Is it possible for the kinetic-energy integral, Tij, to be negative? I was messing around with some HF code and found that the integral became negative on some off-diagonal terms. (This could also be a bug with the code.)

If the kinetic energy integral is negative, what would the physical meaning be?

  • 3
    $\begingroup$ Page 126 of this book and the following pages suggest that negative Tij values are reasonable and shows them occuring for a few example calculations. $\endgroup$
    – Tyberius
    Commented Sep 15, 2019 at 18:36

2 Answers 2


Generally, the off-diagonal elements of matrices, especially in physical systems, are interpreted as the coupling between whatever the $i$ and $j$ elements correspond to. So, in this case, a negative element of $T_{ij}$ corresponds to a negative kinetic coupling between atomic orbital basis functions. In plain language, this means that basis functions $i$ and $j$ tend to mutually lower the kinetic energy of an electron placed in one of those orbitals. I am not sure it really makes sense to give much of an interpretation to this, however, because what one is really interested in is diagonalizing the Fock matrix, of which the kinetic energy is only one part.

Also, it is always possible to choose a basis where the kinetic energy matrix is diagonal, but again, this would not get you anywhere as in solving the HF problem, you will diagonalize the Fock Matrix which will surely put you in a basis in which the kinetic energy matrix is non-diagonal. What might be more interesting would be to take the actual molecular orbitals which are part of the solutions of the Roothan-Hall equations and re-compute the kinetic energy matrix over these orbitals. Looking at the coupling between these orbitals may be more easily interpretable for e.g. aromatic $\pi$-systems.


@jheindel has given an insightful answer. I want to supplement it with some mathematical analysis. Szabo and Ostlund note that the kinetic energy integral over two primitive $s$-Gaussians $A,B$ is equal to$^{[1]}$ $$ \left(A \left| - \frac{1}{2} \nabla^2 \right|B\right) = \frac{\alpha\beta}{\alpha+\beta} \left[ 3 - \frac{2\alpha\beta}{\alpha+\beta} \left| \mathbf{R}_A - \mathbf{R}_B \right|^2 \right] \cdot S_{AB} $$ where $\alpha, \beta$ are the Gaussian exponents, $\mathbf{R}_A, \mathbf{R}_B$ are the centers of the Gaussians, and $S_{AB}$ is the relevant overlap integral, which we know is greater or equal to zero. The term inside the brackets can become negative, rendering the entire integral negative.

For more insight, one may do the following:

  1. Graph a one-dimensional Gaussian. Observe that it is positive everywhere. (The results translate to three dimensions.)
  2. Apply the kinetic energy operator to the Gaussian from step 1 and graph it. Observe that there are negative regions.
  3. Multiply the result from step 2 with a second Gaussian. Observe that depending on the center of the second Gaussian, one can amplify the negative sections and suppress the positive sections of the result of step 2. By taking the integral over all space, one can obtain a negative total value.

[1] A. Szabo, N. Ostlund: Modern Quantum Chemistry, Dover Publications, 1st. edition, revised, 1996, page 412.


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