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In my book (Daniel V.Schroeder - An introduction to thermal physics) in page 267 they introduce the Fermi-Dirac distribution is given by

$$\bar{n}_{FD}=\frac{1}{e^{(\epsilon-\mu)/kT}+1}. \tag{1}$$

But a few pages later they claim that the Fermi-Dirac distribution becomes a step function and then proceed to define the Fermi energy as

$$\epsilon_F=\mu(T=0).$$

Questions:

1) How does $\bar{n}_{DF}$ become a stepfunction at $T=0$? It's not even possible to plug it in the distribution since then we have division by zero.

2) The expression for the chemical potential $\mu$ is given by

$$\mu=-kT\ln{\frac{Z_1}{N}},$$

where $N$ is the number of particles and $Z_1$ is the partition function for any single particle. So setting $T=0$ should just give $\epsilon_F=\mu=0$. But this is apparently not the case. Why?

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    $\begingroup$ Where do you get a division by zero? $\endgroup$
    – user137289
    Oct 21, 2019 at 20:21
  • $\begingroup$ In the exponent in the distribution. $e^{(\epsilon - \mu)/k\cdot 0}.$ $\endgroup$
    – Parseval
    Oct 21, 2019 at 20:22
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    $\begingroup$ Have you attempted taking the limit as $T\to0$ for the cases of $\epsilon>\mu$ and $\epsilon<\mu$? What do you get? And then think like a physicist and just say, "Ah, when $T=0$ we have a step function." If you don't upset some mathematicians with your physics then you are doing physics wrong :P $\endgroup$ Oct 21, 2019 at 21:02

2 Answers 2

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1) Yes. These curves (for example) were computed with Mathematica with $T$ very small, approaching zero but still finite.enter image description here

Discussion of the limit:

$$ \lim_{T\rightarrow 0} \frac{1}{e^{\delta/T}+1},$$ where $\delta = (E-\mu)/k_B$.

If $E>\mu$, then $\delta >0$ and $\delta/T \rightarrow +\infty$, hence $e^{\delta/T} \rightarrow \infty$ and $f=0$.

If $E<\mu$, then $\delta <0$ and $\delta/T \rightarrow -\infty$, hence $e^{\delta/T} \rightarrow 0$ and $f=1$.

2) I don't know exactly where your formula comes from, but usually you do an expansion with $E_F \gg k_BT$ like in here. That is because you never consider a particle at energy $\epsilon = E_F$, but rather at $\epsilon = E_F + k_B T$.

Also, $Z$ for non-interacting fermions is: $$ Z = \sum_{n=0}^1 r^n = 1+r,$$ where $r= \exp \left (-\frac{\epsilon - \mu}{k_B T} \right ) $ which thus also depends on temperature.

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  • $\begingroup$ Great answer, thank you! I guess the physicist notation $T=0$ instead of $T\Rightarrow 0$ confused my mathematical mind. However I have a follow up question for 2) Why don't we consider a particle att energy $\epsilon?$ In my book they do it multiple times. For example they usually consider the groundstate to be att $E_0=0 $eV and the first excited state $E_1=\epsilon.$ $\endgroup$
    – Parseval
    Oct 22, 2019 at 8:36
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    $\begingroup$ Fixed some of the math for you. $\endgroup$ Oct 22, 2019 at 12:07
  • $\begingroup$ @Parseval Well you can, but for fermions anything at $\epsilon \leq E_F$ is "locked". You usually consider particles that are "around" the Fermi level, to within the energy width allowed by the temperature $k_B T.$ $\endgroup$
    – SuperCiocia
    Oct 22, 2019 at 16:52
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The Fermi-Dirac distribution is a statistical expression for the fact that only a single Fermion can occupy a specific state. Then, the probability of a state to be occupied is given by the expression. Now let's think of a situation where we have a system with $N$ states of energy, ordered from the lowest to the highest as $0 < \epsilon_1 < \epsilon_2 < \cdots < \epsilon_N$. And we have $n$ identical fermions in the system. If the temperature is very low $T \ll \epsilon_1$, then of course the 'first' fermion will have to be in the lowest level, and once it is occupied, the next fermion will have to go to the next level etc. until we will fill the $n$ lowest energy levels. This will give the Fermi energy -- the highest energy occupied when the temperature is very very very low. You can think of $T=0$ as 'temperature is much smaller than the lowest energy scale in the system'. The Fermi-Dirac distribution then becomes effectively a $\theta(E_F-\epsilon)$ function.

Note that this is in contrast to bosons, that can all pile up in the lowest energy level when the temperature is lowered, and therefore there is no concept of Fermi energy for them (on the other hand, you get a cool phenomena like the Bose-Einstein condensate).

Regarding your worry about $\mu = -k T \ln(Z/N)$ leading to a zero chemical potential at zero temperature: this is not the case as $Z$ also depends on temperature. Again - you can expand for small $T$ and get the expression.

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  • $\begingroup$ Thank you very much for this answer! The first paragraph is beautifully explained, however the last sentence, can you explain what the $\theta$ function is and show how you get to it by taking the limit of $T\rightarrow 0?$ Also, how does it matter if $Z$ depends on $T$ as well? We are multiplying a logarithm with essentially zero, to me this seems obvious that it should approach zero as well. $\endgroup$
    – Parseval
    Oct 22, 2019 at 8:42
  • $\begingroup$ I almost forgot to ask another important question: You write that "if the temperature is very low $T<<\epsilon_1$, then ofcourse the first fermion will have to go to the lowest level". Why is this? Why does the occupation of a certain energy state-level depend on $T?$ How do I see this? I ask because in the book they say "When a gas of fermions is so cold that nearly all states below $E_F$ are occupied while nearly all states above $E_F$ are unoccupied, it is said to be degenerate". Doesnt this contradict the fact that filling the $n$ energy levels gives us $E_F$? $\endgroup$
    – Parseval
    Oct 22, 2019 at 8:49
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    $\begingroup$ @Parseval To answer your questions: $theta(x)$ is a function that is zero if $x<0$ and 1 if $x>0$. If you look at the Fermi-Dirac distribution and assume $T$ to be very small, then $f(\epsilon)=1$ if $\epsilon-\mu < 0$ (and then you have an exponential of a very large negative number, giving essentially zero) and $f(\epsilon) = 0$ if $\epsilon-\mu >0$, as then the denominator approaches infinity. $Z$ depends on $T$ through an exponent. Think of $x \ln (e^{1/x})$ at the limit of $x\to 0$. It is still finite. $\endgroup$
    – user245141
    Oct 22, 2019 at 9:18
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    $\begingroup$ @Parseval the fundamental thing to understand about temperature and statistical distribution is that the probability of a system to occupy a state with energy $E$ is proportional to $\exp(-E/T)$. Lower temperatures mean that the system will tend to occupy lower energy states, while higher temperature allows the system to also include higher energy states. This is also the explanation for the Fermi energy - at very low temperatures all single-particle states with energy below $E_F$ are occupied. $\endgroup$
    – user245141
    Oct 22, 2019 at 9:21

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