I'm trying to understand how the Kohn-Sham equations arise from the variational principle, failing. I think my problem is the inability to apply the variational principle. Or, I lack some crucial understanding about many body physics. (Background: I've never studied many body physics, but basic quantum mechanics and classical electrodynamics, yes. I run into DFT while searching for ways to model biochemical reactions.)

So the Kohn Sham equations are the time-independent Schrödinger equations (one for each electron) \begin{equation} \big[ -\frac{\nabla^2}{2} + v_{\mathrm{eff}}(r) \big] \phi_i = \epsilon_i \phi_i. \end{equation}

In density functional theory the energy functional can be written (for non-interacting electrons that have the same electron density as the exact system) \begin{align} E_v[n(\vec{r})] =& T_s[n_s(\vec{r})] + \int \mathrm{d}^3r \, v_{\mathrm{eff}}(\vec{r}) n_s(\vec{r}), \end{align} where $n(\vec{r})$ is the (trial) electron density, $T_s$ is the kinetic energy and the integral is the potential energy of the electrons moving in a potential $v(r)$.

With a fixed number of $N$ electrons, the method of Lagrange multipliers ought to have the form (as far as I've understood) \begin{equation} \frac{ \partial E[n] }{ \partial n(\vec{r}) } = \lambda \frac{\partial}{\partial n(\vec{r})} \int \mathrm{d}^3r \, n(\vec{r}) \end{equation} I would start to expand the right hand side as \begin{align} \frac{\partial}{\partial n(\vec{r})} \int \mathrm{d}^3r \, n(\vec{r}) =& \int \mathrm{d}^3r \frac{ \partial n(\vec{r}) }{ \partial n(\vec{r}) } \\ =& \int \mathrm{d}^3 r \end{align} And the left hand side as \begin{align} & \int \mathrm{d}^3r \, \bigg\{ \big(\frac{\partial}{\partial n}v(\vec{r})\big) n(\vec{r}) + v(\vec{r}) \frac{\partial}{\partial n}n(\vec{r}) \bigg\} + \frac{\partial}{\partial n}T_s[n] \\ =& \int \mathrm{d}^3 r \, v(\vec{r}) + \frac{\partial}{\partial n}T_s[n] \end{align}

What am I doing wrong? What should I be doing?

  • 1
    $\begingroup$ 1) The 5000th homework-and-exercises question; 2) Are you entirely concerned with the mathematical part of this? $\endgroup$
    – HDE 226868
    Dec 13, 2014 at 17:10
  • 2
    $\begingroup$ @HDE If I wasn't, I would not have asked. $\endgroup$
    – swirld
    Dec 14, 2014 at 6:09

1 Answer 1


After studying variational calculus for a while, here is what I gathered.

The variation of a functional $J[L(x,f,f')] = \int \, L(x,f,f')\mathrm{d}x$ can be written as \begin{equation} \delta J[L] = \int \frac{ \delta J[f] }{ \delta f(x) } \delta f(x) \mathrm{d}x \end{equation} with \begin{equation} \frac{\delta J}{\delta f(x)} = \frac{\partial L}{\partial f} - \frac{\mathrm{d}}{\mathrm{d}x} \frac{\partial L}{\partial f'} \end{equation}

Here, $L(x,f,f') = E_v[n(r)] - \lambda\big( \int n(r) \mathrm{d}^3 r - N \big)$, a function of r and n(r) (not n').

In my question I tried to apply the required but not sufficient condition: $\nabla_{r,n(r)} E_v[n(\vec{r})] = \lambda \nabla_{r,n(r)} \big(\int n(\vec{r}) \mathrm{d}^3r - N \big)$. Instead, if I take the variation of the Lagrangian $E_v[n] - \lambda \big(\int n(\vec{r}) \mathrm{d}^3r - N \big)$ of the fictious system: \begin{align} &\delta \bigg(E_v[n(r)] - \lambda \big(\int n(r) \mathrm{d}^3r - N \big) \bigg) \\ =& \int \frac{\delta E_v[n]}{\delta n(r)} \delta n(r) \mathrm{d}^3r - \lambda \int \bigg( \frac{\delta (\int n(r) \mathrm{d}^3 r)}{\delta n(r)} \bigg) \delta n(r) \mathrm{d}^3 r \\ =& \int \frac{\delta E_v[n]}{\delta n(r)} \delta n(r) \mathrm{d}^3r - \lambda \int \bigg( \frac{\partial n(r)}{\partial n(r)} \bigg) \delta n(r) \mathrm{d}^3 r \\ \end{align} Furthermore, after expanding $E_v[n(r)] = \int v(r)n(r) \mathrm{d}^3 r + T_s[n(\vec{r})]$ I get \begin{align} & \int \frac{\partial \big(v(r) n(r)\big)}{\partial n(r)} \delta n(r)\mathrm{d}^3 r + \int \frac{\delta T_s[n(r)]}{\delta n(r)} \delta n(r) \mathrm{d}^3 r - \lambda \bigg( \int \delta n(r) \mathrm{d}^3r \bigg) \\ =& \int \big[n(r)\frac{\partial v(r)}{\partial n(r)} + v(r)\frac{\partial n(r)}{\partial n(r)} \big]\delta n(r)\mathrm{d}^3 r + \int \frac{\delta T_s[n(r)]}{\delta n(r)} \delta n(r) \mathrm{d}^3 r - \lambda \bigg( \int \delta n(r) \mathrm{d}^3r \bigg) \\ =& \int \delta n(r) \bigg\{ v(r) + \frac{\delta T_s[n(r)]}{\delta n(r)} - \lambda \bigg\} \mathrm{d}^3 r = 0 \end{align} Which is what I wanted to get.

The next question is why do we arrive at the 1D time-independent Schrödinger equation from this? First, the integral is zero only when the kernel is zero (except for a set of points with zero measure), since the energy functional (which is the expectation value of the system's energy) is positive (by convention) and $\lambda$ is a constant, so that the sign of the integrand is always the same. This means that the integrand must be zero. After that, we can multiply the part in $\{ \}$-brackets with some function $\phi$, and notice that this is the Schrödinger equation we were looking for. After this, we can define the electron density to be $n(r) = \sum_i |\phi_i|^2$. After iterating enough times, you should get the same $n(r)$ which was your initail guess.

Although I must admit that I'm not a 100% satisfied with that last part, yet. I'm not sure if that's the correct mindset, or if it was known from elsewhere that $n(r) = \sum_i \phi_i$.

  • $\begingroup$ Nice attempt! . $\endgroup$
    – Mikkel Rev
    Dec 26, 2019 at 0:35

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