A metric outer measure on a metric space $(X,d)$ is an outer measure satisfyingHere $d(A,B)=\inf \{d(a,b):a\in A,b\in B\}$. Every Borel set is measurable for a metric outer measure. Consequently its measurable sigma-algebra contains the Borel sigma-algebra; it may be strictly larger.

The Caratheodory measurability criterion yields a complete sigma-algebra ${\mathfrak{M}}_{\mu }^{\ast }$, and $\mu$ restricted to it is a countably additive measure. Completeness means that every subset of a measurable null set is measurable.

For the requested algebra proof, $X$ and $\varnothing$ plainly satisfy the criterion. Replacing $A$ by $X\setminus A$ merely exchanges the two terms, proving closure under complements. If $A,B$ are measurable, apply their splitting identities successively to any $E$:Subadditivity applied to the first two pieces impliesThe reverse inequality is outer subadditivity. Hence $A\cup B$ is measurable, proving closure under finite unions and therefore the algebra of sets property. The argument never subtracts infinite values.

An outer measure on $X$ is a function $\mu :\mathcal{P}(X)\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$, monotonicity, and countable subadditivity:Its measurable sets are those satisfying Caratheodory measurability criterion:The test runs over all subsets $E$, including nonmeasurable ones.

The absolute continuity of measures condition isThe Radon-Nikodym theorem says that if both positive measures are sigma-finite measures, then this condition is equivalent to the existence of a nonnegative measurable function $h$, unique up to equality $\kappa$-almost everywhere, such thatOne writes $h=d\nu /d\kappa$, the Radon-Nikodym derivative. Under these hypotheses $h$ is finite $\kappa$-almost everywhere. If $\nu$ is finite, then $h\in L^1(\kappa )$. A finite signed measure absolutely continuous with respect to a sigma-finite positive measure likewise has an integrable signed Radon-Nikodym derivative.

The sigma-finiteness hypothesis on the dominating measure cannot simply be omitted. On the Borel subsets of $[0,1]$, let $\kappa$ be counting measure and $\nu$ Lebesgue measure. Absolute continuity holds because the only $\kappa$-null set is empty. A Radon-Nikodym derivative would satisfy $h(x)=\nu (\{x\})=0$ at every point, contradicting $\nu ([0,1])=1$. Here counting measure is not sigma-finite.

The given angular momentum commutation relations imply $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. By linearity,Similarly,For the remaining commutators, the commutator product rule gives, for each $j$,The last sum vanishes because the Levi-Civita symbol is antisymmetric under $i\leftrightarrow k$, while $L_i{L}_k+L_k{L}_i$ is symmetric. Taking $j=3$ and then linear combinations with $j=1,2$ proves

If $L^2\psi =\lambda {\hslash }^2\psi$ and $L_3\psi =m\hslash \psi$, these identities yieldOne must also prove that this candidate eigenstate is nonzero. Since the angular momentum operators are self-adjoint operators, $L_{+}^{\ast }=L_{-}$, and expanding the product givesThereforeThis is nonnegative and, by the assumed inequality, nonzero. Hence it is positive. The angular momentum raising operator really does produce a nonzero simultaneous eigenstate with the claimed eigenvalues.

To prove that raising terminates, let $v$ be any nonzero state reached after $r$ steps, with projection $q=m+r$. Positivity givesso $q^2\le \lambda$. Thus the sequence cannot remain nonzero for every integer $r\ge 0$. At its last nonzero state $v=(L_{+}{)}^k\psi$, one has $L_{+}v=0$, and the squared norm identity givesUnder the paper's initial inequality the first step is nonzero, so in fact $k\ge 1$. An angular momentum ladder endpoint must therefore be reached: bounded projection follows from positivity, and the final raising norm fixes the total angular momentum eigenvalue. Writing $j=m+k$, the complementary product $L_{+}{L}_{-}=L^2-L_3^2+\hslash L_3$ gives $\Vert L_{-}v{\Vert }^2=2j{\hslash }^2\Vert v{\Vert }^2$, also showing $j\ge 0$.

Put $\gamma =2m\hslash \omega$. If ${\psi }_n$ is a normalized energy eigenstate at $E_n$, the factorization from part (a) givesIt follows inductively that $v_n=(a_{+}{)}^n{\psi }_0$ has squared norm ${\gamma }^{n}n!$. The commutator with $H$ establishes its energy eigenvalue at each step. Henceis normalized and has $E_n=(n+\frac{1}{2})\hslash \omega$. Distinct levels are orthogonal because $H$ is a self-adjoint operator. In particular,

To decompose the specified initial wavefunction, retain this exact phase convention for the creation and annihilation operators. Their difference givesSince $a_{-}{\psi }_0=0$ and $a_{-}{a}_{+}{\psi }_0=\gamma {\psi }_0$,This quadratic excitation of an oscillator ground state, $\Psi (x,0)=C_0^{-1}{X}^2{\psi }_0$, contains only the $n=0$ and $n=2$ energy eigenstates. The printed initial function has an unspecified overall normalization, so the Born rule requires division by its squared norm. After wavefunction normalization the state isThe minus sign is consistent with the paper's convention: its $a_{+}$ is $i$ times the usual real-coordinate raising operator multiplied by $\sqrt{\gamma }$, so its ${\psi }_2$ has the opposite sign to the usual real second harmonic oscillator eigenfunction. The energy measurement probabilities are independent of this choice of phase.

Therefore the complete answer isEvery other energy has probability zero. At any later time the two coefficients only acquire their separate dynamical phases, so these probabilities remain the same. A direct check of the normalization isThis also confirms the factor of three in the squared norm of the two-component expansion.

On a smooth wavefunction $f$, the position operator and momentum operator giveThus the canonical commutation relation is $\boxed{[X,P]=i\hslash I}$. Since $X$ and $P$ are self-adjoint operators and $m,\omega$ are real, the adjoint operator of $a_{-}$ isExpanding the products in the stated order is essential:Subtracting gives $\boxed{[a_{-},a_{+}]=2m\hslash \omega I}$. These identities initially hold on smooth rapidly decaying functions, which contain the quantum harmonic oscillator energy eigenstates used below.

The factorization also proves a lower bound: for a normalized energy eigenstate $H\psi =E\psi$,Therefore $E\ge E_0:=\hslash \omega /2$. The commutator product rule, applied to $H=a_{+}{a}_{-}/(2m)+E_0$, givesSo a nonzero application of the creation operator $a_{+}$ raises the energy eigenvalue by $\hslash \omega$, while the annihilation operator $a_{-}$ lowers it by the same amount.

Begin with any energy eigenstate, whose existence is assumed. Repeated lowering must terminate: otherwise it would eventually produce a nonzero energy eigenstate below $E_0$, contradicting the lower bound. More explicitly, if $(E-E_0)/(\hslash \omega )=N+r$ with integer $N\ge 0$ and $0\le r<1$, the first $N$ lowering steps are nonzero. If $r>0$, one further step is also nonzero by its positive squared norm and has energy $E_0+(r-1)\hslash \omega <E_0$, an impossibility. Thus $r=0$ and the last state satisfies $a_{-}{\psi }_0=0$. This provesEvery level exists as well: $\Vert a_{+}{\psi }_n{\Vert }^2=2m(E_n+\hslash \omega /2)=2m\hslash \omega (n+1)>0$, so repeated raising from ${\psi }_0$ never vanishes. The same factorization bounds the energy expectation of any normalized state in the one-dimensional harmonic oscillator form domain below by $E_0$, proving that the terminated state really is a ground state.

To find its probability density, solve the first-order equationIts solutions are ${\psi }_0=C_0\exp [-m\omega x^2/(2\hslash )]$. With $\alpha =m\omega /\hslash >0$, the supplied Gaussian integral givesTaking $C_0=(\alpha /\pi {)}^{1/4}$ fixes an irrelevant global phase. ConsequentlyThe ground-state energy is positive even though the potential minimum is zero: it is the oscillator's zero-point energy.