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Insert hydrogen energy eigenstates on either side of the commutator. Because $L_z$ is Hermitian and has eigenvalue $\hslash m$,ThusThis is a necessary electric-dipole selection rule, not a sufficient condition for a nonzero matrix element. For example, $⟨200|z|200⟩=0$ by parity even though $m=m^{\prime }=0$. The angular factor $z=r\cos \theta$ additionally requires $l^{\prime }=l\pm 1$: multiplication of a spherical harmonic by $\cos \theta$ gives only harmonics with the same $m$ and adjacent $l$. The commutator alone establishes only the magnetic quantum number restriction; radial integrals and the remaining electric-dipole selection rules for hydrogen must still be considered.

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.

By linearity of the time-independent Schrödinger equation, the even and odd partseach satisfy the same energy eigenvalue equation whenever they are nonzero. They obey $R{\psi }_{+}={\psi }_{+}$ and $R{\psi }_{-}=-{\psi }_{-}$, so they have definite parity. At least one is nonzero, and $\psi ={\psi }_{+}+{\psi }_{-}$. Each nonzero part can be normalized separately.

Therefore a basis of stationary solutions can be chosen to have definite parity, and solving the even and odd sectors loses no energy eigenvalues. This does not assert that every possible quantum superposition has definite parity; a superposition of the two sectors need not. For one-dimensional bound states in the usual regular setting, nondegeneracy of one-dimensional bound states further implies that each individual energy eigenstate has definite parity.

At a finite jump in potential energy, the wavefunction and its first derivative must be continuous. A jump in the derivative would create a Dirac delta function in ${\psi }^{\prime \prime }$, for which the finite potential has no balancing term. For the even solution, matching at $x=a$ givesA nonzero solution cannot have $B=0$. It also cannot have $\cos \beta =0$: the first condition would give $D=0$, while the second would force $B\kappa \sin \beta =0$, a contradiction. Division is therefore valid and yieldsParity makes the conditions at $x=-a$ identical to those already imposed. Finally the definitions of the decay and oscillation constants giveThe circle relation fixes the well strength; the tangent relation selects its even bound states.

For an even function, the interior sine term must vanish, because sine is odd and cosine is even. Comparing reflected exterior tails at any $x>a$ gives $D{e}^{-kx}=A{e}^{-kx}$. Hence the parity condition alone isIt does not yet impose the matching condition on $B$. Continuity at $x=a$, when imposed in the next part, additionally gives $A=D=B{e}^{ka}\cos (\kappa a)$.

In two spatial dimensions there are only two kinetic matrices and one mass matrix. The three Pauli matrices square to identity and anticommute pairwise, sois a two-component Dirac Hamiltonian with $H_{2+1}^2=c^2(p_x^2+p_y^2)+m^2{c}^4$. It therefore reproduces the stated first-order equation and its Klein-Gordon square. Requiring a fourth anticommuting matrix in three spatial dimensions is precisely what prevented this two-component construction above.

Printed angular-momentum qualification. Define the dimensionless spin $s_i=-i{\epsilon }_{ijk}{\alpha }_j{\alpha }_k/4$ and the physical orbital angular momentum ${\ell }_i={\epsilon }_{ijk}{x}_j{p}_k$, with $[x_i,p_j]=i\hslash {\delta }_{ij}$. ThenThe source prints $L_i=i{\ell }_i$ instead. Taken literally, it is anti-Hermitian, has incompatible units with the dimensionless spin when they are added, and givesIt cannot yield the requested usual angular-momentum algebra or conservation law. For a concrete counterexample set $\hslash =1$ and take a wave with momentum along $z$. Sincethe printed $J_i=i{\ell }_i+s_i$ obeys $[H_D,J_x]=(1+i)c{\alpha }_y{p}_z\ne 0$.

The physically consistent operator is ${\mathbf{J}}_{\mathrm{phys}}=\mathbf{\ell }+\hslash \mathbf{s}$. It satisfiesThe cancellation follows directly from the two mixed commutators above. In a spherical basis the free spinor can be labelled by energy, $j=1/2,3/2,\dots$, and $m_j=-j,\dots ,j$, together with a parity or equivalent Dirac angular label distinguishing the two orbital couplings $\ell =j\pm 1/2$. Alternatively one can choose a momentum and helicity basis. One cannot simultaneously assign all components of momentum and all angular labels as independent commuting observables.

Let $K_i$ generate Lorentz boosts with the same exponential convention as the rotation generators of rotations $J_i$. The Lie algebra of the Lorentz group isConsequently non-collinear Lorentz boosts do not close among themselves: the product contains a rotation, the Wigner rotation. For an explicit counterexample, in coordinates $(ct,x,y)$ write $c_a=\cosh a$, $s_a=\sinh a$ and similarly for $b$. ThenA pure rotationless Lorentz boost has a symmetric matrix, whereas this product has spatial entries $(xy)=0$ and $(yx)=s_a{s}_b\ne 0$ for $a,b\ne 0$. It is therefore a concrete product of two pure Lorentz boosts that is not a pure Lorentz boost. Collinear Lorentz boosts are the special case that do form a one-parameter group, with additive rapidities.

Allowing all spatial directions and closure generates both the $K_i$ and, through their commutators, all $J_i$. The smallest connected group containing all these Lorentz boosts is the proper orthochronous Lorentz groupIt contains arbitrary compositions of spatial rotations and Lorentz boosts continuously connected to the identity. It has six continuous parameters. Here “smallest” refers to the subgroup generated by all continuous pure Lorentz boosts, not to an arbitrary collection already including discrete reversals.

The full Lorentz group $O(1,3)$ has four connected components, distinguished by the sign of the determinant and the direction of time. Representatives are the identity, parity $P=\mathrm{diag}(1,-1,-1,-1)$, time reversal $\mathsf{T}=\mathrm{diag}(-1,1,1,1)$, and $P\mathsf{T}=-I$. ThusMultiplication by these discrete transformations relates the components algebraically; no continuous path within the Lorentz group joins different components. Translations are not required for closure here; adding them gives the larger Poincaré group.

The light states have the counting $3+4+1=8$. This suggests approximate chiral symmetry breakingwith a nonzero quark chiral condensate $⟨{\bar{q}}_R^j{q}_L^i⟩\propto {\delta }^{ij}$ for the three light flavours $u,d,s$. The three pions, four kaons and predominantly octet eta meson are the resulting pseudo-Goldstone bosons. Their small nonzero masses reflect explicit chiral symmetry breaking by the light-quark masses; for example, $m_{\pi }^2$ is proportional to $m_u+m_d$ at leading order. The larger strange-quark mass makes the kaons and octet eta meson heavier than the pions, while electromagnetic interactions help split charged and neutral partners.^[ref]

Although the massless strong-interaction terms admit independent rotations of all six flavours, the charm, bottom and top masses are too large to treat six-flavour chiral symmetry as a small explicit breaking at the light-hadron scale. Full approximate $SU(6{)}_L\times SU(6{)}_R$ breaking would suggest 35 light modes, rather than this octet. The classical axial $U(1)$ is also broken by the chiral anomaly, so it does not supply a ninth equally light Goldstone boson.

Terminology qualification: these mesons have spin zero but negative parity: they are pseudoscalar mesons, with $J^P=0^{-}$, rather than positive-parity scalar particles with $J^P=0^{+}$. Calling the listed states “scalar” is acceptable only if it is being used loosely to mean spin zero.

Every finite connected cubic graph with at most two bridges has a perfect matching.

Proof. The handshaking lemma implies that the number $n$ of vertices is an even integer. Fix a nonempty subset $S$ of the vertex set, and let $q$ count the components of $G-S$ of odd integer order. For such a component $C$, counting vertex degrees gives $3|C|=2|E(C)|+d(C)$, where $d(C)$ counts its edges to $S$. Thus $d(C)$ is a positive odd integer. If it is one, the unique outgoing edge is a bridge; this happens for at most two components. For all others $d(C)\ge 3$. Therefore $3q-4\le 3|S|$. Meanwhile $q\equiv n-|S|\equiv |S|(\mathrm{mod}2)$. If $q>|S|$, the same parity would force $q\ge |S|+2$, contradicting the preceding bound. For $S=\varnothing$ there is one component of even integer order and hence $q=0$. Apply Tutte's theorem on perfect matchings.

For a negative muon with rest-frame spin vector $\mathbf{s}$, the leading massless-electron distribution is $d\Gamma /(dxd\Omega )=G_F^2{m}_{\mu }^5{x}^2[(3-2x)+(1-2x)\hat{\mathbf{k}}\cdot \mathbf{s}]/(384{\pi }^4)$, where $x=2E_e/m_{\mu }\in [0,1]$. Its final-spin sum is $64G_F^2(k\cdot q^{\prime })(q\cdot (p-m_{\mu }s))$. The spin vector is axial whereas the electron direction is polar, so their scalar product reverses under parity. Its occurrence demonstrates parity violation by the left-chiral weak charged current.

A missing proton relative to a filled nuclear shell. In a closed-core approximation a single hole in a $j$ orbital has the same total angular momentum $j$ and orbital parity as the removed particle. For example, removing a $p_{1/2}$ proton from a spin-zero closed core gives $J^P=1/2^{-}$.

A meson with zero spin angular momentum and negative parity. The pion and kaon are familiar examples. Between two pseudoscalar mesons, parity conservation in the strong interaction forces the axial current matrix element to vanish, while the vector current has pseudoscalar-to-pseudoscalar form factors. The weak decay as a whole need not conserve parity.

An axially symmetric quadratic potential energy perturbation preserves parity and the $z$ component of orbital angular momentum. For $1s,2s,2p_0,2p_{\pm 1}$, its first-order energy shifts are respectively $(2\alpha +\beta )a_0^2$, $14(2\alpha +\beta )a_0^2$, $(12\alpha +18\beta )a_0^2$, and $(24\alpha +6\beta )a_0^2$. They follow by multiplying the hydrogen radial second moment by the angular average of $\alpha +(\beta -\alpha ){\cos }^2\theta$, whose ${\cos }^2\theta$ averages are $1/3,1/3,3/5,1/5$. Negative $\beta$ makes the globally extended potential unbounded below; the shifts then describe local perturbative resonance poles or a potential stabilized at large distance.

Invariance under a spatial reflection. A one-dimensional Quantum Hamiltonian invariant under $x\mapsto a-x$ admits energy eigenfunctions of definite even or odd midpoint parity. Even and odd functions are orthogonal on a reflection-symmetric interval.

The lowest even quantum harmonic oscillator quantum state has density $\rho (y)=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$, and the lowest odd quantum state has density $2y^2\rho (y)/{\tau }^2$. Since ${\rho }^{\prime }=-2y\rho /{\tau }^2$, integration by parts gives ${\mathbb{E}}_{\mathrm{F}}[f]={\mathbb{E}}_{\mathrm{B}}[f+y{f}^{\prime }]$ whenever the boundary terms vanish. For $f(y)=y^n$, this proves the displayed identity. Odd moments vanish in both quantum states by parity; even moments quantify the larger spatial spread in the odd quantum state. In particular the relative-coordinate variance is three times as large, and the odd density vanishes at particle coincidence.

Opposite parity of same-centre s orbitals and axial p orbitals makes their left and right nearest-neighbour hopping integrals opposite in sign. Their Bloch wave sum is consequently odd in wavevector and vanishes at the zone centre and boundary. Away from those points, orbital hybridization repels coupled levels and can produce an avoided crossing.

The even, spherically symmetric hydrogen atom ground state has no permanent electric dipole, so a uniform electric field gives no first-order energy shift. The electric-dipole selection rules for hydrogen allow odd-parity excited states to mix in at second order, yielding a negative shift quadratic in the field.

Here there is a nondegenerate energy eigenvalue, so second-order nondegenerate perturbation theory applies. With $H^{\prime }=e\mathcal{E}z$, it givesThe notation includes integration over the continuous spectrum, as well as a sum over discrete energy eigenstates. All denominators are negative because $1s$ is the ground state. The positive electric polarizability ${\alpha }_{1s}$ therefore lowers its energy.^[ref]

An independent way to see the coefficient avoids truncating that sum. Define $E_{\mathrm{h}}=e^2/(4\pi {ϵ}_0{a}_0)$ and $f=e{a}_0\mathcal{E}/E_{\mathrm{h}}$. In coordinates $\mathbf{\rho }=\mathbf{r}/a_0$, the dimensionless unperturbed Hamiltonian operator and normalized wavefunction are $h_0=-{\nabla }_{\rho }^2/2-1/\rho$ and ${\psi }_0={\pi }^{-1/2}{e}^{-\rho }$, with energy eigenvalue $-1/2$. Writing the first-order wavefunction correction as $f\chi$, substitution into the time-independent Schrödinger equation givesThe second equality can be checked by differentiation. Its parity makes $\chi$ orthogonal to ${\psi }_0$. The resulting integral issince the normalized radial probability density is $4{\rho }^2{e}^{-2\rho }$, giving $⟨{\rho }^2{⟩}_0=3$ and $⟨{\rho }^3{⟩}_0=15/2$. Converting back to SI units gives the stated electric polarizability. The induced electric dipole moment is $-\partial \Delta E/\partial \mathcal{E}={\alpha }_{1s}\mathcal{E}$ and tends to zero with the electric field, unlike the selected degenerate branches' leading dipoles. Both calculations describe weak-field level shifts; a spatially uniform static electric field also allows eventual quantum tunnelling and ionization, which these real energy corrections do not calculate.

Strong and electromagnetic interactions conserve net strangeness; a decay changing it requires the weak interaction. A strange baryon decay into a less-strange baryon and a pion is an example. The weak interaction does not impose a parity selection rule, so a final orbital partial wave cannot by itself determine the parent's parity. Qualitative lifetime classes do not provide an exact lifetime without an amplitude calculation or measurement.

The symmetric double-delta potentialseparates into even and odd parity channels. At either interaction, the wavefunction is continuous and its derivative has the jump

Let $B_1,B_2,B_3$ be independent random variables, each with the Bernoulli distribution of parameter $1/3$, and let $S=B_1+B_2+B_3$. Multiplying their characteristic functions and expanding by the binomial theorem givesThe coefficients are exactly the probability mass function of the binomial distribution with parameters $3,1/3$: the sum takes $0,1,2,3$ with probabilities $8/27,12/27,6/27,1/27$. Direct outcome counting verifies them: an outcome with $k$ ones has probability $(1/3{)}^k(2/3{)}^{3-k}$, and there are $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ such outcomes. The coefficients sum to one, recovering ${\phi }_S(0)=1$.

At $t=\pi$, ${\phi }_S(\pi )=(1/3{)}^3=1/27$. This is also $\mathbb{P}(S\text{ even})-\mathbb{P}(S\text{ odd})=(8+6-12-1)/27$: the complex exponential now records parity. The moments agree by two routes: linearity of expectation gives $\mathbb{E}[S]=1$, while the variance of a sum of independent random variables gives $\mathrm{Var}(S)=3(2/9)=2/3$. Summing $k^2\mathbb{P}(S=k)$ gives the second moment $5/3$, also recovered as $-{\phi }_S^{\prime \prime }(0)$.