1
$\begingroup$

For a ideal Bose gas in harmonic trap, the total particle number can be written as,

N harmonic, riemanand z is fugacity.

Now I want to find the expression for particle density for excited states. In case of Bose gas in box potential, the expression for particle number was

N box

and the excited state particle density can be calculated moving volume term on left hand side and writing remaining term as thermal de Broglie wavelength,

d box

However, in the case of harmonic trap there seems no term that relates to volume. So my first question is can we relate the volume of harmonic oscillator to any way with trap frequency ?

Secondly, if change my particle number equation to same form as box potential and get a thermal de Broglie cube, then the remaining terms above should give volume (thought),

d harmonic

But now the volume seems to depend on the temperature. How can this be expained ? My initial thought was that with decreasing temperature the non condensate fraction decreases but if we compare to the box potential, it is total volume of system. Any suggestions are appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

The harmonic trap gives an energy dependent effective volume. Think in classical terms: at energy $E$, the accessible region for the particle is a ball of radius $r$ and volume $V$ with: \begin{align} r &= \sqrt{\frac{2E}{m\omega^2}} \\ V &= \frac{4\pi}{3}\left(\frac{2E}{m\omega^2}\right)^{3/2} \end{align} Adding thermal fluctuations, in the canonical ensemble, the spacial distribution is Gaussian with root mean square distance $r_{rms}$ (given by the equipartition theorem): $$ r_{rms} = \sqrt{\frac{3k_BT}{m\omega^2}} $$ which gives an effective accessible ball of volume: $$ V = \frac{4\pi}{3}\left(\frac{3k_BT}{m\omega^2}\right)^{3/2} $$ Up to a numerical factor, this is the effective volume you find in the numerator of your expression. This is because $r_{rms}$ is the only length scale of your problem, so by dimensional analysis a volume will necessarily proportional to it's cube.

In short, you should think of $1/\omega$ as the effective length of your box. In particular, just as for the gas in a box, you have the same extensive scaling. Since $V\sim \omega^{-D}\sim N$, you need $\omega\sim N^{-1/D}$. The temperature dependence is there in some sense to match the dimensions.

For your second question, the temperature dependence is purely classical. As temperature increase, thermal fluctuations increase, the "available" energy increases which increases the effective volume by the first equations.

Note that the purely classical interpretation works above the critical temperature. Below the critical temperature, you'll have a macroscopic number of bosons in the ground state. Their effective occupied volume will be given by the ground state of the harmonic oscillator. It is therefore temperature independent: $$ r_{rms} = \sqrt{\frac{\hbar}{2m\omega}} \\ V = \frac{4\pi}{3}\left(\frac{\hbar}{2m\omega}\right)^{3/2} $$ Thus, in the BEC phase, the classical fraction will still have a temperature decreasing effective volume, but the ground state fraction will have a fixed volume just like for a gas in a box.

Btw, if you wanted a full quantum treatment, this would modify the total particle number to: $$ N = \sum_{n\in\mathbb N^3} \frac{ze^{\beta \hbar \omega(n_x+n_y+n_z)}}{1-ze^{\beta \hbar \omega(n_x+n_y+n_z)}} $$ It is consistent with the classical treatment for all temperature because in the thermodynamic limit, if the temperature is intensive, $\beta\omega\hbar \gg 1$. You just need to treat apart from the ground state. The classical theory applies in particular the equipartition theorem as well.

Hope this helps.

$\endgroup$
6
  • $\begingroup$ Thank you.. That was very helpful. $\endgroup$
    – AdShil00
    Dec 8, 2022 at 7:17
  • $\begingroup$ @Ipz so Does that mean for BEC in harmonic trap, the confinement volume can be given by the thermal energy but If i try to derive the same expression for total particle number N, using the discrete sum of occupancy of each state, i could not derive it. Since we have only have energy of each state h*w but no way to relate to KT. What do you think about it ? $\endgroup$
    – AdShil00
    Dec 13, 2022 at 9:44
  • 1
    $\begingroup$ If I understand correctly, you tried a full quantum treatment using my last formula. The previous effective volumes estimations still work in the quantum case. The idea is to find the relevant length scale giving the radius of the effective sphere, typically by calculating the $r_{rms}$. $\endgroup$
    – LPZ
    Dec 13, 2022 at 17:41
  • $\begingroup$ Ok. So if we use classical treatment, then from Equipartition theorem we can relate total average energy to thermal energy and then relate effective radius to r_rms. Then, for quantum treatment, finding the average energy of oscillator from canonical ensemble for high temperature limit(KT>hw) also seems to give us the same result. Is this a valid way to calculate the effective radius, or do you think there are better ways ? $\endgroup$
    – AdShil00
    Dec 14, 2022 at 8:32
  • $\begingroup$ For large temperature indeed the classical and quantum results agree and equipartition is applicable. However, at low temperature (which interests you since you want to know about the BEC), the results differ and you should rather use the fully quantum version. Intuitively, you have an extensive number of particles in the ground state. The effective volume will be given by the $r_{rms}=\sqrt{\frac{\hbar}{m\omega}}$ of the ground state. $\endgroup$
    – LPZ
    Dec 14, 2022 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.