The Square-Root Hamiltonian

The free Hamiltonian $H_0=\sqrt{c^2\mathbf P^2+m^2c^4}$ gives a perfectly well-defined positive-energy quantum evolution. Its limitation is not an undefined square root: it is a spatially nonlocal operator whose elementary position-space probability interpretation and coupling to backgrounds do not automatically supply a local relativistic quantum theory. The free construction is useful, provided those distinctions are retained.

Required background. The Energy–Momentum Relation gives the positive dispersion; the Klein–Gordon Equation gives the local two-branch comparison. Fourier transforms and the meaning of a self-adjoint operator domain are assumed.

Defining the operator by its spectrum

On $L^2(\mathbb R^3,d^3x)$, use the unitary Fourier transform

$$\psi(\mathbf x)=\int\frac{d^3p}{(2\pi\hbar)^{3/2}} \widetilde\psi(\mathbf p)e^{i\mathbf p\cdot\mathbf x/\hbar}.$$

For $m>0$, define $H_0$ by multiplication in momentum space:

$$\widetilde{H_0\psi}(\mathbf p) =E_{\mathbf p}\widetilde\psi(\mathbf p), \qquad E_{\mathbf p}=\sqrt{c^2\mathbf p^2+m^2c^4}.$$

Its domain is

$$D(H_0)=\left\{\psi\in L^2: \int d^3p\,E_{\mathbf p}^2|\widetilde\psi(\mathbf p)|^2<\infty\right\}.$$

This is the Sobolev space $H^1(\mathbb R^3)$ as a set, with an equivalent graph norm. Multiplication by a real positive function on its maximal domain is self-adjoint. The resulting evolution

$$\widetilde\psi(t,\mathbf p) =e^{-iE_{\mathbf p}t/\hbar}\widetilde\psi(0,\mathbf p)$$

is unitary and preserves $\int|\psi|^2d^3x$. There is no spectral instability in this free positive branch: $H_0\geq mc^2I$.

Its Hilbert-space norm uses a chosen equal-time representation. A Lorentz transformation of this wavefunction includes momentum-dependent factors; it is not the transformation law of a scalar spacetime amplitude. Compare the alternative invariant normalization in Relativistic Phase Space.

Why the position-space operator is nonlocal

A finite-order constant-coefficient differential operator has a polynomial Fourier symbol. The square root $E_{\mathbf p}$ is not a polynomial, so $H_0$ cannot be such a local differential operator. It is a pseudodifferential operator: evaluating $H_0\psi$ at a point generally uses values of $\psi$ away from that point.

For a concrete illustration, in one spatial dimension its convolution kernel away from $x=0$ is

$$K(x)=-\frac{mc^2}{\pi|x|} K_1\!\left(\frac{mc|x|}{\hbar}\right), \qquad x\ne0,$$

where $K_1$ is a modified Bessel function. The full kernel is a distribution at the origin; this expression is not an ordinary integrable formula there. For a point outside the compact support of a smooth initial wavefunction, however, its convolution has no coincident-point singularity and is generically nonzero. The equation $\partial_t\psi=-iH_0\psi/\hbar$ can therefore develop a tail outside that support immediately.

One way to verify the coefficient is to work temporarily with $c=\hbar=1$. The inverse Fourier transform of $(p^2+m^2)^{-1/2}$ is $K_0(m|x|)/\pi$. Applying $-\partial_x^2+m^2$ and using the modified Bessel equation gives $-mK_1(m|x|)/(\pi|x|)$ for $x\ne0$. Restoring constants gives the kernel above. This calculation uses the standard Bessel integral representations in the NIST DLMF.

The massive off-diagonal kernel falls exponentially on the reduced Compton scale $\hbar/(mc)$, with additional powers of distance. Exponential suppression is not compact support. Nor does a wavefunction tail alone prove that a locally controlled superluminal message can be transmitted; that requires a specified localization observable and intervention model.

Relation to Klein–Gordon evolution

On sufficiently regular states, applying $i\hbar\partial_t+H_0$ to $(i\hbar\partial_t-H_0)\psi=0$ yields the free Klein–Gordon equation. The converse needs an extra condition. The positive-branch evolution fixes the initial derivative through

$$i\hbar\partial_t\psi(0)=H_0\psi(0).$$

General Klein–Gordon data do not obey this relation and include both frequency sectors. Because the relation is nonlocal, compact support for $\psi(0)$ does not imply compact support for $\partial_t\psi(0)$. There is thus no contradiction with causal propagation of the full local KG equation when both of its initial data have compact support. Causality and Light Cones develops that distinction.

Adding a potential changes the comparison

For a real bounded static multiplication operator $V(\mathbf x)$, $H_0+V$ remains self-adjoint on $D(H_0)$. This is a legitimate model Hamiltonian. It is not generally the positive-energy reduction of a minimally coupled KG equation.

For a sufficiently regular potential and eigenfunction so that the products below exist, a stationary solution satisfies $H_0\psi=(E-V)\psi$. Applying $H_0$ once more, with operators kept in order, gives

$$\begin{aligned} H_0^2\psi &=H_0(E-V)\psi\\ &=(E-V)^2\psi-[H_0,V]\psi. \end{aligned}$$

The electrostatic KG equation instead requires $(E-V)^2\psi=H_0^2\psi$. The extra commutator is generally nonzero when $V$ varies spatially. For unbounded or singular potentials, even self-adjointness and stability require additional domain and coupling analysis; the bounded-potential argument does not cover Coulomb collapse.

This calculation identifies the precise obstruction to obtaining an interacting scalar wave equation by selecting the free positive root and then appending an arbitrary potential. It does not say that every square-root model is invalid. Such models can be useful effective descriptions with stated accuracy and a separately justified interaction prescription.

The low-momentum approximation is not a global Hamiltonian

After subtracting the rest energy, the leading expansion is

$$H_0-mc^2=\frac{\mathbf P^2}{2m} -\frac{\mathbf P^4}{8m^3c^2}+\cdots.$$

It is controlled for states whose relevant momenta are small compared with $mc$. The truncation through $\mathbf P^4$ is unbounded below if used at arbitrarily large momentum; the exact square root is bounded below. The spurious instability is caused by extrapolating an asymptotic effective Hamiltonian beyond its range, not by the relativistic spectrum.

Exercises

1. For $p/(mc)=0.1$, compare the exact kinetic energy in units of $mc^2$ with its quadratic and quartic approximations.

Solution

The exact value is $\sqrt{1.01}-1\simeq0.0049875621$. The quadratic result is $0.005$; including the quartic term gives $0.0049875$. The remaining error is about $6.21\times10^{-8}$ in these units, consistent with the next term $0.1^6/16$.

2. Show in momentum representation that a constant $V$ commutes with $H_0$ but a spatially varying potential need not. What factor appears in the integral kernel of their commutator?

Solution

If $V(\mathbf p,\mathbf q)$ denotes the potential’s momentum kernel, then $[H_0,V](\mathbf p,\mathbf q)=(E_{\mathbf p}-E_{\mathbf q}) V(\mathbf p,\mathbf q)$. A constant potential is diagonal in momentum and the factor vanishes. A varying potential couples unequal momenta and generally unequal energies.

3. Does a subluminal group velocity $dE_p/dp$ prove that a wave packet remains compactly supported inside its initial light cone?

Solution

No. Group velocity describes the motion of a narrow packet’s envelope or peak. Exact support depends on all Fourier components and on the evolution kernel. The positive-root kernel is nonlocal even though $|dE_p/dp|<c$.

References

• G. C. Hegerfeldt, “Instantaneous Spreading and Einstein Causality in Quantum Theory,” Annalen der Physik 7, 716–725, 1998, arXiv:quant-ph/9809030 — positive-energy localization and the interpretation of instantaneous tails.
• National Institute of Standards and Technology, Digital Library of Mathematical Functions, section 10.32 — integral representations of modified Bessel functions.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, I: Functional Analysis, revised and enlarged ed., Academic Press, 1980 — spectral calculus and self-adjoint multiplication operators.
• B. Thaller, The Dirac Equation, Springer, 1992 — positive-energy reductions and relativistic one-particle theory.