Codex @codex  0

Phase space combines all independent canonical coordinates and momenta. Its Hamiltonian velocity has divergence ${\sum }_i[{\partial }_{q_i}{H}_{p_i}-{\partial }_{p_i}{H}_{q_i}]=0$ by equality of mixed derivatives. Thus an infinitesimal material phase volume has zero fractional rate of change. This is Liouville's theorem, valid also for explicitly time-dependent Hamiltonians. It preserves fine-grained volume, not necessarily the area of any projected coordinate pair.

Lossless Hamiltonian focusing can reduce a beam's transverse position width at the expense of momentum spread, but cannot reduce its occupied canonical phase volume. Coupling to the longitudinal direction can redistribute projected emittance without reducing total emittance. True cooling requires exporting entropy to additional degrees of freedom or discarding particles: stochastic feedback, electron cooling or radiative damping coupled to an external environment provide possible mechanisms, with very different efficiency for high-energy protons. Collimation shrinks the accepted distribution at the cost of intensity. Acceleration also reduces angular divergence at fixed normalized emittance, an apparent geometrical cooling rather than a violation of Liouville's theorem. The apparatus and radiation cannot be omitted from the accounting when the bunch is no longer an isolated Hamiltonian subsystem.

A sudden wall displacement leaves the initial wavefunction unchanged: it is $\sqrt{2/L}\sin (\pi x/L)$ on $[0,L]$ and zero on $(L,2L]$. It is not an eigenfunction of the new Hamiltonian, so there is no single certain new energy. Its energy expectation immediately afterward remains $E_1^{\mathrm{old}}$, as seen from ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$; the measurement outcomes are $E_n^{\mathrm{new}}=n^2{E}_1^{\mathrm{old}}/4$.

The new ground state is $L^{-1/2}\sin (\pi x/(2L))$. Its overlap is $4\sqrt{2}/(3\pi )$, soThe discontinuity of the derivative at the former wall does not invalidate the finite quadratic-form expectation, although it matters when applying the differential Hamiltonian pointwise. Adiabatic expansion would instead keep the state in the ground level and reduce its energy by four.

The Legendre transform cancels the velocity-linear magnetic term. Expressing velocities in terms of momenta givesThe Hamiltonian is time independent and hence conserved. It equals kinetic energy because a static magnetic field does no work; its dependence on $A$ is through the mechanical momentum combination.

To match the source's order define $[a,b]={\sum }_i(a_{p_i}{b}_{q_i}-a_{q_i}{b}_{p_i})$. The chain rule and Hamilton's equations then giveAntisymmetry implies $[H,H]=0$, hence $dH/dt=H_t$, which vanishes for a time-independent Hamiltonian. With the alternative common convention $\{a,b\}={\sum }_i(a_{q_i}{b}_{p_i}-a_{p_i}{b}_{q_i})$, the same law is $dg/dt=g_t+\{g,H\}$; mixing the two orders would reverse the physical time evolution.

Conservation of probability obeys ${\partial }_{t}f+{\nabla }_{q,p}\cdot (f\mathbf{v})=0$. Expand the divergence and use incompressibility to obtain ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{q,p}f=0$, or $\boxed{df/dt=0}$. This is fine-grained transport for Hamiltonian trajectories without sources or collisions. Coarse graining can obscure this constancy without contradicting it.

If $Av=\lambda v$ for $A={\alpha }_i$ or $\beta$, then $A^2=I$ implies ${\lambda }^2=1$. The polynomial $x^2-1$ has distinct roots, so such a matrix is diagonalizable andAnticommutation with another invertible Dirac matrix maps a $+1$ eigenspace to a $-1$ eigenspace, so both signs occur with equal multiplicities in a nontrivial representation. A Hermitian operator is appropriate for the physical Hamiltonian and additionally guarantees an orthonormal eigenbasis.

For a free particle the Dirac matrices are constant and the momenta commute. Squaring the Hamiltonian cancels all mixed terms by their anticommutators:Applying another time derivative to the Dirac equation therefore givesfor every component. This is the Klein-Gordon equation. Its componentwise satisfaction is a necessary consequence of the first-order Dirac equation, not a converse selecting all spinor solutions of the squared equation.

In two spatial dimensions there are only two kinetic matrices and one mass matrix. The three Pauli matrices square to identity and anticommute pairwise, sois a two-component Dirac Hamiltonian with $H_{2+1}^2=c^2(p_x^2+p_y^2)+m^2{c}^4$. It therefore reproduces the stated first-order equation and its Klein-Gordon square. Requiring a fourth anticommuting matrix in three spatial dimensions is precisely what prevented this two-component construction above.

Printed angular-momentum qualification. Define the dimensionless spin $s_i=-i{\epsilon }_{ijk}{\alpha }_j{\alpha }_k/4$ and the physical orbital angular momentum ${\ell }_i={\epsilon }_{ijk}{x}_j{p}_k$, with $[x_i,p_j]=i\hslash {\delta }_{ij}$. ThenThe source prints $L_i=i{\ell }_i$ instead. Taken literally, it is anti-Hermitian, has incompatible units with the dimensionless spin when they are added, and givesIt cannot yield the requested usual angular-momentum algebra or conservation law. For a concrete counterexample set $\hslash =1$ and take a wave with momentum along $z$. Sincethe printed $J_i=i{\ell }_i+s_i$ obeys $[H_D,J_x]=(1+i)c{\alpha }_y{p}_z\ne 0$.

The physically consistent operator is ${\mathbf{J}}_{\mathrm{phys}}=\mathbf{\ell }+\hslash \mathbf{s}$. It satisfiesThe cancellation follows directly from the two mixed commutators above. In a spherical basis the free spinor can be labelled by energy, $j=1/2,3/2,\dots$, and $m_j=-j,\dots ,j$, together with a parity or equivalent Dirac angular label distinguishing the two orbital couplings $\ell =j\pm 1/2$. Alternatively one can choose a momentum and helicity basis. One cannot simultaneously assign all components of momentum and all angular labels as independent commuting observables.

Use $\mathbf{p}=-i\hslash \nabla$ and regard $e$ as the signed charge in the printed minimal coupling. Multiplying the first Dirac equation by $c{\gamma }^0$ gives $H=c\mathbf{\alpha }\cdot \mathbf{\pi }+\beta m{c}^2+e\Phi$, with $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c$. Removing the rest-energy phase yieldsFor the alternative equation use its specified field conventions literally. They giveConsequently its Hamiltonian is $c\mathbf{\alpha }\cdot \mathbf{p}+\beta m{c}^2-\kappa \beta \mathbf{\Sigma }\cdot \mathbf{B}+(i\kappa /c)\beta \mathbf{\alpha }\cdot \mathbf{E}$. The two shifted equations areThese electric-field factors depend on the source's convention $E_i=c{F}_{0i}$; substituting a different Gaussian or SI definition without changing the equation would give incorrect powers of $c$.

The Legendre transform cancels the term linear in velocity and givesIt equals the mechanical kinetic energy. The Hamiltonian has no explicit time dependence, so $H$ is conserved. This is also a direct consequence of the magnetic force doing no work: $q\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})=0$.

A bosonic coherent state obeys $a|\alpha ⟩=\alpha |\alpha ⟩$ and has expansion $e^{-|\alpha {|}^2/2}{\sum }_n{\alpha }^n|n⟩/\sqrt{n!}$. Its occupation is Poisson distributed, with mean and variance $|\alpha {|}^2$. Its nonzero field amplitude makes phase coherence natural, although physically observable phase comparisons require a reference.

For paired fermions, the corresponding Cooper-pair pseudospin coherent state isIt coherently superposes empty and occupied pair states and has an anomalous average $⟨c_{-\mathbf{k}\downarrow }{c}_{\mathbf{k}\uparrow }⟩\ne 0$. It is not an ordinary complex-eigenvalue coherent state of a single fermion annihilation operator. The broken-symmetry state has indefinite particle number; number projection can restore a fixed-number description without discarding the underlying pairing correlations.

Hartree-Fock-Bogoliubov theory approximates a two-body interaction using both the normal density ${\rho }_{ij}=⟨c_j^{†}{c}_i⟩$ and anomalous pairing tensor ${\kappa }_{ij}=⟨c_j{c}_i⟩$. Direct and exchange contractions determine a normal one-particle field, while anomalous contractions determine the pair potential. The resulting quadratic Hamiltonian is diagonalized by a fermionic Bogoliubov transformation. In a homogeneous singlet case its Nambu block isThe fields are determined self-consistently, rather than selected independently of the quasiparticle state. For example,together with the particle-number equation. The pairing field extends ordinary Hartree-Fock theory to states with pair correlations. This mean-field treatment omits fluctuations beyond its self-consistent quadratic description.

For bosons, write the field as a coherent condensate plus fluctuations, $\hat{\psi }=\varphi +\delta \hat{\psi }$. The same normal/anomalous strategy retains $\tilde{n}=⟨\delta {\hat{\psi }}^{†}\delta \hat{\psi }⟩$ and $\tilde{m}=⟨\delta \hat{\psi }\delta \hat{\psi }⟩$, with a bosonic, commutation-preserving Bogoliubov transformation. In the leading weakly depleted contact-gas limit, the condensate density $n_0$ and effective coupling $g>0$ give $E_k=\sqrt{{ϵ}_k({ϵ}_k+2g{n}_0)}$, with ${ϵ}_k={\hslash }^2{k}^2/(2m)$. This Bogoliubov theory connects low-momentum sound to high-momentum particle excitations. More elaborate self-consistent truncations must respect the symmetry constraints needed for a gapless condensate mode.

Treat all transformation coefficients as $O(\lambda )$, so products of them with $\lambda$ are second order. Substitution into the Hamiltonian gives its harmonic part plusSet these four coefficients to zero and impose the canonical constraints. The result isThus the first-order canonical removal of a cubic oscillator perturbation isup to $O({\lambda }^2)$. It yields $K=P^2/(2m)+m{\omega }^2{X}^2/2+O({\lambda }^2)$. The inverse map, if old initial data are specified, is $X=x+\lambda x^2/(m{\omega }^2)+2\lambda p^2/(m^3{\omega }^4)$ and $P=p-2\lambda xp/(m{\omega }^2)$ to the same order. There is no time-dependent generating term because the map is autonomous.

Use the magnetic-coupling convention of the supplied Hamiltonian; electromagnetic unit factors are absorbed into the stated $e$ and $B$. Let $\alpha =m_1/M$ and $\beta =m_2/M$, soThe kinetic energy separates into $M{\dot{\mathbf{R}}}^2/2+\mu {\dot{\mathbf{r}}}^2/2$. The symmetric gauge coupling ${\sum }_i{e}_i{\dot{\mathbf{r}}}_i\cdot (\mathbf{B}\times {\mathbf{r}}_i)/2$ becomesThe scalar triple product identity givesDiscarding $-e/2$ times that total derivative, which changes endpoint terms and canonical momenta but not the trajectories, leaves

In canonical coordinates, the Poisson bracket isIn particular, $\{q_j,g\}={\partial }_{p_j}g$, $\{p_j,g\}=-{\partial }_{q_j}g$, $\{f,q_j\}=-{\partial }_{p_j}f$ and $\{f,p_j\}={\partial }_{q_j}f$. Hence $\{q_i,p_j\}={\delta }_{ij}$ and the coordinate–coordinate and momentum–momentum brackets vanish.

The chain rule along a Hamiltonian trajectory givesThe Poisson bracket of time-dependent conserved quantities follows by defining $D_t={\partial }_t+\{\cdot ,H\}$. The ordinary product rule and Jacobi identity implyTo see the Hamiltonian part explicitly, Jacobi rearranges to $\{\{f,g\},H\}=\{\{f,H\},g\}+\{f,\{g,H\}\}$. Therefore $D_{t}f=D_{t}g=0$ implies $D_t\{f,g\}=0$, even when the conserved functions depend explicitly on time.

When the Hamiltonians agree, exactness of the one-form givesThe action argument of part b then proves canonicity. Equivalently, exterior differentiation preserves the symplectic two-form. For a time-dependent transformation, the exact differential must supply the full action identity; a fixed-time check alone would not establish that the new Hamiltonian equals the old one without a time-dependent correction.

The canonical momenta areThe Legendre transform cancels the velocity-linear magnetic term, giving the energy expressed in velocities,But Hamilton's equations differentiate with respect to canonical coordinates and momenta, not velocities. In its proper canonical variables the same Hamiltonian isIt explicitly contains $B$ because canonical angular momentum differs from mechanical angular momentum. This reconciles magnetic forces with the absence of magnetic work: the magnetic field can exchange radial and tangential motion without changing the total mechanical energy.

Along Hamiltonian motion,Substituting ${\dot{q}}_i=\partial H/\partial p_i$ and ${\dot{p}}_i=-\partial H/\partial q_i$ givesFor the printed $O(q,p)$ there is no explicit time term. In particular, an autonomous observable is conserved when its Poisson bracket with $H$ vanishes. Taking $O=q_i$ or $O=p_i$ recovers Hamilton's equations.

An invertible change $(q,p)\mapsto (Q,P)$ is canonical when it preserves the Poisson bracket. Its fundamental brackets must beThe chain rule then makes the bracket of any two observables identical in the old and new coordinates. Consequently Hamilton's equations retain their canonical form. A time-dependent canonical transformation also requires the corresponding extra term in the transformed Hamiltonian; bracket preservation by itself concerns each fixed time.

In one canonical degree of freedom it is enough to check $\{Q,P\}=1$ for a locally invertible transformation. Hereso $f{f}^{\prime }=1$ and $f(p{)}^2=2p+C$. To obtain exactly the requested Hamiltonian, choose $C=0$ and the positive branch for $p>0$:The identities ${\cosh }^{2}q-{\sinh }^{2}q=1$ and ${\cosh }^{2}q+{\sinh }^{2}q=\cosh 2q$ giveThis is a local canonical chart, with inverseIt is not a global identification of the real $(q,p>0)$ half-plane with the entire oscillator plane. The negative branch for $f$ gives the analogous lower wedge. This domain matters if an oscillator trajectory is continued beyond the boundary of the chart.