Phy5646/Simpe Example of Time Dep Pert: Difference between revisions
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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>, | ||
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'''Solution:''' | '''Solution:''' | ||
The harmonic oscillator eigenstates: <math>|n | 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 | 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 | ||
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 | |||
x= | To calculate <math>\langle n|x|0\rangle</math> we use | ||
and recall that | |||
:<math> | |||
2 | x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger}) | ||
We also see that | </math> | ||
Note the following | |||
1. As | 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 | ||
probability goes to zero; that is, the system adjusts adiabatically to the presence of the field, | |||
without being "jolted" into making a transition. | :<math> | ||
2. The potential, through its | P_{1}=\frac{\pi}{2}\frac{q^{2}E^{2}\tau^{2}}{m\hbar\omega}e^{-\omega^{2}\tau^{2}/2} | ||
other words, we see that there is a selection rule that operates here. If the potential were | </math> | ||
proportional to | |||
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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q} in a one-dimensional harmonic oscillator of characteristic frequency Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega} is placed in an electric field that is turned on and off so that the potential energy is
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda V(t) = qExe^{-t^{2}/\tau^{2}} }
If the particle is initially in the ground state, what is the probability that after time Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} , such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t >> \tau} , 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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle} , with eigen-energies Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h\omega (n + 1/2) } . Then
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}} }
Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau >> T} ,we may take the upper and lower limits in the time integral to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm \infty} . We therefore get
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} }
Therefore,
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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} }
To calculate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|x|0\rangle} we use
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger}) }
and recall that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a|0\rangle = 0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{\dagger}|0\rangle = |1\rangle} . Thus, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\langle n|x|0\rangle |^{2} = \frac{\hbar}{2m\omega}\delta_{n1} } and therefore
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{1}=\frac{\pi}{2}\frac{q^{2}E^{2}\tau^{2}}{m\hbar\omega}e^{-\omega^{2}\tau^{2}/2} }
We also see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_n = 0} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} = 2, 3, ...
Note the following
1. As Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau \to \infty P_{1}\to 0} . 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^{2}} , for example, then transitions to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} = 2 would be allowed.