In the ideal gas model of statistical mechanics one very often assumes the gas is dilute so the average separation between particles is large and hence their mutual interaction is correspondingly small.

In addition, if the gas is sufficiently dilute, the average separation between its particles is much larger than de Broglie wavelength of a particle. In this case, quantum-mechanical effects are of negligible importance and it is permissible to treat the molecules as distinguishable particles moving along classical trajectories (classical approximation).

I am confused about the shaded statement, why we judge the quantum-mechanical effect in terms of the wavelength? I mean if the observable processes are of no disturbance then the particle's dynamics is governed by classical mechanics, so here is there anything I am missing?


1 Answer 1


There are two quantum regimes that one might want to check:

  • The wave-like behaviour of particles which renders the concept of trajectory and definite positions somewhat fuzzy, leading to interferences which in turn lead to discretised values of the momenta. This particular regime can be checked by ensuring that $\Delta x \Delta p \gg \hbar$ thus avoiding what is sometimes called the saturation of Heisenberg inequality. If $L$ is the size of the box then for an ideal gas particle $\Delta x = L$. Furthermore, assuming classical fluctuations (i.e. equipartition) we have that $\Delta p \simeq \sqrt{m k_B T}$ this leads then to the condition that $L \sqrt{m k_B T} \gg \hbar$ which is equivalent to $\Lambda \ll L$ where $\Lambda$ is the thermal de Broglie wavelength $\Lambda = h/\sqrt{2\pi m k_B T}$.

  • The fact that indistinguishable particles may want to obey quantum statistics depending on whether they are likely to be in the same state or not. To determine this regime, you can imagine that you have a system of N particles in an ideal gas. The semi-classical partition function for a single particle is $q = L^3 / \Lambda^3$. Keep in mind that $q$ gives roughly an estimate of the number of states accessible to a particle. Now, if you were to insert an additional particle in the system, you want to know how likely it is that it has the same state as one of the particles already in the gas. The probability for it being in the same state as one of the N others is roughly $N/q$. If you ask this probability to be much smaller than one you get that $N/q \ll 1$ which is equivalent to $\Lambda \ll (V/N)^{1/3}$ where the right hand side is the typical distance between any two particles in the gas.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.