then I thought that the desired transition probability would be:

##P_{01=}|\left <\phi_0 | \phi_1' \right >|^2##

I'm not sure if this is right, so then I expressed ##\left <\phi_0 | \phi_1' \right >## in the x representation which gives me a cumbersome integral:

##\left <\phi_0 | \phi_1' \right >=\displaystyle \int_{-\infty}^{\infty}dx \left <\phi_0 | x \right > \left < x | \phi_1' \right >=\displaystyle \left [\frac{4}{\pi^2}\right ] \frac{m\omega}{\hbar} \int_{-\infty}^{\infty} u e^{-\frac{1}{2}\left [ (u-\frac{qE}{m\omega^2})^2+\frac{m\omega}{\hbar}u^2 \right ] }du##

I've used the change of variables ##u=x+\frac{qE}{m\omega^2}##, at t=0 I have ##\left < x | \psi (t=0) \right >=\phi_0(x)##, and for t>0 my eigeinstates are the ##\left <x | \phi_n' \right > =\phi_n(x+\frac{qE}{m\omega^2})=\phi_n(u)##

Now, I think this procedure is wrong. I believe the way I've solved the schrödinger equation is wrong, because I'm solving the perturbed hamiltonian, but I'm not considering what happens before t=0. Besides, the integral doesn't look like the cosine I've obtained before.

Any help will be appreciated.

*At time ##t=0## a constant and uniform electric field ##\vec E## oriented in the ##\vec x## direction is applied over a charged particle with charge ##+q##. This same particle is under the influence of an harmonic potential ##V(x)=\frac{1}{2}m \omega^2 X^2##, which at time ##t=0## is in it's fundamental state. Compute the transition probability to the first excited state. Solve the problem exactly. Verify that the result obtained is the same for first order (time dependent) perturbation theory. Study what would happen for a transition to the second excited state.*

Well, I computed the transition probability using time dependent perturbation theory at first, cause I thought it would be easier than solving exactly.

I have:

##H=H_0+\lambda \hat W\\

E_n^0=\hbar\omega\left (n + \frac{1}{2}\right ) \\

H_0=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2\\

\ W(t) = \left\{ \begin{array}{ll}

qEX & \mbox{if $t \geq 0$};\\

0 & \mbox{if $t < 0$}.\end{array} \right. \ \\

P_{if}=\frac{1}{\hbar^2} \left | \int_0^t e^{i\omega_{fi}t'}W_{fi}dt' \right |^2 \\ ##

Then using some harmonic oscillator algebra I have:

##W_{10}=qE\sqrt {\frac{\hbar}{2m\omega}}\\

\omega_{10}=\omega##

And the probability of transition using time dependent perturbation theory gave me:

##P_{01}=\frac{q^2E^2}{\hbar m \omega^3}\left (1-\cos(\omega t) \right )##

Then I've tried to solve the Schrödinger equation exactly, which is easy to solve by completing the square.

That gives the eigenfunctions:

##\phi_n'(x)=\phi_n(x+\frac{qE}{m\omega^2})##

The effect of the electric field is just a shift in x.

So then I thought that the desired transition probability would be:

##P_{01=}|\left <\phi_0 | \phi_1' \right >|^2##

I'm not sure if this is right, so then I expressed ##\left <\phi_0 | \phi_1' \right >## in the x representation which gives me a cumbersome integral:

##\left <\phi_0 | \phi_1' \right >=\displaystyle \int_{-\infty}^{\infty}dx \left <\phi_0 | x \right > \left < x | \phi_1' \right >=\displaystyle \left [\frac{4}{\pi^2}\right ] \frac{m\omega}{\hbar} \int_{-\infty}^{\infty} u e^{-\frac{1}{2}\left [ (u-\frac{qE}{m\omega^2})^2+\frac{m\omega}{\hbar}u^2 \right ] }du##

I've used the change of variables ##u=x+\frac{qE}{m\omega^2}##, at t=0 I have ##\left < x | \psi (t=0) \right >=\phi_0(x)##, and for t>0 my eigeinstates are the ##\left <x | \phi_n' \right > =\phi_n(x+\frac{qE}{m\omega^2})=\phi_n(u)##

Now, I think this procedure is wrong. I believe the way I've solved the schrödinger equation is wrong, because I'm solving the perturbed hamiltonian, but I'm not considering what happens before t=0. Besides, the integral doesn't look like the cosine I've obtained before.

Any help will be appreciated.

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