Proof of the Hartree-Fock equations

As per the proof of the Hartree equations, the proof of the Hartree-Fock equations involves the use of an analogue of eq4 to find an expression similar to eq18. Applying the Lagrange method of undetermined multipliers, the resultant expression is then minimised to obtain the non-canonical form of the Hartree-Fock equations.

The analogue of eq4 is:

$E=\int\cdots\int\Psi_a^*\left(-\frac{1}{2}\sum_{i=1}^n\nabla_i^2-\sum_{i=1}^n\frac{Z}{r_i}+\sum_{i=1}^{n-1}\sum_{j=i+1}^n\frac{1}{r_{ij}}\right)\Psi_ad\tau_1\cdots d\tau_n\;\;\;\;\;\;\;\;(93)$

where $\Psi_a$ is given by eq66.

Substituting eq86 and eq89 in eq93

$E=\sum_{i=1}^nH_i+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n(J_{ij}-K_{ij})\;\;\;\;\;\;\;\;(94)$

where
$H_i=\int\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(x_i)dx_i$
$J_{ij}=\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_i(x_i)\phi_j(x_j)}{r_{ij}}dx_idx_j$
$K_{ij}=\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_j(x_i)\phi_i(x_j)}{r_{ij}}dx_idx_j$

Since $\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n(J_{ij}-K_{ij})=\frac{1}{2}\sum_{j= 1}^n\sum_{i=1}^n(J_{ij}-K_{ij})$ (see this article for explanation),

$E=\sum_{i=1}^nH_i+\frac{1}{2}\sum_{j=1}^n\sum_{i=1}^n(J_{ij}-K_{ij})\;\;\;\;\;\;\;\;(95)$

For $\langle\Psi_a\vert\Psi_a\rangle=1$ , we must have $\langle\phi_i\vert\phi_j\rangle=\delta_{ij}$ . To find the minimum value of E subject to the constraints of $\int\phi_i^*(x_i)\phi_j(x_i)dx_i=\delta_{ij}$ , we apply the Lagrange method of undetermined multipliers to form the new functional $F[\phi_i^*,\phi_i]$ , where

$F[\phi_i^*,\phi_i]=E-\sum_{i=1}^n\sum_{j=1}^n\lambda_{ij}\left[\int\phi_i^*(x_i)\phi_j(x_i)dx_i-1\right]\;\;\;\;\;\;\;\;(96)$

Substituting eq95 in eq96,

$F[\phi_i^*,\phi_i]=\int\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(x_i)dx_i+\frac{1}{2}\sum_{i,j=1}^n\biggr(\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_i(x_i)\phi_j(x_j)}{r_{ij}}dx_idx_j$ $-\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_j(x_i)\phi_i(x_j)}{r_{ij}}dx_idx_j\biggr)-\sum_{i,j=1}^n\lambda_{ij}\left[\int\phi_i^*(x_i)\phi_j(x_i)dx_i-1\right]\;\;\;\;\;\;\;\;(97)$

Applying the functional variation method as per the proof of the Hartree equations and noting that the minimum energy corresponding to F is when a small change in the functional’s input yields no change in the functional’s output, i.e. when $dF=F[\phi_i^*+\delta\phi_i^*,\phi_i+\delta\phi_i]-F[\phi_i^*,\phi_i]=0$ or

$dF=\sum_{i=1}^n\int\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\delta\phi_i(x_i)dx_i+\sum_{i=1}^n\int\delta\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(x_i)dx_i$ $+\frac{1}{2}\sum_{i,j=1}^n\int\int\biggr[\delta\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_i(x_i)\phi_j(x_j)}{r_{ij}}+\phi_i^*(x_i)\delta\phi_j^*(x_j)\frac{\phi_i(x_i)\phi_j(x_j)}{r_{ij}}\biggr]dx_idx_j$ $+\frac{1}{2}\sum_{i,j=1}^n\int\int\biggr[\phi_i^*(x_i)\phi_j^*(x_j)\frac{\delta\phi_i(x_i)\phi_j(x_j)}{r_{ij}}+\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_i(x_i)\delta\phi_j(x_j)}{r_{ij}}\biggr]dx_idx_j$ $-\frac{1}{2}\sum_{i,j=1}^n\int\int\biggr[\delta\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_j(x_i)\phi_i(x_j)}{r_{ij}}+\phi_i^*(x_i)\delta\phi_j^*(x_j)\frac{\phi_j(x_i)\phi_i(x_j)}{r_{ij}}\biggr]dx_idx_j$ $-\frac{1}{2}\sum_{i,j=1}^n\int\int\biggr[\phi_i^*(x_i)\phi_j^*(x_j)\frac{\delta\phi_j(x_i)\phi_i(x_j)}{r_{ij}}+\phi_i^*(x_i)\phi_j^*(x_j)\frac{\phi_j(x_i)\delta\phi_i(x_j)}{r_{ij}}\biggr]dx_idx_j$ $-\sum_{i,j=1}^n\lambda_{ij}\int\left[\phi_i^*(x_i)\delta\phi_j(x_i)+\delta\phi_i^*(x_i)\phi_j(x_i)\right]dx_i\;\;\;\;\;\;\;\;(98)$

It can be easily shown, by expanding the Coulomb and exchange terms in eq98, that the two terms within each square bracket are the same. Therefore,

$dF=\sum_{i=1}^n\int\phi_i^*(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\delta\phi_i(x)dx+\sum_{i=1}^n\int\delta\phi_i^*(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\phi_i(x)dx$ $+\sum_{i,j=1}^n\int\int\biggr[\delta\phi_i^*(x)\phi_j^*(x')\frac{\phi_i(x)\phi_j(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}+\phi_i^*(x)\phi_j^*(x')\frac{\delta\phi_i(x)\phi_j(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}\biggr]dxdx'$ $-\sum_{i,j=1}^n\int\int\biggr[\delta\phi_i^*(x)\phi_j^*(x')\frac{\phi_j(x)\phi_i(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}+\phi_i^*(x)\phi_j^*(x')\frac{\delta\phi_j(x)\phi_i(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}\biggr]dxdx'$ $-\sum_{i,j=1}^n\lambda_{ij}\int\left[\phi_i^*(x)\delta\phi_j(x)+\delta\phi_i^*(x)\phi_j(x)\right]dx=0$

where we have changed the dummy variables $x_i$ and $x_j$ to $x$ and $x'$ respectively.

Let’s carry out the following for the above equation:

1) Use the Hermitian property of the operator $-\frac{1}{2}\nabla^2-\frac{Z}{r}$ on the first term.
2) Switch the dummy indices i and j to j and i respectively for the 2^ndexchange integral and the 1^st undetermined multiplier term.
3) Rearrange.

The result is:

$\int\sum_{i=1}^n\delta\phi_i(x)\biggr\{\biggr[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{j=1}^n\int\phi_j(x')\frac{(1-P_{ij})\phi_j^*(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}dx'\biggr]\phi_i^*(x)$ $-\sum_{j=1}^n\lambda_{ji}\int\phi_j^*(x)\biggr\}dx$ $+\int\sum_{i=1}^n\delta\phi_i^*(x)\biggr\{\biggr[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{j=1}^n\int\phi_j^*(x')\frac{(1-P_{ij})\phi_j(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}dx'\biggr]\phi_i(x)$ $-\sum_{j=1}^n\lambda_{ij}\int\phi_j(x)\biggr\}dx=0\;\;\;\;\;\;\;\;(99)$

Question

Show that $\lambda_{ji}^*=\lambda_{ij}=\lambda_{ij}^{\dagger}$ , i.e. the Lagrange undetermined multipliers are elements of a Hermitian matrix.

Answer

Changing the dummy variables $x_i$ and $x_j$ of eq96 to $x$ and $x'$ respectively and taking the complex conjugate throughout, noting that E and hence F is real (recall that the functional F is obtained by subtracting the function $g=0$ , where $g=\sum_{i,j=1}^n\lambda_{ij}\int\phi_i^*\phi_jdx-\delta_{ij}$ from E),

$F=E-\sum_{i,j=1}^n\lambda_{ij}^*\left[\int\phi_i(x)\phi_j^*(x)dx-1\right]\;\;\;\;\;\;\;\;(100)$

Using eq96 again and switching the dummy indices i and j to j and i respectively,

$F=E-\sum_{i,j=1}^n\lambda_{ji}\left[\int\phi_j^*(x)\phi_i(x)dx-1\right]\;\;\;\;\;\;\;\;(101)$

Comparing eq100 and eq101,

$\lambda_{ji}=\lambda_{ij}^*\;\;\;\;\;\;\;\;(101a)$

or equally

$\lambda_{ji}^*=\lambda_{ij}=\lambda_{ij}^{\dagger}\;\;\;\;\;\;\;\;(102)$

Substitute eq101a in eq99,

$dF=\int\sum_{i=1}^n\delta\phi_i(x)A_idx+\int\sum_{i=1}^n\delta\phi_i^*(x)B_idx=0\;\;\;\;\;\;\;\;(103)$

where

$A_i=\left[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{j=1}^n\int\phi_j(x')\frac{(1-P_{ij})\phi_j^*(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}dx'\right]\phi_i^*(x)-\sum_{j=1}^n\lambda_{ij}^*\int\phi_j^*(x)$
$B_i=\left[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{j=1}^n\int\phi_j^*(x')\frac{(1-P_{ij})\phi_j(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}dx'\right]\phi_i(x)-\sum_{j=1}^n\lambda_{ij}\int\phi_j(x)$

Using the same logic described in the steps taken from eq20 to eq21, we have

$\left[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{j=1}^n\int\phi_j^*(x')\frac{(1-P_{ij})\phi_j(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}dx'\right]\phi_i(x)=\sum_{j=1}^n\lambda_{ij}\int\phi_j(x)\;\;\;\;(104)$

and its complex conjugate

$\left[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{j=1}^n\int\phi_j(x')\frac{(1-P_{ij})\phi_j^*(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}dx'\right]\phi_i^*(x)=\sum_{j=1}^n\lambda_{ij}^*\int\phi_j^*(x)\;\;\;\;(105)$

Eq104 and eq105 are the Hartree-Fock equations. However, they are not in the canonical form (eigenvalue form). For the derivation of the canonical Hartree-Fock equations, see the next article.

Proof of the Hartree-Fock equations

Question

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