Let me give a bit of context before asking the actual questions: In the second edition of Condensed Matter Physics, Michael P. Marder derives the specific heat of Fermi liquids in chapter 17.5.4.

He defines $$ \varepsilon_{\vec{k}} = \varepsilon_{\vec{k}}^{\left(0\right)} + \sum_{\vec{k^\prime},\sigma^\prime} u_{\vec{k}\vec{k^\prime}} \delta f_{\vec{k^\prime}} $$ as the energy needed to add a quasi-particle in the state $\vec{k}$ to the considered state characterized by the $\delta f_{\vec{k^\prime}}$ (the spin quantum numbers are suppressed in his notation). $\varepsilon_{\vec{k}}^{\left(0\right)}$ is the energy of said quasi-particle in the non-interacting case, $u_{\vec{k}\vec{k^\prime}}$ describes the interactions, and $\delta f_{\vec{k^\prime}}$ is the difference in occupation number between the considered state and the ground state (so that $\delta f_{\vec{k^\prime}}$ can only be either $-1$, $0$, or $1$.)

The system's total energy is $$ \varepsilon \left[\delta f\right] = \varepsilon_0 + \sum_{\vec{k},\sigma} \varepsilon_{\vec{k}}^{\left(0\right)} \delta f_{\vec{k}} + \frac{1}{2} \sum_{\vec{k},\sigma,\vec{k^\prime},\sigma^\prime} \delta f_{\vec{k}} u_{\vec{k}\vec{k^\prime}}\delta f_{\vec{k^\prime}} $$ with $\varepsilon_0$ being the ground state energy.

The specifc heat is then argued to be (I suppose the auther implicitly assumes that $u_{\vec{k}\vec{k^\prime}} = u_{\vec{k^\prime}\vec{k}}$) $$ C_V = \left.\left(\frac{\partial\varepsilon}{\partial T}\right)\right|_V = \sum_{\vec{k}\sigma}\varepsilon_{\vec{k}} \frac{\partial \delta f_{\vec{k}}}{\partial T} $$ where the notation has been changed so that $\delta f_{\vec{k}}$ now means the expectation value for the occupation number of state $\vec{k},\sigma$. It is shown that $$ \frac{\partial \delta f_{\vec{k}}}{\partial T} = \frac{\exp\left(\beta\left(\varepsilon_{\vec{k}}-\mu\right)\right)} {\left(1+\exp\left(\beta\left(\varepsilon_{\vec{k}}-\mu\right)\right)\right)^2} \left\{ \frac{\varepsilon_{\vec{k}} - \mu}{k_B T^2} - \frac{1}{k_B T} \sum_{\vec{k^\prime},\sigma^\prime} u_{\vec{k}\vec{k^\prime}} \frac{\partial \delta f_{\vec{k}}}{\partial T} + \frac{1}{k_B T} \frac{\partial \mu}{\partial T} \right\} $$ Here, $\mu$ is the chemical potential, and $\beta=\frac{1}{k_B T}$. $k_B$ is Boltzman's constant, $T$ is the temperature.

Up to this point, I think I understood everything. Now, the author claims that at low temperatures, the contribution of the first term in the curly brackets dominates, so the others are neglected. (Is there an easy way to see this is true, and under which conditions other than low $T$ this is true? It is a bit worrisome that the quantity one seeks to calculate, $\frac{\partial \delta f_{\vec{k}}}{\partial T}$, appears on the right hand side under a summation over its index.)

Finally, the author aproximates the summation $\sum_{\vec{k},\sigma}$ by an integration $V\intop\left[\mathrm{d}\vec{k}\right]$ and writes $$ C_V = V\intop\left[\mathrm{d}\vec{k}\right] \frac{1}{k_B T^2} \left(\varepsilon_{\vec{k}}-\mu\right)^2 \frac{\exp\left(\beta\left(\varepsilon_{\vec{k}}-\mu\right)\right)} {\left(1+\exp\left(\beta\left(\varepsilon_{\vec{k}}-\mu\right)\right)\right)^2} $$

This is where I don't understand what hapened. Given the equations above, I would have expected a factor $$ \varepsilon_{\vec{k}}^2-\varepsilon_{\vec{k}}\mu $$ instead of $$ \left(\varepsilon_{\vec{k}}-\mu\right)^2 = \varepsilon_{\vec{k}}^2 - 2 \varepsilon_{\vec{k}}\mu + \mu^2 $$

I would be very happy if someone could explain to me what hapened here, or share their thoughts and educated guesses. Are the expressions somehow equivalent? Are they approximately equivalent, and if yes, under which conditions in the approximation sensible?

The result given in the book, $C_V = V \frac{\pi^2}{3} k_B^2 T D\left(\varepsilon_F\right)$, I have encountered in other sources already, so that sems to be correct.

  • $\begingroup$ In a system of interacting particles I think it is not worrisome that the term appears right hand side in a sum. Taking only the first term is OK if the derivatives of $\delta f$ and $\mu$ stay finite. Then $1/T^2$ always wins over $1/T$. Had no time to look at the main part of your question, yet. $\endgroup$ Feb 12, 2015 at 17:01
  • $\begingroup$ Can you check the integral you show here again, as it only has $1/(k_\mathrm{B}T)$ and not $1/(k_\mathrm{B}T^2)$. The rest is OK? $\endgroup$ Feb 12, 2015 at 17:08
  • $\begingroup$ @mikuszefski: Thanks, that missing $1/T$ was my fault and is now corrected. The rest should be correct. Regarding $\delta f$: If one accepts that the $T^{-2}$ term dominates for low $T$, then one can plug that approximation in again; $\frac{\partial\delta f}{\partial T}$ on the right hand side then is again proportional to $T^{-2}$, but then is multiplied with $T^{-1}$ to give a pole of even higher order. I'd really like to understand what's going on here, but if you only had time to answer this or the main question, I'd be grateful if you'd decide to explain the latter. Thanks again! $\endgroup$
    – RQM
    Feb 12, 2015 at 17:28
  • $\begingroup$ Hi,you don't plug in the equation in itself. Sure, you would generate higher orders then, but the thinking goes the other way: "We assume this has a finite solution. In such a case $\delta f$ may vary with $T$ but stay finite. If so the $T^{-2}$ will win." Actually it is reasonable to assume that $\delta f$ and $\mu$ are finite and smooth functions of $T$ for $T \rightarrow 0$ and far away from a possible phase transition. $\endgroup$ Feb 13, 2015 at 12:39
  • $\begingroup$ I also found a hint in this. If I get it right your $\delta f$ is the $f(\epsilon) D(\epsilon)$ there. They add a zero, i.e. $\mu \times 0$ where $0=\int \partial f/\partial T D$ This is zero because it is $\partial/ \partial T \int f D$, which is somewhat like $\partial / \partial T N$ and $N$ is constant. $\endgroup$ Feb 13, 2015 at 13:13

1 Answer 1


Summarizing my comments:

As we have an interacting system with $u_{kk'}$, it is rather normal that the change of occupation depends on all others, i.e; $\partial_T \delta f_k$ depends on $\sum_k \delta f_k $ or its derivatives. Surely this complicates drastically the overall solution. Luckily we are interested in the low temperature approximation here. In a next step we assume that such a solution exists and that it is somewhat well behaved. In detail we assume that $\delta f_k$ and $\mu$ are smooth and differentiable functions with respect to $T$. If that is the case the according terms in the equation will have a finite value and the term with $T^{-2}$ will be more important than terms with $T^{-1}$, for temperatures below a certain value or, in this case, $T \rightarrow 0$.

Concerning the square the trick seems to be the following (see here):

$\int \mathrm{d}k\; \epsilon_k \partial_T\delta f_k =\int \epsilon_k \partial_T \delta f_k + \mu \times 0$

and with $N$ constant

$0= \partial_T N = \partial_T \int \delta f_k=\int\partial_T\delta f_k$

as the sum or the integral over all occupation probabilities is a constant (the number of particles).

So one gets

$\int \mathrm{d}k\; \epsilon_k \partial_T \delta f_k + \mu \int \mathrm{d}k\; \partial_T \delta f_k =\int \mathrm{d}k\; (\epsilon_k+\mu)\partial_T \delta f_k$

Plugging in the second equation gives the square you were wondering about. I agree that not explaining this step is rather nasty for a textbook.

  • $\begingroup$ I am very sorry that I haven't had time to think about this problem since now. Your answer was very helpful, thank you! $\endgroup$
    – RQM
    Mar 8, 2015 at 20:09

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