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I am reading Relativistic Quantum Mechanics by Bjorken and Drell and on page 5 they present the following attempt at a relativistic Hamiltonian for a free particle

\begin{equation} i\hbar\frac{\partial\psi}{\partial t} = \sqrt{-\hbar^2 c^2\nabla^2+m^2c^4}\psi. \end{equation}

About this equation they say "If we expand it, we obtain an equation containing all powers of the derivative operator and thereby a nonlocal theory". They also mentioned how the equation's time and space components are asymmetrical and would transform differently under Lorentz transformations.

I can see the asymmetry of the equation as a problem for its relativistic ambitions, but how does it relate to non-locality. If feels they are equating being un-transformable under Lorentz transformations with non-locality but I can't quite see the connection. Even if a covariant equation preserves the theory's locality, does it mean a non-covariant equation is necessarily non-local?

Qmechanic
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NeonGabu
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  • The written equation is covariant, in the sense that if $\psi$ is a scalar field, it solves that equation in every reference frame as it follows by passing in Fourier representation. It is simply a positive energy KG field. – Valter Moretti Dec 21 '22 at 07:21
  • Non locality arises from the Paley-Wiener theorem again in fourier representation because $\sqrt{m^2+ p^2}$ is not an entire analyitic function. It does not matter if the Taylor expansion includes infinite terms. – Valter Moretti Dec 21 '22 at 07:24
  • Possible duplicates: https://physics.stackexchange.com/q/13624/2451 and links therein. – Qmechanic Dec 21 '22 at 08:44

2 Answers2

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I think a simple example might help, consider a scalar field $\phi$ in 1D. The standard treatment of this problem would involve starting from a chain of particles with the displacement of the $i^{th}$ particle being equal to $\phi(ia)$ where a is a length scale, when a is very small, the field and 1D chain description become identical. A $\nabla^2$ term will come up in the equation of motion of this field if the coupling is nearest neighbour i.e. $$F=k(\phi_{i+1}-\phi_i)-k(\phi_{i}-\phi_{i-1})$$ Higher derivatives like $\nabla^3$ will come when the coupling(Force on a particle) is of the form $$k'[\phi_{i+2}-\phi_{i-2}-2\phi_{i+1}+2\phi_{i-1}]$$ As the order of the derivative increases, one needs to include couplings to particles which are farther away from a given particle making the theory non-local (see Meharan Kardar's Statisitcal Physics of Fields Ch.1 and 2, these chapters deal with field theory only). Since any problem on fields($\psi$ here) can be treated in this way(atleast mathematically) this might be the logic by which the authors say that higher derivatives of $\psi$ indicates a non-local theory.

amogh waghmare
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The written equation is the Schroedinger equation for a Klein-Gordon particle. The written Hamiltonian is the correct one: it is the restriction of the QFT Hamiltonian to the one-particle space.

The Hilbert space is made of functions which solve the said equation and that can be written as

$$\psi(t, \vec{x}) = \int_{\mathbb{R}^3} \frac{\hat{\psi}(\vec{k})}{\sqrt{m^2+ \vec{k}^2}}e^{i(\vec{k}\cdot \vec{x}- t \sqrt{m^2+\vec{k}^2})} \frac{d\vec{k}}{(2\pi)^{3/2}} \tag{1}$$

with

$$\int_{\mathbb{R}^3} |\hat{\psi}(\vec{k})|^2 d\vec{k}< +\infty$$

Though the equation is relativistically invariant (though it is not co-variant), the equation gives rise to non-local solutions as is well-known.

The active action Poincaré group is the usual one $(U_{(V,\Lambda)}\psi)(x) = \psi (\Lambda^{-1}(x -V))$ and it preserves (1) and the Schroedinger equation.

Non locality can be intepreted as follows in addition to the, in a sense, trivial intepretation that the Hamiltonian is not the linear combination of differential operators. (I am assuming in the rest of this asnswer that $\vec{k} \in \mathbb{R}$ to dela with standard complex-variable function instead of functions of many complex variables, the extension to that case involves some subtleties.)

Suppose that at $t=0$ the initial datum $\psi(0,\vec{x})$ vanishes outside a ball of finite radius $R$ centered at the origin. One expects that after $t>0$ seconds the support of the solution $\psi(t,\vec{x})$ is still bounded in a ball of finite radius $R+ct$. As a matter of fact this is false: after evry arbitrarily short time $t>0$ the support of the solution fulfills the whole space.

The mathematical reason of this sort of nonlocality is due to the fact that the function $$\vec{k} \mapsto e^{-it \sqrt{m^2+\vec{k}^2}}$$ is not the real restriction of an entiere analytic function defined in the whole complex $\vec{k}$ plane for $t>0$ as the Paley-Wiener theorem requires. Notice that instead $$ \frac{\hat{\psi}(\vec{k})}{\sqrt{m^2+ \vec{k}^2}}e^{i\vec{k}\cdot \vec{x}}$$ is such a function by hypothesis (the fourier transform must have a shape such that it cancel the branch points of $\sqrt{m^2+k^2}$ arising from $\vec{k}^2=-m^2$ on the imaginary axis), but when we switch on $t$ the nature of the function changes, bacause of the function $t\sqrt{m^2+k^2}$.

The non-local feature of the written Schroedinger equation is still the object of many discussions in relation to the problem of localisation of a relativistic particle which is a longstandig issue.

Valter Moretti
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