View source for Phy5646/Simpe Example of Transition Probability Calculation

This example taken from "Quantum Physics" 3rd ed., Stephen Gasiorowicz, p. 237-238.

'''Problem:''' 
A particle of charge <math>q</math> in a one-dimensional harmonic oscillator of characteristic frequency <math>\omega</math> is placed in an electric field that is turned on and off so that the potential energy is 

:<math>
\lambda V(t) = qExe^{-t^{2}/\tau^{2}}
</math>

If the particle is initially in the ground state, what is the probability that after time <math>t</math>, such that <math>t >> \tau</math>, 
the particle is found in the first excited state of the harmonic oscillator? What is the probability that 
it is found in the second excited state?

'''Solution:'''
The harmonic oscillator eigenstates: <math>|n\rangle</math>, with eigen-energies <math>h\omega (n + 1/2) </math>. Then 

:<math>
\langle n|\psi(t)\rangle  = \langle n|0\rangle +  \frac{qE}{i\hbar}\int_{-\infty}^{t}dt'e^{\frac{i}{\hbar}(\epsilon_n - \epsilon_0)t'}\langle n|x|0\rangle e^{-t'^{2}/\tau^{2}} 
</math>

Since <math>\tau >> T</math>,we may take the upper and lower limits in the time integral to be <math> \pm \infty</math>. We therefore get
 
:<math>
\begin{align}
\langle n|\psi(t)\rangle & = \frac{qE}{i\hbar}\langle n|x|0\rangle\int_{-\infty}^{\infty}dt'e^{in \omega}e^{-t'^{2}/\tau^{2}}\\ & = \sqrt{\pi}\frac{qE\tau}{i\hbar}\langle n|x|0\rangle e^{-n^{2}\omega^{2}\tau^{2}/4}
\end{align}
</math>

Therefore,

:<math>
P_{n}=\pi\frac{q^{2}E^{2}\tau^{2}}{\hbar^{2}}|\langle n|x|0\rangle|^{2} e^{-n^{2}\omega^{2}\tau^{2}/2}
</math>

To calculate <math>\langle n|x|0\rangle</math> we use

:<math>
x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})
</math>

and recall that <math>a|0\rangle = 0</math> and <math>a^{\dagger}|0\rangle = |1\rangle</math>. Thus, <math> |\langle n|x|0\rangle |^{2} = \frac{\hbar}{2m\omega}\delta_{n1} </math> and therefore 

:<math>
P_{1}=\frac{\pi}{2}\frac{q^{2}E^{2}\tau^{2}}{m\hbar\omega}e^{-\omega^{2}\tau^{2}/2}
</math>
 
We also see that <math>P_n = 0</math> for <math>n</math> = 2, 3, ...
 
Note the following
 
1. As <math> \tau \to \infty P_{1}\to 0</math>. When the electric field is turned on very slowly, then the transition probability goes to zero; that is, the system adjusts adiabatically to the presence of the field, without being "jolted" into making a transition.
 
2. The potential, through its <math>x</math>-dependence, will allow only certain transitions to take place. In other words, we see that there is a selection rule that operates here. If the potential were proportional to <math>x^{2}</math>, for example, then transitions to <math>n</math> = 2 would be allowed.