# Quantising the energy-momentum relation

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?

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.
• 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 $x_0$. – Jannik Pitt May 18 at 9:49
• @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. – Dvij D.C. May 18 at 9:53
• @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/…. – Dvij D.C. May 18 at 11:42