Phy5646/Another example: Difference between revisions

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P_{01} = \frac{p^2}{2 m \hbar w} exp[-\frac{1}{2} (w t)^2]
P_{01} = \frac{p^2}{2 m \hbar w} exp[-\frac{1}{2} (w t)^2]
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Note that for <math>\tau>> \frac{1}{w}</math> the probability for the excitation is extremely small. This is the case of a so-called adiabatic perturbation. On the contrary, for a rapid perturbation <math>\tau<< \frac{1}{w}</math> the probability of excitation is constant.

Latest revision as of 12:57, 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

we have:


Note that for the probability for the excitation is extremely small. This is the case of a so-called adiabatic perturbation. On the contrary, for a rapid perturbation the probability of excitation is constant.