Is there a significance or an important interpretation of the fact that the plots in the Fermi-Dirac distribtion graph, all intersect at the point with coordinates ($\epsilon_F$,$\frac {1}{2}$)?
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The curve shown in the image quoted in OP correspond to different values of temperature, but the same value of the Fermi level. It is a simple mathematical fact that the Fermi function $$ f_{\epsilon_F}(\epsilon)=\frac{1}{\exp\left(\frac{\epsilon-\epsilon_F}{k_BT}\right)+1} $$ takes value $1/2$ when $\epsilon=\epsilon_F$. What changes with temperature is the spread of the function about the Fermi level, which characterizes how many electrons from below the Fermi level have been excited to above the Fermi level.
See also this answer regarding the difference between Fermi energy and Fermi level.
Roger V.
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Is it possible to showcase the spread change as temeperature changes just by considering the FD distribution only? An increase in temperature means that the exponential for fixed energy value decreases in value. This decrease is the consequence why below the fermi level, different plots at the same energy value have lower occupation value as we increase the temperature, and the opposite above the fermi level ? – imbAF Mar 27 '23 at 09:59
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@imbAF For non-interacting particles. You may find this thread (and my answer in it) helpful: Are Distribution functions really the probability or the number of particles? – Roger V. Mar 27 '23 at 10:02
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But the formula I have for the distribution is: $\overline n_i=\frac{1}{e^{\beta(\epsilon_i-\mu)}+1}$. But $\mu$ is dependent on Temperature. In your formula, you are making the assumption that $\mu=\epsilon_F$ why? – imbAF Mar 27 '23 at 21:11
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$\mu\rightarrow\epsilon_F$ is just a change of notation - Fermi level and chemical potential are the same thing (but perhaps you are right, that using $\mu$ would avoid a confusion with Fermi energy.) Your question does not say anything about temperature dependence - in fact, if $\mu$ were a function of temperature, the curves of course would not cross at the same point... but what is more important here is that we are working in the grand canonical ensemble where $\mu$ is among the independent variables (while the number of particles $N$ is a function of temperature.) – Roger V. Mar 28 '23 at 08:10
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In our class we said that $\mu(T=0,V,N)=\epsilon_F$. Meaning that the value that the chemical potential has at absolute zero, is taken as the fermi energy, which is the highest energy level at T=0. But I see we keep the $\mu$ constant – imbAF Mar 28 '23 at 08:28
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To quote Wikipedia: "In band structure theory, used in solid state physics to analyze the energy levels in a solid, the Fermi level can be considered to be a hypothetical energy level of an electron, such that at thermodynamic equilibrium this energy level would have a 50% probability of being occupied at any given time."
States with low energy are always the most likely to be occupied, and states with higher energy are less likely to be occupied. The Fermi energy is the "balancing point" with 50% expected occupation.
Marius Ladegård Meyer
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