Quantum Harmonic Oscillator - What is the Temperature of the system?

1. Dec 2, 2009

obi-wan

1. The problem statement, all variables and given/known data

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. Dec 2, 2009

MaxL

β = 1/(Kb*T)

where Kb is bolzmann's constant.

3. Dec 2, 2009

obi-wan

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

thanks

4. Dec 2, 2009

jdwood983

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 $E=\frac{1}{2}\hbar\omega+0.3\hbar\omega$, you can stick that into the equation and solve for $\beta$:

$$\frac{1}{2}\hbar\omega+0.3\hbar\omega=\frac{1}{2}\hbar\omega\coth[\beta\hbar\omega]$$

5. Dec 2, 2009

MaxL

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.

6. Dec 2, 2009

jdwood983

Hmm...I'll double check my work, but I'm pretty sure that when you take the derivative with respect to $\beta$, you get an exponential in the numerator and a $1-exp[\beta\hbar\omega]$ which can be reduced to the hyperbolic cotangent.

7. Dec 2, 2009

jdwood983

I'm pretty sure it is the same thing,

$$\coth[x]=\frac{\exp[x]+\exp[-x]}{\exp[x]-\exp[-x]}$$

and when you use the proper energy (the OP forgot a factor $\hbar\omega$ in his equation)

$$E=\frac{1}{2}\hbar\omega+\frac{\hbar\omega\exp[-\hbar\omega\beta]}{1-\exp[-\hbar\omega\beta]}=\frac{1}{2}\hbar\omega\left(\frac{1+\exp[-\hbar\omega\beta]}{1-\exp[-\hbar\omega\beta]}\right)$$

which is the same thing as the hyperbolic cotangent.

8. Dec 2, 2009

MaxL

Okay, I double checked my work. My mistake!

Obi-wan, I hope that helps!

9. Dec 3, 2009

obi-wan

thank you for all your help. I've worked out the temperature to be ~ 0.6208ΘE which sounds reasonable