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A harmonic oscillator problem

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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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Answers and Replies

  • #2
fzero
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You have a simple mistake, the coefficients are

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

where [tex]\psi_{n}^{\omega'}[/tex] are the complete set of states for the harmonic oscillator at the new frequency. However to answer the two questions stated in the problem, you only need [tex]c_0[/tex].
 
  • #3
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You have a simple mistake, the coefficients are

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

where [tex]\psi_{n}^{\omega'}[/tex] are the complete set of states for the harmonic oscillator at the new frequency. However to answer the two questions stated in the problem, you only need [tex]c_0[/tex].
ahh you're right. So if I'm interpreting this correctly according to [tex]E_{n}^{\omega}= \left(n+\frac{1}{2}\right)\hbar\omega[/tex], [tex]\hbar\omega/2[/tex] will not be found. thanks!
 
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  • #4
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Could someone give a more conceptual feel for this problem? What happens when the spring constant changes and the angular frequency thus doubles? Does this mean the complete n-states now change? What happens with the energy? Is the energy of the ground state the same before and after frequency change?
 
  • #5
fzero
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Could someone give a more conceptual feel for this problem? What happens when the spring constant changes and the angular frequency thus doubles? Does this mean the complete n-states now change? What happens with the energy? Is the energy of the ground state the same before and after frequency change?
When we suddenly change the potential, we must write the initial state as a linear combination of the new basis states. The ground state energy of the new potential is different (double the old if we double the frequency). The initial state will have a different energy because work was done on the system.
 

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