The energy-momentum relation in special relativity states $m^2 = E^2 - ||p||^2$ (in natural units). So $$ E = \pm\sqrt{\| p \|^2 + m^2}. $$ If we want to find a theory for a relativistic free particle, one could quantise this expression and use the result as a Hamiltonian. This gives: $$ H = \sqrt{-\Delta^2 + m^2}. $$ Since the Laplacian is positive and self-adjoint (on a suitable domain), for $m^2 \geq 0$ this is a well-defined self-adjoint operator. Then one could look to solutions to $$i\sqrt{-\Delta^2 + m^2}\psi = \frac{\partial \psi}{\partial t},$$ for example in $L^2(\mathbb{R^3})$ to describe a relativistic free particle moving in flat space. This equation is Lorentz-invariant so it describes (at least mathematically) a relativistic phenomenon. Why was this theory dismissed for the description of a relativistic free particle? What does it predict and how does this differ from experiment?


1 Answer 1


The most glaring issue with this theory is that it is non local. You can realize this by expanding the square root. The expansion would never end and it will contribute indefinitely high orders of the $\partial$ operator, which means that the value of $\frac{\partial \psi}{\partial t}$ at one point would depend on values of $\psi$ at distant points. This is unacceptable because we expect our theories to be local to be of scientific value.

You can get rid of the non-locality issue by applying the squared Hamiltonian on the state, and then you get rid of the square-root and obtain the famous Klein-Gordan equation, the issues with which are well known, see, for example, the Introduction section of Dyson's lectures on quantum mechanics.

The Klein-Gordon equation doesn't make physical sense as an equation for the wave-function of a single particle for a single relativistic particle cannot be localized in quantum mechanics. However, it can be meaningfully seen as the classical equation of motion of a free scalar field. Its Lagrangian can be quantized to obtain the quantum field theory of a free scalar field.

  • $\begingroup$ But if I'm not mistaken one knows a function locally, then one knows all of its derivatives locally? Or precisely: To know $\frac{\partial\psi}{\partial x^\alpha}(x_0)$ one only needs to know the values of $\psi$ in some ball of arbitrary radius $\epsilon > 0$ which contains $x_0$. $\endgroup$ May 18, 2020 at 9:49
  • $\begingroup$ @JannikPitt Yes, what you're saying is precisely correct but only for finite orders of derivatives. For example, if you know the value of $\psi$ in some ball of infinitesimal small radius as you describe, you can find the hundredth derivative of $\psi$ at the point, no issues. But, here, the expansion involves indefinitely high orders of derivatives. To put it in a crass manner, in order to know $\frac{d^\infty}{dx^\infty}\psi$, you'd need to make the radius of that ball finite. $\endgroup$
    – ACat
    May 18, 2020 at 9:53
  • $\begingroup$ Ah, so "locality" is not preserved by "taking limits" (or rather applying functional calculus in this case). Very interesting, your answer then makes it very clear why this theory of a relativistic particle isn't physically satisfactory. $\endgroup$ May 18, 2020 at 10:54
  • $\begingroup$ should not the expected locality also depend on how steep the ensuing nonlocality be, i.e., how fast the convergence of that "Taylor" series is? I mean if it is very very very fast converging series why would that be a problem? $\endgroup$
    – hyportnex
    May 18, 2020 at 11:32
  • $\begingroup$ @hyportnex What do you mean by fast converging? Simply put, since $\phi(x+a)=\sum_{k=0}^\infty \phi^{(k)}(x) a^k/k!$ where $a$ doesn't need to be infinitesimal, having indefinitely large derivatives at a point $x$ amounts to knowing the function at distant places $x+a$. For a more detailed discussion, see: physics.stackexchange.com/questions/13624/…. $\endgroup$
    – ACat
    May 18, 2020 at 11:42

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