Time dependent perturbation theory, HO subject to electric field

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SUMMARY

This discussion focuses on the application of time-dependent perturbation theory to compute the transition probability of a charged particle in a harmonic potential subjected to a uniform electric field. The initial Hamiltonian is defined as ##H=H_0+\lambda \hat W##, with ##H_0## representing the harmonic oscillator. The transition probability to the first excited state is calculated as ##P_{01}=\frac{q^2E^2}{\hbar m \omega^3}\left (1-\cos(\omega t) \right )##. The user also attempts to solve the Schrödinger equation exactly, leading to a shift in the eigenfunctions due to the electric field, but expresses uncertainty about the correctness of their approach.

PREREQUISITES
  • Understanding of quantum mechanics, specifically harmonic oscillators
  • Familiarity with time-dependent perturbation theory
  • Knowledge of Schrödinger equation and eigenfunction analysis
  • Basic concepts of transition probabilities in quantum systems
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Telemachus
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Hi there. I'm dealing with this problem, which says:

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.
 
Last edited:
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Ok, I think I've realized what I have to do. I think I should use the set of differential equations obtained from the Schrödinger equation in the series expansion. Anyway, I'll appreciate any help or suggestion.
 
Last edited:

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