In the entire book, perturbation theory is used as a qualitative tool to rationalise some chemical phenomena. The authors write that

$$\psi_i = \sum_\mu{T_{ji}\psi_j^\circ}\tag{1}$$

The proof of the derivation for $T_{ji}$ involves the following statement:

$$T_{ji} = \left(\mathbf{C}_j^\circ\right)^\mathrm{T}\mathbf{S}^\circ\mathbf{C}_i \tag{2}$$


$$\left(\mathbf{C}_j^\circ\right)^\mathrm{T} = \left(C_{1j}^\circ \quad C_{2j}^\circ \quad \cdots \quad C_{mj}^\circ\right)\tag{3}$$

(indexing the coefficients for the jth unperturbed MO), and

$$ \mathbf{C}_i = \begin{pmatrix} C_{1i} \\ C_{2i} \\ \vdots \\ C_{mi} \end{pmatrix} \tag{4} $$

(indexing the coefficients for the $i$th perturbed MO); $\mathbf{S},$ the overlap matrix, is defined as usual.

I'm not quite sure how the aforementioned statement is obtained. Could someone please walk me through how

$$T_{ji} = \sum\sum C_{nj}^\circ S_{nm}C_{mi}?$$

Source: Orbital Interactions in Chemistry, 2nd edition, page 794-802

  • $\begingroup$ Please utilize MathJax for typesetting math and chemical formulas whenever possible: this makes them searchable and reduces traffic (especially if images are large). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, see MathJax basic tutorial and quick reference. $\endgroup$
    – andselisk
    Commented Nov 17, 2019 at 15:03
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    $\begingroup$ This equation (as written) is wrong; it's not actually in the book. The (correct) formula is inversely proportionate to $E_j^{(0)} - E_i^{(0)}$. It is a standard result in quantum mechanics and I suggest looking in a QM textbook of choice. Or the Internet. There are lots of good websites with the full derivation - search "perturbation theory". $\endgroup$ Commented Nov 17, 2019 at 15:05
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    $\begingroup$ Also, please add a complete reference for the textbook you are quoting from, and format quoted portions of the text as such with >. $\endgroup$
    – andselisk
    Commented Nov 17, 2019 at 15:05
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    $\begingroup$ @orthocresol it's in Appendix 1, and I've reflected that in an edit. $\endgroup$ Commented Nov 17, 2019 at 15:39
  • $\begingroup$ Ah, ok, I take that back then, sorry. It is confusing because in the actual text they have $t_{ji}$ which has a different meaning... $\endgroup$ Commented Nov 17, 2019 at 22:18

2 Answers 2


Rather than focusing on the individual matrix entries, it may be more intuitive to look at the matrix $\mathbf{T}$ formed by putting together all entries $T_{ij}$. In this case, $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$, where $\mathbf{C}$ is the coefficient matrix with column $k$ denoting how atomic orbitals mix together into molecular orbital $k$.

Now the point is that molecular orbitals, or the columns of $\mathbf{C}$, are orthonormal and are eigenfunctions of $H_{eff}$ (see page 34). This orthonormality makes sense if you think of the Hamiltonian matrix, which is certainly self-adjoint, and solving the eigenproblem is equivalent to diagonalizing the matrix with an orthonormal basis of eigenvectors.

Hence, $\mathbf{C}$ is unitary, and $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$ can be simply seen as a change-of-basis transformation done on the overlap matrix from atomic basis to the molecular orbital basis.

I hope this rather mathematical perspective may be of help to you. Thanks!


It look like the expression is just the equivalent way of expression the matrix product. I've attached two pictures that might help.maytrix-pic2

matrix pic1

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    $\begingroup$ I think you've misunderstood my question; I don't understand the rationalise behind the matrix expression itself, and had simply restated it in terms of the double summation. Thanks for the help, though! $\endgroup$ Commented Nov 18, 2019 at 12:58

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