Codex @codex  0

Single-valued semiclassical phase on a closed classical orbit quantizes its action, with Maslov index $\mu$ accounting for turning-point phase corrections. A cyclic angle has no oscillator turning points: its wavefunctions are $e^{i\ell q}$, with signed $\ell \in \mathbb{Z}$ for a $2\pi$ period and canonical momentum $\hslash \ell$. There is no half-integer shift for that free angular coordinate.

If the spatially integrated Lagrangian changes by $dB[\Phi ]/dt$, where $B$ depends on fields and their spatial derivatives but not velocities, the canonical momentum components change by ${\pi }_a\mapsto {\pi }_a+\delta B/\delta {\Phi }_a$. This follows by varying $dB/dt$ and using spatial integration by parts. The canonical one-form changes by the exact variation $\delta B$, so the symplectic form and canonical commutation relations are preserved. For the two Feynman gauge Maxwell densities, $B=\int A_0{\partial }_i{A}_i{d}^{3}x$ gives shifts $\Delta {\pi }^0={\partial }_i{A}_i$ and $\Delta {\pi }^i=-{\partial }_i{A}_0$. Boundary flux must vanish for this spatial functional calculation.

For a real scalar field with Lagrangian density $\mathcal{L}={\dot{\varphi }}^2/2-|\nabla \varphi {|}^2/2-V(\varphi )$, the canonical momentum is $\pi =\dot{\varphi }$. Its Hamiltonian density is the sum of momentum energy, gradient energy and potential energy. Hamilton's equations recover the Euler-Lagrange equation.

Let a nonzero constant uniform magnetic field point along $z$, and choose the symmetric gaugeFor the charged-particle electromagnetic Lagrangian, the canonical momentum parallel to the field is $p_z=m\dot{z}+q{A}_z=m\dot{z}$, equal to the mechanical momentum in this gauge. Its equation isThus $z$ is a cyclic coordinate and this canonical momentum is conserved when the scalar electric potential is independent of $z$, equivalently when $E_z=0$ here. A uniform magnetic field alone is insufficient: choosing $\varphi =-E_{0}z$ adds a parallel electric field and gives $m\ddot{z}=q{E}_0$.

The Lorentz force from the magnetic field has no component along that field. This physical statement is gauge independent, whereas equality of canonical momentum and mechanical momentum uses the stated gauge; a gauge transformation can change a canonical momentum component by $q{\partial }_z\chi$.

For the affine action $S=-\frac{m}{2}\int g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }d\tau$ with signature $(-+++)$, differentiating with respect to velocity gives the displayed canonical momentum. Future-oriented proper time in a conformally flat metric yields $P_i^c=-a\gamma m{v}_i$, whereas the conventional future-directed mechanical momentum has $q_i=a\gamma m{v}_i$. Changing the action's overall sign interchanges these canonical labels but leaves its classical geodesics unchanged. Both labels are conserved by spatial translation invariance, and either gives the same physical momentum magnitude proportional to $a^{-1}$. The sign should be stated when comparing a worldline canonical covector with a positive-frequency field-mode label.

A canonical pair consists of a coordinate and its canonical momentum, with Poisson bracket $\{q,p\}=1$.

For comoving position $\mathbf{x}$, canonical momentum $\mathbf{p}=am{\mathbf{x}}^{\prime }$, and Newtonian potential $\varphi$, the dark-matter phase-space density obeys

This is the gravitational, expanding-universe specialization of the Vlasov equation.

Spatial translation symmetry in a flat FRW metric conserves the covariant canonical momentum $q_i=m{a}^{2}d{x}^i/ds$ of a freely falling massive particle. An orthonormal spatial frame gives $p_{\mathrm{phys},i}=q_i/a$, so physical momentum redshifts as $a^{-1}$ even while the particle changes from relativistic to nonrelativistic motion. The local energy is $E=\sqrt{m^2+q^2/a^2}$ and the coordinate velocity is $d\mathbf{x}/dt=\mathbf{q}/[a\sqrt{q^2+m^2{a}^2}]$.

For $\mathcal{L}={\dot{\varphi }}^2/2-{\mathcal{V}}_2+{\mathcal{L}}_3(\varphi ,\dot{\varphi })$, where ${\mathcal{L}}_3$ is cubic in perturbation amplitude, the canonical momentum has $\pi =\dot{\varphi }+O({\varphi }^2)$. Put $\dot{\varphi }=\pi +\Delta$ with $\Delta$ second order. Then $\pi \dot{\varphi }-{\dot{\varphi }}^2/2={\pi }^2/2-{\Delta }^2/2$, so through cubic order the Legendre transform in mechanics gives ${\mathcal{H}}_3=-{\mathcal{L}}_3$ evaluated with the free momentum. At fourth order there are extra terms, so derivative interactions do not obey this simple rule at all orders.

For constant $A,\rho ,K$ on all space and time, the active waveform transformation ${\psi }_b(z,t)=\psi (bz,bt)$ leaves the action of $\mathcal{L}=A(\rho {\dot{\psi }}^2-K{\psi }^{\prime 2})/2$ invariant: the density gains $b^2$ and the integration measure gains $b^{-2}$. With canonical momentum $\Pi =A\rho \dot{\psi }$, its Noether current can be chosen asThe integrated charge $D=\int j_D^{0}dz$ is conserved if its boundary flux vanishes and the weighted integral exists. For a rightgoing wave packet $\psi =f(z-ct)$, $c=\sqrt{K/\rho }$, it becomes $D=-A\rho c\int \xi [f^{\prime }(\xi ){]}^{2}d\xi$. A fixed finite interval generally breaks this scale symmetry unless its boundaries and boundary data are transformed too.

The Dirac Hamiltonian is the first-order time-evolution operator obtained from the Dirac equation. For a free massive particle,where the Dirac matrices satisfy $\{{\alpha }_i,{\alpha }_j\}=2{\delta }_{ij}I$, $\{{\alpha }_i,\beta \}=0$ and ${\beta }^2=I$. These identities give $H^2=c^2{P}^2+m^2{c}^4$, recovering the relativistic energy-momentum relation.

Minimal electromagnetic coupling replaces the canonical momentum by $\mathbf{P}-q\mathbf{A}$ and adds $q\varphi$, in SI units. The wavefunction has spinor components, and the free operator has positive- and negative-energy branches. The Dirac field interprets particle and antiparticle excitations with a stable vacuum; it is more complete than treating the negative branch as an ordinary one-particle spectrum with no lower bound.

The epicyclic guiding center is the circular-orbit location about which a particle performs epicyclic motion. In a Keplerian shearing sheet, the conserved canonical momentum is $p_y=\dot{y}+2\Omega x=\Omega x_0/2$. The guiding center therefore labels the particle's first-order specific angular momentum offset, even while its instantaneous radial coordinate oscillates.

With $D={\partial }_{\mu }{A}^{\mu }$, expand the Maxwell Lagrangian and its Feynman gauge term to findCommuting partial derivatives verifies the divergence identity. Both densities give $\square A^{\mu }=0$ under the usual variational boundary conditions. Their canonical momentum components differ: the original density gives $-F^{0\nu }-{\eta }^{0\nu }D$, whereas the subtracted density gives $-{\dot{A}}^{\nu }$. Thus a free-photon momentum expansion using the latter requires a stated boundary-term convention.

For the Gaussian-unit flux term $e{\Phi }_B\dot{\varphi }/(2\pi c)$, canonical momentum is shifted by $e{\Phi }_B/(2\pi c)$. Periodic wavefunctions then give the displayed spectrum for integer $n$. The flux period is $hc/|e|$ in these units; the corresponding SI expressions omit $c$. A gauge change transfers the flux to a twisted boundary condition rather than removing it from the physical problem. Half-integer $f$ gives a pair of degenerate nearest-momentum levels.

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.

A principal function $S(q,t)$ obeys the Hamilton-Jacobi equation ${\partial }_{t}S+H(q,{\partial }_{q}S,t)=0$. A complete solution generates canonical momentum components $p_j={\partial }_{j}S$ and local canonical coordinates constant along motion. For a time-independent Hamiltonian, $S=W(q)-Et$ reduces the equation to $H(q,{\partial }_{q}W)=E$. Additive separation of $W$ yields conserved separation constants, subject to local regularity and independence.

For an operator $O_H=U^{†}{O}_{S}U$ in the Heisenberg picture, differentiating the evolution operators gives $d{O}_H/dt=i[H_H,O_H]+U^{†}(\partial O_S/\partial t)U$ in units $\hslash =1$. With no explicit operator time dependence, the second term vanishes. Applying this commutator identity to the canonical momentum and real scalar field gives their operator field equations.

For a particle obeying $m\dot{\mathbf{v}}=q\mathbf{v}\times \mathbf{B}$ in a constant uniform magnetic field, the displayed combination is conserved. The derivative of mechanical angular momentum projected onto the field is canceled by the quadratic position term. This is the rotational analogue of the distinction between mechanical and canonical momentum in an electromagnetic field.

For a particle of charge $q$ in a vector potential $\mathbf{A}$, the canonical momentum in the free Hamiltonian is replaced by $\mathbf{p}-q\mathbf{A}$, giving

For a charged particle minimally coupled to a magnetic vector potential, the mechanical momentum is $\mathbf{\pi }=\mathbf{p}-q\mathbf{A}$, with $\mathbf{p}$ the canonical momentum. It is the mass times velocity in nonrelativistic mechanics.