If I want to minimize the energy of a Slater determinant subject to the constraint that the spin orbitals are orthonormal (as in the Hartree-Fock approximation), I can use Lagrange's method of undetermined multiplier, i.e.

$$L[\{\chi_{a}\}] = E_{0}[\{\chi_{a}\}]-\sum_{a=1}^{N}\sum_{b=1}^{N}\varepsilon_{ba}([a|b]-\delta_{ab})$$

where $\{\chi_{a}\}$ are the spin orbitals, $E_{0}$ is the ground state energy, $[a|b]$ is the overlap integral between spin-orbitals $\chi_{a}$ and $\chi_{b}$ and $\varepsilon_{ba}$ is a Langrange multiplier. By setting the first variation of $L$ to zero I can proceed to minimize the energy of a single determinant $|\Psi_0\rangle=|\chi_{1}\chi_{2}\cdots\chi_{a}\chi_{b}\cdots\chi_{N}\rangle$.

A problem in Szabo and Ostlund's "Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory" asks me to prove that the Lagrange multipliers are elements of a Hermitian matrix, i.e. $\varepsilon_{ba} = \varepsilon_{ab}^{*}$, given that $L$ is real and $[a|b] = [b|a]^{*}$.

How would one go about proving this?

Where I am starting at the moment is that since $L$ is real, then $L = L^{*}$. If I find $L^{*}$ then making this equal to $L$ will show that it is necessary for the Lagrange multipliers to be elements of a Hermitian matrix. However, following through with this doesn't seem to get me to a point where I can say that $\varepsilon_{ba} = \varepsilon_{ab}^{*}$.


1 Answer 1


The Szabo-Ostlund problem (v1) can be recast in purely mathematical terms as follows:

Assume that:

  1. $C$ and $H$ are two complex $n\times n$ matrices,

  2. that $H=H^{\dagger}$ is Hermitian, and

  3. that the trace ${\rm tr}(CH)~\in~\mathbb{R}$ is real.

Prove that $C$ is Hermitian.

(Here the entries of the $C$ matrix are the Lagrange multipliers $\varepsilon_{ab}$, and the entries of the $H$ matrix are the overlap constraints $[a|b]-\delta_{ab}$.)

OP is correct that the above exercise is not possible to solve with no further information, but one can say the following:

  1. The complex matrix $C$ can always be written uniquely as a sum $C=A+B$ of a Hermitian matrix $A=A^{\dagger}$ and an anti-Hermitian matrix $B=-B^{\dagger}$. (In fact, the choices $A=\frac{C+C^{\dagger}}{2}$ and $B=\frac{C-C^{\dagger}}{2}$ are the only possibility.)

  2. The trace ${\rm tr}(AH)~\in~\mathbb{R}$ is real.

  3. The trace ${\rm tr}(BH)~=0$ is zero.

Thus Szabo and Ostlund should instead have asked to prove that one might as well assume that $C$ is Hermitian, since its anti-Hermitian part is anyway always zero by the three assumptions, and thus plays no role in the minimization process.

The logic of the exercise is somewhat artificial. If one goes beyond the exercise: More realistically, one would argue that an $n\times n$ Hermitian matrix $H$ of $n^2$ real constraints requires $n^2$ real Lagrange multipliers. These Lagrange multipliers can be neatly organized in an $n\times n$ Hermitian matrix $C$. And in this way, the term ${\rm tr}(CH)$ in the Lagrangian becomes manifestly real.

  • $\begingroup$ Thanks for your answer. As I understand it, in your generalization, $C$ is equivalent to the matrix of Lagrange multipliers and $H$ is equivalent to the matrix of the overlap integrals $[a|b]$. The second term on the right of the Lagrange function equation is equivalent to the trace over $CH$, so showing $C$ is Hermitian is equivalent to showing that $\varepsilon_{ab} = \varepsilon_{ba}^{*}$. I was not aware that $C = A + B$ - could you point me to somewhere where this is explained? $\endgroup$ Jul 11, 2012 at 11:38
  • 1
    $\begingroup$ I updated the answer (v3). $\endgroup$
    – Qmechanic
    Jul 11, 2012 at 12:52

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