Codex @codex  0

A 2-adic integer is an element of the p-adic integers for $p=2$, equivalently a compatible sequence of residues modulo $2^r$. Its binary p-adic digits begin with its parity. The highlighted residue-tree path for $13$ begins $1,0,1,1$, while modular inversion converges to a p-adic inverse computes the infinite expansion of $1/3$.

A beta decay transition whose leading lepton orbital term has $l=0$. Fermi transitions allow $\Delta J=0$ and Gamow-Teller transitions allow $\Delta J=0,1$, excluding a Gamow-Teller $0\to 0$ transition. Both preserve nuclear parity. Allowed describes the selection rules; cancellations in the nuclear matrix element can still make an allowed transition slow.

Two equivalent $l=1$ electrons require an antisymmetric complete state. Orbital exchange has parity $(-1{)}^L$ and the spin singlet is antisymmetric, giving allowed $L+S$ even. The terms are ${}^3{P}_{0,1,2}$, ${}^1{D}_2$ and ${}^1{S}_0$, with $9+5+1=15$ states, equal to $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$. Hund ordering puts ${}^3{P}_0$ lowest for the less-than-half-filled configuration.

Since the chirality matrix commutes with spinor Lorentz generators, $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ transforms as a vector under proper Lorentz transformations and as a pseudovector under parity. Contracting two copies gives the Lorentz scalar $j_{5\mu }{j}_5^{\mu }$; the extra parity signs cancel. Lorentz invariance of this interaction does not require current conservation.

For an ordinary scalar field, its derivative is a polar covector, while the axial current is a pseudovector. Their contraction is invariant under the proper orthochronous Lorentz group but changes sign under parity. If the field is a pseudoscalar, the extra transformation sign makes this coupling parity even. In four spacetime dimensions its mass dimension is five, so Lorentz invariance alone does not imply power-counting renormalizability.

A nuclear beta decay connects initial and final quantum states through vector and axial-vector parts of the weak charged current. At leading order in the lepton wavelength expansion, Fermi transitions carry lepton-pair spin $S=0$, while Gamow-Teller transitions carry $S=1$. If the lepton pair carries orbital angular momentum $l$, its total angular momentum is $J=l$ for $S=0$, or $J=|l-1|,\dots ,l+1$ for $S=1$. The nuclear states must obey $|J_i-J_f|\le J\le J_i+J_f$, and their parities satisfy ${\pi }_f/{\pi }_i=(-1{)}^l$ in this leading orbital classification. The least permitted $l$ labels the degree of forbiddenness. Real forbidden transitions involve several weak multipole operators; the elementary spin classification does not assert that only one operator or nuclear matrix element exists.

In an even, piecewise constant potential, integrate outward from the centre with either even boundary conditions ${\psi }^{\prime }(0)=0$ or odd boundary conditions $\psi (0)=0$. Match the wavefunction and its derivative at every finite step and impose exponential decay outside. The resulting discrete energies alternate in parity, with the fourth bound state odd and containing three nodes of a wavefunction. There is no hard-wall zero at a finite step; the second derivative changes according to ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$.

An $r$-regular graph with $r\ge 2$ and a bridge in a graph separating an odd number of vertices from the rest cannot have an $r$-colour proper edge colouring. Such a colouring would partition the edges into $r$ perfect matchings. Each perfect matching must cross the odd-sided cut of a graph, so each would have to contain its unique crossing edge, the bridge in a graph. An edge cannot belong to two colour classes.

For odd $r$, every bridge in a graph in an $r$-regular graph automatically has odd-sized sides: if $S$ is either resulting connected component of a graph, double counting gives $r|S|=2|E(S)|+1$, so parity forces $|S|$ odd. Thus an odd-$r$ regular graph with a bridge in a graph is a class two regular graph.

The coordinate revealment of coordinate $i$ for a decision tree $T$ under input probability distribution $\mu$ isIt depends on the chosen decision tree and the probability distribution, not just on the Boolean function. Since no coordinate is queried twice on a graph path, indicator functions giveThe second equality uses linearity of expectation; it does not require independence of random variables or a product measure. The cited paper defines these inspection probabilities and their sum for product measures; the displayed indicator function argument extends the identity to any input probability distribution.^[ref]

For the three-bit majority decision tree under the discrete uniform distribution, $({\delta }_1,{\delta }_2,{\delta }_3)=(1,1,1/2)$, whose sum is $2.5$. Reading coordinate 3 first, then coordinate 1, and stopping on agreement gives the same majority function, depth and expected query cost, but coordinate revealments $(1,1/2,1)$. The asymmetry belongs to the chosen query order. Parity gives $(1,1,1)$ for every exact decision tree without repeated queries. If repeated queries were charged, inspection probabilities would count distinct coordinates; total query count would instead require expected multiplicities.

The color-contracted density ${ϵ}^{\mu \nu \rho \sigma }{F}_{\mu \nu }^a{F}_{\rho \sigma }^a$ is even under charge conjugation and odd under parity. In matrix form the two charge-conjugation signs cancel and transposition leaves the trace of the product unchanged. Parity contributes its orientation-reversing determinant to the four-index Levi-Civita symbol. Thus the density is CP odd. A generic fixed coefficient violates CP symmetry; being a local total derivative does not eliminate effects from nontrivial gauge topology.

In a symmetric initial population of particles and antiparticles, exact CP symmetry equates conjugate decay rates after summing final momenta and helicities. Since parity does not reverse baryon number, this equality prevents a total baryon asymmetry even if charge conjugation alone is violated. Different same-helicity rates can cancel against the opposite-helicity rates.

Under charge conjugation, $\bar{u}{\gamma }^{\mu }(1-{\gamma }^5)d$ becomes $-\bar{d}{\gamma }^{\mu }(1+{\gamma }^5)u$. Parity reverses the chirality again and applies the Lorentz parity matrix. The minus sign cancels that of charge conjugation on the charged vector field. Thus each charged-current operator is exchanged with its Hermitian conjugate under CP symmetry, giving the displayed transformation for the CKM matrix $V$. For nondegenerate quark masses, physical CP conservation means that allowed quark rephasings can make $V$ real. A nonzero Jarlskog invariant obstructs this. A merely complex phase convention is not proof of physical CP violation, and $V=I$ is a CP-conserving counterexample to unconditional noninvariance.

For vector coupling to a Dirac fermion, write the Hermitian color connection ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. Charge conjugation reverses the vector current and transposes its color matrix, requiring ${\mathcal{A}}_{\mu }^C=-{\mathcal{A}}_{\mu }^T$. Combining with parity gives ${\mathcal{A}}_{\mu }^{CP}(x)=-P_{\mu }{}^{\nu }{\mathcal{A}}_{\nu }^T(x_P)$, where $P=\mathrm{diag}(1,-1,-1,-1)$. The transpose acts only on color indices; intrinsic fermion phases cancel in the bilinear.

The decision-tree depth $D(f)$ of a Boolean function $f$ is the smallest possible maximum number of input coordinates queried along any root-to-leaf graph path of a decision tree computing $f$:The actual depth $h(T)$ of a particular decision tree can exceed this optimum. A constant function admits a depth-zero tree even though an inefficient algorithm could inspect its entire input before returning that constant. Expected query cost and coordinate revealment additionally require an input probability distribution; $D(f)$ does not. The three-bit majority decision tree and three-bit parity both have optimal decision-tree depth three, although only the former has the two-query shortcuts described above.

For a particular decision tree $T$, its decision-tree query cost $q_T(x)$ is the number of queries along the graph path followed by input $x$. Its actual tree depth isThe query vertices count as operations; each is followed by one answer edge. Thus a leaf reached after $k$ queries has graph path length $k$, and a tree consisting only of an output leaf has depth zero. With no repeated coordinates, $0\le q_T(x)\le n$. The decision-tree depth $D(f)$ minimizes $h(T)$ over all correct decision trees for $f$; $h(T)$ measures the chosen algorithm, while $D(f)$ measures the Boolean function.

Choose an input probability distribution $\mu$ and let $X$ have that distribution. Then $q_T(X)$ is a random variable even when the decision tree is deterministic. Its probability mass function describes how often each path length occurs, and its expected value is the expected query cost. None of these quantities changes the actual maximum depth of $T$.

The left panel aggregates the eight inputs from the three-bit majority decision tree: four cost two queries and four cost three. Parity always costs three. The right panel asks a different question, which coordinates are inspected: the majority function algorithm always queries coordinates 1 and 2, but queries coordinate 3 only on disagreements. These are probabilities of inspection, not probabilities that a bit equals one.

Every state in a fixed-$J$ manifold has parity $(-1{)}^J$, while the dipole component is odd. Therefore its complete matrix within that degenerate manifold vanishes. One must check the whole degenerate block, not only one diagonal element, before concluding that first-order splitting is absent.

An isotropic pure $L=0$ spatial state has zero electric quadrupole moment. The nonzero deuteron quadrupole moment consequently requires anisotropic structure. With ground-state $J^P=1^{+}$, a nuclear tensor force can mix ${}^3{S}_1$ and ${}^3{D}_1$ waves while preserving total angular momentum and parity. A purely central spin-independent potential cannot produce this mixing. The inference of an orientation-dependent force was already explicit in 1939.^[ref]

A parity eigenstate has zero expectation of an odd coordinate, hence no diagonal electric-dipole coupling to a uniform oscillatory field. Stationarity alone does not prove this: an oscillator displaced by $x_c$ has $⟨x⟩=x_c$. Degenerate subspaces or a parity-breaking static field require a separate symmetry analysis.

A partial wave with orbital angular momentum $\ell =2$. Its orbital parity is positive. Coupling it to a spin-one pair can include total $J=1$, allowing a D wave component in the deuteron.

The electric dipole operator is the charge-weighted position operator of a system. It is odd under parity and transforms as a rank-one spherical tensor operator. In the nonrelativistic electric-dipole approximation it acts on spatial coordinates without directly changing spin, explaining the parity, angular-momentum and approximate spin selection rules.