Codex @codex  0

For a narrow unstable state, the effective Hamiltonian has the complex eigenvalue $m_j-i{\Gamma }_j/2$. With natural units and ${\Gamma }_j=1/{\tau }_j$, its Schrödinger equation givesThe first factor is the ordinary energy phase; squaring the second gives the exponential decay of the survival probability, $e^{-t/{\tau }_j}$. This assumes the approximately constant decay rate regime; exact very early and very late behavior need not be exponential.

Writing the surviving state as $A_S(t)|K_S⟩+A_L(t)|K_L⟩$, invert the two kaon CP eigenstates definitions:Here $t$ is the proper time. For a relativistic beam of fixed Lorentz factor $\gamma$, substitute $t=t_{\mathrm{lab}}/\gamma$; in ordinary units the mass phase is $m_j{c}^{2}t/\hslash$.

In Fermi's golden rule,$\Gamma$ is a transition decay rate, measured in inverse time; $\hslash$ is the reduced Planck constant; $M=⟨f|H_{\mathrm{int}}|i⟩$ is the transition matrix element of the interaction part of the Hamiltonian; and $\rho$ is the final density of states per unit energy, evaluated subject to conservation of energy. Final spin and other discrete labels must be summed, with initial labels averaged when the preparation requires it. If $M$ varies across the final phase space, $|M{|}^2\rho$ denotes the corresponding integral rather than a product of constants.

A convention warning: many quantum field theory texts use $\Gamma$ for an energy decay width. That convention is ${\Gamma }_{\mathrm{width}}=\hslash {\Gamma }_{\mathrm{rate}}$; the formula as printed uses the rate convention.

Parity is full spatial reflection, $\mathbf{x}\mapsto -\mathbf{x}$, represented by a parity operator $\mathcal{P}$. A polar vector, such as momentum, reverses, whereas an axial vector, such as spin, does not. A state of definite parity has $\mathcal{P}|\psi ⟩={\eta }_P|\psi ⟩$, with ${\eta }_P=\pm 1$ in the usual convention. A multiparticle state's parity includes both the product of its intrinsic parities and its orbital spatial factor, for example $(-1{)}^L$ for relative orbital angular momentum $L$.

An interaction conserves parity when its dynamics are invariant under this transformation, equivalently $[H,\mathcal{P}]=0$ for a time-independent Hamiltonian. Parity violation means that a process and its fully reflected counterpart need not have the same probability. It is more than a left-right asymmetry in an unreflected laboratory apparatus.

The electromagnetic and strong sectors of the Standard Model conserve parity to the accuracy relevant here, taking the strong-sector Yang-Mills theta term to vanish. The weak charged current selects left-chiral fermions and right-chiral antifermions, as defined by fermion chirality; the reflected right-chiral charged current is not equally present. The neutral weak current also generally has different left- and right-chiral couplings. Consequently the weak sector violates parity; this does not mean that every individual weak decay must have opposite initial and final parity. For a massive particle, helicity and fermion chirality should be distinguished.

Write $\bar{\chi }=({\chi }_1+{\chi }_2)/2$ and $\delta \chi ={\chi }_2-{\chi }_1$. Since the supplied rotation matrix is real and orthogonal, $U^{-1}=U^T$. Direct matrix multiplication gives the flavor-basis HamiltonianThe common term $\bar{\chi }I$ produces only an overall quantum phase. The off-diagonal elements $H_{e\mu }^{\prime }=H_{\mu e}^{\prime }=(\delta \chi /2)\sin 2\theta$ couple the two interaction eigenstates in the Schrödinger equation. Flavor change requires both $\delta \chi \ne 0$ and $\sin 2\theta \ne 0$. For ultrarelativistic neutrinos in vacuum, $\delta \chi \simeq \Delta m^2/(2E)$, so a nonzero squared-mass difference is needed. Observable neutrino oscillation also requires coherent production and propagation of the two components and a resolvable nonzero relative phase; special propagation times can return the initial flavor even when these Hamiltonian conditions hold.

A useful distinction for the next part is that $H_{ee}^{\prime }$ is an energy, not a time-dependent survival probability amplitude. The latter is the $ee$ element of $S(t)=\exp (-i{H}^{\prime }t)$.

For a positive electron neutrino matter potential and the normal ordering implicit in positive $K$, choose $0\le {\theta }_m\le \pi /2$ by continuous diagonalization of the matter Hamiltonian. With $D=[(\cos 2\theta -K\rho {)}^2+{\sin }^{2}2\theta {]}^{1/2}$, the quadrant is fixed byAt high density $\cos 2{\theta }_m\to -1$, soTaking a principal-value arctangent alone and concluding ${\theta }_m\to 0$ would select the wrong branch. The production state is the second matter propagation eigenstate, rather than literally the vacuum mass eigenstate while still inside the dense medium.

Assume the solar density decreases slowly enough for adiabatic matter-enhanced neutrino conversion, and neglect subsequent Earth matter effects. The state follows ${\nu }_{2m}$ continuously into the vacuum mass eigenstate ${\nu }_2$. Since $⟨{\nu }_e|{\nu }_2⟩=\sin \theta$,This is substantially below the phase-averaged vacuum result $0.75$. There is no interference between two populated vacuum mass eigenstates in this ideal adiabatic limit. More generally, after phase averaging, $P_{ee}=\frac{1}{2}+\frac{1}{2}\cos 2\theta \cos 2{\theta }_m$ at production if the evolution is adiabatic. In the high-density limit, a nonadiabatic level-jump probability $P_c$ instead gives $P_{ee}=(1-P_c){\sin }^2\theta +P_c{\cos }^2\theta$. Thus the supplied core-density condition alone does not guarantee the numerical Earth result: the usual solar adiabatic assumption is also needed. These are the stipulated two-flavor parameters, not a precision prediction for the measured three-flavor solar spectrum.

For an infrared-regulated phase covariance, fix the uniform order-parameter phase zero mode and introduce ultraviolet and infrared cutoffs before taking limits. With the Fourier conventionthe dimensionless quadratic Hamiltonian diagonalizes asCompleting the square with a source gives $Z[J]/Z[0]=\exp [\frac{1}{2}\int JGJ]$, with inverse kernelThere is no additional $k_{B}T$ factor: it is already included in $\bar{K}$ because the supplied Hamiltonian is $\beta H$.

It is simplest to evaluate the real-space dependence by its defining Green equation, $-\bar{K}{\nabla }^{2}G={\delta }^{(d)}$. Rotational invariance gives $r^{d-1}{G}^{\prime }(r)=$ constant for $r>0$. Integrating over a small sphere fixes $-\bar{K}{S}_d{r}^{d-1}{G}^{\prime }(r)=1$, where $S_d=2{\pi }^{d/2}/\Gamma (d/2)$. Integrating once more,In two dimensions, take the limit after absorbing the divergent $r$-independent piece into $C$:For $d\le 2$, $G$ itself has an infrared-divergent constant in the infinite system; it is not a finite thermodynamic covariance without a regulator. The physical phase-difference variance is finite at fixed separation and cancels that constant. These are smooth spin wave correlations; compact-field vortex sectors are considered separately below.

Spontaneous symmetry breaking means that a pure thermodynamic phase is less symmetric than the Hamiltonian even when the selecting external field is removed. For example,The direction of the infinitesimal field is held fixed in this limit. Reversing the limits in a finite symmetric sample normally gives zero. For the XY model, an ordered state chooses one direction from a circle of energetically equivalent directions.

A Goldstone mode is a long-wavelength fluctuation along this continuously degenerate family. A uniform angle rotation costs no energy; a slowly varying one costs only gradient energy. In the stated static phase-only model the quadratic kernel is $\bar{K}{q}^2$, tending to zero as $q\to 0$. There is one independent soft order-parameter phase coordinate for broken $O(2)$ rotation. The static Hamiltonian does not specify a dynamical dispersion $\omega (q)$; that requires a dynamical law. Magnitude fluctuations instead have a restoring local curvature and have been excluded from this low-energy model.

Renormalization compares the same long-distance physics at different observation scales. The starting object is a microscopic or cutoff statistical Hamiltonian ${\mathcal{H}}_g$ with every symmetry-allowed coupling $g$. Because a critical correlation length is much larger than a lattice spacing, repeatedly eliminating short-distance detail can reveal scale-invariant behavior.

In a real-space transformation choose a normalized blocking kernel $B(S|s)$ for retained variables. DefineSumming over $S$ proves exact preservation of $Z$ before truncation. The transformed Hamiltonian generally gains longer-range and many-body interactions and a field-independent constant. Observables must be blocked consistently. In momentum-shell language split $\varphi ={\varphi }_{<}+{\varphi }_{>}$, integrate $\Lambda /b<|q|<\Lambda$, then use $x^{\prime }=x/b$ and a field rescaling to restore cutoff and kinetic normalization. Coarse-graining, rescaling and tracking generated interactions are distinct steps.

Repeated transformations generate a flow $g^{\prime }=R_b(g)$ with composition $R_b{R}_c=R_{bc}$ when the same exact coarse-graining family is used. Losing degrees of freedom generally makes this a semigroup, despite its conventional group name. A renormalization-group fixed point is invariant after rescaling. At a continuous transition the large-distance theory approaches a critical renormalization-group fixed point; finite correlation length instead eventually flows into an ordinary thermodynamic phase regime.

Linearize in scaling coordinates $u_i$:For increasing observation length, $y_i>0$ is relevant and must be tuned, $y_i<0$ is irrelevant, and $y_i=0$ is marginal and needs nonlinear analysis. Discrete multipliers $b^{y_i}$ must be compared with one, not with zero. The stable manifold forms the critical surface after relevant directions have been set to zero.

After subtracting regular additive contributions, ordinary hyperscaling givesChoosing $b=|t{|}^{-1/y_t}$ or choosing the field as scale yieldsFor a critical two-point function $G(r)\sim r^{-(d-2+\eta )}$, the order parameter scaling dimension is $x_m=(d-2+\eta )/2$, and its conjugate field has $y_h=d-x_m=(d+2-\eta )/2$. This recovers the Q1 identities while distinguishing the gap exponent $\Delta$ from the field's scaling dimension $y_h$.

As a concrete dimensional check, start from a scalar gradient action. Restoring its coefficient uses ${\varphi }^{\prime }(x^{\prime })=b^{(d-2)/2}{\varphi }_{<}(b{x}^{\prime })$, so at the Gaussian fixed pointQuartic interactions are relevant below four dimensions. They can produce an interacting Wilson-Fisher fixed point, whose exponents differ from naive Gaussian counting. Above four, the quartic coupling is irrelevant at the Gaussian fixed point but dangerous for ordered observables: ${\varphi }_0^2=-r/u$ is singular when $u$ is dropped. Thus the simple two-variable hyperscaling formula is not universal above the upper critical dimension.

Different microscopic systems can flow to the same long-distance renormalization-group fixed point, explaining universality of exponents and normalized scaling forms. Critical temperatures, metric factors and regular free energies remain nonuniversal. Dimension, order parameter symmetry, interaction range and additional soft variables constrain the universality class. Truncations, perturbative expansions and numerics approximate the flow; they must not be confused with the exact elimination definition.^[ref]

For a ferromagnetic nearest-neighbour Ising model on a chain let $s_i=\pm 1$, $K=J/(k_{B}T)>0$ and $h=H/(k_{B}T)$. The dimensionless Hamiltonian is $-K{\sum }_i{s}_i{s}_{i+1}-h{\sum }_i{s}_i$. In one dimension, decimating every other spin closes exactly within the nearest-neighbour family, including its additive constant. This is the exact chain building block of the Migdal-Kadanoff approach; bond moving becomes an additional approximation in higher-dimensional lattices.

For exact zero-field Ising-chain decimation, sum over an intermediate spin $\sigma$ between two retained spins:Parallel and antiparallel end spins give $A{e}^{K^{\prime }}=2\cosh 2K$ and $A{e}^{-K^{\prime }}=2$. HenceFor an even periodic chain, $Z_N(K)=A^{N/2}{Z}_{N/2}(K^{\prime })$. Rescaling the lattice spacing $2a$ back to $a$ completes the renormalization step. Decimation of $b$ successive links gives $\tanh K^{\prime }=(\tanh K{)}^b$, as follows by multiplying the transfer matrix $b$ times.

Let $v=\tanh K$. Then $v^{\prime }=v^2$ on $0\le v\le 1$. The infinite-temperature renormalization-group fixed point $v=0$ attracts every $0\le v<1$; the zero-temperature renormalization-group fixed point $v=1$ is unstable to heating, with slope two. There is no intermediate ferromagnetic renormalization-group fixed point. Equivalently $K^{\prime }\simeq K^2$ for small $K$, while $K^{\prime }\simeq K-\frac{1}{2}\log 2$ for large finite $K$. Thus there is no finite-temperature ordered thermodynamic phase in the short-range chain.

The domain-wall scaling coordinate of an Ising chain near $T=0$ is the domain-wall Boltzmann factor $w=e^{-2K}$. Its recursion and the zero-field correlation length areSince $\xi (K^{\prime })/a=[\xi (K)/a]/2$, the recursion has the correct length scaling. Its exponent is one with respect to $w$; it is not a finite power of physical temperature about $T_c=0$. In the thermodynamic limit, a bond is a domain wall with probability $(1-\tanh K)/2\sim e^{-2K}$, giving the same exponentially large length scale. Their finite energy allows entropy to proliferate walls at any positive temperature.

The Ising-chain decimation constant matters for thermodynamics. With dimensionless free energy per spin $f=-\lim N^{-1}\log Z_N$,The second expression follows independently from transfer eigenvalues $2\cosh K$ and $2\sinh K$ and satisfies the recursion. Multiplying it by $k_{B}T$ restores the physical free energy. Dropping $A$ would preserve normalized retained-spin probabilities but lose the eliminated-spin entropy.

Finite-field Ising-chain decimation is also exact. Use the symmetric transfer matrixand read the new couplings fromHere $W_{++}=e^{2K+2h}+e^{-2K}$, $W_{--}=e^{2K-2h}+e^{-2K}$ and $W_{+-}=2\cosh h$. The latter off-diagonal entry, the field convention split between adjacent bonds and the constant must all be tracked consistently. Near the zero-temperature zero-field renormalization-group fixed point $h^{\prime }\simeq 2h$, while at $K=0$ the independent-spin field is unchanged.

In the higher-dimensional Migdal-Kadanoff approximation, moving bonds creates chains that can be decimated. With bond moving first on a $d$-dimensional hypercubic lattice and length factor $b$, the strengthened links have $b^{d-1}K$, giving the conventionFor $d=1$ no transverse bonds move, so this reduces to the exact chain result. In higher dimensions it discards geometric correlations and is generally approximate on the original lattice; moving and decimating in another order yields another approximation, not a contradictory exact chain formula.^[ref]

Allowing azimuthal motion makes the endpoint speed satisfy $v^2=l^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\phi }}^2)$. The light-rod spherical pendulum therefore hasThe azimuth $\phi$ is a cyclic coordinate, since the Lagrangian is independent of it. Its conjugate canonical momentum is the conserved vertical angular momentum:For a finite uniform rod, replace $m{l}^2$ by $I$ and $mgl$ by $A$ from i. The same azimuthal symmetry then gives $p_{\phi }=I{\sin }^2\theta \dot{\phi }$.

The Hamiltonian is the Legendre transform of this Lagrangian. With $p_{\theta }=I\dot{\theta }$,There is no explicit time dependence, so $dH/dt=\partial H/\partial t=0$. Here $H$ also equals the total mechanical energy: the constraint is stationary, the kinetic energy is quadratic in generalized velocities, and gravity is conservative. The apparent singularity of the momentum expression at $\sin \theta =0$ is a spherical-coordinate singularity; trajectories reaching a pole must be described consistently, and nonzero $J_z$ prevents reaching it at finite energy.

Choose zero potential energy at the equilibrium position. For the simple harmonic oscillator, the kinetic energy and elastic potential energy give the LagrangianIts canonical momentum is $p=\partial L/\partial \dot{x}=m\dot{x}$. The Legendre transform defining the Hamiltonian consequently givesThe Euler-Lagrange equation is $m\ddot{x}+kx=0$. Thus along the motionThis proves conservation of energy. Equivalently, the Lagrangian has no explicit time dependence, so $dH/dt=-\partial L/\partial t=0$. The Lagrangian itself need not be constant: the kinetic energy and potential energy exchange periodically.

For ordinary conservative classical mechanics, the Lagrangian is $L(q,\dot{q},t)=T-V$, where $T$ is the kinetic energy and $V$ the potential energy. More generally it is the function whose action $\int Ldt$ satisfies the principle of stationary action. The canonical momenta and Hamiltonian areFor a regular Lagrangian, the first equations can be inverted to express the generalized velocities in terms of $q,p,t$; this defines the Legendre transform on phase space.

Along a solution of the Euler-Lagrange equations, ${\dot{p}}_i=\partial L/\partial q_i$. Taking a total time derivative givesThus an autonomous Lagrangian hasFor the usual time-independent mechanical coordinates and a velocity-independent potential energy, the kinetic energy is quadratic in the generalized velocities, so ${\sum }_i{p}_i{\dot{q}}_i=2T$ and $H=T+V$. The conservation proof itself uses the Euler-Lagrange equations, rather than assuming this special expression for energy.

The massless inextensible string has hanging length $y=l-r$, measured downward from the table. Thus $\dot{y}=-\dot{r}$. In polar coordinates, the tabletop mass has squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$. Adding the two kinetic energies and setting the hanging gravitational potential energy to $-Mgy$ givesand the Lagrangian isThe canonical momenta are $p_r=(M+m)\dot{r}$ and $p_{\theta }=m{r}^2\dot{\theta }$. Performing the Legendre transform gives the HamiltonianIt equals the total mechanical energy and is conserved because $L$ has no explicit time dependence. Also $\theta$ is a cyclic coordinate, so $J=p_{\theta }=m{r}^2\dot{\theta }$ is conserved angular momentum about the vertical through the hole. The table and the string guide supply external forces; conservation of the system's total linear momentum does not follow. These are the two independent generic integrals needed for the reduced tabletop mass coupled to a hanging mass.

Write the lattice reciprocal vector as $G=2\pi /a$ in one dimension. The nearly-free electron model starts from ${\epsilon }_k={\hslash }^2{k}^2/(2m_e)$ and couples plane waves separated by G. Near a zone boundary, the two-wave Hamiltonian isAt $k=G/2$ the free waves are degenerate and the periodic potential opens a gap $2|V_G|$. A reduced-zone drawing folds k modulo G into $[-G/2,G/2]$, displaying multiple bands at each reduced wavevector. The original plot shows free folded parabolas, avoided crossings and an allowed vertical optical transition. ^[ref]

Including two spin states, filling a free Fermi sphere gives $n=2(4\pi k_F^3/3)/(2\pi {)}^3=k_F^3/(3{\pi }^2)$. A monovalent BCC cell contributes two electrons, soThe sphere therefore lies inside every first-zone face, since the supplied distance is the nearest boundary.

For the shortest BCC reciprocal vector, $G=2\sqrt{2}\pi /a$. In the zero-potential folded-band limit, the energy of a vertical transition from k along G to the neighbouring free branch is ${\hslash }^2(G^2-2Gk)/(2m_e)$. It decreases over the occupied interval and is smallest at $k=k_F$. Thus the monovalent BCC direct optical threshold isThe final state is empty because $G-k_F>k_F$. Finite weak $V_G$ changes the energies and makes optical matrix elements nonzero; an exactly free electron gas by itself has no interband absorption. The expression is a leading weak-potential threshold, not a claim that every peak equals the onset.

Using the supplied lattice constants and bare-electron free bands gives $E_F=(3.24,2.12,1.85)\mathrm{eV}$ and thresholds $(2.06,1.35,1.18)\mathrm{eV}$ for Na,K,Rb. The source plot's broad higher-frequency features shift toward lower energy in the same order as a increases, qualitatively consistent with the predicted $a^{-2}$ scale. Their peak shapes, heights and exact positions need matrix elements, band corrections and scattering; low-frequency intraband conductivity is a separate contribution. No measured trace is reproduced.

Apply the radial momentum operator to a smooth radial test function. More generally $[f(r),p_r]=i\hslash f^{\prime }(r)$, giving the first two labelled identities. Expanding the square gives the third:The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A central potential commutes with $L^2$ and $L_z$, so choose simultaneous angular eigenfunctions $Y_{lm}$, with $L^2=l(l+1){\hslash }^2$ and $l=0,1,\dots$. The radial Hamiltonian consequently has the stated centrifugal term. The radial inner product uses measure $r^{2}dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial momentum is not thereby asserted to be a self-adjoint observable on a half-line.

Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i{p}_r/\hslash$. Since $D^{†}=-D$ formally and $[D,W]=W^{\prime }=q/r^2$,The hydrogen radial Hamiltonian factorization follows by matching the Coulomb coefficient, which fixes the Bohr radiusFor the second pair of labelled relations, the opposite product changes the sign of $W^{\prime }$, henceUsing $[A,A^{†}A]=[A,A^{†}]A$ in the factorization then givesThis Coulomb angular-momentum intertwining means applying $A_l$ to an energy eigenstate either gives zero or an eigenstate of $H_{l+1}$ with the same energy. For a normalized negative-energy state,At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The bound-state ladder termination by norm positivity occurs at some $L$ with $A_L\psi =0$, givingThe terminating radial function is proportional to $r^L{e}^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy scattering states are not subject to this terminating negative-energy argument. Here $m$ is the mass in the supplied Hamiltonian; including nuclear motion replaces it by the reduced mass.

The remaining cyclic commutators are $[S_y,S_z]=i\hslash S_x$ and $[S_z,S_x]=i\hslash S_y$. In a spin-$s$ irreducible representation,For spin one-half, $S_i=\hslash {\sigma }_i/2$, appropriate to an electron's spin. Each $S_i^2={\hslash }^{2}I/4$, so this Hamiltonian is $H=A{\hslash }^{2}I/4$. The orthonormal vectors $(1,0{)}^T$ and $(0,1{)}^T$ are eigenstates with the same energy $A{\hslash }^2/4$; every normalized superposition is also an eigenstate. The $B$ term vanishes identically because spin-one-half quadratic anisotropy is constant.

For the supplied three-dimensional basis one consistent choice isThese Hermitian matrices obey the cyclic commutators with the given $S_x$, and $S^2=2{\hslash }^{2}I$, representing spin one. Their squares giveThus the three normalized Cartesian basis vectors $e_1,e_2,e_3$ have energies ${\hslash }^2(A-B)$, ${\hslash }^2(A+B)$ and $0$, respectively; degeneracies permit arbitrary orthonormal choices in the corresponding eigenspaces.

The parallel and antiparallel $S_x$ states are $|+x⟩=(0,1,-i{)}^T/\sqrt{2}$ and $|-x⟩=(0,1,i{)}^T/\sqrt{2}$. Evolution from the first givesThis spin-one reversal under quadratic anisotropy is complete whenprovided $A+B\ne 0$. If $A+B=0$, its initial state is stationary up to phase and complete reversal never occurs. Assume real $A,B$ for a Hermitian Hamiltonian.

The anisotropic quantum harmonic oscillator is a sum of three commuting one-coordinate oscillator Hamiltonians. Substituting $\psi =X(x)Y(y)Z(z)$ and dividing by $XYZ$ gives a sum of three functions of independent coordinates, so each equals a separation constant. ConsequentlyFor a time-independent operator, conservation for arbitrary states follows from $[H,Q]=0$, since $d⟨Q⟩/dt=(i/\hslash )⟨[H,Q]⟩$. The first labelled operator is $i{L}_z/\hslash$; the second is $-L^2/{\hslash }^2$. The kinetic energy is rotationally invariant. The potential is invariant under rotations about $z$ precisely when (i) ${\omega }_x^2={\omega }_y^2$; full rotational invariance requires (ii) ${\omega }_x^2={\omega }_y^2={\omega }_z^2$. For positive oscillator frequencies these are equality of the frequencies themselves. Acting with the commutators on arbitrary test functions excludes anisotropic potentials in case (ii).

In the isotropic case let $b=m\omega /\hslash$. ThenThe polynomial $x$ is the first excited oscillator factor and the other two factors are ground-state Gaussians. The first excited spatial level has three independent choices, proportional to $x,y,z$. Odd parity gives $⟨x⟩=0$, and Gaussian integration yieldsFor oscillator uncertainty and allocation of classical energy, the $x$ mode has energy $3\hslash \omega /2$, whose classical amplitude is $A_x=\sqrt{3\hslash /(m\omega )}$, giving $\Delta x=A_x/\sqrt{2}$. If “same energy” instead means assigning the entire three-dimensional energy $5\hslash \omega /2$ to one classical coordinate, its amplitude is $\sqrt{5\hslash /(m\omega )}$ and the ratio is $\sqrt{3/10}$. A general three-dimensional classical orbit's $x$ amplitude is not fixed by total energy without specifying its allocation among modes.

The capacity of an isotropic oscillator shell, including the two electron spin states, is $(N+1)(N+2)$ electrons. Nine non-interacting electrons occupy shells as $2,6,1$ for $N=0,1,2$. ThusThe least positive excitation promotes an electron by one oscillator quantum. Rearrangements within the partly filled degenerate shell remain ground states and cost no energy.

A continuous symmetry is represented by a unitary family $U(ϵ)=e^{-iϵG/\hslash }$ with a self-adjoint conserved generator of a continuous quantum symmetry $G$. If $UH{U}^{†}=H$ for all $ϵ$, differentiating gives $[H,G]=0$. Thus, for no explicit time dependence,translation invariance conserves the corresponding momentum component; rotational invariance about one axis conserves that angular-momentum component. A central potential conserves all components and $L^2$, but the components need not be simultaneously measurable because $[L_x,L_y]=i\hslash L_z$. Time-translation invariance of a closed system gives a time-independent Hamiltonian and energy conservation.

A discrete symmetry need not have a continuously varying generator. For parity, $[H,P]=0$ lets stationary states be chosen even or odd, and preserves parity sectors. symmetry does not force every state to be an eigenstate of its conserved observable. Explicit time dependence adds $⟨\partial G/\partial t⟩$ to the conservation equation, and an external symmetry-breaking potential generally permits exchange with the environment. Degenerate eigenspaces require choosing a compatible common basis rather than assuming every eigenvector already diagonalizes every commuting operator.

In spherical coordinates, the angular Laplacian has spherical harmonics as eigenfunctions, $L^2{Y}_{lm}={\hslash }^{2}l(l+1)Y_{lm}$ and $L_z{Y}_{lm}=\hslash m{Y}_{lm}$. Separation leavesRegularity and normalizability give principal quantum number $n=1,2,\dots$, orbital quantum number $l=0,\dots ,n-1$, and magnetic quantum number $m=-l,\dots ,l$. Here the $m$ in the kinetic coefficient is the electron mass, distinct from the magnetic label.

For the proposed angular factor, $\cos \theta \propto Y_{10}$, so $l=1$ and magnetic quantum number zero. Direct differentiation gives ${\nabla }^2(r{e}^{-\alpha r}\cos \theta )=({\alpha }^2-4\alpha /r)\psi$. The $1/r$ term in the Schrödinger equation vanishes when $2{\hslash }^2\alpha /m=C$. With Bohr radius $a_0={\hslash }^2/(mC)$,wavefunction normalization uses $\int d\Omega {\cos }^2\theta =4\pi /3$ and ${\int }_0^{\infty }{r}^4{e}^{-2\alpha r}dr=24/(2\alpha {)}^5$, givingAn arbitrary constant phase is allowed. This is the $2p_z$ state. Under spatial inversion, $r$ is unchanged and $\cos \theta$ changes sign, so its parity eigenvalue is $-1$, also given by $(-1{)}^l$.

Rotational invariance means rotated states have the same energy; it does not require each member of a degenerate energy eigenspace to be invariant. rotations mix the three $l=1$ magnetic substates. An orbital oriented along $z$ can therefore have an anisotropic probability density while the Hamiltonian remains spherically symmetric.

The angular momentum commutation relations are $[L_x,L_y]=i\hslash L_z$ and cyclic permutations. The product rule yields $[L_z,L_x^2]=i\hslash (L_y{L}_x+L_x{L}_y)$ and $[L_z,L_y^2]=-i\hslash (L_x{L}_y+L_y{L}_x)$, so $[L_z,L^2]=0$. Simultaneous eigenstates have eigenvalues ${\hslash }^{2}l(l+1)$ and $\hslash m$; orbital angular momentum has integer $l\ge 0$ and $m=-l,\dots ,l$.