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  Difference between proliferation and condensation of bosons

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It is usually seen in literature the terminology that certain bosons proliferate or condense. What is the difference between them? Is it that proliferation just means a large number of bosons occupy certain state, and condensation further means there is also off-diagonal long range order besides that?

For example, in three dimensions free bosons can condense at low temperatures, signaled by a macroscopic number of bosons in the single particle ground state. There is a phase transition going from the high temperature uncondensed phase to the low temperature condensed phase, and this can be seen in singularities in various thermodynamic quantities.

But in two dimensions, as long as the temperature is low enough, there can also be a macroscopic number of bosons in the single particle ground state. However, we do not say this is a Bose-Einstein condensate, because there is no phase transition. Then in this case should we say bosons proliferate but not condense?

asked Apr 11, 2017 in Theoretical Physics by Mr. Gentleman (255 points) [ no revision ]
recategorized Apr 11, 2017 by Dilaton

1 Answer

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Bose-Einstein condensation is a different phase of matter, meaning that you need to go through a phase transition to reach it. Hence, some thermodynamics quantities (or their derivatives) will exhibit a discontinuity  the usual one for the BEC is the heat capacity.

Since it's a phase transition, you want to find the critical value of the parameter which is driving it. <br> This is the temperature, and it is $T_c$.

The physical meaning of the BEC is a saturation effect. <br> For non-interacting bosons, the mean occupancy state $j$ is given by:

$$ f(E_j) = \frac{1}{e^{(E_j - \mu)/kT}-1} .$$

Now you see that the ground state $E_0$, the occupancy is infinity. This is because for the $E_0$ state the chemical potential $\mu$ also needs to be zero, in order to guarantee $f$ to still be positive. Physically, the chemical potential is defined as $\partial U/\partial N$, i.e. the energy added when you add one particle to the system. But if you add it to the $E=0$ state, then the extra energy is 0...

Bose-Einstein condensation begins when you saturate the excited states and start macroscopically occupying the ground state, which has infinite occupancy.

Below $T_c$, $f(E_0)$ starts blowing up so it does not make sense using the above distribution anymore, since the atoms start amassing into the ground state. <br>
So $T_c$ is extracted from when your total $N$ is equal to the number of atoms in the excited states, $N_{ex} = \int_0 ^{\infty}dE \, g(E) \, f(E)$ where $g(E)$ is the *density of states*, i.e. the number of states in a given interval $[E, E+dE]$. The sum should have been over the *states*, but I changed it to the *energy* $E$ just by introducing this density of states term.  

The density of states $g(E)$ scales with the number of dimensions $d$. For a free $d$ dimensional system it goes as $g(E) \propto E^{d/2  -1}$, while for $d$ dimensional harmonic potential it scales as $g(E) \propto E^{d-1}$.

In general you can write:

$$ g(E) \propto E^{\alpha  -1},$$

with $\alpha$ being the number of degrees of freedom in the system divided by 2. For free particles in $d$ dimensions, $\alpha = d/2$, and for a $d$ dimensional harmonic potential, the degrees of freedom are $2d$ ($d$ translations and $d$ oscillations) so $\alpha = d$. All agree with the above.

The integral above can be rewritten as:

$$N_{ex} = \int_0 ^{\infty}dE \, g(E) \, f(E) \propto  (k T_c)^{\alpha} \int_0 ^{\infty} dx\frac{x^{\alpha-1}}{e^x - 1} $$

where I defined $x$ as $E/k T_c$.

The intregral

$$ \int_0 ^{\infty} dx \frac{x^{\alpha-1}}{e^x - 1} = \Gamma(\alpha) \zeta(\alpha),  \qquad \alpha > 1$$

with $\Gamma$ being the gamma function, $\zeta$ being the Riemann zeta function.

Which gives you:

$$ k T_c \propto \frac{1}{[\Gamma(\alpha) \zeta(\alpha)]^{1/\alpha}}. $$

To have a BEC transiton, you want $T_c \neq 0$, i.e. a non-trivial solution.

In free space, $d=2,3,4$ have $\alpha = 1, 3/2, 2$:

$$
\begin{array}{ccc}
\alpha & \Gamma(\alpha) & \zeta(\alpha) \\
\hline
1/2 & \text{integral does not converge}      \\
1 & 1    & \infty      \\
3/2 & \sqrt{\pi}/2  & 2.612      \\
2 & 1    & \pi^2/6      \\
\dots & \dots & \dots
\end{array} $$

So in a free system with $d = 1,2$ the only solution is $T_c = 0$, but for $d>2$, $T_c$ is finite.
 
Hence, in 2D you may well have a lot of bosons in the ground state. But they are not forced to be there because of saturation, therefore it is not a distinct phase of matter. And the heat capacity (or whatever) will not have discontinuities.

answered Dec 1 by Matteo S [ no revision ]

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