In a paper describing a Kohn-Sham Density Functional Theory implementation, the authors describe the use of the density matrix for e.g. the calculation of the electronic density and for efficiency improvements. With a basis expansion of the Kohn-Sham states $$ |\Psi_n^\mathrm{KS}\rangle = \sum_i c_{in}\,|\phi_i\rangle $$ the density operator and matrix are \begin{align*} \hat{\rho} &= \sum_n f_n |\Psi_n\rangle\langle\Psi_n| =\sum_{ij} \rho_{ij}\, |\phi_i\rangle\langle\phi_j| \\ \rho_{ij} &= \sum_n f_n\, c^*_{in}\,c_{jn} \end{align*} with the occupation numbers $f_n$ - the electron density is then obtained via $$ \langle\mathbf{r}|\hat{\rho}|\mathbf{r}\rangle = \rho = \sum_{ij} \phi_i\,\rho_{ij}\,\phi_j. $$ When comparing this with the standard definition of the density operator $$ \hat{\rho} = \sum_n w_n |\Psi_n\rangle\langle\Psi_n|, $$ the weights $w_n$ have a different meaning, being the probabilities to find a subsystem in the state $|\Psi_n\rangle$ and fulfilling $\sum_n w_n = 1$. In my understanding, this has nothing to do with the density operator in above context where $f_n$ denote whether a single-particle orbital is occupied or not and where $\sum_n w_n \geq 1$ is possible.

Now to my question: is the density matrix in the context of DFT as explained above just a tool to ease technical/computational operations or can still any meaning in the 'classical' quantum-mechanical view be given to it?


1 Answer 1


While the Kohn-Sham states $|n^{\rm KS}\rangle$ are really only an auxiliary construct to compute the non-interacting kinetic energy, one often nevertheless goes ahead and interprets the $|n^{\rm KS}\rangle$ as single-particle states and the Kohn-Sham eigenvalues $\epsilon_n$ as quasi-particle energies (e.g. one uses the Kohn-Sham band structure as a first approximation to an experimentally observed quasi-particle band structure with such famous problems of density functional theory typically underestimating band gaps significantly).

If one does accept the interpretation of Kohn-Sham states as single-/quasi-particle states, the eigenvalues $\lambda_n$ of the density matrix $\hat \rho = \sum_n \lambda_n |n^{\rm KS}\rangle\langle n^{\rm KS}|$ are the probabilities to find a particle in state $|n^{\rm KS}\rangle$ (or simply the occupation of that state). $\hat \rho$ is therefore called one-particle reduced density matrix: $N-1$ particle degrees of freedom have been traced out. The eigenvectors of one-particle reduced density matrices are called natural orbitals (which in the case of Kohn-Sham density functional theory coincide with the Kohn-Sham states).

Since Kohn-Sham density functional theory only uses a single Slater determinant, at $T=0$K the $\lambda_n$ are either $0$ or $1$ (or $0$ or $2$ if spin degeneracy is factored into the occupation numbers); correlation beyond a single Slater determinant would lead to fractional occupation even at $0$K.

Besides this single-particle like interpretation of the above density matrix, a potential computational advantage comes about in the case of actually not decomposing $\hat \rho$ into its representation via the Kohn-Sham states, but expressing it in real space as $\rho(\vec r, \vec r^\prime)$. One can then take advantage of the fact that $\rho(\vec r, \vec r^\prime)$ typically decays 'relatively' quickly with $|\vec r - \vec r^\prime|$ and attempt to simply truncate $\rho(\vec r, \vec r^\prime)$ beyond a cutoff radius. This truncation can be used to construct linear-scaling approaches, avoiding the computational disadvantage of contructing orthogonal Kohn-Sham states which accounts for the general cubic scaling with system size of Kohn-Sham density functional theory (there are also basis set choice related techniques based on Kohn-Sham orbitals that allow for linear scaling computational complexity such as atom-centered atomic-orbital-like basis sets, where orbital overlap is truncated beyond a cutoff).

Finally a note on the trace of $\hat \rho$: unlike the trace of the pure density matrix of a single particle, which is $1$, the trace of the one-particle reduced density matrix $\hat \rho$ of an $N$-particle system is $N$; each particle is found with 100% probability somewhere, and there are $N$ particles. Hence this is not in contradiction to the eigenvalues of $\rho$ coinciding with single-particle occupation numbers: the sum of the single-particle occupation numbers must indeed be $N$.

  • $\begingroup$ Thank your for this excellent and enlightening answer! $\endgroup$
    – Lukk
    Aug 26, 2019 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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