# Confusing about Maxwell-Boltzmann distribution in non-classical region

1. Jan 26, 2012

### KFC

Hi there,
I am reading some materials on statistical mechanics and for the section on momentum distribution, there mention for some non-interacting system, the momenta distribution are satisfying the Maxwell-Boltzmann distribution, which is summarized in http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

If we consider the problem as 1 dimensional, the momentum distribution becomes

$$f(p) = [m/(2\pi k_B T)]^{-1/2}\exp[-p^2/(2mk_BT)]$$

This look good to me. But in some books or materials, for getting into quantum region, the author simply considers the non-interacting system by introducing the relation with planck const. as

$$p = \hbar k$$

such that

$$f(k) = \hbar[m/(2\pi k_B T)]^{-1/2}\exp[-(\hbar k)^2/(2mk_BT)]$$

My question is, is that really the correct way to go from classical into quantum mechanics by just introducing the $$\hbar$$ in the momentum relation to the wavenumber for the non-interacting system? As I have noticed in other book for statistical mechanics, the Maxwell distribution can only be applied for the system with the hypothesis that the particles are non-interacting, identical but distinguishable. But for quantum case, I don't know how this condition is hold. Since the author trying to apply the above distribution to a bosonic gas at really low temperature (call BEC?), should this condition hold for this case? But even that's true, why don't we use the boson statistics instead?

For my second question, if f(k) is the correct distribution, how come this quantity $$\hbar[m/(2\pi k_B T)]^{-1/2}$$ has the unit of wave number? I try to do the math but I get meter square instead.

2. Jan 27, 2012

### Rap

Something is wrong with the first equation. Just doing dimensional analysis, we have to have f(p)dp be dimensionless so f(p) has to have units of 1/(mv). It has units of the square root of kT/m. kT is an energy so the units are v^2, not 1/(mv). The correct statement is the square root of $1/(2\pi m k T)$

3. Jan 28, 2012

### KFC

Ok. Thanks. So what's the right expression for the quantum case when the wavenumber distribution being consindered? I try

$$f(k) = \hbar\sqrt{\frac{1}{2\pi m k_B T}} \exp[-(\hbar k)^2/(2mk_B T)]$$

but the dimension is not correct.

4. Jan 28, 2012

### Rap

For a distribution of any variable x, you have to have f(x)dx be dimensionless. So you have to use f(p)dp=f(k)dk, which means f(k)=f(p) dp/dk, that should give you the right answer.

5. Jan 28, 2012

### KFC

Thanks. But that's what I did, not working. since $$p=\hbar k$$ and $$dpdk = \hbar$$, plug that back to the f(p) with p written in k, I got the one I wrote before but the dimension is not correct. The right dimension should be in $$meter^{-1}$$

6. Jan 28, 2012

### Rap

Since k has dimensions 1/meter, and f(k)dk is dimensionless, f(k) must have dimensions of meters.