# One-electron reduced density matrix: Argument for positive semidefiniteness

I cannot follow an argument for positive-semidefiniteness of the one-electron density matrix given in "Molecular Electronic-Structure Theory" by Helgaker/Jorgensen/Olsen. First some definitions: $F(M)$ denotes the Fock Space spanned by a basis of $M$ orthonormal spin orbitals (which are not further specified). $F(M,N)$ denotes its subspace of $N$-electron states. Note that because we are considering a finite one-electron basis, the Fock-Space only includes states with a maximum number of $N$ electrons. The authors further define the one-electron density matrix for an arbitrary $N$-electron state $|x\rangle$ by $\bar D_{PQ} = \langle x| a_P^\dagger a_Q |x\rangle$. Now comes the sentence I don't understand: "The one-electron density matrix is positive semidefinite since its elements are either trivially equal to zero or inner products of states in the subspace $F(M, N-1)$" (given in section 1.7.1).

I'm assuming that the authors mean that the matrix elements are inner products whenever $N\geq 1$ and are equal to zero when $N=0$ (vacuum state). I also know that a matrix is positive-semidefinite when it is the matrix representation of a positive-semidefinite bilinear form, but I don't see a connection to the above argument, especially since the inner product is positive definite (without "semi").

Please do not provide any alternate arguments why the density matrix is positive-semidefinite (I think I can find these myself). I would like to understand the specific argument given above.

The matrix elements $D_{PQ}$ can either be zero if the orbital $P$ or $Q$ are not occupied -- then $a_Q\vert x\rangle=0$ or $\langle x \vert a_P^\dagger = 0$ -- and otherwise, they are overlaps of $N-1$ electron states $a_P\vert x\rangle$.
If you define a matrix $X$ whose columns are $a_P\vert x\rangle$, then you can write $D=X^\dagger X$, and a matrix of this form is always positive semi-definite.
• Could you make the argument a little bit more explicit? In $a_P|x\rangle$ there is only one index, so how is this a matrix? – LLang Aug 16 '15 at 12:19
• For each $P$, $a_P\vert x\rangle$ is ket, i.e. a (column) vector. By taking those column vectors for all $P$ next to each other, I obtain a matrix $X$. It is immediate to see that for this matrix, $D=X^\dagger X$. – Norbert Schuch Aug 16 '15 at 12:35
• Thanks, that helped a lot! I like to distinguish kets and their representations in some basis. If $|i\rangle$ form a complete orthonormal basis of $F(M, N-1)$ one can define $X_{iP} := \langle i | a_P | x \rangle$. Then the rest works exactly as you explained. – LLang Aug 16 '15 at 13:36
• @LLang If you don't want to talk about representations of the kets, you can also argue that the expectation value of $Q$ w.r.t. any vector $\vec c=(c_P)$ is given by $\vec c^\dagger Q\vec c$ = $\langle y\vert y\rangle\ge0$, where $\vert y\rangle = \sum_P c_P a_P\vert x\rangle$. – Norbert Schuch Aug 16 '15 at 15:11