de Broglie wavelength ($\lambda =h/p$)

The de Broglie wavelength is the distance over which a matter wave accumulates one full spatial quantum phase cycle in its propagation direction:Here $h$ is the Planck constant, $\mathbf{p}$ is the momentum, and $\mathbf{k}$ is the angular wavevector. The scalar $p$ in $\lambda =h/p$ means the positive momentum magnitude, not a signed component. Reversing a momentum reverses the wavevector without making the wavelength negative. At zero momentum there is no finite spatial phase period: the limiting wavelength diverges.

This relation makes microscopic diffraction accessible with particles: increasing the momentum shortens the wavelength until it is comparable with the spacings in a crystal scattering problem. The electron accelerating-voltage wavelength gives a practical way to tune this scale, while the Davisson–Germer experiment tests it by comparing electron diffraction directions with independently known crystal geometry. The same relation applies to neutral particles in a neutron interferometer; electric charge is not a requirement for a matter wave.

The wavelength belongs to the wavefunction, not to a sinusoidal path of a particle. A free-particle momentum convention makes the distinction especially clear: a plane wave has a rotating complex quantum phase but a constant probability density. A localized wave packet contains a range of momenta and therefore generally has a range of wavelengths, rather than one exact spatial period.

A material particle carries a quantum phase whose wave vector and angular frequency obey $\mathbf{p}=\hslash \mathbf{k}$ and $E=\hslash \omega$. Interference and diffraction test this relation; a localized particle requires a wave packet.

For a free particle of mass $m>0$ in nonrelativistic quantum mechanics, choose the potential energy zero and writeApplying the momentum operator $-i\hslash \nabla$ returns $\mathbf{p}\psi$; this fixes the sign convention. The angular frequency is measured relative to this chosen energy zero. Adding a constant energy changes only a time-dependent global phase, leaving the spatial de Broglie wavelength unchanged. No rest-energy term is included in this nonrelativistic quantum mechanics convention.

The plane wave has $|\psi {|}^2=|A{|}^2$ and probability current $\mathbf{j}=(\mathbf{p}/m)|A{|}^2$. Its real part can be drawn as a sine or cosine, but that graph is a component of a probability amplitude, not a transverse displacement. On infinite space this is a generalized momentum eigenstate, not a state satisfying wavefunction normalization. A narrow wave packet instead moves with group velocity ${\nabla }_{\mathbf{k}}\omega =\mathbf{p}/m$; the phase velocity $\omega /|\mathbf{k}|=|\mathbf{p}|/(2m)$ is not its transport speed.

The equality of canonical momentum and kinetic momentum here assumes a field-free region with the magnetic vector potential chosen zero. For a charged particle in a magnetic field, the kinetic momentum is $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$. An arbitrary canonical phase gradient must not then be substituted into $m\mathbf{v}$; the free particle formula needs its stated assumptions.

Let an electron start with negligible kinetic energy and move through an increase $U>0$ in electric potential. Its electric charge is $q=-e$, where $e>0$ is the elementary charge, so the decrease in potential energy is $eU$. In a field-free region after acceleration,This is an electron formula, not a universal conversion from voltage to wavelength. It assumes negligible initial kinetic energy, negligible losses, and $eU\ll m_e{c}^2$, where $c$ is the speed of light. It describes the de Broglie wavelength outside the crystal; the potential energy inside a material can change the local kinetic momentum.

For $U=54\mathrm{V}$, take $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, and $m_e\simeq 9.1093837\times 10^{-31}\mathrm{kg}$. The rounded electron mass is consistent with the 2022 recommended constants.^[ref]The electronvolt is an energy unit: $54\mathrm{eV}$ must be converted to joules before combining it with $m_e$ in SI units. Dimensional analysis checks both steps:The resulting velocity is about $4.3584\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, or $0.01454c$, so the nonrelativistic assumption is consistent. For a larger accelerating voltage, the relativistic energy-momentum relation gives $pc=\sqrt{K(K+2m_e{c}^2)}$, with $K=eU$, while $\lambda =h/p$ still holds. At $54\mathrm{V}$ the relativistic wavelength is shorter by about $0.00264\%$.

Explicitly, the exact special relativity formula and its nonrelativistic quantum mechanics limit areHere $K=eU$ is the gained kinetic energy; the total energy is $E=m_e{c}^2+K$. At $80\mathrm{kV}$, the nonrelativistic quantum mechanics expression gives $4.336\mathrm{pm}$, about $3.84\%$ larger than the exact $4.176\mathrm{pm}$. This difference matters when using electron diffraction ring positions to infer lattice plane spacings.