# MCSCF Hessian transformation

I have tried doing the transformation of the Hessian in MCSCF following the notation that is used in Molecular Electronic-Structure Theory (MEST). Here I try to show what steps I have taken (to get the wrong result).

In MCSCF we have that the second order contribution to the electronic Hessian is given as:

$$\overline{\overline{K}}=\left(\begin{array}{cc} ^{\mathrm{cc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{co}}\overline{\overline{K}}^{(2)}\\ ^{\mathrm{oc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{oo}}\overline{\overline{K}}^{(2)} \end{array}\right)\tag{12.2.40, MEST}$$

With,

$$^{\mathrm{cc}}K_{i,j}^{(2)}=2\left\langle i\left|\hat{H}-E^{(0)}\right|j\right\rangle\tag{12.2.41, MEST}$$

and,

$$^{\mathrm{co}}K_{pq,i}^{(2)}=^{\mathrm{oc}}K_{pq,i}^{(2)}=2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle\tag{12.2.42, MEST}$$

and,

$$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \frac{1}{2}\left(1+P_{pq,rs}\right)\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle\tag{12.2.43, MEST}$$

For the last equation the permutation operator $P_{pq,rs}$ can be expanded out, giving:

$$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}P_{pq,rs}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle$$ $$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{rs}^{-},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle$$ $$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{rs}^{-},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle\tag{1}$$

Now for the transformation of the Hessian we can write:

$$\left(\begin{array}{c} ^{\mathrm{c}}\overline{\sigma}\\ ^{\mathrm{o}}\overline{\sigma} \end{array}\right)=\left(\begin{array}{cc} ^{\mathrm{cc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{co}}\overline{\overline{K}}^{(2)}\\ ^{\mathrm{oc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{oo}}\overline{\overline{K}}^{(2)} \end{array}\right)\left(\begin{array}{c} \overline{c}\\ \overline{\kappa} \end{array}\right)=\left(\begin{array}{c} ^{\mathrm{cc}}\overline{\overline{K}}^{(2)}\overline{c}+{}^{\mathrm{co}}\overline{\overline{K}}^{(2)}\overline{\kappa}\\ ^{\mathrm{oc}}\overline{\overline{K}}^{(2)}\overline{c}+{}^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa} \end{array}\right)\tag{12.5.17, MEST}$$

Now I want to determine the expression of the right hand side. The first three terms I seem to get correctly, only the last part gives me trouble. Look for equation (Problem). Lets start with:

$$\left[^{\mathrm{cc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{i}=\sum_{j}2\left\langle i\left|\hat{H}-E^{(0)}\right|j\right\rangle c_{j}$$

The definition:

$$\left|\overline{c}\right\rangle =\sum_{j}c_{j}\left|j\right\rangle$$

can be used to find:

$$\left[^{\mathrm{cc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{i}=2\left\langle i\left|\hat{H}-E^{(0)}\right|\overline{c}\right\rangle \tag{12.5.18, MEST}$$

Now the next term:

$$\begin{array}{ccc} \left[^{\mathrm{co}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{i} & = & \sum_{p>q}2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle \kappa_{pq}\\ & = & 2\left\langle i\left|\sum_{p>q}\kappa_{pq}\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle \end{array}$$

Now by the definition:

$$\hat{\kappa}=\sum_{p>q}\kappa_{pq}E_{pq}^{-}\tag{2}$$

Giving:

$$\begin{array}{ccc} \left[^{\mathrm{co}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{i} & = & 2\left\langle i\left|\left[\hat{\kappa},\hat{H}\right]\right|0\right\rangle \\ & = & 2\left\langle i\left|\hat{H}_{\kappa}\right|0\right\rangle \end{array}\tag{12.5.19, MEST}$$

with,

$$\hat{H}_{\kappa}=\left[\hat{\kappa},\hat{H}\right]\tag{12.5.22, MEST}$$

Now the next term:

$$\left[^{\mathrm{oc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{pq}=\sum_{i}2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle c_{i}$$

Since it can be written that:

$$2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle =\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle +\left\langle 0\left|\left[E_{pq}^{-},\hat{H}\right]\right|i\right\rangle$$

$$\left[^{\mathrm{oc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{pq}=\left\langle \overline{c}\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle +\left\langle 0\left|\left[E_{pq}^{-},\hat{H}\right]\right|\overline{c}\right\rangle\tag{12.5.21, MEST}$$

And the last term. This is here where I cannot figure it out, I tried the following, by first applying equation (1):

$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\sum_{r>s}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle \kappa_{rs}+\frac{1}{2}\sum_{r>s}\left\langle 0\left|\left[E_{rs}^{-},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle \kappa_{rs}\tag{Problem}$$

This can be rewritten as, by applying equation (12.5.22, MEST) and (2):

$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[\hat{\kappa},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle$$

Now by using the Jacobi identity:

$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle -\frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[\hat{H},\hat{\kappa}\right]\right]\right|0\right\rangle -\frac{1}{2}\left\langle 0\left|\left[\hat{H},\left[\hat{\kappa},E_{pq}^{-}\right]\right]\right|0\right\rangle$$

Now by using the relation, $\left[\hat{A},\left[\hat{B},\hat{C}\right]\right]=-\left[\left[\hat{B},\hat{C}\right],\hat{A}\right]$ and $\left[\hat{A},\hat{B}\right]=-\left[\hat{B},\hat{A}\right]$:

$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[\hat{\kappa},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[\left[\hat{\kappa},E_{pq}^{-}\right],\hat{H}\right]\right|0\right\rangle$$

Now:

$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\frac{3}{2}\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[\left[\hat{\kappa},E_{pq}^{-}\right],\hat{H}\right]\right|0\right\rangle$$

Now by $\left\langle 0\left|\left[\left[\hat{\kappa},E_{pq}^{-}\right],\hat{H}\right]\right|0\right\rangle =\left[^{\mathrm{o}}\overline{E}^{(1)},\overline{\kappa}\right]_{pq}$:

$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\frac{3}{2}\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left[^{\mathrm{o}}\overline{E}^{(1)},\overline{\kappa}\right]_{pq}$$

But this is NOT the right result in MEST the right result is giving as:

$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\left[^{\mathrm{o}}\overline{E}^{(1)},\overline{\kappa}\right]_{pq}\tag{12.5.21, MEST}$$

I cannot seem to figure out where I did my maths wrong and ended up with a factor 1/2 and factor 3/2.

As a side note the book explain the above math as:

where te elements of $\overline{\overline{K}}^{(2)}$ are defined in (12.2.41)-(12.2.43). Inserting these expressions in (12.5.17) and carrying out some simple algebra, we obtain

Since it is supposed to be simple, I might have forgot about something basic.