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I have got so far the 2D density of states as $g(\epsilon)=\frac{Am}{\pi\hbar^2}$ where $A$ is the area of the "square" and $m$ is the the electron mass. Then I have found an expression for the the chemical potential for the gas using: $N={\int_0^\infty}g(\epsilon)n_Fd\epsilon$, where $n_F$ is the fermi distribution function. This spits out $N=\frac{Am}{\pi\hbar^2}k_BT\ln(1+e^{\frac{\mu}{k_BT}})$, and taking the limit of $T$ goes to zero we get $\epsilon_F=\frac{N\pi\hbar^2}{Am}$, so $\mu=k_BT\ln(e^{\frac{\epsilon_F}{k_BT}}-1)$. This is where I start to get a little stuck. I am thinking I should next work out $<E>=\int_0^\infty{n_F}{\epsilon}g(\epsilon){d}\epsilon$ However I think this is going down the wrong route (or I can't get the algebra right) as I have a hint to use the result $\int_{-\infty}^{\infty}dx\frac{x^2e^x}{(e^x+1)^2}=\frac{\pi^2}{3}$. If anyone has a hint on where I should be going then it'd be much appreciated!

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Do you want this "at" T = 0, or at T small, but nonzero? If it's the latter, you want to look up "Sommerfeld expansion". –  Lagerbaer Apr 9 '13 at 17:30
    
Low T limit, couple of things confusing me if you could take a quick look below –  Dmist Apr 10 '13 at 13:29
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The procedure to obtain the specific heat of a free electron gas in arbitrary dimension is related to the so-called Sommerfeld expansion that is applied to integrals of the form

$$I(\mu,T)=\int_{-\infty}^{+\infty} d\epsilon \, f(\epsilon) n_F(\epsilon) $$

where $n_F(\epsilon)$ is the Fermi-Dirac distribution and $f(\epsilon)$ is an arbitrary function which vanishes at $\epsilon \rightarrow -\infty$ and diverges as a power law when $\epsilon \rightarrow +\infty$.

This technique uses the fact that the derivative of the Fermi function is peaked around the chemical potential $\mu$ and broadened by $k_BT$ [and indeed collapses to $-\delta(\epsilon-\mu)$ at $T=0$]. Therefore, if the function behaves as above, which is true for $f(\epsilon)=\epsilon g(\epsilon$), we can integrate by parts to get

$$\int_{-\infty}^{+\infty} d\epsilon \, f(\epsilon) n_F(\epsilon)=-\int_{-\infty}^{+\infty} d\epsilon \, F(\epsilon) \dfrac{\partial n_F}{\partial \epsilon}$$ where $$ F(\epsilon)=\int_{-\infty}^{\epsilon}d\epsilon' f(\epsilon')$$ and expand $F(\epsilon)$ around $\epsilon=\mu$ in Taylor series $$F(\epsilon)=F(\mu)+\sum_{k=1}^{+\infty} \dfrac{F^{(k)}(\mu)}{k!}(\epsilon-\mu)^k$$ with $F^{(k)}(\mu)$ being the $k$-th derivarive of the function $F$. Observing that $(-\partial n_F/\partial \epsilon)$ is normalized to unity and that is an even function (so that only terms even in $k$ contribute) we can go back to the original function $f(\epsilon)$ and using $x=\beta(\epsilon-\mu)$ $$I(\mu,T)=\int_{-\infty}^{\mu}d\epsilon f(\epsilon)+\sum_{k=1}^{+\infty} \dfrac{f_k}{\beta^{2k}}\dfrac{d^{2k-1}}{d\epsilon^{2k-1}}f(\epsilon)\Bigg|_{\epsilon=\mu}$$ with coefficients $$f_k=\int_{-\infty}^{+\infty} dx \dfrac{x^{2k}}{(2k)!}\left(-\dfrac{d}{dx} \dfrac{1}{e^x+1}\right)dx$$ Using this you should not have any problem to find the mean kinetic energy for the free 2DEG and from there the specific heat.

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Okay, after looking at it for quite a while I'm pretty happy bar a couple of things, firstly the function $f(\epsilon)$ should vanish for epsilon goes to minus infinity, however in this case $f(\epsilon)=g(\epsilon)\epsilon$ and the density of sates is just a constant, so f diverges at minus infinity? Also, I'm confused why the integral's upper limit can be changed to $\mu$ (or even to $\epsilon_F$ as I've seen in some places). Is it that the fermi function goes quickly to zero within a couple of $k_BT$ of the Fermi energy (or chemical potential) and T is small so limit is pretty much $\mu$ –  Dmist Apr 10 '13 at 12:23
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The DOS is not constant everywhere. It's zero below the single particle ground state energy, which we set to 0. So $f(\epsilon) = 0$ for $\epsilon <0 $. It's only constant above that. Next, in the limit of very low $T$, if you do the expansion, you'll find that $\mu = \epsilon_F$. I think there's a typo in the post in the definition of $F(\epsilon)$. I think it should read $F(\epsilon) := \int_{-\infty}^\epsilon d\epsilon' f(\epsilon')$. –  Lagerbaer Apr 10 '13 at 15:47
    
@Lagerbaer: Yep, there is a typo. I fixed it. Thanks! –  DaniH Apr 10 '13 at 17:07
    
@Dmist: I forgot to mention that the integral you give as hint appears at first order in the Sommerfeld expansion. –  DaniH Apr 10 '13 at 21:32
    
Ahh okay, so we set it to zero for $\epsilon$ below zero but that doesn't matter anyway as we can neglect any terms which don't depend on T as that's all we're interested in for the heat capacity. I get out $C/N=\frac{(\pi{k_B})^2N}{6\epsilon_F}T$ which gives the total heat capacity of a metal the required form of $AT^3+BT$ (I am kind of assuming this dependence on T is general for d dimensions) –  Dmist Apr 11 '13 at 10:48
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