Phy5646/Another example: Difference between revisions
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P_{nk} = \frac{1}{\hbar^2} |\int^\infty_\infty <k|V|n> e^{i t (E^{(0)}_k - E^{(0)}_n )/\hbar} dt| | P_{nk} = \frac{1}{\hbar^2} |\int^\infty_{-\infty} <k|V|n> e^{i t (E^{(0)}_k - E^{(0)}_n )/\hbar} dt| | ||
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V(x,t) = -e x \epsilon(t) - x | V(x,t) = -e x \epsilon(t) - x | ||
</math> | </math> | ||
The oscillator is in the ground state (n=0), so the non vanishing elements of the perturbation matrix are | |||
:<math> | |||
V_{01}=V_{10}= - \frac{p}{\sqrt{\pi} \tau} \sqrt{\frac{\hbar}{2mw}} exp[-(\frac{t}{\tau})^2] | |||
</math> | |||
In the first approximation a uniform field can produce a transition of the oscillator only to the first excited state, then: | |||
:<math> | |||
P_{01} = \frac{p^2}{2 \pi \tau^2 m t w} |\int^\infty_{-\infty} exp[i w t - (\frac{t}{\tau})^2] dt| | |||
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Using the identity |
Revision as of 00:35, 22 April 2010
(Submitted by Team 1)
This example was taken from "Theory and Problems of Quantum Physics", SCHAUM'S OUTLINE SERIES, p. 190-192.
Problem: Consider a one dimensional harmonic oscillator embedded in a uniform electric field. The field can be considered as a small perturbation and depends on time according to
where A is constant. If the oscillator was in ground state until the field was turned on at t=0, compute in the first approximation, the probability of its excitation as a result of the action of the perturbation.
Solution:
The probability of a transition from the state n to the state k is given by
Let e, m and w denote the charge, mass and natural frequency of the oscillator, respectively, where x denotes its deviation from its equilibrium position. In the case of an uniform field, the perturbation is given by
The oscillator is in the ground state (n=0), so the non vanishing elements of the perturbation matrix are
In the first approximation a uniform field can produce a transition of the oscillator only to the first excited state, then:
Using the identity