Nonrelativistic Limit

A nonrelativistic limit compares a family of relativistic states and dynamics with a lower-energy theory. It must say which scales are held fixed, how the state spaces are identified, and over what time interval the comparison is made. For free positive-energy particles, the rest-energy-subtracted propagator converges on every fixed square-integrable state over bounded time intervals. It does not converge uniformly over all momentum states. This distinction makes precise why a successful low-energy approximation need not work for every state.

Required background. Energy–Momentum Relation supplies the dispersion and rest energy; Metric and Units fixes the dimensional conventions.

Helpful background. Foldy–Wouthuysen Transformation provides the free positive-energy identification; Klein–Gordon to Schrödinger gives explicit cutoff error bounds; Poincaré Group fixes the boost algebra used below.

A family of low-energy comparisons

Restore $c$ and keep $m>0$ and $\hbar$ fixed. A simple free family holds physical momentum scales and observation times fixed while $c\to\infty$. Then

$$\frac{|\mathbf p|}{mc}\longrightarrow0, \qquad \frac{E_{\mathbf p}-mc^2}{mc^2}\longrightarrow0.$$

For backgrounds one must additionally control field strengths, potential-energy differences, spatial variation, and frequencies relative to the mass scale. Those assumptions are separate from the free limit proved below. In natural units the same content is a hierarchy of dimensionless ratios; one does not literally vary the numeral in the convention $c=1$.

Different choices of what to hold fixed define different families. For a Coulomb problem, holding $q^2/\epsilon_0$ fixed gives $\alpha=q^2/(4\pi\epsilon_0\hbar c)\to0$. Holding $\alpha$ fixed instead changes the interaction strength as $c$ grows; a bound state’s characteristic $v/c\sim Z\alpha$ then need not become small. When restoring SI constants, also respect $\epsilon_0\mu_0c^2=1$ rather than varying them independently.

The rest phase $e^{-imc^2t/\hbar}$ has no pointwise limit for generic nonzero $t$. Remove it before comparing envelopes:

$$\Psi_c(t)=e^{-imc^2t/\hbar}\,\widetilde\Psi_c(t).$$

Removing that phase alone does not eliminate the negative-energy branch or prepare a positive-energy state. The scalar initial-data requirement is treated in Klein–Gordon to Schrödinger.

Identifying the free particle state space

For a free Dirac particle, the exact Foldy–Wouthuysen construction supplies an isometric embedding of a two-component momentum wavefunction $\chi(\mathbf p)$ into the positive-energy subspace:

$$J_c(\mathbf p)\chi =\sqrt{\frac{E_{\mathbf p}+mc^2}{2E_{\mathbf p}}} \begin{pmatrix} \chi\\[2pt] \dfrac{c\,\boldsymbol\sigma\cdot\mathbf p} {E_{\mathbf p}+mc^2}\chi \end{pmatrix},$$ $$J_c^\dagger J_c=I_2.$$

Here $E_{\mathbf p}=\sqrt{m^2c^4+c^2\mathbf p^2}$. The domain is the ordinary two-component $L^2$ momentum space with its common measure on both sides. This is a norm-preserving comparison map, not the instruction to delete two components and leave their norm unaccounted for.

After this identification, exact free positive-energy envelope evolution acts on $\chi$ by the scalar multiplier generated by

$$h_c(\mathbf p) =\sqrt{m^2c^4+c^2\mathbf p^2}-mc^2.$$

The candidate nonrelativistic generator is $h_0(\mathbf p)=\mathbf p^2/(2m)$. The spin index is carried unchanged in this free comparison. For external fields, a dressed comparison map and its transformed observables require their own approximation analysis; the free map cannot establish that result.

Fixed-state convergence on finite time intervals

Define $U_c(t)=e^{-ih_ct/\hbar}$ and $U_0(t)=e^{-ih_0t/\hbar}$ as momentum-space multipliers. For each fixed $\mathbf p$, $h_c(\mathbf p)\to h_0(\mathbf p)$. Let $f$ be any fixed square-integrable two-component state. Then

$$\begin{aligned} \|(U_c(t)-U_0(t))f\|^2 ={}&\int d^3p\,|f(\mathbf p)|^2\\ &\times\left| e^{-ih_c(\mathbf p)t/\hbar} -e^{-ih_0(\mathbf p)t/\hbar}\right|^2. \end{aligned}$$

The phase difference is bounded by two. Dominated convergence therefore proves that this norm tends to zero. More strongly, for $|t|\leq T$,

$$\left|e^{-ih_ct/\hbar}-e^{-ih_0t/\hbar}\right| \leq \min\left(2,\frac{T|h_c-h_0|}{\hbar}\right).$$

Applying the same argument to this bound gives

$$\sup_{|t|\leq T}\|(U_c(t)-U_0(t))f\| \longrightarrow0$$

for every fixed finite $T$. No finite fourth momentum moment is needed for this convergence statement. A quantitative rate does require additional control of the state, such as a cutoff or suitable moments. The explicit cutoff estimate belongs to the scalar reduction owner.

This result compares evolution on prepared particle states. It is neither a theorem about every time-dependent background nor a statement that the unbounded Dirac operator can simply be evaluated at $c=\infty$. Thaller’s resolvent treatment gives a distinct operator framework for more general Dirac limits.

Why the all-momentum operator norm does not converge

Fix any finite $c$ and nonzero $t$. Put

$$d(\mathbf p)=h_0(\mathbf p)-h_c(\mathbf p).$$

As a function of $|\mathbf p|$, $d$ is continuous, starts at zero, and is unbounded above. Hence some momentum satisfies $d(\mathbf p)|t|/\hbar=\pi$. At that momentum the two phase multipliers are opposite, and their distance is two. Continuity gives arbitrarily nearby momenta with distances arbitrarily close to two, so the essential supremum is also two:

$$\|U_c(t)-U_0(t)\|_{\rm op}=2 \qquad(c<\infty,\ t\ne0).$$

This is compatible with the preceding strong convergence. The state that nearly saturates the operator norm is allowed to move to a different momentum region as $c$ changes. Strong convergence holds the state fixed first.

An explicit check sets $d_*=\pi\hbar/|t|$ and chooses

$$\frac{\mathbf p^2}{2m} =d_*+\sqrt{2mc^2d_*}.$$

Substitution gives $h_0-h_c=d_*$. Although the selected ratio $|\mathbf p|/(mc)$ tends to zero as $c\to\infty$, the momentum itself and the nonrelativistic kinetic energy grow. A small relative dispersion error can therefore still produce an order-one absolute phase error. The physical scales being held fixed matter.

Similarly, even for a fixed low-momentum packet, a small energy error can accumulate over a sufficiently long time. A finite-time estimate cannot be promoted to a uniform all-time statement.

The Galilei mass term from the boost algebra

There is also a symmetry check on the limiting procedure. Use the established passive convention $K_i=M^{0i}$, with $[K_i,P_j]=-i\hbar\delta_{ij}H/c$ and $[K_i,H]=-i\hbar cP_i$. On a fixed-mass particle family define

$$G_i=-\frac{K_i}{c},\qquad h=H-mc^2I.$$

The Poincaré commutators become

$$\begin{aligned} [G_i,P_j]&=i\hbar\delta_{ij} \left(mI+\frac{h}{c^2}\right),\\ [G_i,h]&=i\hbar P_i,\\ [G_i,G_j]&=-\frac{i\hbar}{c^2}\epsilon_{ijk}J_k. \end{aligned}$$

On states with controlled residual energy and angular momentum, the limiting boosts commute and the first bracket retains $i\hbar m\delta_{ij}I$. This is the mass central term in the Galilei algebra. A small passive boost has $U_B=e^{i\boldsymbol\xi\cdot\mathbf K/\hbar}$ with $\boldsymbol\xi\simeq\mathbf v/c$, so its limit is $e^{-i\mathbf v\cdot\mathbf G/\hbar}$ in the same convention. The corresponding wavefunction phase and its interpretation are derived in Galilean Boosts.

This is a contraction on a specified mass and scale sector. It does not by itself establish a general prohibition on relativistic superpositions of different masses.

Exercises

Two momentum families. Compare fixed physical momentum $p=p_0$ with $p=\lambda mc$, where $\lambda>0$ is fixed. Which family approaches nonrelativistic kinematics?

Solution

For fixed $p_0$, $p_0/(mc)\to0$ and $h_c\to p_0^2/(2m)$. For $p=\lambda mc$, $h_c=mc^2(\sqrt{1+\lambda^2}-1)$ while $h_0=mc^2\lambda^2/2$. Their relative difference remains a function of $\lambda$; it does not vanish by increasing $c$ alone. The velocity ratio is $v/c=\lambda/\sqrt{1+\lambda^2}$.

A nearly worst-case state. Why is a narrow normalizable packet around the momentum with $d|t|/\hbar=\pi$ enough to prove the operator-norm result, even though a momentum eigenstate is not normalizable?

Solution

The phase distance is continuous in a neighborhood of that momentum. For every $\varepsilon>0$, a small region has distance greater than $2-\varepsilon$. A normalized packet supported there has evolution difference at least $2-\varepsilon$. The universal upper bound is two, giving the norm without using an improper state.

A controlled observation time. A momentum-restricted subspace has $\sup|h_c-h_0|\leq\delta E$. Give a sufficient time condition for a state-norm error at most $\varepsilon$.

Solution

For a normalized state in that subspace, the multiplier bound gives error at most $|t|\delta E/\hbar$. Thus $|t|\leq\varepsilon\hbar/\delta E$ is sufficient. It is a conservative bound, not a claim that every packet actually attains the maximum error.

References

• Foldy, Leslie L., and Siegfried A. Wouthuysen. “On the Dirac Theory of Spin 1/2 Particles and Its Non-Relativistic Limit.” Physical Review 78, 29–36 (1950). doi:10.1103/PhysRev.78.29. The representation change that separates free particle and antiparticle components.
• İnönü, Erdal, and Eugene P. Wigner. “On the Contraction of Groups and Their Representations.” Proceedings of the National Academy of Sciences 39, 510–524 (1953). doi:10.1073/pnas.39.6.510. Group contraction as a symmetry-limit construction.
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0. See also the author’s Nonrelativistic Limit of the Dirac Equation, a version of the book treatment using shifted resolvents.