The Scalar Nonrelativistic Limit in Practice

A scalar packet can follow Schrödinger evolution accurately when its frequency sector is prepared, its kinetic momenta are small, and the neglected energy correction has not accumulated an appreciable phase. This page applies the full derivation in Klein–Gordon to Schrödinger to a Gaussian packet. The calculation distinguishes a typical-momentum estimate from a strict compact-support error bound.

Required background. Klein–Gordon to Schrödinger derives the expansion and its conserved-norm matching.

A Gaussian packet with a declared normalization

Use the positive-frequency free sector with $m>0$. Let $\chi$ be the flat-$L^2$ amplitude obtained by the norm transformation in the canonical derivation, and prepare

$$\chi(0,\mathbf x)=\frac{1}{(\pi\sigma^2)^{3/4}} \exp\left(-\frac{|\mathbf x|^2}{2\sigma^2}\right).$$

Each spatial coordinate has variance $\sigma^2/2$ and each momentum component has variance $\hbar^2/(2\sigma^2)$. The mean momentum is zero, but the packet is not at rest in every momentum component:

$$\langle\mathbf P^2\rangle=\frac{3\hbar^2}{2\sigma^2}, \qquad \langle(\mathbf P^2)^2\rangle=\frac{15\hbar^4}{4\sigma^4}.$$

Let $\lambda_C=\hbar/(mc)$ denote the reduced Compton wavelength. The small parameter is $\lambda_C/\sigma$, rather than $|\langle\mathbf P\rangle|/(mc)$, which vanishes for every width here. The Gaussian has nonzero tails at all momenta, so a theorem assuming compact momentum support cannot be applied to it without estimating the tail separately.

Leading relativistic energy correction

The owner page gives the free correction $\delta H=-\mathbf P^4/(8m^3c^2)$. Its expectation in this packet is

$$\Delta E=-\frac{15\hbar^4}{32m^3c^2\sigma^4}.$$

The leading kinetic energy is $3\hbar^2/(4m\sigma^2)$, so

$$\frac{|\Delta E|}{\langle H_{\rm Schr}\rangle} =\frac58\left(\frac{\lambda_C}{\sigma}\right)^2.$$

For $\sigma=10\lambda_C$, the first correction is about $0.00625$ of the leading kinetic energy. This is an asymptotic energy estimate, not a uniform claim about every observable for all times. In particular, a small error in each momentum’s frequency can build a substantial relative phase over a long evolution.

Because the Gaussian has finite higher moments, one can also bound the free state error without an artificial sharp cutoff. The pointwise dispersion bound from the owner yields

$$\|\chi_{\rm exact}(t)-\chi_{\rm Schr}(t)\|_{L^2} \leq \frac{|t|}{8\hbar m^3c^2} \sqrt{\langle(\mathbf P^2)^4\rangle} =\frac{\sqrt{945}\,|t|\hbar^3} {32m^3c^2\sigma^4}.$$

Here $\langle(\mathbf P^2)^4\rangle=945\hbar^8/(16\sigma^8)$. This bound concerns the common normalized initial amplitude and the two free multiplier evolutions; it is useful when its right side is small. The difference between the rest-phase-removed scalar envelope $\psi$ and $\chi$ is a separate normalization effect.

Preparing the scalar Cauchy data

Writing down the Gaussian shape is not enough to prepare the scalar solution. Recover the initial scalar amplitude by

$$\widetilde\phi(0,\mathbf p) =\sqrt{\frac{mc^2}{E_{\mathbf p}}}\, \widetilde\chi(0,\mathbf p),$$

and choose $\partial_t\phi(0)=-iH_0\phi(0)/\hbar$. Then remove the rest-energy phase to obtain the slow envelope. Setting $\partial_t\phi(0)=0$ instead produces equal frequency-sector amplitudes and fails the assumed preparation.

For an external field, a Gaussian width alone is even less decisive. Kinetic rather than canonical momentum enters the expansion, backgrounds can drive transitions, and field derivatives enter higher orders. Use the covariant residual-energy criterion in the full derivation and the application limits in External Potentials.

Exercises

1. Derive the fourth momentum moment from independent centered Gaussian components with variance $v=\hbar^2/(2\sigma^2)$.

Solution

Expand $(P_x^2+P_y^2+P_z^2)^2$. Each of the three diagonal terms contributes $\langle P_i^4\rangle=3v^2$, while the three cross pairs, each counted twice, contribute $v^2$. The sum is $15v^2=15\hbar^4/(4\sigma^4)$.

2. How do the fractional kinetic correction and the state-error bound change if $\sigma$ doubles at fixed time?

Solution

The fractional correction falls by a factor of four. The absolute energy correction and the displayed state-error bound fall by a factor of sixteen. These are different quantities because the leading kinetic energy also falls by four.

References

• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5 — scalar low-momentum expansion.
• D. Tong, Quantum Field Theory, University of Cambridge lecture notes, 2006–2007, §2.8, Non-Relativistic Fields — the slowly varying scalar sector.