26 May 2025
Matter Waves, Phase and Group Velocities, and Electron Diffraction
For a photon,
\[E=h\nu=\hbar\omega,\qquad p=\frac{h}{\lambda}=\hbar k,\]where $\omega=2\pi\nu$, $k=2\pi/\lambda$, and $\hbar=h/(2\pi)$. De Broglie proposed that the same frequency-energy and wavelength-momentum relations apply to material particles:
\[\boxed{\lambda=\frac{h}{p}},\qquad \boxed{k=\frac{p}{\hbar}},\qquad \boxed{\omega=\frac{E}{\hbar}}.\]The first relation has the required dimension because
\[\left[\frac hp\right] =\frac{\mathrm{J\,s}}{\mathrm{kg\,m\,s^{-1}}} =\mathrm m.\]For a non-relativistic particle, $p=mv$ and $K=p^2/(2m)$, so
\[\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}.\]An electron accelerated from rest through a potential difference $V$ gains kinetic energy $eV$, provided losses are negligible. Consequently,
\[\boxed{\lambda=\frac{h}{\sqrt{2m_e eV}}}.\]This approximation assumes $eV\ll m_ec^2$. In convenient units it becomes $\lambda(\text{angstrom})\simeq12.27/\sqrt{V(\mathrm V)}$. Relativistically, $E_{\rm tot}=m_ec^2+eV$ and
\[p c=\sqrt{E_{\rm tot}^2-m_e^2c^4} =\sqrt{eV(eV+2m_ec^2)},\]so $\lambda=hc/\sqrt{eV(eV+2m_ec^2)}$, which reduces to the non-relativistic expression when $eV/(m_ec^2)$ is neglected.
Phase and group velocities
A single plane component has the form
\[\psi(x,t)=A e^{i(kx-\omega t)}.\]A surface of constant phase satisfies $kx-\omega t=\text{constant}$, and differentiation gives the phase velocity
\[\boxed{v_p=\frac{dx}{dt}=\frac{\omega}{k}=\frac{E}{p}}.\]A localized particle is represented by neighboring wave numbers, not one infinite plane wave. For two close components, use
\[\cos(k_1x-\omega_1t)+\cos(k_2x-\omega_2t) =2\cos\!\left(\frac{\Delta k\,x-\Delta\omega\,t}{2}\right) \cos(\bar kx-\bar\omega t).\]The slowly varying first factor is the envelope. In the limit $\Delta k\to0$, its speed is
\[\boxed{v_g=\frac{d\omega}{dk}=\frac{dE}{dp}}.\]For the relativistic dispersion relation
\[E^2=p^2c^2+m^2c^4,\]differentiation gives
\[2E\frac{dE}{dp}=2pc^2,\qquad v_g=\frac{pc^2}{E}=v.\]Here $p=\gamma mv$ and $E=\gamma mc^2$. Therefore
\[v_p=\frac Ep=\frac{c^2}{v},\qquad \boxed{v_pv_g=c^2}.\]The phase velocity may exceed $c$, but a plane-wave phase carries neither a localized particle nor a signal; the envelope and its information move at $v_g=v<c$.
If the rest-energy phase is removed and the non-relativistic Schrödinger energy $K=p^2/(2m)$ is used, then
\[\omega=\frac{\hbar k^2}{2m},\qquad v_p=\frac{\hbar k}{2m}=\frac v2,\]while $v_g=d\omega/dk=\hbar k/m=v$. There is no contradiction: adding the constant rest energy $mc^2$ changes the phase rotation but not the group velocity or any probability density.
Davisson-Germer experiment
Davisson and Germer accelerated electrons through a known voltage and scattered them elastically from a nickel crystal. The crystal planes form a diffraction grating. Waves reflected from adjacent planes separated by $d$ travel an extra distance $2d\sin\theta$, so constructive interference requires
\[\boxed{2d\sin\theta=n\lambda},\qquad n=1,2,\ldots .\]Here $\theta$ is the glancing angle measured from the crystal plane. If the apparatus reports the angle $\phi$ between incident and detected beams, the reflection geometry gives $\theta=(\pi-\phi)/2$; the distinction prevents a sign or angle-convention error.
For the well-known $54\,\mathrm V$ peak from nickel, $d\simeq0.091\,\mathrm{nm}$ and $\phi\simeq50^\circ$, hence $\theta\simeq65^\circ$. First-order Bragg diffraction gives
\[\lambda_{\rm Bragg}=2d\sin\theta\simeq0.165\,\mathrm{nm}.\]The de Broglie prediction is
\[\lambda_{\rm dB}=\frac{12.27}{\sqrt{54}}\,\text{angstrom} \simeq1.67\,\text{angstrom}=0.167\,\mathrm{nm},\]in agreement within the experimental and crystal-spacing precision.
The detector records localized electron impacts, yet many impacts form the diffraction distribution of a wave amplitude. Wave-particle duality therefore does not mean that an electron is alternately a classical particle and a classical wave. Propagation and interference are described by amplitudes, whereas detection transfers localized quanta of energy and momentum.
Solved Problems
1. Relativistic de Broglie wavelength at 100 kV
An electron is accelerated from rest through $100\ \mathrm{kV}$. Calculate its relativistic de Broglie wavelength and compare it with the non-relativistic value.
Solution. The electron gains kinetic energy $K=eV>0$. Relativistic energy and momentum obey
\[(pc)^2=(K+m_ec^2)^2-m_e^2c^4 =K(K+2m_ec^2).\]Therefore
\[\lambda_{\rm rel} =\frac{hc}{\sqrt{eV(eV+2m_ec^2)}} =\boxed{3.701\ \mathrm{pm}}.\]The non-relativistic expression gives
\[\lambda_{\rm nr} =\frac{h}{\sqrt{2m_e eV}} =\boxed{3.878\ \mathrm{pm}}.\]Thus the approximation overestimates the wavelength by
\[\frac{\lambda_{\rm nr}-\lambda_{\rm rel}}{\lambda_{\rm rel}}\times100\% =\boxed{4.78\%}.\]Both formulas have dimensions of length because momentum occupies the denominator of $h/p$. Relativity increases the momentum associated with a given accelerating voltage, so the corrected wavelength is shorter. When $eV\ll m_ec^2$, the factor $eV+2m_ec^2$ approaches $2m_ec^2$ and the relativistic expression reduces to the non-relativistic one.
2. Predicting a first-order electron-diffraction maximum
Electrons accelerated through $150\ \mathrm V$ strike crystal planes separated by $d=0.100\ \mathrm{nm}$. Find the first-order Bragg glancing angle, using the non-relativistic approximation.
Solution. Since $150\ \mathrm{eV}\ll511\ \mathrm{keV}$, the non-relativistic wavelength is adequate:
\[\lambda=\frac{h}{\sqrt{2m_e eV}} =0.100137\ \mathrm{nm}.\]With $\theta$ measured from the planes and $n=1$, constructive interference requires
\[2d\sin\theta=\lambda.\]The positive glancing angle is therefore
\[\theta=\sin^{-1}\!\left(\frac{0.100137}{2(0.100)}\right) =\boxed{30.05^\circ}.\]The ratio $\lambda/(2d)$ is dimensionless and is less than one, so a real first-order maximum exists. Increasing $V$ decreases $\lambda$ and hence decreases this glancing angle. In the limiting case $\lambda>2d$, no first-order solution is possible because it would require $\sin\theta>1$.
Descriptive Questions
- Why is the de Broglie wavelength a property of momentum rather than of particle charge?
- Distinguish phase velocity from group velocity and explain which one follows the motion of a localized free particle.
- How does the Davisson-Germer experiment establish wave-particle duality without implying that an electron is a classical material wave?
- Why must the reported detector angle be converted carefully before it is inserted into a Bragg-law convention?
Numerical Problems
- Find the non-relativistic de Broglie wavelength of a proton with kinetic energy $1.00\ \mathrm{keV}$.
Final answer: $\boxed{\lambda=0.905\ \mathrm{pm}}$. - A neutron has de Broglie wavelength $0.180\ \mathrm{nm}$. Find its speed.
Final answer: $\boxed{v=2.20\times10^3\ \mathrm{m\,s^{-1}}}$. - For a non-relativistic free particle moving at $2.00\times10^6\ \mathrm{m\,s^{-1}}$, find the phase and group velocities after removal of the rest-energy phase.
Final answer: $\boxed{v_p=1.00\times10^6\ \mathrm{m\,s^{-1}},\quad v_g=2.00\times10^6\ \mathrm{m\,s^{-1}}}$. - A relativistic particle moves at $0.800c$. Find its matter-wave phase and group velocities.
Final answer: $\boxed{v_p=1.25c,\quad v_g=0.800c}$.
Every added wavelength, velocity, and diffraction value is checked in the MJ-11 problem-verification worksheet; every printed residual and check is zero.
Discussion