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

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(19 intermediate revisions by the same user not shown)
Line 4: Line 4:
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  
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>,  
Line 13: Line 15:
The harmonic oscillator eigenstates: <math>|n\rangle</math>, with eigen-energies <math>h\omega (n + 1/2) </math>. Then  
The harmonic oscillator eigenstates: <math>|n\rangle</math>, with eigen-energies <math>h\omega (n + 1/2) </math>. Then  


<math>\begin{align} \langle n|\psi(t)\rangle & = \langle n|0\rangle +  \frac{1}{i\hbar}\int_{-\infty}^{t}dt'\langle n|V(t')|0\rangle\\ &  \end{align}</math>
:<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>


cn{t) = %r \ dt' <?"■*'(n|jc|0)c"''
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
in J
Vf
:<math>
Since (»T,we may take the upper and lower limits in the time integral in (15-10) to be ±oo. We  
\begin{align}
therefore get  
\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}
(°°) = ^-<«W0> I dt' e-'^'e-'"'72
\end{align}
in J
</math>
— <x
 
= ~<w|jt|0> J dt
Therefore,
■' e-""'e-''2h2
 
= \^T^{n\x\0)e-w«
:<math>
in
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}
This means that
</math>
q2Ey
 
h2
To calculate <math>\langle n|x|0\rangle</math> we use
Pn{^ = ^~r-\(n\x\0)\2e^Wl2
 
To calculate |jc|0) we use (6-35), to get
:<math>
x= l^{A+A+)  
x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})
and recall that A|0> = 0 and A*|0) = |1>. Thus |<nbc|0>|2 = -^- S„. and therefore  
</math>
2ma)
 
2 mtuo
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  
We also see that Pn = 0 for n = 2, 3,  
 
Note the following  
:<math>
1. As t —> oo Pt —> 0. When the electric field is turned on very slowly, then the transition  
P_{1}=\frac{\pi}{2}\frac{q^{2}E^{2}\tau^{2}}{m\hbar\omega}e^{-\omega^{2}\tau^{2}/2}
probability goes to zero; that is, the system adjusts adiabatically to the presence of the field,  
</math>
without being "jolted" into making a transition.  
2. The potential, through its jc-dependence, will allow only certain transitions to take place. In  
We also see that <math>P_n = 0</math> for <math>n</math> = 2, 3, ...
other words, we see that there is a selection rule that operates here. If the potential were  
proportional to x2, for example, then transitions to n = 2 would be allowed.
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.