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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.

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.

For a charged particle in a uniform magnetic field, the symmetric gauge $\mathbf{A}=\mathbf{B}\times \mathbf{x}/2$ makes the displayed magnetic pseudomomentum a conserved quantity when no other position-dependent potential breaks translation symmetry. It differs from the kinetic momentum $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}=m\dot{\mathbf{x}}$. With $\mathbf{B}=B{\mathbf{e}}_3$, Hamilton's equations for $H=|\mathbf{p}-e\mathbf{A}{|}^2/(2m)$ give ${\dot{p}}_1=(eB/2){\dot{x}}_2$ and ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, proving conservation of $K_1=p_1-eB{x}_2/2$ and $K_2=p_2+eB{x}_1/2$. The charge factor is necessary unless it has already been absorbed into the field convention. In quantum mechanics the transverse components need not commute, even though each commutes with this Hamiltonian operator; magnetic translation algebra records that distinction.

Let $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$ be the kinetic momentum. The quantum Hamiltonian isAssume $qB>0$ and zero longitudinal momentum, as in the displayed states. Sincewe have $[a,a^{†}]=1$ andFor $w=x+iy$,which annihilates everywith $f$ holomorphic. Hence all these states haveThe Landau levels are $E_n=\hslash {\omega }_B(n+1/2)$ for $n=0,1,\dots$; the arbitrary holomorphic function exhibits the degeneracy of the lowest level. For $qB<0$, the normalizable lowest-level states are anti-holomorphic instead.

In Cartesian components, $\partial L/\partial v_i=m{v}_i+e{A}_i$ and $\partial L/\partial x_i=e{v}_j{\partial }_i{A}_j-e{\partial }_i\varphi$. The Euler-Lagrange equation therefore givesThus the minimally coupled particle Lagrangian yields the Lorentz force.

For an arbitrary smooth gauge function $\chi$, take $\mathbf{A}\to \mathbf{A}+\nabla \chi$ and $\varphi \to \varphi -{\partial }_t\chi$. Both fields are unchanged, whileThe action changes only by an endpoint term, so fixed-endpoint variation gives the same motion. Canonical momentum shifts by $e\nabla \chi$, whereas kinetic momentum $m\mathbf{v}$ and the Lorentz force are gauge invariant. This is the relevant gauge invariance of the classical particle dynamics.

Work in temporal gauge, with zero scalar potential, and use the charge sign exactly as in the given covariant spatial derivative. A spatially uniform vector potential acts on a Fourier mode asThere are no spatial derivatives of $\mathbf{A}$. Substitution into the field equation and multiplication by $c^2$ therefore givesThe canonical wavevector $\mathbf{k}$ is conserved by spatial translation symmetry; the kinetic momentum is $\hslash \mathbf{k}-e\mathbf{A}(t)$. The absence of a term involving $\dot{\mathbf{A}}$ in this mode equation follows from temporal gauge and the spatial minimal-coupling operator. The electric field still affects evolution through the changing oscillator frequency.

In temporal gauge, a spatially uniform step $\mathbf{A}(t)={\mathbf{A}}_0\Theta (t)$ gives $\mathbf{E}=-{\mathbf{A}}_0\delta (t)$. A charged scalar Fourier mode retains its canonical wavevector while its kinetic momentum changes from $\hslash \mathbf{k}$ to $\hslash \mathbf{k}-e{\mathbf{A}}_0$. The resulting frequency jump mixes positive- and negative-frequency modes.