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The thermal de Broglie wavelenght is often defined by the formula

$$\lambda=\frac{h}{\sqrt{2\pi mkT}}$$

but equally frequently is it defined as de Broglie wavelength for a free ideal gas of massive particles in equilibrium, but in this case we obtain

$$\lambda=\frac{h}{\sqrt{3 mkT}}$$

Even though the second version has some theoretical predictions, in the literature the first one is used.

I also find papers where authors used approximation $\pi\approx 3$, and claim that $E_K=3kT$ instead of $\frac{3}{2}kT$.

My question is : What is the reason ? Why we used $2\pi$ instead of $3$ ? Is it motivated historically or we can define this value within some other theory using its formalism ?

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  • $\begingroup$ a well known statistical integral is $$ \int e^{x^{2}} = \sqrt{\frac{\pi}{2}} $$ $\endgroup$ – user97261 Apr 28 '16 at 20:59
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I see two ways in which the thermal de Broglie wavelength is defined. In both cases we can get it from the probability distribution and partition function of an ideal gas. We will consider a 3D ideal gas with non-relativistic dispersion.

First Way

Consider the partion function of an ideal gas:

\begin{align} Z &= \int_{p_x=-\infty}^{+\infty}\int_{p_y=-\infty}^{+\infty}\int_{p_z=-\infty}^{+\infty} e^{-\frac{1}{2mkT}(p_x^2+p_y^2+p_z^2)}dp_xdp_ydp_z\\ &= \left(2\pi mkT \right)^{\frac{3}{2}} \end{align}

Note that this has dimensions of momentum cubed. Noticing this, we can define the characteristic thermal momentum

\begin{align} p_T = \sqrt{2\pi mkT} \end{align}

We can consider the de Broglie wavelength of a particle with this characteristic momentum to get the first definition of the thermal de Broglie wavelength:

$$\lambda_T = \frac{h}{p_T} = \frac{h}{\sqrt{2\pi mkT}}$$

Second Way

Consider the average energy of an ideal 3D gas. This can be found from the equipartion theorem to be

\begin{align} \langle E \rangle = \frac{3}{2}kT \end{align}

A relation for the de Broglie wavelength is

$$ \lambda = \frac{h}{\sqrt{2mE}} $$

Considering the de Broglie wavelength of a particle with energy equal to the average thermal energy of a 3D ideal gas we get the second definition of the thermal de Broglie wavelength:

$$ \lambda_T = \frac{h}{\sqrt{3mkT}} $$

A Note on a Possible Third Way

A third way which would make sense would be to calculate the average de Broglie wavelength of all of the particles in an ideal gas:

$$ \langle \lambda \rangle = \iiint \frac{h}{p} e^{-p^2/2mkT} dp^3 = \frac{2h}{\sqrt{2\pi m kT}} $$

We see that this is within a factor of 2 of the first definition.

Summary

The three approaches differ by factors of order unity so they all refer to similar length scales. In the end the thermal de Broglie wavelength is largely a notational convenience so we don't need to carry around factors of $\frac{h^2}{mkT}$ all over the place so we shouldn't worry too much about the pre-factor. But it is nice to know where the different conventions come from. Though it is largely a notational convenience it does clearly have physical significance since it is related to $\langle \lambda \rangle$.

I have never really seen the third way presented. I have seen the first way presented by far the most often. I think this is because the partition function appears all over the place so very commonly the specific factor $2\pi mkT$ arises so it is nice to give this quantity a name. The second approach may be presented more often in introductory approaches to statistical mechanics. This convention is most problematic because it very clearly relies on the specific problem of a 3 dimensional ideal gas.

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