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In my statistical mechanics intro class, we are taught that at sufficiently high temperatures, the quantum treatment of things becomes unnecessary. Why is this? Can this be shown using certain equations?

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    $\begingroup$ Too heuristic to be a full answer: in QM statistical mechanics you have two types of uncertainty, statistical uncertainty quantified by temperature $T$ and quantum uncertainty. If you make statistical uncertainty large then the quantum uncertainty is not the dominant force so can't matter so may as well be zero (ie classical). $\endgroup$ – jacob1729 Apr 13 at 1:43
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    $\begingroup$ By 'the quantum treatment of things breaks down' do you mean 'the quantum treatment is arbitrarily close to the classical treatment so is unnecessary'. Quantum mechanics always gives the right answers, just sometimes the corrections are small. $\endgroup$ – jacob1729 Apr 13 at 1:44
  • $\begingroup$ @jacob1729 yes this is what I mean $\endgroup$ – NoThangButtaChknWang Apr 13 at 1:49
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Qualitatively, quantum effects can be ignored if the interchange properties of boson or fermions can be ignored, which is to say if the system is dilute. When will a system be dilute? A single particle occupies a volume of its thermal de Broglie wavelength cubed (in three dimensions). For massless particles, the thermal de Broglie wavelength must go as $\lambda_{th}\sim1/T$ by dimensional analysis. For massive particles, the thermal de Broglie wavelength goes as $\lambda_{th}\sim1/\sqrt{mT}.$ Thus for both massless and massive particles, as the temperature increases, the volume occupied by any one particle decreases and quantum effects become less and less important.

To flesh this out further, I'll essentially follow 8.4 and 8.5 of Baierlain's Thermal Physics book.

The number distribution of particles is either Fermi or Bose, $$n(\varepsilon)\propto \frac{1}{e^{(\varepsilon-\mu)/T}\pm1},$$ where, for simplicity, I'm using units in which Boltzmann's constant $k_B=1$.

The "quantumness" of the distributions is in the $\pm1$, which is to say that the quantumness can be ignored when $e^{(\varepsilon-\mu)/T}\gg1$. Then these two distributions can be well approximated by the classical number distribution of particles, the Maxwell distribution, $$n(\varepsilon)\propto e^{-(\varepsilon-\mu)/T}.$$

In the case that $e^{(\varepsilon-\mu)/T}\gg1$ we see that $n(\varepsilon)\ll1$, which is to say that the system has low occupation number.

When will a system have low occupation number? Qualitatively, when the system is dilute. Baierlein derives the following expression for the average energy of a semi-classical system, i.e. one in which quantum effects are only a small correction. In three dimensions, $$\langle E \rangle = \frac{3}{2}NT\Big(1\pm\frac{1}{2^{5/2}}\frac{N \lambda_{th}^3}{(2s+1)V}\Big),$$ where $N$ is the number of particles, $s$ is their spin, $V$ is the volume of the system, and $\lambda_{th} = 1/\sqrt{2\pi mT}$.

One can see, then, that quantum effects can be ignored when $\frac{1}{2^{5/2}}\frac{N \lambda_{th}^3}{(2s+1)V}\ll1$, i.e. when $$T\gg\frac{1}{2\pi m}\Big( \frac{N}{2^{5/2}(2s+1)V} \Big)^{2/3}.$$ The denser the system---the large is $N/V$---the higher the temperature must be for one to ignore quantum effects.

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