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Why the derivative of Fermi-Dirac distribution function at absolute zero temperature becomes negative of Dirac_Delta function. The Fermi-Dirac distribution function is \begin{equation} f_{0}(E)=\frac{1}{e^\frac{{E-E_{F}}}{k_{B}T}+1}, \end{equation} As $T\rightarrow0$, the Fermi-Dirac distribution becomes a step function \begin{equation} f_{0}(E)=\Theta({E}-E_{F}). \end{equation} and \begin{equation} \frac{\partial f_{0}}{\partial E}=\frac{-2}{\left(2\pi\hbar\right)^{3}} \delta\left(E-E_{F}\right), \end{equation} how we can get this and why the derivative becomes negative? should it not be \begin{equation} \frac{\partial f_{0}}{\partial E}=\frac{2}{\left(2\pi\hbar\right)^{3}} \delta\left(E-E_{F}\right), \end{equation}

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    $\begingroup$ Actually $\lim_{T\to 0} f_0(E) = \Theta(E_F-E)$. This means that only states below the Fermi energy are occupied. The derivative should be negative since the distribution goes from 1 to 0 when $E$ is increased above $E_F$. You should be able to convince yourself that the derivative must be negative at any temperature. It just means that lower-energy states are more likely to be occupied. $\endgroup$ – Mark Mitchison Dec 7 '15 at 4:40
  • $\begingroup$ It's not clear how you define $\Theta(E)$, but you may have it inverted. Notice $f_0(E)\to{}1$ for $E\ll{}E_F$ and $f_0(E)\to{}0$ for $E\gg{}E_F$. Does your definition of $\Theta(E)$ do that? $\endgroup$ – The Photon Dec 7 '15 at 5:01
  • $\begingroup$ @MarkMitchison that sounds like it should be an answer $\endgroup$ – David Z Dec 7 '15 at 7:17
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The Fermi-Dirac distribution is $$ f_T(E)=\frac{1}{\exp\left(\frac{E-E_F}{k_{\text B}T}\right)+1}$$ when $T\to0$, the denominator of $\frac{E-E_F}{k_{\text B}T}$ goes to zero and this ratio goes to $+\infty$ if $E>E_F$ and to $-\infty$ if $<E_F$. Therefore the exponential $\exp\left(\frac{E-E_F}{k_{\text B}T}\right)$ goes to $0$ if $E<E_F$ and to $+\infty$ if $E>E_F$. Thus the distribution $f_0$ goes to $1$ if $E<E_F$ and to $0$ if $E>E_F$. This is expressed by the limiing expression $$f_0=\Theta(E_F-E)$$ (and not $\Theta(E-E_F)$ as you wrote).

So the distribution $f_0$ is constant except at the point $E=E_F$ where it decreases from $1$ to $0$. That is why the derivative is negative.

By the way, we have $\frac{\partial f_0}{\partial E} =-\delta(E_F-E)$.

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  • $\begingroup$ Why i cannot write \begin{equation} f_{0}(E)=\Theta({E}-E_{F}). \end{equation} Delta function is symmetric. $\endgroup$ – sara Dec 7 '15 at 10:15
  • $\begingroup$ The letter $\Theta$ is called theta. Obviously, if $$\Theta(x)=\left\{\begin{array}{cc}0&\text{if}\;x<0\\1&\text{if}\;x\geq 0\end{array}\right.$$ then it is not symmetric. $\endgroup$ – Tom-Tom Dec 7 '15 at 10:31

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