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Let $J_k=\int {ϵ}_{k\ell j}{x}_{\ell }{g}_j{d}^{3}x$. Local momentum conservation and integration by parts, with the boundary term vanishing for a localized field, givebecause ${\sigma }_{ij}$ is symmetric. Hence the total electromagnetic angular momentum is independent of time.

With $x^0=ct$, so that ${\partial }_0=c^{-1}{\partial }_t$, the $\nu =0$ component of
${\partial }_{\mu }{T}^{\mu \nu }=0$ isand the $\nu =j$ component isThese are exactly the energy and momentum conservation laws from parts (a) and (b), now combined into conservation of the electromagnetic stress-energy tensor.

Differentiate $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, use the vacuum Maxwell equations, and use $\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{B}=0$. Componentwise vector identities then givewhereThis is the negative of the conventional Maxwell stress tensor; with the sign used here, its divergence is the outward momentum flux.

Dot the vacuum Ampère-Maxwell equation with $\mathbf{E}$ and Faraday's law with $\mathbf{B}/{\mu }_0$. The vector identity
$\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{B}$
gives the Poynting theoremThusis the Poynting vector, the electromagnetic energy-flux density.

For relativistic one-particle energy
$ϵ(p)=\sqrt{p^2{c}^2+m^2{c}^4}$, replace $q_1$ byClassical particle counting still givesSince $q_1^{\mathrm{rel}}$ is proportional to $V$, $\Phi =-k_{B}T⟨N⟩=-pV$, and thereforeThe classical ideal-gas equation of state is independent of the dispersion relation.

Extensivity makes $\Phi (T,V,\mu )$ proportional to $V$. Comparing its differential
$d\Phi =-SdT-pdV-⟨N⟩d\mu$
with Euler scaling gives

The grand potential is

A grand canonical ensemble describes a system that exchanges both energy and particles with a reservoir, at fixed $T,V,\mu$. If microstate $n$ has occupation $N_n$ and one-particle energy $E_n$, thenwith the allowed occupations chosen for the particle statistics.

PutAn even solution is proportional to $\cosh (\kappa x)$ inside the potential barrier. Normalize the right exterior region asContinuity of $\psi$ and ${\psi }^{\prime }$ at $x=a/2$ gives, with
$\gamma =\kappa \tanh (\kappa a/2)$,Solving,Its modulus is one, as expected for the even-parity eigenvalue of the unitary S-matrix.

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.

For incidence from the left, normalize the asymptotic wavefunction byFor incidence from the right, define $r^{\prime },t^{\prime }$ analogously with the directions reversed. Conservation of the probability current givesApplying the same conserved sesquilinear current to the two scattering solutions givesThese are exactly the column orthonormality relations $S^{†}S=I$ for the stated One-dimensional S-matrix, so $S$ is unitary.

The quantum harmonic oscillator Hamiltonian is
$H=\hslash \omega (N+\frac{1}{2})$. Thereforeso the evolved state remains an $A$-eigenstate, with eigenvalue
$\boxed{z{e}^{-i\omega t}}$.

The coherent-state overlap givesFor the classical period $T=2\pi /\omega$, this probability equals one whenever $t=kT$.

The number probabilities area Poisson distribution of parameter $|z{|}^2$. HenceThe relative uncertainty is $1/|z|=1/\sqrt{⟨N⟩}$, which tends to zero as the mean occupation tends to infinity.

Orthogonality of the energy eigenstates givesUsing the lowering-operator action from part (a) and shifting the summation index,Thus this coherent state is a normalized eigenstate of $A$ with eigenvalue $z$.

From the number operator commutator $[N,A]=-A$,so $A|n⟩$ is proportional to $|n-1⟩$. Its squared norm isWith the conventional phase choice,

Differentiating the first auxiliary equation in $t$, the second in $x$, and equating ${\psi }_{xt}$ and ${\psi }_{tx}$ leaves preciselyso the Airy equation is their compatibility condition.

Multiplication of ${\psi }_x-ik\psi =q$ by $e^{-ikx}$ givesHenceFor $\mathrm{Im}k\ge 0$, the kernel $e^{ik(x-y)}$ is bounded when $y\le x$; dominated differentiation makes ${\psi }_{+}$ analytic for $\mathrm{Im}k>0$. Rapid decrease gives

For real $k$, multiply the time equation by $e^{-ikx}$ and let $x\to +\infty$. The rapidly decreasing terms on its right vanish, leavingThereforeand the Fourier inversion theorem gives

Since $L^T=L$ and $B^T=-B$,so the commutator is symmetric. If $Lv=\lambda v$ with $\Vert v\Vert =1$, symmetry permits a differentiable orthonormal eigenbasis locally, andThus a Lax pair evolution is isospectral.

For the harmonic oscillator,With the stated $L$, takeDirect calculation givesThe eigenvalues of $L$ areTheir constancy proves conservation of the oscillator energy $H={\lambda }_{+}^2/2$.