Phy5646/Another example: Difference between revisions

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:<math>
:<math>
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|
</math>
</math>


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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|
</math>
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