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Homework Help: Quantum Harmonic Oscillator - What is the Temperature of the system?

  1. Dec 2, 2009 #1

    A quantum mechanical harmonic oscillator with resonance frequency ω is placed in an environment at temperature T. Its mean excitation energy (above the ground state energy) is 0.3ħω.
    Determine the temperature of this system in units of its Einstein-temperature ΘE = ħω/kB

    2. Relevant equations

    Partition-Function= E= -δ/δβ ln[Z(β)]

    3. The attempt at a solution

    I have used the Partiton-Function, E= -δ/δβ ln[Z(β)] = -δ/δβ ln[(x^1/2)/1-x]
    .where x= exp^(ħω/2)

    E= -[δ/δβ ln(exp^-ħωβ / 1-exp^-ħωβ] = Eo - exp^-ħωβ / -1 + exp^-ħωβ
    . where Eo= ħω/2

    E= Eo + εħω
    .where ε = 0.3ħω

    I need find β in order to work out the temperatureof the system but stuck at this point. Really appreciate if some one can guide me in the right direction.
  2. jcsd
  3. Dec 2, 2009 #2
    β = 1/(Kb*T)

    where Kb is bolzmann's constant.
  4. Dec 2, 2009 #3
    MaxL yes I know β= 1/kBT and have substituted β intentionally but what im asking is how can i get a numerical value for β to work out the temperature? or do i need to? I'm just confused

  5. Dec 2, 2009 #4
    When I did this, I got

    E=\frac{\partial}{\partial \beta}\ln[z]=\frac{1}{2}\hbar\omega\coth[\beta\hbar\omega]

    which looks pretty similar to what you have (you didn't take your exponentials to the hyperbolic functions).

    Since you know that [itex]E=\frac{1}{2}\hbar\omega+0.3\hbar\omega[/itex], you can stick that into the equation and solve for [itex]\beta[/itex]:

  6. Dec 2, 2009 #5
    Sorry! I didn't mean to be condescending, I just read the question too fast.

    Okay, so what you're ultimately after is an answer that looks like T=Teinstein*(Some number). It's pretty easy to show that Teinstein=ħωβ*T, so what you really need is to numerically determine ħωβ. You could get that using a calculator or mathematica to numerically solve an equation like the one jdwood just dropped.

    Although, jdwood, I did not get that same equation. I think you may have made a chain rule mistake.
  7. Dec 2, 2009 #6
    Hmm...I'll double check my work, but I'm pretty sure that when you take the derivative with respect to [itex]\beta[/itex], you get an exponential in the numerator and a [itex]1-exp[\beta\hbar\omega][/itex] which can be reduced to the hyperbolic cotangent.
  8. Dec 2, 2009 #7
    I'm pretty sure it is the same thing,


    and when you use the proper energy (the OP forgot a factor [itex]\hbar\omega[/itex] in his equation)


    which is the same thing as the hyperbolic cotangent.
  9. Dec 2, 2009 #8
    Okay, I double checked my work. My mistake!

    Obi-wan, I hope that helps!
  10. Dec 3, 2009 #9
    thank you for all your help. I've worked out the temperature to be ~ 0.6208ΘE which sounds reasonable