In Peskin & Schröder QFT, just before equation 7.93, he writes in passing,

Next let us examine how $\Pi _2 (q^2)$ modifies the electromagnetic interaction, as determined by Eq. (7.77). In the nonrelativistic limit it makse sense to compute the potential V(r). For the interaction between unlike charges, we have, in analogy with Eq. (4.126),

$$V(\vec{x}) = \int \frac{d^3 q}{(2 \pi)^3} e^{i \vec{q}\cdot \vec{x}} \frac{-e^2}{|q|^2(1 - \Pi_2(-|q|^2))}.$$

Is it easy to see why potentials cannot exist in a fully relativistic quantum mechanical formalism? I see that with the usual coulomb potential, the potential changes instantly at all x when a charge moves. However, in general this need not be an issue; in electrodynamics it does not cause causality to be violated. See for ex: Griffiths electrodynamics section 10.1.3

There is a peculiar thing about the scalar potential in the Coulomb gauge; it is determined by the distribution of charge right now. If I move an electron in my lab, the potential V on the moon immediately records this change. That sounds particularly odd in the light of special relativity, which allows no message to travel faster than the speed of light. The point is that V by itself is not a physically measurable quantity - all the mann in the moon can measure is $\vec{E}$, and that involves $\vec{A}$ as well. Somehow it is built into the vector potential, in the coulomb gauge, that whereas V instantaneously reflects all changes in $\rho$, the combination $-\nabla V - \frac{\partial \vec{A}}{\partial t}$ does not, $\vec{E}$ will change only after sufficient time has elapsed for the "news" to arrive.

The above statement is for the electromagnetic potential, not a potential energy, but this differs only by a factor of charge.

In QM the "offense" would be if a measurable change in probability occurs with spacelike separation. Does this occur with a potential? Is it easy to see how, mathematically?


The problem with potentials is the same as the problem with position.

Relativistic quantum field theories (RQFTs) require a number of conceptual changes as compared to their non-relativistic counterpart, non-relativistic quantum mechanics (NRQM) due to the limit on the speed of information transmission. Effectively, RQFT treats the domain where both the Universe's natural informational limits: on transmission ($c$) and on content ($\hbar$), meet.

Most notably, one of these is the disappearance of the position operator in its ordinary, NRQM sense. To understand why, consider the following thought experiment. In NRQM, agents possess reduced information about a particle's position, which is described by its positional wave function


which, as you know, maps to a probability distribution. Consider now the following problem: A "position measurement", as it is often called, is really more a series of yes-no queries of the form "Is the particle located within a connected region $R$"? (Ideally, we can make them in arbitrarily fast succession.) If the answer is "yes", then the agent updates its positional wave function accordingly; i.e.

$$\psi(\mathbf{x}) \stackrel{\mbox{saw a "1"}}{\to} \psi(\mathbf{x}) \cdot 1_R(\mathbf{x})$$

where $1_R(\mathbf{x})$ is the "indicator function" on region $R$. (Of course, there's also a need to normalize afterward.) This is the famous "von Neumann projection axiom" that causes a lot of controversy (but which makes perfect sense in this understanding.). Conversely, if the answer is "no", it updates via

$$\psi(\mathbf{x}) \stackrel{\mbox{saw a "0"}}{\to} \psi(\mathbf{x}) \cdot (1 - 1_R(\mathbf{x}))$$


Now note something here: in the first case, what we have is a zeroing out of all probabilities outside the region $R$, because the answer "Yes", by definition, cateegorically excludes the particle as being anywhere outside. The integrated total probability is still $1$ (after re-normalization), but all the "support" - the nonzero "mass" of probability - is entirely, and perfectly, contained in $R$.

But now consider the move up to RQFT. Can you (in the role of the agent) make that kind of categorical exclusion, logically in a relativistic setting? Note that by saying "it is definitively outside $R$", you are making a statement about the entire Universe, all the way to billions of light-years and beyond, from $R$. That should smell funny.

Think about the time it takes you to do the query, because it must be a physical interaction after all. If it takes 5 seconds to query the particle position, do you think it'd be "kosher" in a relativistic setting to be able to claim a definitive, due to results therefrom, about something at a distance of, say, $150\ \mathrm{Gm}$, or 1 AU, from you? Think about this.

This is the intuitive essence of the Reeh-Schlieder no-go theorem in relativistic quantum field theory, and it requires us to make the following concession:

It is not possible to categorically answer a question of the form "Is the particle within the region $R$?" as long as that region is finite.

In other words, while such a query is always "worth" one bit, here it must only be ever worth strictly less than one bit.

And the results of this are significant. One is that it means that any putative "positional wave function" for a particle must not decay to actually zero outside a local region, to always leave finite probability outside the region.

But what this does is even worse! In NRQM, you see, the position operator $\hat{\mathbf{x}}$ only makes sense if you can, in theory, have a basis of positional "eigenstates", which must here be delta functions in the position - perfect localization. But the above just blew that out of the water! A position operator, in the same sense of NRQM, doesn't, and can't, fly in the RQFT.

But what about potential? Well, here's the thing. Potential, when written in the form $V(\mathbf{x})$, is actually an operator on a positional wave function. That's why in the non-relativistic Schroedinger equation, the potential $V(x)$ multiplies $\psi$:

$$-\frac{\hbar^2}{2m} \nabla^2 \psi + V(\mathbf{x}) \psi = i\hbar \frac{\partial \psi}{\partial t}$$

. You should have noted something was up when that this is a classical function giving a classical numerical quantity for the potential energy, and there it is. But now in RQFT, as we just saw before, there is no positional wave function in the sense we had before, because the needed operator formalism fell apart from underneath our feet!

Here's another way to look at it. Let's suppose you want to be insistent that nonetheless, we should be able to put down such a function anyways. Then consider the one-dimensional NRQM case. Suppose you had a potential like

$$V(x) = e^{cx}$$

i.e. a ramping potential. Suppose we query the particle potential energy, and get an exact value $V_\mathrm{obs}$. Or even suppose we ask enough yes-no questions to categorically exclude certain potential energies outside of a range.

What now is the signification of the quantity $\frac{1}{c} \log(V_\mathrm{obs})$, by the above equation? What does that mean making measurements (querying) of the potential energy also lets us do here?

What did we just talk about earlier?

Does something smell? Do you think it should be sensible to be able to do this in RQFT? Do you think "querying the potential energy" will run into any issues?

This is why the ordinary potential from NRQM doesn't work in RQFT any more than the usual idea of position does.

  • $\begingroup$ well-defined spatial probability densities should still be recoverable for particle fields, even after all finite-speed accounting is taken care of $\endgroup$
    – lurscher
    Jul 24 '19 at 23:29
  • $\begingroup$ @lurscher : Perhaps - but you will not be able to use the usual Hermitian operator formalism because position eigenstates will not be valid RQFT states. Moreover, you'll need some modification of the conceptualization (seems to become a bit "counterfactual", i.e. it represents information that only an impossible process could acquire) of what a "probability distribution" means given that it is in this case no longer possible to ever reduce it to a single possibility or a well-defined cut-out (i.e. with sharp edges). $\endgroup$ Jul 25 '19 at 2:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.