Fix $t$ and let $F_t(u)=(u-t{)}_{+}^{k-1}$. Initially assume distinct spline knots, so the ordinary polynomial interpolation definitions are unambiguous. The two interpolants ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree with $F_t$ at their $k-1$ shared nodes $t_{i+1},\dots ,t_{i+k-1}$. Their difference has degree at most $k-1$ and all these roots, henceFor $k=1$ the product is empty and the same statement concerns constants. The leading coefficient of an interpolating polynomial is its highest-order divided difference, so the divided difference recurrence givesThis proves Lee's formula:Summing over $i$ telescopes to ${\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. If $t_k<t<t_{n+1}$, the first interpolation nodes $t_1,\dots ,t_k$ are all smaller than $t$, so their truncated-power data vanish and ${\ell }_1=0$. The last nodes $t_{n+1},\dots ,t_{n+k}$ are larger than $t$, so their data are values of the polynomial $(u-t{)}^{k-1}$ and uniqueness gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the telescoped identity isthe Marsden identity, valid for every real $x$.

If repeated spline knots are allowed, ordinary value interpolation at repeated nodes does not specify ${\ell }_i$ uniquely. The standard extension uses confluent divided differences and Hermite interpolation, or limits of distinct knots: the same polynomial identity then persists wherever these quantities are defined. At a maximally repeated knot where the truncated power lacks the required derivative, the appropriate one-sided spline convention must be fixed; an unspecified classical derivative is not part of the proof. For distinct knots, the printed identity requires no such qualification.

Density operator controllability cannot defeat unitary invariants: it grants access to the unitary orbit of a density operator, whose members have the same eigenvalues, rather than to every density operator. In the Hilbert-Schmidt inner product orthonormal normalization of this question,Quantum Hamiltonian evolution preserves this norm, so the target is not reachable from the stated initial state. Equivalently, their values of purity of a density operator are $1/2+1/4=3/4$ and $1/2+1/2=1$: the initial state is a mixed state and the target a pure state. Dissipation, an appropriate measurement protocol or an auxiliary system with additional resources would be needed to change the spectrum; freely controlling the closed-system Quantum Hamiltonian alone does not suffice.

For a Hermitian Quantum Hamiltonian, $[{\sigma }_m,{\sigma }_n{]}^{†}=-[{\sigma }_m,{\sigma }_n]$. ConsequentlySwapping $m,n$ changes the sign of the commutator, so $L_{nm}=-L_{mn}$. The generator is therefore real and antisymmetric. Its last column vanishes because the identity commutes with $H$, so $c=0$ and the traceless block satisfies $A^T=-A$.

For constant $H$, the Bloch vector evolves by $s(t)=R(t)s(0)$ with $R(t)=e^{At}$. Direct differentiation givesHence $R^{T}R=I$. Its determinant is continuously connected to one and cannot change between the two possible values $\pm 1$, so $\det R=1$. This proves that the evolution is a rotation in the special orthogonal group. A time-dependent Hermitian Quantum Hamiltonian has the same conclusion from $\dot{R}=A(t)R$.

These Hamiltonian rotations of Bloch vectors preserve the Euclidean norm and purity of a density operator. For $N>2$, not every rotation of the ambient sphere is a physical unitary conjugation: the higher-dimensional positivity constraints and full spectrum of the density operator impose further restrictions.

Let $\mathcal{G}=-i[H,\cdot ]+{\sum }_d\mathcal{D}[V_d]$ denote the Lindbladian. Substituting $\rho ={\sum }_n{r}_n{\sigma }_n$ and projecting with the Hilbert-Schmidt inner product givesFor the Quantum Hamiltonian term, cyclicity of the trace yieldsFor each Lindblad dissipator, the same cyclic rearrangement givesThese are the required coefficient matrices, so $\dot{r}=(L+{\sum }_d{D}^{(d)})r$. They are real because both generators map Hermitian matrices to Hermitian matrices, whose coordinates are real by (a).

The trace of a commutator vanishes. AlsoThus the last row of $L+{\sum }_d{D}^{(d)}$ is zero and $\boxed{{\dot{r}}_{N^2}=0}$. It is trace preservation, not preservation of the purity of a density operator, that produces this constant coordinate.

Write $G=L+{\sum }_d{D}^{(d)}$ and $n=N^2-1$. Since $r_{N^2}=1/\sqrt{N}$, the remaining coordinates obey the Affine Bloch equationBecause $[H,I]=0$ and $\mathcal{D}[V](I)=[V,V^{†}]$, its offset can also be writtenIt vanishes for unital dynamics, but need not vanish for a general dissipative evolution.

For $N=2$, take the normalized Pauli operators as the traceless basis. The matrix associated with a vector $s$ is $\rho =I/2+s\cdot \sigma$, where these basis matrices are the usual Pauli matrices divided by $\sqrt{2}$. Their multiplication relations giveAt radius $1/\sqrt{2}$, these eigenvalues are one and zero. Every sphere point therefore gives a positive rank-one projector. Any other orthonormal Hermitian basis merely changes the real coordinates by an orthogonal transformation, so the result is basis independent.

For every $N>2$, choose a rank-one projector $P$ and consider the Hermitian trace-one matrixIts eigenvalues are $2/N-1$ once and $2/N$ with multiplicity $N-1$. They sum to one and satisfyBy orthonormality its traceless coordinate vector has squared length $1-1/N$, so it is on the specified sphere. But $2/N-1<0$, so $A$ is not a density operator. Thus for $N>2$, the pure states occupy a proper subset of the sphere; positivity excludes some of its points. In fact, this example is the antipodal Bloch vector to that of $P$.

The usual assertion presumes $N\ge 2$. If the trivial dimension $N=1$ is admitted, the radius-zero sphere consists of the single coordinate point, also corresponding to the single pure state, so that dimension is an additional trivial exception.

Let $\mathcal{P}$ be the pinching map determined by the projectors, and put $\sigma =\mathcal{P}(\rho )$. This is a nonselective projective measurement; its projectors are its Kraus operators. We will prove the exact relative entropy of a pinched state identity.

First the support of $\rho$ is contained in that of $\sigma$. Indeed, if $v$ lies in the kernel of $\sigma$, positivity givesEvery summand is zero. Since ${\sum }_i{P}_i=I$, also ${\rho }^{1/2}v=0$, and thus $v$ is in the kernel of $\rho$. Therefore the logarithm of $\sigma$ may be used on its support without an infinite relative-entropy term.

The operator $\sigma$ is block diagonal, so each $P_i$ commutes with $\log \sigma$ on that support. Cyclicity of the trace givesExpand the quantum relative entropy:Its nonnegativity now proves the entropy increase under nonselective projective measurement:The equality case in nonnegativity of quantum relative entropy says equality holds exactly when $\rho =\sigma$. Equivalently,Thus equality means the state already has no coherence between distinct measurement blocks. It need not be diagonal inside a block of rank greater than one.

Use position eigenstates normalized by $⟨q|q^{\prime }⟩=\delta (q-q^{\prime })$, and the free Hamiltonian operator $\hat{H}={\hat{p}}^2/(2m)$. The time-sliced path integral is the matrix element of the time-evolution operator:Insert momentum eigenstates, with $⟨q|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipq/\hslash }$. The free-particle propagator becomesComplete the square in the exponent:The Fresnel integral, defined by the damping prescription $T\to T-i0$, gives, for $T>0$,The square-root branch is $i^{-1/2}=e^{-i\pi /4}$; it is fixed by the regulated Gaussian integral and the short-time Dirac delta distribution limit, rather than an arbitrary phase.

The Euler-Lagrange equation for the free particle is $m\ddot{q}=0$. Its unique classical path satisfying the endpoint boundary conditions is $q_{\mathrm{cl}}(t)=q^{(i)}+\Delta qt/T$. Therefore its classical action isSubstituting into the operator result yields $\mathcal{I}=C(T)e^{i{S}_{\mathrm{cl}}/\hslash }$, with $C(T)=(m/(2\pi i\hslash T){)}^{1/2}$ independent of the endpoint positions.

The same independence is transparent directly in the path integral. Write $q=q_{\mathrm{cl}}+\eta$, with $\eta (0)=\eta (T)=0$. The cross term in the action is $m{\dot{q}}_{\mathrm{cl}}{\int }_0^T\dot{\eta }dt=0$, so $S[q]=S_{\mathrm{cl}}+(m/2){\int }_0^T{\dot{\eta }}^{2}dt$. The fluctuation integral depends only on $m,T,\hslash$. Because the action is quadratic, the factorization is exact; the semiclassical propagator already equals the full answer.

At $m^2=0$ and $h=0$ the model is at a Gaussian fixed point. Its connected correlation function has the power-law critical form $r^{-(D-2)}$ rather than a finite exponential decay length. Approaching from $m^2>0$, the uniform susceptibility ${\tilde{G}}_0(0)=1/m^2$ and the correlation length both diverge. If the thermal scaling field is linear in $m^2$ with $\alpha$ finite and positive, thenThe same result follows from the mass multiplier $b^2$: its logarithmic thermal eigenvalue is two, so $\nu =1/2$.

The determinant also shows a thermodynamic singularity. Two derivatives of the free-energy density with respect to $m^2$ give, up to sign and a nonzero normalization,as $m^2\downarrow 0$. Rescaling $p=\sqrt{\alpha m^2}q$ establishes the power, and the dimensionless integral is ultraviolet-convergent for $D<4$. Integrating back gives a nonanalytic free-energy term proportional to $(m^2{)}^{D/2}$ after regular backgrounds are removed. This is critical behavior of the quadratic model.

There is a stability qualification: a pure Gaussian potential with $m^2<0$ is unbounded below in the uniform mode, so it does not by itself provide a stable ordered phase on the other side. The ordered-phase obstruction in a Gaussian scalar model is removed by adding a positive quartic interaction. In $2<D<4$ that interaction is relevant, so the resulting generic scalar critical point need not retain the Gaussian exponents. Thus the calculation proves the Gaussian critical singularity and $\nu =1/2$, without claiming that the unmodified quadratic integral defines both phases of a physical ordered transition.

Split the Fourier modes into $|p|<\Lambda /b$ and the shell $\Lambda /b<|p|<\Lambda$. The source is uniform, so it couples only to the zero mode, which belongs to the retained part. The Gaussian field theory has no couplings between the shell and retained modes. Integrating out the shell therefore contributes only a source-independent determinant to the free energy, without changing the quadratic coefficients of the retained Hamiltonian.

Restore the cutoff by $x=b{x}^{\prime }$ and choose the field normalization ${\varphi }_{<}(b{x}^{\prime })=b^{-(D-2)/2}{\varphi }^{\prime }(x^{\prime })$. Counting the measure, derivative and field factors explicitly givesThus in the convention that preserves the kinetic normalization, the Gaussian momentum-shell scaling isThe field has engineering dimension $(D-2)/2$ and zero anomalous dimension. The measure Jacobian and shell determinant supply the additive constant described in Q2. A different choice of field rescaling can redistribute factors into $\alpha$, but must give the same physical transformed correlation length ${\xi }^{\prime }=\xi /b$.

Use the product-limit estimate in (b). Between consecutive event times there are no mortality jumps, soThus the estimated relative survivor curve rises between events when background mortality is positive, and is constant if it is zero. Censoring changes the risk set and can change the slope, but supplies no downward event jump. At the next event time the curve is multiplied by $1-d_k/r_k$.

Provided the curve is still positive, the ratio of the post-event estimates at the two consecutive event times isFor $d_k<r_k$, the later value is larger precisely whenFor example, suppose there are 100 individuals in the risk set after the earlier event, no intervening censoring, a 20-day gap, and one death at its end. If every background hazard is $0.001$ per day, then the integrated background contribution is $0.02$ andThe individual background hazards are small, but the interval contains less observed mortality than expected from that background. This can occur by chance in a small event sample or when the cohort is healthier than the life-table reference on other dimensions.

A nonnegative true excess hazard makes $F_E$ nonincreasing, but this unconstrained estimator need not inherit that restriction. An increase is therefore not proof of a clinical survival benefit. If the exponential cumulative-hazard version is chosen instead, its corresponding condition is ${\int }_{t_j}^{t_k}{\overline{h}}_B>d_k/r_k$; the qualitative explanation remains the same. If the risk set is exhausted by deaths, the product-limit factor is zero and no later increase is possible.

Define the individual background survivor function and common excess component byIntegrating the additive hazard function decomposition givesThe relative survivor function therefore compares actual survival with the survival expected from background mortality, and is common across individuals under this model. If $h_E\ge 0$, it is the survivor function of a hypothetical process having only the excess hazard. If excess mortality can be negative, the ratio may exceed one and lacks an ordinary survival-probability interpretation.

To estimate it, let $Y_i(u)$ indicate that individual $i$ is in the risk set, let $Y(u)={\sum }_i{Y}_i(u)$, and let $N(u)$ count all observed deaths. Assume appropriate independent censoring. The aggregate event intensity isConsequently the background correction must use the current risk-set averageonly on the observed range where $Y>0$. Dividing the event count by $Y$ estimates the total hazard there. Subtracting the known average background gives the offset-adjusted cumulative hazard estimatorHere $d_j$ is the event count at $t_j$ and $r_j=Y(t_j)$; under an untied scheme $d_j=1$. One possible estimate is ${\hat{F}}_{E,\exp }=\exp [-{\hat{H}}_E]$.

A product-limit version retains the full event fraction, which is especially relevant when the excess mortality is not small. Estimate the relative-survival differential equation $d{F}_E/F_E=-d{H}_E$ byAt an event time it multiplies the curve by $1-d_j/r_j$. Between events it solves ${\hat{F}}_E^{\prime }={\overline{h}}_B{\hat{F}}_E$. Starting at one therefore gives the risk-set-adjusted relative survivor estimatorThe Kaplan–Meier estimator factor accounts for observed all-cause mortality; the continuous multiplier corrects for known background mortality. The exponential cumulative-hazard version instead has event jumps $\exp (-d_j/r_j)$; it is close when event fractions are small but is not identical to this product-limit version.

Even if each background hazard is small, its accumulated contribution over long follow-up cannot simply be discarded. Nor should the excess contribution be linearized merely because the background is small. Using the known background integral inside the exponential avoids both errors. The average is over the evolving risk set: replacing it by an initial-cohort mean, or dividing by an arbitrary average of individual background survivor functions, is generally wrong when the background hazards differ. Censoring and deaths change which individuals contribute to the background correction.

A standard setting is relative survival in cancer registries. Patients' all-cause mortality is compared with mortality expected from population life tables matched for factors such as attained age, sex and calendar year. These provide the known individual background hazards $h_B^{(i)}$; the remaining common excess hazard $h_E$ describes mortality above that expected background.

This is useful when causes of death are incomplete or unreliable: all observed deaths can be used without deciding which were caused by the cancer. It also separates changes in background population mortality from excess mortality in the diseased cohort. The reference life tables must be appropriate, and a common excess hazard is a modeling assumption, not automatic. A relative survivor function can then describe the cohort's excess survival, but interpreting it as the effect of a causal elimination of the disease requires further assumptions.