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The Lorenz gauge is ${\partial }_{\mu }{A}^{\mu }=0$. Substituting the electromagnetic four-potential into the covariant Maxwell equations then removes the mixed derivative and givesEquivalently,

Let $g(E)$ denote the total density of states, including both spin directions. The field lowers one spin energy by ${\mu }_{B}B$ and raises the other by the same amount. To first order in $B$, each spin population changes by the one-spin density $g(E_F)/2$ times the displacement of its Fermi edge. The population imbalance is thereforeMultiplying by the Bohr magneton carried by each excess aligned electron givesThus, with this total-density convention,The zero-temperature magnetic susceptibility isThis weak positive response is Pauli paramagnetism. If susceptibility per unit volume is required, both $M$ and $g$ are divided by $V$.

Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.

Choose the $z$-axis along $\mathbf{E}$, so the perturbation isup to an irrelevant sign convention for the electron charge. By second-order nondegenerate perturbation theory,Parts (a) and (b) restrict the contributing bound states toFor hydrogen,and henceThereforeEvery denominator is negative, so the discrete-state contribution lowers the ground-state energy.

The hydrogen ground-state charge cloud is spherically symmetric and has even parity, so it has no permanent electric dipole moment. A uniform electric field couples through a perturbation proportional to $\mathbf{E}\cdot \mathbf{r}$, which is odd under parity. Its expectation in the ground state therefore vanishes, giving no first-order energy correction.

The field can nevertheless polarize the atom: it admixes odd-parity excited states and creates an induced dipole. Such mixing contributes from second order onward, producing the quadratic Stark effect of the hydrogen ground state.

WriteTaking the stated double-commutator identity between the two eigenstates gives, if the matrix element is nonzero,For nonnegative integers $\ell ,{\ell }^{\prime }$, this equation has precisely the possibilitiesIndeed, ${\ell }^{\prime }=\ell$ forces $\ell =0$, while for unequal values the difference
$({\ell }^{\prime }-\ell )({\ell }^{\prime }+\ell +1)$ satisfies the equation only when its first factor has magnitude one.

Under parity, $z\mapsto -z$, while a hydrogen state of orbital angular momentum $\ell$ has parity $(-1{)}^{\ell }$. Therefore a nonzero matrix element requiresso ${\ell }^{\prime }$ and $\ell$ must have opposite parity. This excludes every case ${\ell }^{\prime }=\ell$, including ${\ell }^{\prime }=\ell =0$. Combining the results gives

In position representation,which differentiates only with respect to $x$ and $y$. Multiplication by $z$ therefore commutes with it:Taking the matrix element of this commutator between angular-momentum eigenstates givesThus the matrix element can be nonzero only whenThis is the magnetic part of the electric-dipole selection rules for hydrogen.

The Poincaré-Bendixson theorem states that a nonempty compact $\omega$-limit set of a planar $C^1$ flow which contains no equilibrium is a periodic orbit.

The origin is the only equilibrium: at a nonzero equilibrium, putting $q=x^2+y^2-a$ would requirebut its determinant $q^2+3$ is positive. By local asymptotic stability, choose a small simple closed Lyapunov level curve on which the original vector field points inward. On a sufficiently large circle,so the original vector field points outward.

Reverse time. The annulus between these two curves is then a compact positively invariant trapping region: the reversed field points into it at both boundaries. It contains no equilibrium. The Poincaré--Bendixson theorem applied to any reversed trajectory in the annulus produces a periodic orbit. Reversing time does not change its image, so the original system also has a periodic orbit.

For $k=1$, the interval in part (b) contains one exactly when $a>1$. With $V=x^2+y^2=r^2$,because $2xy\le r^2$. Hence $\dot{V}<0$ throughoutEvery trajectory starting in the disc $x^2+y^2<a-1$ remains there and approaches the origin. Thus this disc is contained in the basin of attraction of the origin.

A Lyapunov function for an equilibrium is a continuously differentiable function that is positive definite there and whose derivative along nonconstant trajectories is negative definite in a neighbourhood.

For $V=x^2+k{y}^2$, positivity requires $k>0$, andThe quartic term is negligible sufficiently close to the origin, so $V$ is a local Lyapunov function exactly whenis positive definite. Its determinant condition isThe two roots of equality areTherefore

The Jacobian at the origin iswhose eigenvalues areTheir real parts are negative because $a>0$. The linearization stability theorem therefore makes the origin a locally asymptotically stable hyperbolic sink.

PutOn $[0,2]$, its global maximum is the endpoint $t=2$, whereand the amplitude $\log t$ has value $\log 2$. The endpoint form of Laplace's method therefore givesIndeed, with $u=2-t$, Taylor expansion gives $\varphi (2-u)=2-5u+O(u^2)$ and $\log (2-u)=\log 2+O(u)$; the scale contributing to the integral is $u=O(x^{-1})$. Outside any fixed neighbourhood of $2$, $\varphi$ is bounded above by $2-\delta$, so that part is exponentially smaller. This also controls the integrable logarithmic singularity at zero.

Statement (iii) is false. On the unit disc, takeThe function $f$ is continuous on the closed disc, but $\varphi$ has its strict maximum $1$ at the interior point $(0,0)$ and equals zero on the boundary.

Statement (ii) is true. Since $\Delta (-\varphi )=0$, applying the maximum principle to $-\varphi$ gives

Because $\overline{\Omega }$ is compact, $\varphi$ attains a maximum. If a maximizer lay in $\Omega$, its Hessian matrix would be negative semidefinite, so its trace would satisfy $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. Hence every strictly subharmonic function under the stated hypotheses attains a maximum on $\partial \Omega$.

Statement (i) is true. For $ϵ>0$, the functionhas $\Delta {\varphi }_{ϵ}=4ϵ>0$, soLetting $ϵ\downarrow 0$ givesThis is the maximum principle for harmonic functions.