I've been reading about virtual particle exchanges in physics books and in Physics SA posts, where a particle interpretation of gravity and Coulomb interaction is established. The Feynman Diagram picture (combined with the snowball picture that a fellow contributor provided in a related Phys.SE post) serves to account for the repulsion part: grosso modo two particles in motion, a virtual photon/graviton, momentum change induces opposite motions ($\Delta \vec p \ $ and $ \ -\Delta \vec p$). Still, how can we account for the attraction (as a layman's analogy if not in formalism)?


2 Answers 2


The short answer is the Heisenberg uncertainty principle allows for the attraction. Suppose you have two opposite charges, and the one on the left emits a virtual photon with momentum directed leftwards. The left charge begins to move towards the charge on the right. Now, where's the virtual photon? It's momentum is some exact value directed left, so classically we expect that at a later time we will find the virtual photon at a new position further to the left of the original position of the charge that emitted it.

But because of the HUP, if it is in a state of precise momentum, then it's position becomes infinitely uncertain. Hence, there's an equal chance of finding the new position of left-momentum photon to be on the right of where it was emitted (i.e., where the other charge is located) as there is of finding it to the left. And that happens, the right charge can absorb that virtual photon and it's leftwards momentum, and lo the two charges are moving towards each other.

Now there's a new, weirder, problem. What is it then that distinguishes between attraction and repulsion? It seems like two particles have an equal chance of doing either, regardless charge. The resolution of this in QFT defies analogy to anything we are familiar with, and hinges on the fact that quantum particle states are described by a wave-function instead of a set of coordinates like we do with classical particles. The parity of the wave-function for our two charges will be either even or odd, depending on whether the charges have equal or opposite sign, respectively. When the wave-function of the virtual photon interferes with that of the charges, the pattern of constructive and destructive interferences that results depends crucially on whether the charges' wave-function is even or odd. In the even case, there is more destructive interference between the two (like) charges and constructive interference outside this region, meaning the charges are now more likely to be found in positions with a wider separation. The opposite happens if the particles have opposite charge.

The quantum mechanical phenomenon underlying the macroscopic effect we observe and describe as attraction repulsion is so much farther removed from the classical conceptions of the process than any sane physicist ever would have credited. For me, my first encounter with the ideas described above was when I really started to believe Bohr who famously said, " Those who are not shocked when they first come across quantum theory cannot possibly have understood it." In QM you don't understand things. You just get used to them.

  • 5
    $\begingroup$ "you just get used to them" i.e. you develop an intuition. For classical mechanics we have been developing our intuitions from the time we were babies. It does take time to reach a stage where one can "guess" a quantum mechanical behavior. $\endgroup$
    – anna v
    Oct 8, 2013 at 3:28

Let us reformulate OP question as follows:

If the force between 2 charges $q_1$ and $q_2$ in QFT is mediated by a virtual force-carrying particle going back and forth between the charges, cf. Fig. 1, wouldn't the momentum conservation in each collision/vertex always lead to repelling charges? How can such theory ever explain attraction?

 --------------------------- q_1
     |      /    |   \
     |     /     |    \
     |    /      |     \
     |   /       |      \
 --------------------------- q_2

$\uparrow$ Fig. 1. A typical ladder diagram. The 2 sides constitutes 2 charges $q_1$ and $q_2$ and the rungs are virtual force-carrying particles.

That's a good question!

TL;DR: The somewhat oversimplified answer is that there are also virtual force-carrying (anti)particles running backwards in time that pull rather than push [1]. Moreover, QFT treats attractive and repulsive interactions in a unified manner in the sense that the Feynman rule for an interaction vertex is proportional to the charge, and this works in a similar manner for each sign of the charge.

Let us try to illustrate this for a simple QFT model:

  1. We will for simplicity assume that the force-carrying field is spin-0/a scalar. [It turns out that this will reverse what is repulsive and what is attractive relative to the ordinary E&M spin-1 interaction, i.e. opposite charges repel and like changes attract [2], but the important point is here that there will still be an attractive case to explain!]

  2. We will for simplicity model charges as point particles rather than fields.

  3. We will Wick rotate to Euclidean spacetime to avoid singular propagators.

The partition function becomes$^1$

$$\begin{align} e^{-\frac{1}{\hbar}W_c[J]}~=~~& Z[J]\cr ~=~~&\int \! {\cal D}\phi~\exp\left\{\frac{1}{\hbar} \int\! d^4x \left(-\frac{1}{2}\phi(-\partial_t^2 -\vec{\nabla}^2+m^2)\phi +J\phi\right)\right\} \cr ~\stackrel{\begin{array}{c}\text{Gauss.}\cr\text{int.}\end{array}}{\sim}& \exp\left\{\frac{1}{2\hbar}\int\! d^4x\int\! d^4x^{\prime}~J(x)G(x\!-\!x^{\prime})J(x^{\prime}) \right\}\tag{1}\cr ~\stackrel{\begin{array}{c}\text{Four.}\cr\text{transf.}\end{array}}{\sim}& \exp\left\{\frac{1}{2\hbar}\int\! dt \frac{d^3k}{(2\pi)^3} \int\!dt^{\prime}\frac{d^3k^{\prime}}{(2\pi)^3}~\tilde{J}(\vec{k},t)\tilde{G}(\vec{k}\!-\!\vec{k}^{\prime},t\!-\!t^{\prime})\tilde{J}(\vec{k}^{\prime},t^{\prime}) \right\} \tag{2}\cr ~\stackrel{\begin{array}{c}\text{Four.}\cr\text{transf.}\end{array}}{\sim}& \exp\left\{\frac{1}{2\hbar}\int\! \frac{d\omega}{2\pi}d^3r \int\!\frac{d\omega^{\prime}}{2\pi}d^3r^{\prime}~\hat{J}(\vec{r},\omega)\hat{G}(\vec{r}\!-\!\vec{r}^{\prime},\omega\!-\!\omega^{\prime})\hat{J}(\vec{r}^{\prime},\omega^{\prime}) \right\},\tag{3} \end{align} $$

where the 2-point correlation function/Greens function is

$$\begin{align} G(x)~=~&\frac{m}{4\pi^2|x|} K_1(m|x|), \cr (-\partial_t^2 -\vec{\nabla}^2+m^2)G(x)~=~&\delta^4(x),\end{align}\tag{1}$$

$$\begin{align}\tilde{G}(\vec{k},t)~=~& -\frac{\sinh(|t|\sqrt{k^2+m^2})}{2\sqrt{k^2+m^2}}, \cr (-\partial_t^2 +k^2+m^2)\tilde{G}(\vec{k},t)~=~&\delta(t), \end{align}\tag{2}$$

$$\begin{align}\hat{G}(\vec{r},\omega) ~=~&\frac{e^{-r\sqrt{\omega^2+m^2}}}{4\pi r}, \cr (-\vec{\nabla}^2+\omega^2+m^2)\hat{G}(\vec{r},\omega)~=~&\delta^3(\vec{r}), \end{align}\tag{3}$$

We can already make 2 observations:

  1. From eq. (2) we see that 3-momentum is indeed conserved at each vertex.

  2. From eq. (3) we start to recognize the shape of a Yukawa potential, cf. e.g. my Phys.SE answer here.

Both observations become clearer in the following.


$\uparrow$ Fig. 2. Only 1 Feynman diagram contributes to our simple QFT model. It consists of 2 sources and 1 propagator. The 2 charges $q_1$ and $q_2$ share ALL their 3-momenta with a virtual force-carrying particle.

  • Given 2 Dirac delta function point-sources in position space $$\begin{align} J(\vec{r},t)~=~&q_1\delta^3(\vec{r}\!-\!\vec{r}_1(t)) +q_2\delta^3(\vec{r}\!-\!\vec{r}_2(t)),\tag{1}\cr \tilde{J}(\vec{k},t)~=~&q_1 e^{i\vec{k}\cdot\vec{r}_1(t)} +q_2e^{i\vec{k}\cdot\vec{r}_2(t)},\tag{2} \end{align}$$ the time-integrated free energy (excluding self-interactions) is $$\begin{align} -W_c[J]~\sim~&\frac{1}{2}\int\! d^4x\int\! d^4x^{\prime}J(x)G(x\!-\!x^{\prime})J(x^{\prime}) \cr ~=~&q_1q_2 \int\! dt_1\int\! dt_2~G(\vec{r}_1(t_1)\!-\!\vec{r}_2(t_2),t_1\!-\!t_2) \tag{1}\cr ~\approx~& q_1q_2 \int\! d\bar{t}~\underbrace{\hat{G}(\vec{r}_1(\bar{t})\!-\!\vec{r}_2(\bar{t}),\omega\!=\!0)}_{\text{Yukawa potential}},\tag{3} \end{align}$$ where in the last equality we (i) assumed that the 2 point charges are heavy/immobile, and we (ii) changed time-coordinates $$ \bar{t}~:=~\frac{t_1+t_2}{2}, \qquad \Delta t~:=~t_1-t_2. $$

Note that a point-charge localized in position space is completely delocalized in momentum space, and vice-versa, in accordance with the HUP.

  • Given 2 Dirac delta function point-sources in momentum space $$\begin{align} \tilde{J}(\vec{k},t)~=~&q_1(2\pi)^3\delta^3(\vec{k}\!-\!\vec{k}_1(t)) +q_2(2\pi)^3\delta^3(\vec{k}\!-\!\vec{k}_2(t)),\tag{2}\cr J(\vec{r},t)~=~&q_1 e^{-i\vec{r}\cdot\vec{k}_1(t)} +q_2e^{-i\vec{r}\cdot\vec{k}_2(t)},\tag{1} \end{align}$$ the time-integrated free energy (excluding self-interactions) is $$\begin{align} -W_c[J]~\sim~&\frac{1}{2}\int\! d^4x\int\! d^4x^{\prime}J(x)G(x\!-\!x^{\prime})J(x^{\prime}) \cr ~=~&q_1q_2 \int\! dt_1\int\! dt_2~\tilde{G}(\vec{k}_1(t_1)\!-\!\vec{k}_2(t_2),t_1\!-\!t_2) \tag{2}\cr ~\approx~& q_1q_2 \int\! d\bar{t}~\frac{1}{(\vec{k}_1(\bar{t})-\vec{k}_2(\bar{t}))^2+m^2}. \end{align}$$

So what are the conclusions? Well, as predicted the sign of the charges (and hence attractive and repulsive interactions) enters in a unified manner. Also momentum conservation is respected.


  1. T. Lancaster & S.J. Blundell, QFT for the Gifted Amateur, 2014; sections 17.2-17.4.

  2. A. Zee, QFT in a nutshell, 2010; section I.5.


$^1$ Here we work in units where $c=1$.


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