# Non-locality of pre-Klein-Gordon equation

In Relativistic Quantum Mechanics, Bjorken and Drell state that expanding the square root in the equation $$-\hbar^2\frac{\partial^2\psi}{\partial t^2}=\sqrt{-\hbar^2c^2\boldsymbol{\nabla}^2+m^2c^4}\psi$$ would result in a non-local wave equation. What exactly do they mean by this? One could argue that in order to compute $$\psi$$'s spatial derivatives at a certain point in space-time up to arbitrary order, one has to know the behavior of $$\psi$$ in points arbitrarily far from the point of interest, even if there is no world line linking those points. But if we were to solve a local equation such as the Dirac equation, we could do that nonetheless. Accordingly, I think there must be a deeper reason for the non-locality that Bjorken and Drell warn for.

• Possible duplicates: physics.stackexchange.com/q/156124/2451 , physics.stackexchange.com/q/346780/2451 and links therein. Feb 25, 2021 at 7:29
• Those threads don't contain any answers to my liking! No one really answers the question - the Peskin and Schroeder answer is not what I'm looking for. Feb 25, 2021 at 7:38
• What do you mean by "if we were to solve a local equation, we could do that nonetheless"? Feb 25, 2021 at 9:27
• Well, when we find a well-behaved solution to the Dirac equation or perhaps even the Schrödinger equation, we can compute its derivatives at any point in space at any time up to arbitrary order. What then is the fundamental problem with such derivatives appearing in a wave equation? Feb 25, 2021 at 15:07
• Nice question. Taylor expansion of $f(x+y)$ around $y=0$ allows you to write $f(x+y)$ in terms of $f(x)$ and all its (infinite) derivatives. Here it's the same: having an equation of infinite order amounts to have something non-local. Jul 15, 2023 at 22:04

## 1 Answer

I think you're confusing different things. The "non-local" problem with the equation in your question is not that the expansion of the square root leads to infinitely many derivatives, although that's definitely not desirable, it's that it mixes values of those derivatives from different points in spacetime. So to get the value of the field at some specific point, I not only need the value of all of the derivatives at that point, I need the value of the derivatives at all other points too.

You also seem to have something else backward. As you correctly state in the comments, a solution to, say, the Dirac equation will typically be smooth so that means that you can compute as many derivatives as you want after you have a solution. If you cannot write down a PDE with a finite number of derivatives in it, however, you will have have technical problems in how to frame a solution. For example, how do you specify boundary conditions or initial value conditions in this case? You can probably get around those problems with additional assumptions, but it's not elegant to say the least.