As the wikipedia article for the exchange interaction so aptly notes, exchange "has no classical analogue."

How wonderful.

Exchange shows up essentially while enforcing the condition that two electrons should not be distinguishable.

That makes sense as a condition to enforce, but I don't understand how this should influence the energetics of the system.

For instance, when two neon atoms interact, there is a very small amount of energy contribution which is attributed to exchange. I understand that exchange is not a force because there is no force carrier, but there is indeed an interaction taking place which affects the energy of the system. What then am I meant to believe is going on? Am I supposed to believe that two electrons spontaneously switch places so as to maintain their indistinguishability? This seems ridiculous. Especially because such a process ought to happen over a finite period of time and thus be observable, but this doesn't seem to be the case (at least as far as I've heard).

What is exchange then? I understand this may not even be an answerable question beyond the fact that it is an effect which drops out of the math and simply is.

Hopefully someone has some interesting insight for me here.

Also, I have seen this question which I believe is referencing the same effect because exchange is known to play a role in ferromagnetism, but nothing ever really caught hold there probably because the question referenced a specific application and so was very difficult to answer well.

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    $\begingroup$ partly related chemistry.stackexchange.com/q/58625/16683 If I'm not wrong the interaction between Ne atoms can be understood as an instantaneous correlation of electrons - ie van der Waals interaction $\endgroup$ Oct 18, 2016 at 7:59
  • $\begingroup$ This is really a quantum physics question; there is a difference between the energy of a system with two distinguishable elements and a system with two indistinguishable elements, simply because there are fewer combinations of the latter. This wouldn't mean anything in a classical system, but that's just how the world works - our classical assumptions are simply wrong. There is no classical analogue because in the classical models, there is no true indistinguishability. In reality, two particles can be in a configuration that isn't factorizable - and this has a (predicted!) measurable effect. $\endgroup$
    – Luaan
    Oct 19, 2016 at 7:50

2 Answers 2


In quantum chemistry, probably the easiest way to understand the "exchange interaction" is within the context of the Hartree-Fock model. $ \newcommand{\op}{\hat} \newcommand{\el}{_\mathrm{e}} \newcommand{\elel}{_\mathrm{ee}} \newcommand{\elnuc}{_{\mathrm{en}}} \newcommand{\core}{^{\mathrm{core}}} \newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle} $

To reduce the complexity of the electronic Schrödinger equation $$ \op{H}\el \psi\el(\vec{q}\el) = E\el \psi\el(\vec{q}\el) \, , $$ where $$ \op{H}\el = \op{T}\el + \op{V}\elnuc + \op{V}\elel = - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2} - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}} + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, , $$ we would like to separate the electronic coordinates from each other by writing down the many-electron wave function as a product of one-electron ones. Unfortunately, such separation won't work due to the presence of $\op{V}\elel$ term in the electronic Hamiltonian. But if instead of $\op{V}\elel$ potential, which prevents the separation of electronic coordinates, a model potential of the form $\sum\nolimits_{i=1}^{n} v_{\mathrm{MF}}(\vec{r}_{i})$ had entered the electronic Schrödinger equation, it would be reduced to a set of $n$ one-electron Schrödinger equations with the many-electron wave function being just a simple product of their solutions, one-electron wave functions $\psi_{i}(\vec{r}_{i})$.

More precisely, since we have to take spin of electrons and the principle of antisymmetry of the electronic wave function into account, the electronic wave function would be an antisymmetric product of one-electron wave functions $\psi_{i}(\vec{q}_{i})$ termed the Slater determinant, $$ \Phi = \frac{1}{\sqrt{n!}} \begin{vmatrix} \psi_{1}(\vec{q}_{1}) & \psi_{2}(\vec{q}_{1}) & \cdots & \psi_{n}(\vec{q}_{1}) \\ \psi_{1}(\vec{q}_{2}) & \psi_{2}(\vec{q}_{2}) & \cdots & \psi_{n}(\vec{q}_{2}) \\ \vdots & \vdots & \ddots & \vdots \\ \psi_{1}(\vec{q}_{n}) & \psi_{2}(\vec{q}_{n}) & \cdots & \psi_{n}(\vec{q}_{n}) \end{vmatrix} \, , $$ where the one-electron wave functions $\psi_{i}$ of the joint spin-spatial coordinates of electrons $\vec{q}_{i} = \{ \vec{r}_{i}, m_{si} \}$ are referred to as spin orbitals. Physically the model potential of the above mentioned form represent the case when electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons and given by $v_{\mathrm{MF}}(\vec{r}_{i})$.

Minimization of the electronic energy functional $$ E\el^{\mathrm{HF}} = \bracket{ \Phi }{ \op{H}\el }{ \Phi } = \bracket{ \Phi }{ \op{T}\el }{ \Phi } + \bracket{ \Phi }{ \op{V}\elnuc }{ \Phi } + \bracket{ \Phi }{ \op{V}\elel }{ \Phi } $$ with respect to variations in spin orbitals subject to constraint that spin orbitals remain orthonormal eventually results in the set of the so-called canonical Hartree-Fock equations which define canonical spin orbitals $\psi_i$ together with the corresponding orbital energies $\varepsilon_i$ $$ \op{F} \psi_{i}(\vec{q}_{1}) = ε_{i} \psi_{i}(\vec{q}_{1}) \, , \quad i = 1, \dotsc, n \, , $$ where $\op{F} = \op{H}\core + \sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$ is the Fock operator with $$ \op{H}\core \psi_{i}(\vec{q}_1) = \Big( - \frac{1}{2} \nabla_{1}^2 - \sum\limits_{\alpha=1}^{\nu} \frac{Z_{\alpha}}{r_{\alpha 1}} \Big) \psi_{i}(\vec{q}_1) \, , \\ \op{J}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{j}(\vec{q}_2) } \psi_{i}(\vec{q}_1) \, , \\ \op{K}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{i}(\vec{q}_2) } \psi_{j}(\vec{q}_1) \, . $$

Here, the first part of the Fock operator, $\op{H}\core$, is known as the one-electron core Hamiltonian and it is simply the Hamiltonian operator for a system containing the same number of nuclei, but only one electron. The second part of the Fock operator, namely, $\sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$, plays the role of the mean-field potential $v_{\mathrm{MF}}$ that approximates the true potential $\op{V}\elel$ of interactions between the electrons. The first operator $\op{J}_{j}$ rewritten in the following form $$ \op{J}_{j}(\vec{q}_{1}) = \int \psi_{j}^*(\vec{q}_{2}) r_{12}^{-1} \psi_{j}(\vec{q}_{2}) \,\mathrm{d} {\vec{q}_{2}} = \int \frac{ |\psi_{j}(\vec{q}_2)|^{2} }{ r_{12} } \mathrm{d} \vec{q}_2 \, , $$ can be clearly interpreted as the the Coulomb potential for electron-one at a particular point $\vec{r}_{1}$ in an electric field created by electron-two distributed over the space with the probability density $|ψ_{j}(\vec{q}_2)|^{2}$, and for this reason it is called the Coulomb operator. The second operator $\op{K}_{j}$ has no simple physical interpretation, but it can be shown to arise entirely due to anti-symmetry requirement, i.e. if instead of a Slater determinant one uses a simple product of spin orbitals, termed the Hartree product, there will be no $\op{K}_{j}$ terms in the resulting equations (the so-called Hartree equations). It is for that reason that $\op{K}_{j}$ is called the exchange operator.

To quickly understand why exchange terms appear when using a Slater determinant instead of a simple product of spin orbitals look no further than at the Slater rules. For $\op{V}\elel$ which is a two-electron operator we have

$$ \bracket{ \Phi }{ \op{V}\elel }{ \Phi } = \sum\limits_{i=1}^{n} \sum\limits_{j>i}^{n} \Big( \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{i}(1) \psi_{j}(2) } - \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{j}(1) \psi_{i}(2) } \Big) \, , $$ where the second (exchange) part would be absent if $\Phi$ were a simple product of spin orbitals, rather than a Slater determinant.1)

To the question on why do we interpret exchange terms in the Hartree-Fock as a manifestation of some sort of interaction. On the one hand, it must be clear from exchange terms $\op{K}_{j}$ being part of potential energy terms $v_{\mathrm{MF}}$, that they are indeed related to some type of interaction between electrons. However, it must be said that, unlike for the four fundamental interactions, there exists no force based on exchange interaction. Strictly speaking, I don't even think that the term "interaction" has a definite meaning in physics, unless it simply stands for the "fundamental interaction" in which case it is merely a synonym for "fundamental force".

Exchange interaction is not one of the four fundamental interactions, so the meaning of the word "interaction" here is a bit different than that for fundamental interactions. Oxford Dictionary defines interaction as follows,

1.1 Physics A particular way in which matter, fields, and atomic and subatomic particles affect one another, e.g. through gravitation or electromagnetism.

And exchange interaction indeed is a way by which electrons (of the same spin) "affect one another". This can be readily seen as follows. Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events:

  • $r_1$ - the event of finding electron-one at a point $\vec{r}_{1}$;
  • $r_2$ - the event of finding electron-one at a point $\vec{r}_{2}$.

Probability theory reminds us that in the general case the so-called joint probability of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by $$ \Pr(r_1 \cap r_2) = \Pr(r_1\,|\,r_2) \Pr(r_2) = \Pr(r_1) \Pr(r_2\,|\,r_1) \, , $$ where

  • $\Pr(r_1)$ is the probability of finding electron-one at point $\vec{r}_{1}$ irrespective of the position of electron-two;
  • $\Pr(r_2)$ is the probability of finding electron-two at point $\vec{r}_{2}$ irrespective of the position of electron-one;
  • $\Pr(r_1\,|\,r_2)$ is the probability of finding electron-one at point $\vec{r}_{1}$, given that electron-two is at $\vec{r}_{2}$;
  • $\Pr(r_2\,|\,r_1)$ is the probability of finding electron-two at point $\vec{r}_{2}$, given that electron-one is at $\vec{r}_{1}$.

The first two of the above probabilities are referred to as unconditional probabilities, while the last two are referred to as conditional probabilities, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are independent of each other.

If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts $$ \Pr(r_1\,|\,r_2) = \Pr(r_1) \, , \quad \Pr(r_2\,|\,r_1) = \Pr(r_2) \, , $$ and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities, $$ \Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2) \, . $$ In reality, however, the events $\vec{r}_{1}$ and $\vec{r}_{2}$ are not independent of each other, because electrons "affect one another", $$ \Pr(r_1\,|\,r_2) \neq \Pr(r_1) \, , \quad \Pr(r_2\,|\,r_1) \neq \Pr(r_2) \, , $$ and consequently the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ is not equal to the product of unconditional probabilities, $$ \Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2) \, . $$

Inequalities above hold true for two reasons, i.e. there are two ways electron "affect one another".

First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons $\Pr(r_1\,|\,r_2) < \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) < \Pr(r_2)$, while at large distances $\Pr(r_1\,|\,r_2) > \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) > \Pr(r_2)$. In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.

Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above. This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the latter, it is still noticeable when electrons are close to each other and not just at the very same location in space.

Now, for a Hartree product wave function $$ \psi_{\mathrm{HP}}(\vec{q}_{1}, \vec{q}_{2}) = \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) \, , $$ one can show that $\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ if two electrons are in different spin states as well as when they are in the same spin state2). But for a Slater determinant $$ \Phi(\vec{q}_{1}, \vec{q}_{2}) = \frac{1}{\sqrt{2}} \Big( \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) - \psi_{1}(\vec{q}_{2}) \psi_{2}(\vec{q}_{1}) \Big) \, , $$ $\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ will hold only for electrons of unlike spin, while when both electrons are in the same spin state $\Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2)$2). And this inequality is a clear indication of exchange interaction.

1) I leave it to OP as an exercise. Derivation for a simple product of spin orbitals is pretty trivial, while for the case of a Slater determinant OP can consult, for instance, Szabo & Ostlund, Modern Quantum Chemistry, Section 2.3.3.

2) I leave it as another exercise.

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    $\begingroup$ Very nice summary on HF theory : ) But one thing that does not come clear to me yet is how one can understand the exchange 'interaction' between systems as an interaction. Part of the question was if there is some way in which we can kind of "visually" understand the process that is responsible for this energy. Is that possible? $\endgroup$ Oct 18, 2016 at 14:13
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    $\begingroup$ @ketbra, exchange interaction is not an interaction in the same sense as the four fundamental interaction are: there is no exchange force, so that the word "interaction" in "exchange interaction" must not be though as a synonym to "force". Rather, I think, "interaction" here is a synonym to "interdependence" which can be demonstrated by working out the probabilities as I did in the answer. $\endgroup$
    – Wildcat
    Oct 18, 2016 at 16:04
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    $\begingroup$ @ketbra You're trying to understand this from a classical perspective. There is no analogue. You simply have to look at the math and say, yes, there is a contribution from this term, and this term is the exchange piece. $\endgroup$
    – Zhe
    Oct 18, 2016 at 17:19

There are three different integrals commonly called the ‘exchange’ integral, which are the resonance integral, the exchange integral itself and the exchange operator which is also an integral. These integrals are related to the Coulomb integral and are conventionally one is called J the other K or vice versa; textbook authors differ. Almost all authors are coy about giving an explanation as to what the exchange integral represents. Those few that do are confusing so you are not in alone in wanting to understand this.

Resonance integral. This occurs in the description of $\ce{H_2^+}$ and is also sometimes called the exchange integral. In the variational solution to this problem the equations to be solved are

$$\begin{vmatrix} H_{11} - E & H_{12} - ES \\ H_{12} - ES & H_{22}-E \end{vmatrix} $$

where E is the energy and $H_{ij} $ and S are integrals. The $\ce{H_2^+}$ wavefunctions $\psi$ are found as linear combinations of 1s atomic orbitals $\psi = c_1\phi_1 \pm c_2 \phi_2$ where $c_1$ and $c_2$ are coefficients.

The overlap integral is $S=\int {\phi_1 \phi_2 ~d\tau} $ and $H_{11} = \int\phi_1~H ~\phi_1 ~d\tau$ and similarly for $H_{22}$. H is the operator $H=p^2/(2m) +V_{en} + V_{nn}$ where p is momentum and the V are electron-proton (e-p) and proton-proton (p-p) interactions and are of the form $e^2/r$ where r is the appropriate separation, e-p or p-p.

Expanding the $H_{11}$ integral has a term that produces the Coulomb integral $$C = \int \phi_1 \frac{e^2}{r_2}\phi_1 ~d\tau$$ which we interpret as the electrostatic potential between proton 2 and the charge distribution of the electron associated with the 1s orbital around proton 1. The integral $H_{22}$ is explained similarly.

Expanding the $H_{12}$ integral produces the exchange or resonance integral,

$$A=\int \phi_1 \frac{e^2}{r_1} \phi_2 ~d\tau$$

which like the Coulomb integral is positive and is where the fun/confusion begins.

If the $H_{12}$ terms were zero, the ground state would be doubly degenerate because by symmetry $H_{11} = H_{22}$. But as the $H_{12}$ are not zero the states based on orbitals $\phi_1$ and $\phi_2$ are not stationary states of the molecule and thus a lower energy is obtained because of the coupling $H_{12}$ between $H_{11}$ and $H_{22}$. A linear superposition of $\phi_1$ and $\phi_2$ has this lower energy.

Some authors (Cohen-Tannoudji, Diu & Laloe , Quantum Mechanics) call this ‘Quantum Resonance’ and describe the fact that $H_{12}$ is not zero as the possibility of the electron ‘jumping’ from the neighbourhood of one proton to that of the other. ‘The electron ‘oscillates’ in time between the two sites under the influence of $H_{12}$ which is responsible for quantum resonance’.

Coulson (Valence) also describes resonance in this way but then warns that it is easy to take the analogy too far and states that there is no possible physical mechanism whereby the electron could oscillate. He then goes on to say that integrals of this form are a consequence of making a linear combination to describe the molecular orbital. So it seems that it has no physical meaning other than it describes that coupling between the two isolated states $\phi_1$ and $\phi_2$ that we started with, and which by forming a molecular orbital lowers the energy compared to these states.

In multi-electron atoms and molecules similar integrals occur but now are mathematically a more complicated but the interpretation, such as it is, remains the same. In the simplest case of two electrons a and b in i and j spatial orbitals where each orbital can accommodate two electrons, the Coulomb integral is

$$ J=\int \int \phi_i (a) \phi_j (b) \frac{e^2}{r_{ab}} \phi_i (a) \phi_j (b) ~d\tau_a d\tau_b = \left\langle ij \middle |ij\right\rangle $$ the Coulomb operator is the integral $$\int \phi_j(b)\frac{e^2}{r_{ab}}\phi_j(b)~d\tau_b $$ The exchange integral, as its name suggests, swaps electrons a and b between orbitals i and j and is $$ K=\int \int \phi_i (a) \phi_j (b) \frac{e^2}{r_{ab}} \phi_j (a) \phi_i (b) ~d\tau_a d\tau_b = \left\langle ij\middle |ji\right\rangle $$ where the exchange operator is $$\int \phi_j(b)\frac{e^2}{r_{ab}}\phi_i(b)~d\tau_b $$ and operates on one of the $\phi_i(a)$ functions.

Coulomb and exchange integrals are encountered elsewhere for example in the determining the energy of spectral terms $\mathrm{^1S}$ and $\mathrm{^3S}$ arising from the 1s2s configuration in Helium. The energy of the two levels is $J\pm K$. The Coulomb term J raises the energy of both energy levels and K splits them. The underlying reason for the energy splitting resides in the respective symmetry of the spatial and spin wavefunctions.

  • $\begingroup$ I was just rereading this and I really like the view of exchange as a coupling between states. That has a very nice interpretation for at least something like $\ce{H2^+}$. Thanks for that insight. $\endgroup$
    – jheindel
    Oct 7, 2018 at 23:59

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