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arenaninja

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## Homework Statement

A particle in the ground state of the harmonic oscillator with classical frequency [itex]\omega[/itex], when the spring const quadruples (so [itex]\omega^{'}=2\omega[/itex]) without initially changing the wave function. What is the probability that a measurement of the energy would still return the value [itex]\frac{\hbar\omega}{2}[/itex]? And the probability for [itex]\hbar\omega[/itex]?

## Homework Equations

The initial ground state of a harmonic oscillator is [tex]E_{0}^{\omega}=\frac{\hbar\omega}{2}[/tex]

The time-independent part of the wavefunction: [tex]\psi_{0}^{\omega} = \left(\frac{m\omega}{\hbar\pi}\right)^{\frac{1}{4}} e^{\frac{-m \omega x^{2}}{2\hbar}}[/tex]

## The Attempt at a Solution

After the spring const change, [tex]E_{0}^{\omega'}=\hbar\omega[/tex]

And [tex]\psi_{0}^{\omega'}=\left(\frac{2m\omega}{\hbar\pi}\right)^{\frac{1}{4}}e^{\frac{-m \omega x^{2}}{\hbar}}[/tex]

To build our wavefunctions, we will need our coefficients, which can be done via

[tex]c_{n}=<\psi_{0}^{\omega'}|\psi_{0}^{\omega}>[/tex]

When I evaluated the integral, I got the answer [itex]c_{n}=\frac{2^{1/4}}{\sqrt3}[/itex]

But now I'm stuck, because there must be a mistake somewhere. You see, the problem states that the probability for measuring the energy and getting [itex]\hbar\omega[/itex] should be 0.943. The probability of the nth energy is easily given by [itex]|c_{n}|^2[/itex], but my coefficient gave me only a discrete value, which I plugged in to wolframalpha to square it and it does seem to be 0.943 (it's ~0.9428). My issue is that I didn't get any [itex]c_{n}[/itex] for any other nth energy, but I know the sum of the square of all these coefficients should equal one. By my crude instincts (this stuff is terribly hard), my result says that the probability for getting the ground state measurement should be zero, but this is obviously wrong since the square of my coefficients do not add up to one.

Any ideas and/or suggestions would be welcome.

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