About the definition of chemical potential in the context of Density Functional Theory (DFT)

Question

I am preparing some slides on DFT.

The second Hohenberg-Kohn theorem states that the exact ground state electron density, $\rho(r)$, minimizes the Energy functional, E[$\rho$], subject to the constraint $\int \rho(r) dr - N =0$, being N the number of electrons.

So, we use Lagrange multipliers to find the solution:

$$\frac{\delta \{E[\rho]-\mu [\int \rho(r) dr - N]\}}{\delta \rho(r)}=0$$

Now, every textbook I have read states the Lagrange multiplier is:

$$\mu = \frac{\delta E}{\delta \rho(r)} $$

I do not understand that, because in my opinion it should be something like:

$$\mu = \frac{\delta E}{\delta ( \int \rho(r) dr - N)} $$

and I do not see a way to transform one denominator into another, or am I doing something wrong?

Thanks.

Answer 1

The first functional derivative can be rearranged to $$\frac{\delta E}{\delta \rho(r)}=\mu \frac{\delta}{\delta \rho(r)}\left(\int \rho(r)dr-N\right)\implies \mu =\frac{\delta E}{\delta \rho(r)}\Big/\frac{\delta}{\delta \rho(r)}\left(\int \rho(r)dr-N\right).$$

Were we working with the usual one-variable derivative, one could use the chain rule to formally 'cancel' the two denominators and obtain the stated result. But that property doesn't hold for the functional derivative, so that jump isn't valid.

That being the case, the above form will agree with the textbook result if $\dfrac{\delta}{\delta \rho(r)}\left[\int \rho(r)dr-N\right]=1$. This is readily verified using the usual formula for such a functional derivative (see here).

Answer 2

$\delta E = \int \frac{\delta E}{\delta\rho(r)}\delta\rho(r) dr$, while $\mu\,\delta\int \rho(r) dr = \int \mu\,{\delta\rho(r)}dr$. The condition that these two integrals are equal for any variation $\delta\rho(r)$ implies that the coefficients of $\delta\rho(r)$ must be equal, i.e., $\frac{\delta E}{\delta\rho(r)}=\mu$.