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Heat capacity at fixed volume is $C_V=(\partial E/\partial T{)}_{V,N}=T(\partial S/\partial T{)}_{V,N}$. From the Helmholtz free energy differential $dF=-SdT-PdV$ at fixed particle number, equality of mixed derivatives gives $(\partial S/\partial V{)}_T=(\partial P/\partial T{)}_V$. Differentiating once more givesFor the canonical partition function, $F=-k_{B}T\log Z$, $E=k_B{T}^2{\partial }_T\log Z$, and $S=k_B(\log Z+T{\partial }_T\log Z)$. Thus, for $V>aN$,andNo Stirling approximation is needed. The heat capacity is independent of volume, while the pressure is linear in temperature, so both sides of the Maxwell-derived identity vanish.

For a periodic potential, the Hamiltonian commutes with every lattice translation. The three primitive translations are commuting unitary operators, so within each energy eigenspace one can choose simultaneous translation eigenstates. Their eigenvalues have modulus one and can be written $e^{i\mathbf{k}\cdot {\mathbf{a}}_i}$. Translation by any lattice vector then givesThis proves Bloch's theorem in its eigenbasis form; arbitrary superpositions inside a degenerate eigenspace need not themselves have a single Bloch wavevector.

The reciprocal lattice consists of vectors $\mathbf{g}$ with $\mathbf{g}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$. Changing $\mathbf{k}$ to $\mathbf{k}+\mathbf{g}$ and $u$ to $e^{-i\mathbf{g}\cdot \mathbf{r}}u$ gives the same physical wavefunction with a periodic new $u$. The eigenvalue problem on one cell has a discrete set of levels, giving bands indexed by $n$, and this equivalence yields $E_n(\mathbf{k}+\mathbf{g})=E_n(\mathbf{k})$. A reciprocal-lattice fundamental domain, conventionally a Brillouin zone, therefore labels inequivalent wavevectors.

Periodic boundary conditions in physical volume $\mathcal{V}$ give one allowed wavevector per reciprocal-space volume $(2\pi {)}^3/\mathcal{V}$. Including the two spin states, there are $2\mathcal{V}{d}^{3}k/(2\pi {)}^3$ states; per unit physical volume the count is the printed $2d^{3}k/(2\pi {)}^3$. Each occupied state carries charge $-e$ and velocity ${\hslash }^{-1}{\nabla }_k{E}_n$, soFor a full band, the integral of ${\nabla }_k{E}_n$ is a boundary integral. Opposite faces of a reciprocal fundamental cell have equal energies by periodicity and opposite normals, so their contributions cancel. Hence a full band carries zero electric current.

With $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$, the matrix of $S_z\cos \theta +S_x\sin \theta$ is $(\hslash /2)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right)$. Multiplication by $(c,s{)}^T$ and $(-s,c{)}^T$ gives eigenvalues $+\hslash /2$ and $-\hslash /2$, using the double-angle identities.

The one-particle transformation is the matrix $R=\left(\begin{array}{cc}c& -s\\ s& c\end{array}\right)$, with determinant one. The antisymmetric tensor $|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩$ transforms by this determinant under $R\otimes R$, so the singlet is unchanged. The total $S_z$ kills it. Its invariance under these rotations implies that total $S_z\cos \theta +S_x\sin \theta$ also kills it for every $\theta$, so total $S_x$ kills it. The spin commutator then gives total $S_y$ acting as zero. Therefore $S^2|\chi ⟩=0$ and it has total spin zero.

Rotating only particle B producesThe measurement probabilities are thereforeEach arrow represents the corresponding value $\pm \hslash /2$. The sequential measurements commute because they act on different particles. The two parallel outcomes leave triplet states with total spin one and $S^2=2{\hslash }^2$. Each antiparallel product state is a superposition of the spin-zero singlet and the spin-one zero-projection triplet, and is not an $S^2$ eigenstate. Outcomes of zero probability at a special angle do not actually occur.

Orient the circle counterclockwise and define its Cauchy transformThe inside and outside limits satisfy ${\Phi }_{+}-{\Phi }_{-}=\varphi$ and ${\Phi }_{+}+{\Phi }_{-}=(\pi i{)}^{-1}\mathrm{PV}{\int }_C\varphi (\tau )/(\tau -t)d\tau$. Also ${\Phi }_{-}(z)=O(z^{-1})$ at infinity. Applying these formulas to the two polynomial coefficients and dividing by $t^2$ gives the Riemann–Hilbert conditionThe homogeneous jump is $G=t/(t^2-1)$. A canonical factorization of a rational scalar Riemann-Hilbert jump isThe inside factor is analytic and nonvanishing; the exterior factor has no zeros or poles outside the circle and grows like $z$. The jump index is $1-2=-1$, consistently with this growth.

Put $F_{+}={\Phi }_{+}$ and $F_{-}={\Phi }_{-}/X_{-}$. Their additive jump is $R(t)=((A-1)t+1)/(t^2-1)$. Since $R$ is analytic outside the circle and decays there, the functions $F_{+}$ inside and $F_{-}+R$ outside glue to an entire function. This function tends to zero at infinity, so Liouville's theorem gives $F_{+}=0$, $F_{-}=-R$. ConsequentlyThe required decay of a Cauchy transform forces $A=1$. Then ${\Phi }_{-}=-1/z$, and the jump givesFor the homogeneous problem, ${\Phi }_{+}/X_{+}$ and ${\Phi }_{-}/X_{-}$ glue to an entire function vanishing at infinity, so only the zero solution exists; this proves uniqueness. Direct verification uses the principal-value integral $-\pi i/t$: the two terms on the left combine to $(t^4+t^3-t^2-t^4+t^3+t^2)/(2t)=t^2$, exactly the right side at $A=1$.

A fundamental solution of a constant-coefficient operator $P(D)$ is a distribution $E$ satisfying $P(D)E={\delta }_0$. The proposed $N$ is locally integrable in three dimensions and harmonic away from zero. For a smooth compactly supported test function $\phi$, apply Green's identity outside the ball of radius $\eta$. On the inner boundary the outward normal is $-\hat{r}$, so ${\partial }_{\nu }N=1/(4\pi {\eta }^2)$. The outer boundary contributes nothing. ThusThe first term tends to $\phi (0)$ and the second is $O(\eta )$. Local integrability permits taking the limit on the left. Therefore $-\Delta N={\delta }_0$ in distributions.

For a harmonic function on a ball, define its spherical mean $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(x_0+r\omega )d\omega$. Differentiation and the divergence theorem give $M^{\prime }(r)=(4\pi r^2{)}^{-1}{\int }_{B_r(x_0)}\Delta u=0$. Its limit at zero is $u(x_0)$, soThe ball formula follows by integrating spherical means in the radius. If two solutions of the prescribed Poisson equation tend to zero at infinity, their difference is globally harmonic and also tends to zero. Apply the spherical mean at an arbitrary $x_0$ and let $r\to \infty$; every point of that sphere has distance at least $r-|x_0|$ from the origin, so the mean tends to zero. Hence the difference is zero everywhere, proving uniqueness.

Let $V(x)=F(\pi ,x)$ satisfy the Bellman optimality equation $V(x)={\sup }_u\{r(x,u)+\beta E[V(x_1)\mid x,u]\}$. It is nonnegative because the rewards are. Under any competing policy, iterate its Bellman inequality to obtain $V(x)\ge E[{\sum }_{t=0}^{N-1}{\beta }^{t}r(x_t,u_t)+{\beta }^{N}V(x_N)]$. Drop the nonnegative terminal term and let $N\to \infty$. Monotone convergence gives $V(x)\ge F(\tilde{\pi },x)$ for every competitor. Since $V$ is the actual return of $\pi$, this proves optimality without a transversality assumption.

Assume the income process remains nonnegative, as required by the feasible-control interval; for unrestricted feasible controls a sufficient condition is ${ϵ}_t\ge -1$. If $\beta \le 1/(1+\theta )<1$, spending everything gives constant income $x$ and value $V(x)=x/(1-\beta )$. Its Bellman expression for general $u$ isThe coefficient of $u$ is nonnegative, so the maximum over $[0,x]$ occurs at $u=x$ and equals $V(x)$. Therefore spending all income at every step is optimal.

If $1/(1+\theta )<\beta <1$ and $x>0$, invest all income for $T$ steps and thereafter spend everything. Its expected return is $x[\beta (1+\theta ){]}^T/(1-\beta )$, unbounded as $T\to \infty$. In fact a sufficiently small fixed positive consumption fraction gives infinite expected discounted reward if $\beta [1+\theta (1-c)]\ge 1$. Thus the value is infinite and the problem has no finite-valued optimum. For $\beta =1$, spending everything already gives infinite value when $x>0$. If $x=0$, the value is zero. A mean condition alone, without nonnegative income or equivalent admissibility restrictions, would not make the model well-defined.

A statistic is sufficient when the conditional distribution of the observation given it does not depend on the parameter. It is minimal sufficient when it is, up to null sets, a measurable function of every sufficient statistic. For a dominated family, the likelihood-ratio criterion identifies two observations precisely when their likelihood ratio is independent of the parameter.

The Rao–Blackwell theorem says that conditioning a square-integrable unbiased estimator on a sufficient statistic preserves unbiasedness and cannot increase variance. The Cramér–Rao bound for an unbiased estimator of scalar $\theta$ is ${\mathrm{Var}}_{\theta }\hat{\theta }\ge 1/I(\theta )$. It requires parameter-independent support, differentiability, interchange of differentiation and integration, and finite positive Fisher information. More generally the numerator is the square of the derivative of the estimand. The bound follows from $E_{\theta }[\hat{\theta }{\partial }_{\theta }\log f]=1$ and Cauchy–Schwarz.

For the uniform sample, the joint likelihood is ${\theta }^{-n}{1}_{\{0<x_i<\theta \forall i\}}$. Its dependence on the observations is through $T={\max }_i{X}_i$, proving sufficiency. Two such likelihoods have parameter-independent ratio exactly when their maxima coincide, proving minimality. The density of $T$ is $n{t}^{n-1}/{\theta }^n$ on $(0,\theta )$. Unbiasedness of $h(T)$ saysDifferentiating this absolutely continuous identity gives $h(t)=(n+1)t/n$ almost everywhere. Conversely that function is unbiased. Every finite-variance unbiased estimator, after Rao–Blackwellization, must therefore give this same function of $T$. Its variance cannot exceed that of any such estimator. ThusThe Cramér–Rao argument does not apply because the support depends on $\theta$; in particular differentiation of the support boundary invalidates the regular score identity.