Write the four cyclic factors as $A=M_{ij}$, $B=M_{jk}$, $C=M_{kl}$ and $D=M_{li}$. The Gaussian moment pairing theorem gives three terms, corresponding to $(AB)(CD)$, $(AC)(BD)$ and $(AD)(BC)$. We sum each term over $i,j,k,l$.

For $(AB)(CD)$, the entry covariance givesIndeed the four products leave respectively three, two, two and one free indices. A cyclic relabelling gives exactly the same sum for $(AD)(BC)$.

For the crossing pairing $(AC)(BD)$,The first product allows $i=k$ and $j=l$ freely, whereas each other product forces all four indices to coincide. Combining the three pairings and the trace normalization,

The spectral theorem for real symmetric matrices identifies the sum of the $k$th powers of the eigenvalues with the trace of the $k$th matrix power. Therefore the moment of the empirical spectral measure isFor odd $k$, every summand has zero expectation by part (a)(ii), so $\boxed{\mathbb{E}[m_k(n)]=0}$ for every $n$.

For the second moment, use $M_{ji}=M_{ij}$. There are $n$ diagonal entries with variance two and $n(n-1)$ ordered off-diagonal entries with variance one. Their means vanish, givingHence $\boxed{\underset{n\to \infty }{\lim }\mathbb{E}[m_2(n)]=1}$.

The probabilistic form of Wick's theorem is Isserlis's theorem. If $(Y_1,\dots ,Y_k)$ is a centered random vector with a multivariate Gaussian distribution and covariance matrix $C$, its odd product moments vanish, and, for even $k$,Here ${\mathcal{P}}_2(k)$ is the set of perfect matchings of the $k$ labelled positions. The formula allows repeated coordinates and singular covariance matrices. To see the pairing rule, differentiate the moment-generating function $\exp (t^{T}Ct/2)$ once in each coordinate and set $t=0$: only terms pairing all differentiated coordinates survive, each with the displayed product of covariances.

Take $Y_a=M_{i_a{i}_{a+1}}$, with $i_{k+1}=i_1$. These variables are linear combinations of the original independent normal random variables, hence are jointly Gaussian. They have zero means, soSubstituting the entry covariance from part (a)(i) identifies each parenthesis with $\mathbb{E}[M_{i_a{i}_{a+1}}{M}_{i_b{i}_{b+1}}]$, giving the requested identity.

The joint probability distribution of all entries of $X$ is unchanged by $X\mapsto -X$, so $M$ and $-M$ have the same probability distribution. For the specified cyclic product $P=M_{i_1{i}_2}\cdots M_{i_k{i}_1}$, this transformation multiplies $P$ by $(-1{)}^k$. Every such product is an integrable random variable, since normal random variables have finite moments of all orders. Thus, for odd $k$,Repeated indices cause no difficulty: all factors change sign simultaneously. It would be incorrect to assume that these factors are independent.

Taking the transpose gives $M^T=(X^T+X)/\sqrt{2}=M$, so $M$ is a symmetric matrix. For $i<j$, $M_{ij}=(X_{ij}+X_{ji})/\sqrt{2}$ has the standard normal distribution, whereas $M_{ii}=\sqrt{2}{X}_{ii}$ has the normal distribution $N(0,2)$. Distinct upper-triangular entries depend on disjoint sets of the original independent random variables, so they remain independent.

Consequently the joint probability density function, relative to ${\prod }_{i\le j}d{m}_{ij}$, isHere $\mathrm{tr}(m^2)={\sum }_i{m}_{ii}^2+2{\sum }_{i<j}{m}_{ij}^2$ because $m$ is a symmetric matrix. This is the Gaussian orthogonal ensemble, with off-diagonal variance one and diagonal variance two; the rescaled matrix $M/\sqrt{n}$ has the corresponding $1/n$ off-diagonal variance.

We will also use its entry covariance. Expanding in the independent entries of $X$, for which $\mathbb{E}[X_{ij}{X}_{kl}]={\delta }_{ik}{\delta }_{jl}$, givesThe symbols ${\delta }_{ab}$ are Kronecker deltas. This formula includes the diagonal case, where the two terms both contribute.

Diagonalizing the $J_1$ matrix gives the normalized eigenvectorswith eigenvalues $+\hslash ,0,-\hslash$, respectively. The first projective measurement prepares $v_{+}$ up to global phase, regardless of the earlier state, conditional on obtaining $+\hslash$.

The unitary time evolution under $H=\omega J_3$ gives, with $\theta =\omega t$,The probability amplitudes for the second projective measurement areThe Born rule therefore gives the general spin-one transverse measurement after axial precession:At the specified times $t_k=\pi k/\omega$, $\theta =\pi k$, soFor even $k$, the result is certainly $+\hslash$; for odd $k$, it is certainly $-\hslash$. Although zero belongs to the spectrum of the observable, it never occurs at these times. Indeed, for even $k$ the evolved vector is $v_{+}$, while for odd $k$ it is $-v_{-}$, where the minus sign is a global phase.

The Hamiltonian operator $H=\omega J_3$ is already diagonal in the angular momentum eigenstate basis of part (b). Its three distinct energy eigenvalues and stationary states areThus the three column eigenvectors are $(1,0,0{)}^{\mathsf{T}}$, $(0,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$, with energies $+\hslash \omega$, zero and $-\hslash \omega$, respectively. Each is nondegenerate since $\omega >0$. Any nonzero scalar multiple gives the same stationary state after normalization; a quantum superposition of different energies is generally not stationary.

Recover the angular momentum operators from $J_1=(J_{+}+J_{-})/2$ and $J_2=(J_{+}-J_{-})/(2i)$. The spin-one angular momentum matrices in the chosen descending-$J_3$ orthonormal basis areThey are self-adjoint operators, satisfy $[J_i,J_j]=i\hslash {\epsilon }_{ijk}{J}_k$, and give $J^2=2{\hslash }^{2}I$, as required for $j=1$.

The phases of the three basis vectors can be chosen so that $A$ and $B$ are nonnegative real numbers. Matrix multiplication then givesThe angular momentum commutation relations require this to equal $2\hslash J_3=2{\hslash }^2\mathrm{diag}(1,0,-1)$. Equating diagonal entries yieldsHenceThese are also the $j=1$ special case of the normalized spin ladder operator coefficients $\hslash \sqrt{j(j+1)-\mu (\mu \pm 1)}$.

Choose the ordered angular momentum eigenstates $(|1,1⟩,|1,0⟩,|1,-1⟩)$, so $J_3=\hslash \mathrm{diag}(1,0,-1)$. Since $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$, a nonzero matrix element of $J_{+}$ must increase the magnetic quantum number by one, and one of $J_{-}$ must decrease it by one. Thus only entries $(1,2),(2,3)$ of $J_{+}$ and $(2,1),(3,2)$ of $J_{-}$ can be nonzero. The adjoint operator identity $J_{-}=J_{+}^{†}$ givesFor the $j=1$ irreducible spin representation, both adjacent transitions are present; the following calculation determines their sizes.

In the ordered orthonormal basis $(|\frac{1}{2},\frac{1}{2}⟩,|\frac{1}{2},-\frac{1}{2}⟩)$, the spin one-half angular momentum operators are $J_i=(\hslash /2){\sigma }_i$, with Pauli matricesThe ladder operators areExplicit multiplication gives $J_{+}{J}_{-}={\hslash }^2\mathrm{diag}(1,0)$ and $J_{-}{J}_{+}={\hslash }^2\mathrm{diag}(0,1)$. Their commutator is therefore

The angular momentum commutation relations arewhere ${\epsilon }_{ijk}$ is the Levi-Civita symbol. The angular momentum eigenstates havewith angular momentum quantum number $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and magnetic quantum number $\mu =-j,-j+1,\dots ,j$. There are $2j+1$ such states in one irreducible spin representation, so the $J^2$ eigenspace has dimension $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the magnetic quantum number with the particle mass in the earlier questions.

For completeness, the ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfy $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$ andThe upper and lower ends are $\mu =\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet dimension. Pure orbital angular momentum permits only integer $j$, whereas general spin allows half-integers. If a larger Hilbert space contains several copies of the same irreducible spin representation, the full $J^2$ eigenspace includes all those copies: its dimension is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.

Write $\alpha =m\omega /\hslash$. The first Hermite polynomials give ${\psi }_{0,0}\propto e^{-\alpha (x_1^2+x_2^2)/2}$ and ${\psi }_{1,0},{\psi }_{0,1}$ proportional, with the same constant, to $x_1{e}^{-\alpha (x_1^2+x_2^2)/2}$ and $x_2{e}^{-\alpha (x_1^2+x_2^2)/2}$. Thus in Cartesian coordinates the new wavefunctions areUsing polar coordinates $x_1=r\cos \phi$, $x_2=r\sin \phi$ givesThe orbital angular momentum is $L_3=-i\hslash {\partial }_{\phi }$, which immediately verifies its eigenvalues $0,\pm \hslash$. Multiplying these spatial wavefunctions by $e^{-i\omega t}$ for ${\chi }_0$ and $e^{-2i\omega t}$ for ${\chi }_{\pm }$ gives the corresponding full stationary states.

Choose the first three energy eigenstates consistently as ${\psi }_{0,0}$, ${\psi }_{1,0}=a_{1,+}{\psi }_{0,0}$ and ${\psi }_{0,1}=a_{2,+}{\psi }_{0,0}$. The Fock vacuum is annihilated by both lowering operators, soFor example $a_{1,-}{a}_{1,+}{\psi }_{0,0}={\psi }_{0,0}$, while $a_{2,-}{\psi }_{1,0}=0$, giving the middle formula. Thus the simultaneous eigenstates can be chosen asTheir pairs of eigenvalues $(H,L_3)$ are respectively $(\hslash \omega ,0)$, $(2\hslash \omega ,+\hslash )$ and $(2\hslash \omega ,-\hslash )$. The factor $1/\sqrt{2}$ normalizes the combinations when the original three states are normalized. Mixing only inside the degenerate first-excited energy eigenspace preserves the energy eigenvalue while diagonalizing $L_3$.

Set ${\omega }_1={\omega }_2=\omega$. Inverting the creation and annihilation operators givesSubstitute these expressions in $L_3=X_1{P}_2-X_2{P}_1$. Terms with two creation operators or two annihilation operators cancel because different modes commute. The remaining orbital angular momentum in oscillator ladder operators isEquivalently it is $-i\hslash (a_{1,+}{a}_{2,-}-a_{2,+}{a}_{1,-})$, consistent with the circular ladder operators of a two-dimensional oscillator.

For an arc represented by $0\le x\le a$, take $0\le a\le 2\pi R$. The Born rule usesIntegrate using ${\cos }^{2}u=(1+\cos 2u)/2$:The result is zero at $a=0$ and one at $a=2\pi R$, checking the wavefunction normalization. At $s=0$ it reduces to $a/(2\pi R)+\sin (4a/R)/(8\pi )$, the stationary position probability. Longer intervals on the periodically identified coordinate require specifying the corresponding subset of the circle; integrating multiple turns would count the same positions repeatedly.

Put $\tau =t-T$ and retain the global phase from the preceding Lüders rule update. Since $P{\varphi }_n=(\hslash n/R){\varphi }_n$, the same orthonormal basis diagonalizes the new Hamiltonian operator, withIn particular, ${\tilde{E}}_{\pm 2}=E_{\star }+m{s}^2/2\mp 2\hslash s/R$. Applying unitary time evolution from $T$ therefore givesThis is rigid translation from a momentum term in the Hamiltonian: the position probability density moves with velocity $-s$. The plus sign in $x+s\tau$ follows directly from the negative sign of $-sP$. At $\tau =0$ the expression agrees exactly with the preceding post-measurement state.

Let ${\Pi }_{\star }=|{\varphi }_2⟩⟨{\varphi }_2|+|{\varphi }_{-2}⟩⟨{\varphi }_{-2}|$ be the orthogonal projection for $E_{\star }$. The Lüders rule gives the normalized post-measurement stateThe factor $-e^{-i{E}_{\star }T/\hslash }$ is a global phase, so $\cos (2x/R)/\sqrt{\pi R}$ is an equivalent representative. The projective measurement preserves the quantum coherence between the two momentum eigenstates; it does not select one of them. This matters for the subsequent position probabilities.

By the Born rule, the probability for a measured energy eigenvalue is the squared norm of the orthogonal projection onto its whole energy eigenspace. ThusEvery other energy eigenvalue has probability zero. These probabilities are independent of $T$: unitary time evolution under the unchanged Hamiltonian operator changes the dynamical phases but not the squared expansion coefficients. The two equal-energy terms describe one measurement outcome.

Under unitary time evolution, each energy eigenstate acquires its own dynamical phase. Since $E_2=E_{-2}=E_{\star }$, the wavefunction isAt $t=0$, the identity $1-\cos (2x/R)=2{\sin }^2(x/R)$ recovers the normalized initial wavefunction.