Phy5646/Simpe Example of Time Dep Pert: Difference between revisions

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This example taken from "Quantum Physics" 3rd ed., Stephen Gasiorowicz, p. 237-238.
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  
'''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>
:<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>,  
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  
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?
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.

Latest revision as of 05:47, 28 March 2010

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

Problem: A particle of charge in a one-dimensional harmonic oscillator of characteristic frequency is placed in an electric field that is turned on and off so that the potential energy is

If the particle is initially in the ground state, what is the probability that after time , such that , 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: , with eigen-energies . Then

Since ,we may take the upper and lower limits in the time integral to be . We therefore get

Therefore,

To calculate we use

and recall that and . Thus, and therefore

We also see that for = 2, 3, ...

Note the following

1. As . 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 -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 , for example, then transitions to = 2 would be allowed.