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I have been reading Neamen's "Semiconductor Physics and Devices" Chapter 3. He derives in Eq. (3.79) the Fermi-Dirac distribution function $$f_F(E) = \frac{1}{1 + \exp(\frac{E-E_F}{kT})}, $$ where $T$ is the temperature, $E_F$ is the Fermi energy, i.e. the Fermi level at $T = 0 K$, and $k$ is the Boltzmann constant. He then states that for $T > 0 K$, $f_F (E = E_F) = \frac{1}{2}$, i.e. the probability that the energy state at $E = E_F$ is populated is always $\frac{1}{2}$, regardless of temperature. Here is the excerpt and plot from the book:

enter image description here

However, it is my understanding that the correct formula for the Fermi-Dirac distribution for any temperature $T$ would replace $E_F$ with the chemical potential $\mu$, which is dependent on temperature and not constant (it is only equal to $E_F$ for $T = 0 K$).

So am I correct in assuming that the author makes an approximation in assuming that the chemical potential is constant and equal to the Fermi energy $E_F$ even for $T > 0 K$ (he writes so in the last line above the plot)? If so, under what assumptions? Or is it a mistake?

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  • $\begingroup$ The article referenced in my answer at physics.stackexchange.com/questions/487998/… goes into gory detail on the difference between the Fermi level and the chemical potential. $\endgroup$
    – Jon Custer
    Commented Sep 11, 2023 at 14:24
  • $\begingroup$ @JonCuster I understand that the Fermi level seems to refer to the chemical potential in the case of semiconductors. However, how does that relate to the Fermi energy, which I understand to be the Fermi level (the chemical potential) at zero temperature. Does Neamen also use "Fermi energy" interchangeably with "Fermi level" (the chemical potential)? Do all references to one of those terms in the case of semiconductors mean the chemical potential, i.e. a quantity that is in general temperature dependent? $\endgroup$
    – shortwhile
    Commented Sep 11, 2023 at 14:33
  • $\begingroup$ In a metal, where there is a non-zero density of states at the fermi energy, the chemical potential at zero temperature is equal to the Fermi energy. In a semiconductor, unlike a metal, there is a gap between highest occupied and lowest unoccupied states, so it is ambiguous/vague as to where any "fermi energy" would be. But, for a semi-conductor, the chemical potential at zero temperature can be found to be midway between the highest occupied and lowest unoccupied states. $\endgroup$
    – hft
    Commented Sep 11, 2023 at 16:20
  • $\begingroup$ Anyways, you are misinterpretting what the author has written. The author is not trying to explain to you what happens when the physical temperature changes. The author is trying to illustrate how the function $\frac{1}{1+e^{\alpha x}}$ looks different for different values of $\alpha$ when plotted as function of $x$. $\endgroup$
    – hft
    Commented Sep 11, 2023 at 19:55
  • $\begingroup$ @hft The correct formula for the Fermi-Dirac distribution involves the chemical potential $\mu$ instead of the Fermi energy (chemical potential at $T = 0 K$), correct? $\endgroup$
    – shortwhile
    Commented Sep 12, 2023 at 8:24

3 Answers 3

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The Fermi energy is indeed defined to be the chemical potential at $T = 0$, and for higher temperatures it is no longer true that $\mu = E_F$. However, the difference between the two is often negligible. For example, for a 3D gas of free fermions we have the famous Sommerfield expansion, which says that $$ \mu = E_F \left( 1 - \frac{\pi^2}{12} \left( \frac{kT}{E_F} \right)^2 + \cdots \right) $$ and as long as the thermal energy scale is much less than the Fermi energy (which is often the case in realistic situations) then we have $\mu \approx E_F$.

In a more general situation, the deviation between $\mu$ and $E_F$ as a function of temperature will differ quantitatively (depending on a different power of $k T$, or even non-analytically on $kT$.) But so long as $T$ is "sufficiently small" (a context-dependent criterion, to be sure) we can get away with assuming that $\mu \approx E_F$ even when $T \neq 0$.

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So am I correct in assuming that the author makes an approximation in assuming that the chemical potential is constant and equal to the Fermi energy EF even for T>0K

It would generally be an approximation to take the chemical potential as equal to the Fermi energy at finite temperature. But the author does not make such as approximation. I think you are misreading or misinterpreting what the author has written.

(he writes so in the last line above the plot)?

No, the author writes: "assuming the Fermi energy is independent of temperature." The author is about to present a diagram regarding how the Fermi-Dirac distribution function changes when the literal $T$ parameter in the function is changed. (Changing $T$ as a parameter in the fermi distribution and plotting the result is not the same as changing the physical temperature.) The author presumably understands that the chemical potential is not always equal to the Fermi energy. The author seems to just be saying, hey this "$T$" thing is a parameter in the Fermi-Dirac distribution and I'm going to show you how the Fermi-Dirac function changes when the $T$ parameter changes.

If so, under what assumptions?

To get a sense of how the chemical potential changes with temperature, you can think of the chemical potential as being "repelled" by the higher density of states region. For example, in the free electron gas case, the density of states is higher at higher energy and the chemical potential decreases with temperature (see the negative sign in the $-\pi^2 k^2 T^2/(12E_F^2)$ term in Michael Seifert's answer).

In a real solid (as opposed to a free electron gas) it is possible that there could be a larger density of states just below the Fermi energy and then the Chemical potential would move up with temperature.

If there is some solid wherein the density of states is exactly constant near the Fermi energy then the chemical potential may not even move at all to second order.

But, generally, in real life, unless the density of states is completely flat (e.g., 2d free electron gas, which is not a real-life situation) the chemical potential at finite temperature will not be equal to the Fermi energy.

Or is it a mistake?

I would not call it a mistake. The author is using the diagram to show how a fixed function's shape changes when the $T$ parameter is adjusted. The author is indicating to you that changing the $T$ parameter is not the same as changing the physical temperature, and the author indicates this by their use of the word "assuming..."

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The correct formula for the Fermi-Dirac distribution is $$f_F(E) = \frac{1}{1+\exp(\frac{E-\mu}{kT})}, $$ where $\mu$ is the chemical potential, a quantity which is in general temperature-dependent (see, e.g., Ashcroft & Mermin 1976 Chapter 2, Eq. (2.48)). At $T = 0 K$, we have $\mu = \mu_0 \equiv E_F$, which is called the Fermi energy.

The author cited in the question seems to use the term "Fermi level" as well as the notation $E_F$ to refer to the chemical potential.

This is illustrated in Chapter 4 of his book (4th ed.), Eq. (4.26b), where a formula for the Fermi level of an intrinsic semiconductor is derived: $$E_{Fi} - E_{midgap} = \frac{3}{4} k T \ln(\frac{m^*_p}{m^*_n}), $$ where $E_{Fi}$ is the Fermi level for an intrinsic semiconductor, and $m^*_n$ and $m^*_p$ are the effective masses of electrons (in the conduction band) and holes (in the valence band) respectively. Clearly, the quantity is temperature-dependent and can therefore plausibly only refer to the chemical potential. Comparing this expression to Ashcroft & Mermin 1976, Chapter 28, Eq. (28.22), $$\mu_i = E_v + \frac{1}{2} E_g + \frac{3}{4} k_B T \ln(\frac{m_v}{m_c}), $$ where $\mu_i$ is the chemical potential of the intrinsic semiconductor, $E_v$ is the top of the valence band, $E_g$ is the bandgap energy, and $m_c$ and $m_v$ are the effective masses of the electrons in the conduction band and the holes in the valence band, respectively, we find that the two formulae are equivalent if $E_{Fi}$ and $\mu_i$ refer to the same thing.

Furthermore, Ashcroft & Mermin note in Chapter 28 of their book that "[it] is the widespread practice to refer to the chemical potential of a semiconductor as 'the Fermi level,' a somewhat unfortunate terminology. [...] The term 'Fermi level' should be regarded as nothing more than a synonym for 'chemical potential' in the context of semiconductors."

We therefore conclude that Neamen, who is cited in the question, uses the term "Fermi level" and the notation $E_F$ to refer to the chemical potential of a semiconductor.

As correctly noted in Michael Seifert's answer, there are conditions under which the chemical potential may be assumed to be constant, i.e. equal to the Fermi energy (Sommerfeld expansion). However, this is not what Neamen refers to when he, in general, uses the term "Fermi level".

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