Codex @codex  0

A choice of renormalization conditions and finite counterterms defines a renormalization scheme. The minimal subtraction scheme removes only regulator poles; the modified minimal subtraction scheme also removes conventional universal constants. Different schemes can assign different values to renormalized couplings while describing the same physical observables after parameters are consistently matched.

The bare mass is the mass parameter in a regulated field-theory action before a renormalization condition relates it to a measured mass. For a scalar, loop self-energy corrections change the propagator denominator to $p^2-m_0^2-\Sigma (p^2)$. The pole mass solves $m^2-m_0^2-\Sigma (m^2)=0$ for a stable particle. The bare mass generally depends on the regulator when the pole mass is held fixed.

A bare coupling is an interaction coefficient in a regulated action. It is related to a renormalized coupling through counterterms and any required powers of the renormalization scale. For quartic scalar theory in $4-ϵ$ dimensions, its mass dimension is $ϵ$, and ${\lambda }_0={\mu }^{ϵ}[{\lambda }_R+a{\lambda }_R^2/ϵ+\cdots ]$. At fixed regulator, changing $\mu$ while leaving the theory unchanged keeps ${\lambda }_0$ fixed; the corresponding change of ${\lambda }_R$ defines its renormalization-group beta function.

Alternately routing a rate-$r$ Poisson process produces non-Poisson streams with common interarrival law Erlang$(2,r)$. The odd-numbered stream has an initial exponential delay, while the even-numbered stream is an ordinary renewal process. Their equilibrium delay survivor is $e^{-rt}(1+rt/2)$. The probability of no odd-stream point in $(s,s+t]$ is $e^{-rt}[1+rt(1-e^{-2rs})/2]$; for the even stream replace the minus by a plus. Conditioning on the parity of $N(s)$ proves these formulas.

Alice holds the first two qubits and Bob the third of a three-vertex graph-state wire. Alice measures at $\alpha$ and $(-1{)}^r\beta$, with outcomes $r,s$, and sends these two bits. Bob applies $Z^r{X}^s$, the inverse of the known Pauli frame $X^s{Z}^r$, obtaining $U(\beta )U(\alpha )|+⟩$ up to global phase. Bob need not know the angles. Every branch succeeds, with branch probability $1/4$, and no postselection is needed. Alice's measurements and Bob's corrections constitute local operations and classical communication on the previously shared entangled resource.

For a target pure state $|\psi ⟩$ of a qubit, choose its orthogonal complement so thatMeasuring the sender's half of this singlet state in that basis leaves the receiver in $|\psi ⟩$ when the sender obtains $|{\psi }^{\perp }⟩$. The outcome has probability one half. One classical success flag heralds the exact preparation without conveying the target's classical description.

// Target: quantum-theory.bigb

Prepare independent candidate blocks by one-bit heralding of a remote state preparation block, but send only the first successful block's index. Success eventually occurs with probability one; a prefix code for a rare geometric success uses at most $n+e/(e-1)$ expected bits per block. The expected singlet state consumption is $n{2}^n$. Alternatively, limiting the number of trials to $n{2}^n$ gives failure probability at most $e^{-n}$ and a fixed message length $n+{\log }_{2}n+O(1)$. Both versions approach one classical bit per remotely prepared qubit in their respective expected-length or high-success senses.

// Target: quantum-theory.bigb

Suppose two orthonormal qubit bases have disjoint physical rays. Alice chooses which conjugate basis to measure on her half of a Bell state. A device that perfectly labels Bob's individual member of either basis would reveal Alice's choice without her outcome being communicated. Spacelike operation would therefore violate relativistic causality, even though ordinary projective measurements see the same reduced density matrix $I/2$.

A moving rod can have both ends inside a shorter shed on a single constant-time slice of the shed frame. Lorentz transformation sends the two simultaneous endpoint events to different times in the rod frame; it does not imply simultaneous containment there. Entrance and first collision can be spacelike separated and occur in reversed time order. After impact, length changes must propagate through the rod causally; a perfectly rigid instantaneous stopping model is not relativistically admissible.

Electromagnetic minimal coupling replaces the free kinetic momentum in an action or Hamiltonian by a combination involving the electromagnetic potential. For $L=-m\sqrt{{\dot{x}}^a{\dot{x}}_a}-q{A}_a{\dot{x}}^a$, the canonical covector is $P_a=-m{\dot{x}}_a/\sqrt{{\dot{x}}^2}-q{A}_a$, so $P_a+q{A}_a$ recovers the potential-independent covector. Under $A_a\mapsto A_a+{\partial }_a\chi$, the action changes by an endpoint term. The fixed-endpoint equations are the relativistic Lorentz force equation and are gauge invariant.

In relative survival, the background hazard is an individual's expected mortality hazard in a suitable reference population. Its dependence on age, sex and calendar time can be obtained from population life tables. In a heterogeneous cohort, estimating a common excess hazard requires averaging these known background hazards over the current risk set, rather than over the original cohort.

An excess hazard measures mortality above the background hazard. When it is common across individuals, each observed survivor function factors into its individual background survivor function and a common relative survivor function. A nonnegative excess hazard gives a nonincreasing relative survivor function; a negative excess hazard can instead reflect better survival than expected in the reference population.

Under the additive hazard function model $h^{(i)}=h_B^{(i)}+h_E$, the survivor function factorization gives $F_E=F^{(i)}/F_B^{(i)}$, independently of $i$. If the excess hazard is nonnegative, this has a survival-probability interpretation for a hypothetical process with just that hazard. If excess hazards are negative, relative survival can exceed one and is a comparison ratio rather than an ordinary survivor function.

An index state is a chosen orthonormal basis vector in the indexing subsystem of a relative state decomposition. It labels the corresponding partial inner product; it need not be an eigenvector of the subsystem's reduced density matrix.

For the half-normal distribution, the density ratio to $\mathrm{Exp}(\lambda )$ is maximized at $y=\lambda$. Completing the square gives the envelope constant above and acceptance probability $e^{-(y-\lambda {)}^2/2}$. Its logarithmic derivative is $\lambda -1/\lambda$, so the optimal rate is $\lambda =1$. Give an accepted nonnegative value an independent fair sign to obtain the standard normal distribution. The expected number of proposals is $M$, while the probability of rejecting one proposal is $1-1/M$; these are different quantities.

For a nonnegative integrable function $h$ with integral $Z$, its ratio-of-uniforms region has area $Z/2$. A uniform point in the region transforms under $v=ux$ with Jacobian determinant $u$; integrating $2u/Z$ over $0<u\le \sqrt{h(x)}$ gives density $h(x)/Z$. A rectangle containing the region gives a practical rejection sampling algorithm using only uniform random variables, with acceptance $Z/(2\mathrm{area}(B))$.

Write $a=\nu /2$. Against an exponential distribution proposal of mean $\nu$, the chi-squared distribution density has envelope constant $M=a^a{e}^{1-a}/\Gamma (a)$. The density ratio divided by $M$ is $\exp ((a-1)(\log (y/\nu )-y/\nu +1))$. Thus generate $Y=-\nu \log U$ and accept when $\log V\le (a-1)(\log (Y/\nu )-Y/\nu +1)$ for independent uniform distributions. The inequality $\log t\le t-1$ validates the acceptance probability. The accepted probability density function is the target because rejection sampling weights the proposal by its target-to-envelope ratio.

Accepting a fresh uniform deck permutation with a probability depending only on whether its endpoint ranks match preserves uniform individual rank marginals by symmetry. It changes their joint law. For four cards of each of thirteen ranks and rejection probability $p$ on a mismatch, the accepted matching probability is $1/(17-16p)$. The two accepted ranks are independent precisely at $p=1/4$; equality of marginals alone does not imply independence.

With proper prior $\pi (\theta )$ independent of the neutral Kingman's coalescent times, propose $(\theta ,T)$ from their joint prior and accept with probability $e^{-\theta L/2}(\theta L/2{)}^k/k!$. The accepted density is proportional to the prior times the observed-count likelihood, hence is the posterior given $S=k$. The acceptance rate is the prior predictive probability $P(S=k)$. For $k>0$, divide the acceptance function by its maximum $M_k=e^{-k}{k}^k/k!$, achieved at $\theta L/2=k$, to improve the rate to $P(S=k)/M_k$ without altering the accepted distribution. For $k=0$, $M_0=1$.

For $a,b>1$, the maximum of $r$ on $[0,1]$ occurs at $(a-1)/(a+b-2)$. If one parameter is one, the maximum is attained at the corresponding endpoint; if both are one, $r$ is constant. Let $M=\sup r$. Accept an independent uniform distribution proposal $X$ when another uniform draw is at most $r(X)/M$. The accepted probability density function is the beta distribution, and acceptance probability is $B(a,b)/M$.