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Average Potential Energy/ Oscillator

  1. Apr 3, 2013 #1
    i regard a single harmonic oszillator: $$H_{1}=\frac{p^{2}}{2m} + \frac{m \omega^{2}}{2} x^{2}$$
    I know the partition function of the oszillator is: $$Z=\frac{kT}{\hbar \omega}$$
    so the probability function is: $$F_{1}(x,p)=\frac{1}{Z}\exp{\frac{-H_{1}(x,p)}{kT}}$$

    Now I want to callculate the average kinetic energy. So, can i do this? :$$
    <\frac{p^2}{2m}>=\int\limits_{-\infty}^\infty dx \int \limits_{-\infty}^\infty dp~~~ F_{1}(x,p) \frac{p^2}{2m}
  2. jcsd
  3. Apr 3, 2013 #2
  4. Apr 3, 2013 #3

    I am not sure if I took the right partition function?
    Is [itex]Z_ {Classical}[/itex] or [itex]Z[/itex] the right function for callculating the average value?

    <\frac{p^2}{2m}> &=\int dp \int dx ~~~\frac{1}{Z} ~~e^{-\beta~\frac{p^2}{2m}}~\cdot ~
    e^{-\beta \frac{m\omega^{2} x^{2}}{2}} ~~\frac{p^2}{2m}\\
    &= \frac{\beta\hbar \omega }{2m} \int dx
    e^{-\beta \frac{m\omega^{2} x^{2}}{2}}
    \int dp~~ p^{2} e^{-\beta~\frac{p^2}{2m}}\\
    &=\frac{\beta\hbar \omega }{2m} ~~\sqrt{\frac{2\pi}{\beta m \omega^{2}}}~~\cdot ~~\sqrt{\frac{2\pi m}{\beta} }\frac{m}{\beta}\\
    &=\frac{\hbar \pi}{\beta}\\
    For the patition function I have seen in a book:
    Z=\frac{Z_{Classical}}{h} =\frac{2 \pi kT}{h \omega} =\frac{kT}{\hbar \omega}
    ~~~~~~~~~~~~~~~~~~~~~~~~~~;Z_{Classical}= \int \int dx dp ~e^{-\frac{H_{1}(x,p)}{kT}}
    Last edited: Apr 3, 2013
  5. Apr 4, 2013 #4
    Well, obviously you don't want that factor of h in the average k.e.
    The resolution of this is that you have to use a consistent integration measure and partition function. If you use what you've called Z as the partition function, you need to integrate over x and p/h rather than x and p. To see why this is true you have to calculate the partition function using that particular convention, i.e. the partition function itself is an integral over x and p/h rather than x and p, as per the fourth equation in the section http://en.wikipedia.org/wiki/Partition_function_(statistical_mechanics)#Definition .
    The reason i say convention is because this factor due to changing the integration measure has no observable consequences, because the probability of finding the particle in a particular infinitesimal volume of phase space is unchanged.
    Similarly, if you were to work out the quantum HO partition function, you would find that redefining the zero of energy is equivalent to multiplying the partition function by a constant - which is consistent with the idea that the choice of zero of energy is arbitrary.
  6. Apr 4, 2013 #5
    Of course, the energy shift thing isn't particular to the SHO - you can see it straight from the definition of Z for any system with discrete energies.
  7. Apr 4, 2013 #6
    Go on

    I have seen that the probability density function [itex]F_{1}(x,p)[/itex] I used, is not 1.

    \int \limits_{-\infty}^{\infty} dx~dp~~F_{1}(x,p)= h

    So is that an error?
    Do I have to divide my result with h?
  8. Apr 4, 2013 #7
    Yes, that's equivalent to using the correct Z.
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