Klein–Gordon to Schrödinger

The Schrödinger limit of the Klein–Gordon equation is a limit of prepared solutions, their conserved norm, and their evolution time scale. Removing the rest-energy phase is necessary but does not remove a negative-frequency component. For a free positive-frequency packet with small momentum, the expansion is particularly transparent and admits an explicit error bound.

Required background. Plane-Wave Solutions fixes frequency preparation; the Klein–Gordon Inner Product fixes normalization; External Potentials supplies the ordered covariant operator used in the final extension.

Preparing the slow scalar envelope

Take $m>0$ and start with a free positive-frequency solution, so $i\hbar\partial_t\phi=H_0\phi$ and $H_0=\sqrt{m^2c^4+c^2\mathbf P^2}$. Equivalently, its initial derivative is $-iH_0\phi(0)/\hbar$. Factor

$$\phi=e^{-imc^2t/\hbar}\psi.$$

The exact first-order equation on this subspace is $i\hbar\partial_t\psi=(H_0-mc^2)\psi$. Direct substitution into the second-order equation gives the equivalent envelope equation

$$i\hbar\partial_t\psi =-\frac{\hbar^2}{2m}\nabla^2\psi +\frac{\hbar^2}{2mc^2}\partial_t^2\psi.$$

To see the cancellation, the second time derivative of the full amplitude is the rest phase times $-\!m^2c^4\psi/\hbar^2-2imc^2\partial_t\psi/\hbar+\partial_t^2\psi$. The first term cancels the mass term in the KG equation; the cross term produces the Schrödinger derivative.

An occupied negative-frequency mode would instead give an envelope proportional to $e^{+i(E_{\mathbf p}+mc^2)t/\hbar}$. Even at small momentum this oscillates on the rest-energy scale. Dropping the second time derivative on such data is inconsistent. The approximation concerns spectral or norm estimates on the solution, not a pointwise ratio $|\partial_t\psi|/|\psi|$, which is ill-defined at nodes.

Expanding the free evolution and bounding its error

Suppose the momentum support lies in $|\mathbf p|\leq\Lambda$, with $\eta=\Lambda/(mc)\ll1$. The residual energy is

$$\begin{aligned} \varepsilon_{\mathbf p} &=E_{\mathbf p}-mc^2\\ &=\frac{\mathbf p^2}{2m} -\frac{\mathbf p^4}{8m^3c^2} +O\!\left(\frac{|\mathbf p|^6}{m^5c^4}\right). \end{aligned}$$

The leading Schrödinger Hamiltonian is $\mathbf P^2/(2m)$; the first free relativistic correction is $-\mathbf P^4/(8m^3c^2)$. The negative sign means the nonrelativistic kinetic energy overestimates the exact residual energy. Taylor’s theorem for $\sqrt{1+z}$, $z\geq0$, gives

$$0\leq\frac{\mathbf p^2}{2m}-\varepsilon_{\mathbf p} \leq\frac{|\mathbf p|^4}{8m^3c^2}.$$

For a normalized initial packet in either the flat momentum norm or the positive KG norm, the two Fourier multiplier evolutions therefore satisfy

$$\|\psi_{\rm exact}(t)-\psi_{\rm Schr}(t)\| \leq\frac{|t|\Lambda^4}{8\hbar m^3c^2}.$$

The proof is the elementary inequality $|e^{-ix}-e^{-iy}|\leq|x-y|$ applied pointwise in momentum and then integrated in the chosen norm. The bound is useful when its right side is small; it is not uniform for arbitrarily long times at fixed cutoff. The fourth-order Hamiltonian improves the bound to $|t|\Lambda^6/(16\hbar m^5c^4)$. Its energy is unbounded below if used at unrestricted momentum, so it is an expansion on the low-momentum sector, not a new fundamental Hamiltonian.

For packets without compact momentum support, control the high-momentum tail separately. Small mean momentum alone does not ensure small relativistic error.

Matching the conserved norm

For the free positive-frequency envelope,

$$Q=\int d^3p\,\frac{E_{\mathbf p}}{mc^2} |\widetilde\psi(\mathbf p)|^2.$$

It approaches $\|\psi\|_{L^2}^2$ at low momentum, with relative difference at most $\sqrt{1+\eta^2}-1$ on the stated support. To identify an exactly normalized Schrödinger-space amplitude, define

$$\widetilde\chi(\mathbf p) =\sqrt{\frac{E_{\mathbf p}}{mc^2}}\, \widetilde\psi(\mathbf p), \qquad \|\chi\|_{L^2}^2=Q.$$

The multiplier commutes with free evolution. Its expansion is $\chi=[1+\mathbf P^2/(4m^2c^2)+O(\eta^4)]\psi$ on the restricted sector. Thus the same free effective Hamiltonian acts on $\chi$, but the relation of the original scalar amplitude to a normalized wavefunction also receives corrections. Exact integrated norm matching does not turn the original local KG density into $|\chi(\mathbf x)|^2$.

Leading coupling to a prescribed electromagnetic field

Now let $\mathcal E=i\hbar\partial_t-q\Phi$ and $\boldsymbol\pi=-i\hbar\nabla-q\mathbf A$. The rest-phase substitution gives the exact ordered identity

$$\mathcal E\psi =\frac{\boldsymbol\pi^2}{2m}\psi -\frac{\mathcal E^2}{2mc^2}\psi.$$

For states and backgrounds for which the covariant residual energies and kinetic momenta are small compared with $mc^2$ and $mc$, and neglected sector mixing is controlled, the leading equation is

$$i\hbar\partial_t\psi =\left[\frac{(-i\hbar\nabla-q\mathbf A)^2}{2m} +q\Phi\right]\psi.$$

Unlike the free case, replacing the final $\mathcal E^2$ term by a commuting numerical square is generally invalid. Covariant derivatives do not commute with varying potentials; systematic higher orders require their commutators and a consistent normalization transformation. The free $-\mathbf P^4$ term alone is not a complete external-field expansion. The free error bound above also does not apply unchanged to an arbitrary background.

The physical conditions must be expressed through kinetic energies, field scales, variation scales, and transition probabilities. An absolute constant electrostatic potential can be removed by a gauge phase and does not by itself invalidate the expansion.

Exercises

1. For a narrow packet with $|\mathbf p|=\eta mc$, find the leading fractional error of the nonrelativistic kinetic energy.

Solution

The omitted energy is $mc^2\eta^4/8$, while the leading kinetic energy is $mc^2\eta^2/2$. Their ratio is $\eta^2/4+O(\eta^4)$. For $\eta=0.1$ this is about $0.0025$. The ratio uses kinetic energy, not the much larger total rest energy.

2. Estimate the time for the omitted leading correction at cutoff $\Lambda=\eta mc$ to accumulate one radian of phase.

Solution

$t\,mc^2\eta^4/(8\hbar)\sim1$ gives $t\sim8\hbar/(mc^2\eta^4)$. Small instantaneous energy error can therefore matter in a long coherent evolution. Different momenta acquire different errors, so the effect is not generally a removable global phase.

3. A rest-mode scalar has both frequencies: $\phi=Ae^{-imc^2t/\hbar}+Be^{+imc^2t/\hbar}$. Find its envelope and test the omission of the second time derivative.

Solution

$\psi=A+Be^{2imc^2t/\hbar}$. For the second term, $i\hbar\partial_t\psi=-2mc^2Be^{2imc^2t/\hbar}$ and $\hbar^2\partial_t^2\psi/(2mc^2)$ is exactly the same quantity. The supposedly small term supplies the entire evolution of that sector. One must suppress $B$ in the prepared data, not just remove a rest phase.

References

• H. Feshbach and F. Villars, “Elementary Relativistic Wave Mechanics of Spin 0 and Spin 1/2 Particles,” Reviews of Modern Physics 30, 24–45, 1958, doi:10.1103/RevModPhys.30.24 — scalar sectors and low-energy reduction.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5 — nonrelativistic scalar limits.
• D. Tong, Quantum Field Theory, University of Cambridge lecture notes, 2006–2007, §2.8, Non-Relativistic Fields — separating the rest-frequency scale.