In Section 2.3.6 of Szabo & Ostlund's Modern Quantum Chemistry, the exchange integral has the form of

$$\int \mathrm{d}\mathbf{r}_1\,\mathrm{d}\mathbf{r}_2\, \psi_a^*(\mathbf{r}_1) \psi_b(\mathbf{r}_1)\frac{1}{r_{12}} \psi^*_a(\mathbf{r}_2) \psi_b(\mathbf{r}_2) $$

or $\langle ab|ba\rangle$ in physicists' notation. According to this textbook (and many other books), $\langle ab|ba\rangle$ is positive. The conclusion seems obvious but I just cannot find a proof. Is there any simple reason that $\langle ab|ba\rangle$ must be positive?


6 Answers 6


The conclusion seems obvious but I just cannot find a proof.

No, I don't think it is obvious. At least I could not think about a simple algebraic proof that the exchange integral $\langle ab \vert ba \rangle$ defined as follows, $$ \langle ab \vert ba \rangle := \iint \overline{\psi}_a(\vec{r}_{1}) \overline{\psi}_b(\vec{r}_{2}) r_{12}^{-1} \psi_b(\vec{r}_{1}) \psi_a(\vec{r}_{2}) \mathrm{d}\vec{r}_{1} \mathrm{d}\vec{r}_{2} $$ is always positive. But I could think about relatively simple physical argument that proves that at least the sum of all the exchange integrals is positive. The argument is based on the very notion of a variation principle and goes as follows.

  1. We know that exchange integrals arise only if we use an antisymmetric product of orbitals as our trial wave function. If we use just a simple product of them there will be just Coulomb integrals in the energy expression.
  2. Now since the true wave function is antisymmetric, if we use an antisymmetric product of orbitals as our trial wave function instead of a simple product of them, we are guaranteed to get the lower energy since we use the qualitatively right wave function.
  3. Exchange integrals enter the energy expression with the negative sign, thus, they have to be positive, otherwise we won't get the lower energy by using an antisymmetric product of orbitals instead of a simple product.

Again, it does not prove that each and every exchange integral is positive, only that their sum is.

  • 2
    $\begingroup$ Cool. If we apply this argument to a two-electron system, e.g. He or H2, then we can "prove" the exchange integral is positive? $\endgroup$
    – Molec
    Nov 7, 2015 at 19:51
  • 1
    $\begingroup$ @Molec, well, yes. :D $\endgroup$
    – Wildcat
    Nov 7, 2015 at 20:25
  • 3
    $\begingroup$ conceptually it makes sense. That means individual exchange term may be positive or negative but the sum must be positive as coulomb integral is obviously repulsive (or positive). $\endgroup$ Jul 13, 2016 at 2:42
  • $\begingroup$ This is completely incorrect because point 2 is completely incorrect. The variational principle applies to only to actual solutions to the Hamiltonian. Arbitrary functions that are not can take any value. (In fact the 2RDM method converges to the correct answer from below, a sort of anti-variational principle, as the constraints of the true wave function are better satisfied) $\endgroup$
    – levineds
    Nov 11, 2017 at 1:53
  • $\begingroup$ @levineds, no, I dont think that you are correct. A Hartree product is basically a diagonal Slater determinant, so the space of Hartree products is just the subspace of the space of Slater determinants, and, as a result, variationally you can't get energy lower with a Hartree product than with a Slater determinant. Can you quote any source which states otherwise? $\endgroup$
    – Wildcat
    Nov 11, 2017 at 22:55

Pilar offers this explanation in his book Elementary Quantum Chemistry (paraphrased):

$\psi_a$ and $\psi_b$ are orthogonal so the function $f(\vec{r})=\psi_a(\vec{r})\psi_b(\vec{r})$ must have positive and negative regions with equal volume.

If $f(\vec{r}_1)$ and $f(\vec{r}_2)$ have the same sign, $f(\vec{r}_1)r^{-1}_{12}f(\vec{r}_2)$ will make a positive contribution to the integral

The smaller $r_{12}$ is, the more $f(\vec{r}_1)r^{-1}_{12}f(\vec{r}_2)$ will contribute to the integral

The smaller $r_{12}$ is, the more likely it is that $f(\vec{r}_1)$ and $f(\vec{r}_2)$ will have the same sign.

So positive contributions to the integral will be larger than the negative contributions, resulting in a positive exchange integral.

  • $\begingroup$ This answer is also not strictly correct. See Theo Jacobson's answer below for a counterexample to this argument. Points 3 and 4 require much more careful treatment to make this argument valid. $\endgroup$
    – levineds
    Nov 11, 2017 at 1:56
  • $\begingroup$ yes, $V$ probably has be to be some inverse function of $r$ ($V(r) = 1/r^x$ where $x > 0$) for this to work. $\endgroup$
    – Jan Jensen
    Nov 11, 2017 at 14:50

In a nut-shell: exchange integrals are two-electron integrals, and two-electron integrals yield positive values. Note that the "kind" or "meaning" of the input functions is irrelevant, because in practice, you will always have linear combinations of primitives, and in most cases gaussians. For the proof of the claim about positive values, I will defer to the experts, [1, HJO] who cite previous work. [2] As taken from the book:

The two-electron intergrals can be viewed as a matrix with the electron distributions [($\Omega_{ab}, \Omega_{cd}$)] as row and column labels [using AO labels $a,b,c,d$, see above] $$ g_{abcd} = \int \int \frac{\Omega_{ab}(\mathbf{r}_1) \Omega_{cd}(\mathbf{r}_2)}{r_{12}} \mathrm{d}\mathbf{r}_1 \mathrm{d}\mathbf{r}_2 $$ Assuming that the orbitals are real, we shall demonstrate that this matrix is positive definite [2]. Let us consider the interaction between two electrons in the same distribution $\rho(\mathbf{r})$: $$ I[\rho] = \int \int \frac{\rho(\mathbf{r}_1) \rho(\mathbf{r}_2)}{r_{12}} \mathrm{d}\mathbf{r}_1 \mathrm{d}\mathbf{r}_2 $$ Inserting the Fourier transform of the interaction operator $$ \frac{1}{r_{12}} = \frac{1}{2\pi^{2}} \int k^{-2} \exp[\mathrm{i}\mathbf{k} \cdot(\mathbf{r}_1 - \mathbf{r}_2)] \mathrm{d}\mathbf{k} $$ and carrying out the integration over the Cartesian coordinates, we obtain $$ I[\rho] = \frac{1}{2\pi^{2}} \int k^{-2} \vert \rho(\mathbf{k}) \vert^2 \mathrm{d}\mathbf{k} \quad\quad \text{(eq. 4)} $$ where we have introduced the distributions $$ \rho(\mathbf{k}) = \int \exp(-\mathrm{i}\mathbf{k}\cdot\mathbf{r}) \rho(\mathbf{r}) \mathrm{d}\mathbf{r} $$ Since the integrand in [(eq. 4)] is always positive or zero, we obtain the inequality $$ I[\rho] > 0 $$

HJO go on to expand the charge distribution $\rho$ in one-electron orbital distributions and get back to the original $g_{abcd}$, noting afterwards that two-electrons thus satisfy the conditions for inner products, in a metric defined by $r^{-1}_{12}$. Therefore, Schwarz-style inequalities hold and are used extensively in integral screening to throw out insignificant integrals before evaluating them.

[1] T Helgaker, P Jørgensen, J Olsen, Molecular Electronic-Structure Theory, Wiley (2002), p. 403f.

[2] CCJ Roothaan, Rev. Mod. Phys., 23, 69 (1951).


I think neither Wildcat's nor Jan Jensen's answers are satisfactory. In Wildcat's answer, point 2 essentially assumes the result. In Jan Jensen's answer, the role of the magnitude of $f(\vec r)$ is not controlled for, so the conclusion doesn't follow by the logic given.

I found myself at this page because the only proofs of positivity I could find refer to the particular form of the potential, which puzzled me. I thought perhaps the result should be true if we were to replace $1/r_{12}$ by any positive function that is monotonically decreasing as $r_{12}$ grows. However I've constructed a simple finite dimensional example that leads me to doubt that even that is true.

In the finite dimensional analogy, the possible points of space are replaced by just three points. Then consider the following for the analog of the potential $V$ and the function $f=\psi_a(i)\psi_b(i)$: $$V=\left(\begin{array}{lll} 1 &1&1/2\\1&1&1\\1/2&1&1\end{array}\right), \qquad f=\left(\begin{array}{l} 1 \\-2\\1\end{array}\right).$$ Note that this $f$ has the required property $\sum_i f_i = 0$, which follows from the orthogonality of the two wave functions (which could be taken as $(1,\pm\sqrt{2},1)$). Also, the potential has the property of decreasing away from the diagonal. And, yet, we have $f^T V f = -1 <0$. Also, while the potential $V$ above isn't strictly decreasing away from the diagonal, we could replace the off diagonal 1's by 0.9 and again we'd get a negative value, $f^T V f = -0.2 <0$. I think this counterexample shows that the explanations given above are not valid, and it strongly suggests that, to some degree, the specific form of $1/r_{12}$ is important for the result.


If you combine the fact that the exchange integral is a sum of terms denoted by $G^k$, with positive coefficients,1 and Racah's proof that the $G^k$ are positive,2 then the positivity of the exchange integral follows. That the exchange integral is unlikely to be negative was suggested by Bacher in 1933.3


  1. Condon, E. U.; Shortley, G. H. The Theory of Atomic Spectra. Cambridge University Press, Cambridge, U.K., 1935; p 176.

  2. Racah, G. Phys. Rev. 1942, 62 (9–10), 438–462. DOI:10.1103/PhysRev.62.438. The proof is on p 460.

  3. Bacher, R. F. Phys. Rev. 1933, 43 (4), 264–269. DOI:10.1103/PhysRev.43.264.


I will ofer just a simple, qualitative, resoning. In the expresion $\langle ab \vert ba \rangle := \iint \overline{\psi}_a(\vec{r}_{1}) \overline{\psi}_b(\vec{r}_{2}) r_{12}^{-1} \psi_b(\vec{r}_{1}) \psi_a(\vec{r}_{2}) \mathrm{d}\vec{r}_{1} \mathrm{d}\vec{r}_{2}$

The term $\overline{\psi}_a(\vec{r}_{1}) \psi_b(\vec{r}_{1})$ represents the "overlap" of $\psi_a$ and $\psi_b$, while, $\overline{\psi}_b(\vec{r}_{2})\psi_a(\vec{r}_{2}) $ also represents the "overlap" of $\psi_a$ and $\psi_b$. Thus, we get a square of the "overlap" of $\psi_a$ and $\psi_b$, which has to be positive and smaller then Coulomb integral ("overlap" with itself). Also, if you are wondering why it even has to be real, $\overline{\psi}_a\psi_b$ and $\overline{\psi}_b\psi_a$ are Hermitian conjugates of each other.


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