2
$\begingroup$

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}

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 7, 2015 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, 2015 at 5:01
  • $\begingroup$ @MarkMitchison that sounds like it should be an answer $\endgroup$
    – David Z
    Dec 7, 2015 at 7:17

1 Answer 1

5
$\begingroup$

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 $\frac{\partial f_0}{\partial E} =-\delta(E_F-E)$.

$\endgroup$
2
  • $\begingroup$ Why i cannot write \begin{equation} f_{0}(E)=\Theta({E}-E_{F}). \end{equation} Delta function is symmetric. $\endgroup$
    – sara
    Dec 7, 2015 at 10:15
  • 1
    $\begingroup$ 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, 2015 at 10:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.