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The orbital angular momentum $L_3=X_1{P}_2-X_2{P}_1$ generates rotations in the plane. The canonical commutation relations giveUse the commutator product rule $[A,B^2]=[A,B]B+B[A,B]$. Since different coordinates commute,Thus, for arbitrary positive angular frequencies,At ${\omega }_1={\omega }_2$, this becomes $\boxed{[L_3,H]=0}$. The rotational symmetry of the two-dimensional isotropic harmonic oscillator allows a basis of simultaneous energy eigenstates and orbital angular momentum eigenstates.

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 $L_3$.

A horizontal Stern-Gerlach spin measurement measures $S_x$. The inputs are its orthogonal eigenstates, soThe output beam identifies the input with certainty in a single ideal measurement.

For a normalized quantum state, $|\alpha {|}^2+|\beta {|}^2=1$, and the Born rule gives $\boxed{p(z+)=|\alpha {|}^2,p(z-)=|\beta {|}^2}$. If the displayed coefficients are not normalized, divide both by $|\alpha {|}^2+|\beta {|}^2$. A quantum superposition is a coherent linear combination: relative phase affects other spin measurements, unlike an incoherent statistical mixture with the same $z$-measurement probabilities.

Every pure qubit quantum state is an eigenstate of spin along some axis. Its Bloch vector isFor real normalized $\alpha ,\beta$, choose $\theta$ modulo $2\pi$ with $\alpha =\cos (\theta /2)$ and $\beta =\sin (\theta /2)$, allowing an irrelevant global sign. ThenEquivalently, $S_{\mathbf{n}}=(\hslash /2)\left(\begin{array}{cc}{\alpha }^2-{\beta }^2& 2\alpha \beta \\ 2\alpha \beta & {\beta }^2-{\alpha }^2\end{array}\right)$. A Stern-Gerlach spin measurement on this axis yields its positive result with certainty, so the state is not a nontrivial quantum superposition in that measurement basis.

Let ${\theta }_j=j\pi /N$ and retain the positive spin beam at every stage. The overlap of successive positive eigenstates isThus the unconditional survival probability, relative to the original intensity, isThe last retained quantum state is $|{\psi }_{\pi }^{+}⟩=|\downarrow ⟩$, so its final negative-$z$ result is certain conditional on survival and adds no further factor. In particular $p_2=1/4$, $p_{10}\simeq 0.7805$, and $p_{100}\simeq 0.9756$. Since $\log \cos x=-x^2/2+O(x^4)$,This is a changing-measurement-axis form of the quantum Zeno effect: frequent selections steer the surviving spin through a continuous succession of nearby eigenstates.

In the source's Gaussian electromagnetic units, set $\mathbf{\Pi }=-i\hslash \nabla -e\mathbf{A}/c$ and take $\Phi =0$. ThenThe chosen gauge is independent of $x,z,t$, so simultaneous energy and canonical $p_x,p_z$ eigenstates justify the exponential ansatz. On the remaining spinors use ${\Pi }_x=p_x+eBy/c$, ${\Pi }_y=-i\hslash d/dy$, ${\Pi }_z=p_z$. The coupled equations are $(E-m{c}^2)\phi =c\sigma \cdot \Pi \chi$ and $(E+m{c}^2)\chi =c\sigma \cdot \Pi \phi$.

Treat $\psi ,{\psi }^{\ast }$ independently in the Legendre transform in mechanics. Their canonical momenta areThe time-derivative terms cancel in ${\pi }_{\psi }\dot{\psi }+{\pi }_{{\psi }^{\ast }}{\dot{\psi }}^{\ast }-\mathcal{L}$, giving the Hamiltonian densityThe first-order time dependence makes the momenta constrained; this direct calculation still gives the canonical density. In physical terms this is the grand-canonical Hamiltonian of a Bose field, $K=H-\mu N$. The physical energy density is obtained by adding $\mu |\psi {|}^2$ back. A stationary field in this chemical-potential frame has the usual factor $e^{-i\mu t/\hslash }$ in the physical frame.

Complete the square:For uniform or periodic boundary conditions, or in the homogeneous bulk, both nonconstant terms are nonnegative and vanish simultaneously atThe ground states form a circle. A particular nonzero ${\psi }_0$ is changed by every nontrivial phase rotation, although the full family is invariant. Selecting one such mean-field theory state therefore exhibits spontaneous symmetry breaking of $U(1)$. This describes phase selection in the thermodynamic limit; an exact finite-system particle-number eigenstate need not itself have a nonzero field expectation.

Define $\Omega =\sqrt{{\Delta }^2+{\Omega }_R^2}$. Squaring the two-by-two matrix gives $H_R^2={\hslash }^2{\Omega }^{2}I/4$, soHere $|\pm {⟩}_R$ is the corresponding normalized eigenstate at $t=0$. In the rotating frame only its quantum phase changes. In the laboratory frame its spin direction also rotates about $z$ at angular velocity $\omega$: $⟨\mathbf{S}{⟩}_{\pm }=\pm \hslash ({\Omega }_R\cos \omega t,{\Omega }_R\sin \omega t,\Delta )/(2\Omega )$. An arbitrary initial state is the same fixed linear combination of these two solutions. If $\Omega =0$, the rotating-frame Hamiltonian operator vanishes and the entire doublet is degenerate; evolution is just $R(t)$.

Take $\omega >0$ and $T=2\pi /\omega$. A full spin one-half rotation gives $R(T)=-I$, an essential sign. The cyclic phase of a spin in a rotating field has factors and total quantum phasesTo separate the adiabatic dynamical phase, set $B=\sqrt{{\Omega }_0^2+{\Omega }_R^2}>0$ and $\cos \theta ={\Omega }_0/B$. Then $\Omega =B-\omega \cos \theta +O({\omega }^2/B)$. The instantaneous laboratory energies are $\pm \hslash B/2$, so the adiabatic dynamical phases are $\mp BT/2$. Subtracting them from ${\Phi }_{\pm }$ givesThe oriented solid angle $\mathcal{A}$ is swept out by the field direction. These are the Berry phases for the eigenstates parallel and antiparallel to that direction. Retaining the spinor sign makes the quantum phase vanish modulo $2\pi$ when the field does not trace a nontrivial loop. The assumption $\omega \ll B$ suffices for adiabatic following; near degeneracy it must not be silently dropped.^[ref]

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

The occupation state is an eigenstate of the projected Quantum Hamiltonian. Since $a_j^{†}{a}_j^{†}{a}_j{a}_j=N_j(N_j-1)$,Equivalently, the bracket is $N(N-1)+2N_0{N}_1$, where $N=N_0+N_1$; there is no self-interaction for a single particle.

Put $g=k\hslash /(2m\omega )$, which has units of energy. In the original oscillator basis,Write $A=\tilde{a}$, $B=\tilde{b}$, $C_2=\cosh 2\theta$ and $S_2=\sinh 2\theta$. The transformed number and pair terms obeyHence the coefficient of $AB+A^{†}{B}^{†}$ is $g{C}_2-\hslash \omega S_2$, which vanishes for

This cancels the requested pair terms but does not diagonalize the printed Hamiltonian. Its number-conserving coupling also transforms:After the pair cancellation, the last creation terms have coefficient $-g{S}_2/2$, nonzero for $k\ne 0$. They create two quanta from the $A,B$ vacuum, so that vacuum is not an eigenstate of the literal Hamiltonian. This is an inconsistency in the ground-state assertion used by the next part, present in the printed PDF.

For the actual coordinate-coupled system, the normal coordinates $q_{\pm }=(x\pm y)/\sqrt{2}$ and their conjugate momenta diagonalize $H$ into independent oscillators withA normalizable ground state requires $|k|<m{\omega }^2$. Eliminating just the two indicated cross-pair terms is not sufficient to establish that ground state.

For one spin channel, let $X_s=a_s^{†}{b}_s$ and $Y_s=X_s^{†}$. The number-operator commutators giveMoving a boson from $B$ to $A$ raises the number difference by two. On a number-difference eigenstate $|d⟩$, $X_s\mathcal{E}(\Delta )|d⟩=\mathcal{E}(d)X_s|d⟩$, while $\mathcal{E}(\Delta -2)X_s|d⟩=\mathcal{E}(d)X_s|d⟩$. ThusSubtracting the reversed products yieldsThis number-difference shift identity fixes the ordering. The placement of the coefficient to the left matters: it acts on the state after the hop. This spectral argument applies to any function defined on the accessible number-difference eigenvalues, without requiring a power-series expansion.

A normalized oscillator coherent state is an eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding it in Fock states givesSince $x=\sqrt{\hslash /(2M\omega )}(a+a^{†})$, $⟨a^2⟩={\alpha }^2$, and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$, the variance is $⟨x^2⟩-⟨x{⟩}^2=\hslash /(2M\omega )$. Henceindependently of the complex amplitude $\alpha$.

For $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$, the commutator is the scalar $[A,B]=|\alpha {|}^{2}I$. The Baker--Campbell--Hausdorff formula therefore terminates and gives the displacement operatorBecause $a|0⟩=0$, applying it to the Fock vacuum yields exactly the Fock state expansion above:For a general complex $\alpha$, its complex conjugate is required in the annihilation term. The expression with $-\alpha a$ printed in the paper is valid when $\alpha$ is real.

The fixed-particle-number sector is spanned by $|1⟩=c_1^{†}|0⟩$ and $|2⟩=c_2^{†}|0⟩$. In this basis,The relevant quantum Gibbs state is normalized within this two-dimensional sector, not over all particle numbers. With $\beta =(k_{B}T{)}^{-1}$ and $x=\beta t$, the exponential of a Pauli vector givesThereforeIts eigenstates are $|\pm ⟩=(|1⟩\pm |2⟩)/\sqrt{2}$, with energies $\mp t$ and probabilities $p_{\pm }=e^{\pm x}/(2\cosh x)$. These are nonnegative and sum to one, so the matrix is Hermitian and has trace one. Its purity of a density operator isFor $t\ne 0$, the zero-temperature limit is the pure ground-state projector,with purity one. As $T\to \infty$, $x\to 0$ and $\rho \to I/2$, with purity $1/2$. At every strictly positive finite temperature and finite $t$, both eigenvalues are nonzero, so the state is mixed. If $t=0$, the two levels are degenerate and the Gibbs state is $I/2$ at all temperatures, including its zero-temperature limit.

Using dimensionless Von Neumann entropy, $S=-\mathrm{Tr}(\rho \ln \rho )$, givesFor nonzero $t$, $S\to 0$ as $T\to 0$, while $S\to \ln 2$ as $T\to \infty$. More precisely, the high-temperature expansion is $S=\ln 2-x^2/2+O(x^4)$, and the low-temperature form is $S\sim (1+2|x|)e^{-2|x|}$. The thermodynamic entropy includes an overall factor $k_B$.

For $H=\mathbf{d}\cdot \mathbf{\sigma }$ with $\mathbf{d}\ne 0$, the eigenvalues are $\pm |\mathbf{d}|$ and the projectors are $P_s=(I+s\hat{\mathbf{d}}\cdot \mathbf{\sigma })/2$. A normalized eigenstate has $⟨\mathbf{\sigma }⟩=s\hat{\mathbf{d}}$. At $\mathbf{d}=0$ the spectrum is degenerate and that direction is undefined.

The two eigenstate rays of the Pauli Y gate have the displayed normalized representatives and respective eigenvalues $+1$ and $-1$. They form an orthonormal basis. Multiplication by $Y$ changes only the global phase of each of these states; it does not exchange their eigenvalue labels.

If a condensate order parameter stays nonzero along a closed loop, its phase changes by an integer multiple of $2\pi$. Broken rotational symmetry does not destroy this single-valuedness constraint. In an inhomogeneous ring the winding integer need not label an angular-momentum eigenstate.

A positive operator-valued measure is a family of positive operators $E_i$ satisfying ${\sum }_i{E}_i=I$. Each $E_i$ is called an effect and determines a probability $p_i=\mathrm{Tr}(E_i\rho )$ for input density operator $\rho$. The effects need not be orthogonal projections or commute. This permits measurements with more outcomes than the dimension of the Hilbert space, such as the trine qubit POVM. Coupling to a quantum ancilla and measuring the larger system supplies a physical route to generalized measurements. ^[ref].

For an elementary unsharp $Z$ example on a qubit, takeOn the eigenstate $|0⟩$, $p_{\pm }=(1\pm \eta )/2$; at $\eta =1/2$ they are $3/4$ and $1/4$. At $\eta =0$ the outcome is a fair coin independent of the input, and at $\eta =1$ these effects specify a sharp $Z$ projective measurement. Choosing $K_{\pm }=\sqrt{E_{\pm }}$ specifies one quantum instrument, but $U_{\pm }\sqrt{E_{\pm }}$ for outcome-dependent unitary operators has the same effects and different state updates. Thus POVM does not determine the post-measurement state.

Choose eigenphase representatives $0\le {\varphi }_j<1$ for a unitary operator $U$ and a positive integer $M$. Its positive-phase fractional power acts on each eigenstate by $e^{2\pi i{\varphi }_j/M}$. This branch uses arguments in $[0,2\pi )$ and can differ from the usual complex principal branch using $(-\pi ,\pi ]$. For exactly dyadic phases, exact quantum phase estimation stores each label in a coherent phase register; bitwise phase gates multiply it by the required phase, and inverse phase estimation removes the labels. Keeping the phase register unmeasured preserves arbitrary eigenstate superpositions and resets the ancilla qubits. With only controlled-$U$ and controlled-$U^{-1}$ primitives, the direct method uses $2(2^n-1)$ oracle calls for $n$ exact phase bits.