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The number operator is $\hat{N}=a^{†}a$. The number state expansion gives $a|\alpha ⟩=\alpha |\alpha ⟩$ by shifting the summation index; taking the adjoint gives $⟨\alpha |a^{†}={\alpha }^{\ast }⟨\alpha |$. Thus the expectation value isFor the second moment, use $a{a}^{†}=a^{†}a+I$:Its expectation value in the coherent state is $⟨{\hat{N}}^2⟩=|\alpha {|}^4+|\alpha {|}^2$. ThereforeThe variance equals the mean, exactly as for the Poisson distribution found in b. For $\mu >0$ the fractional standard deviation is $1/\sqrt{\mu }$, the usual photon shot noise scaling. For the vacuum state, both the mean photon number and its variance vanish; this does not imply zero optical quadrature fluctuations. At $\mu =0$ the fractional fluctuation is undefined.

There is a genuine periodicity mismatch in the source. Write $x_s=sb+u_s$. The intended periodic displacements in a harmonic chain satisfy $u_N=u_0$, equivalently $x_N=x_0+Nb$ for unwrapped positions. The printed $x_N=x_0$ instead makes the potential $\frac{1}{2}m{\lambda }^2{\sum }_s(x_{s+1}-x_s{)}^2+\frac{1}{2}Nm{\lambda }^2{b}^2$, with equilibrium at equal positions, not $x_s=sb$. The following lattice derivation uses periodic displacements, as required by the requested position operator.

Put $k_r=2\pi r/(Nb)$, $r=0,\dots ,N-1$, and Fourier transform the displacements and momenta with normalization $N^{-1/2}$. The spring quadratic form is diagonal, with factor $|e^{i{k}_{r}b}-1{|}^2=4{\sin }^2(k_{r}b/2)$. Thus each nonzero mode is a harmonic oscillator withCreation and annihilation operators obey $[a_r,a_{r^{\prime }}^{†}]={\delta }_{r{r}^{\prime }}$. Inverting the oscillator-coordinate expansion givesThe conjugate momentum is $p_s=-i\sqrt{m\hslash /(2N)}{\sum }_{r\ne 0}\sqrt{{\omega }_r}(a_r{e}^{i{k}_{r}sb}-a_r^{†}{e}^{-i{k}_{r}sb})$. Fourier orthogonality diagonalizes the Hamiltonian as ${\sum }_{r\ne 0}\hslash {\omega }_r(a_r^{†}{a}_r+1/2)$. The normalized oscillator vacuum is annihilated by every $a_r$; applying creation operators gives number states, each added phonon contributing $\hslash {\omega }_r$. The omitted zero mode is a free centre-of-mass coordinate, which must be fixed or treated separately rather than assigned an oscillator vacuum. Accordingly the reduced displacement commutator is $[u_s,p_t]=i\hslash ({\delta }_{st}-1/N)$.

For $iq{u}_s$, separate creation and annihilation parts. Their commutator is a scalar, so the supplied Baker--Campbell--Hausdorff formula and vacuum expectations giveSumming the geometric lattice phases yieldsThe ratio is interpreted by continuity at its removable singularities. The interference factor has its maximum $N^2$ at $q=2\pi n/b$, where all lattice phases coincide. The source's Fourier-sum hint should read $r\equiv 0(\mathrm{mod}N)$, rather than $r=Nb$.

The claim about the full matrix element also needs qualification: its Debye–Waller factor envelope varies with $q$. The absolute global maximum is $|M(0){|}^2=N^2$; at nonzero reciprocal-lattice points the value is $N^2{e}^{-2W(2\pi n/b)}<N^2$. Moreover the derivative of the envelope there is nonzero, whereas that of the interference factor is zero, so those points are not exact local maxima of the full finite-chain intensity. The reciprocal-lattice condition maximizes the interference factor, which is the usual constructive-scattering interpretation of the requested result.

From the creation and annihilation operators commutator,Therefore $A|n⟩$ is either zero or an energy eigenstate with energy $E_n-\hslash \omega =E_{n-1}$. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate, $A|n⟩=c_n|n-1⟩$. Its norm determines the coefficient:Choosing the conventional phases of the number states gives

Solving the given relation for the position operator givesPut $Q=A+A^{†}$. Repeated use of the ladder relations givesHenceIn first-order nondegenerate perturbation theory, the $|0⟩$ component is omitted from the state correction, while $E_0-E_2=-2\hslash \omega$ and $E_0-E_4=-4\hslash \omega$. It follows thatwhich is the first-order ground state of the quartic oscillator.

A coherent pump has a Poisson distribution of photon numbers. Second-harmonic conversion couples a pump number state $|n⟩$ to a state with two fewer pump photons with a matrix element proportional to $\sqrt{n(n-1)}$. Different number components therefore evolve at different rates, and the pump becomes correlated with the harmonic mode. Its reduced state is generally neither pure nor exactly Poisson coherent. An undepleted classical-pump approximation suppresses precisely this backaction; it cannot prove that the exact finite-depletion quantum state stays coherent. Vacuum input or zero interaction are trivial exceptions.

Let $A=a^{†2}$, $B=a^2$ and $N=a^{†}a$. The oscillator algebra gives $[A,B]=-4N-2$, $[N,A]=2A$ and $[N,B]=-2B$. Substituting into the definitions proves closure:The factors depend on the stated normalization: $K_3$ has an extra factor one-half relative to $N+1/2$. The Robertson uncertainty relation consequently givesFor a coherent state with $n=|\alpha {|}^2$, normal ordering givesThus $⟨K_1⟩=\mathrm{Re}{\alpha }^2$ and $⟨K_2⟩=-\mathrm{Im}{\alpha }^2$. In either variance, the fourth-power and $n^2$ terms cancel against the square of the mean, leavingThe variance product is exactly $4⟨K_3{⟩}^2$, so the coherent state saturates the stated bound.

Amplitude-squared squeezing in either of these axes meansThe other variance must then rise sufficiently to respect the product inequality. A rotated field-squared quadrature can reveal squeezing in another direction; the criterion concerns the declared quadrature and reference, not merely a reduction in mean optical intensity.

For stationary photodetection, the normalized correlation in second-order optical coherence is $g^{(2)}(\tau )=⟨E^{-}(t)E^{-}(t+\tau )E^{+}(t+\tau )E^{+}(t)⟩/⟨E^{-}{E}^{+}{⟩}^2$. Normal ordering expresses the absorption of two photons. In a single mode,A coherent state has $g^{(2)}(0)=1$. A single-mode thermal field has $g^{(2)}(0)=2$ and bunches, while a number state with $n>0$ gives $1-1/n$; a single-photon state has zero ideal coincidence probability at zero delay. Antibunching, a zero-delay suppression relative to nearby delays, is distinct from simply reporting an intensity variance. Multimode averaging and finite detector response reduce observed contrast.

The Hanbury Brown and Twiss effect is measured by splitting the light onto two detectors and histogramming coincidences versus delay. Background counts, detector timing, dead time and afterpulsing need calibration. Two detectors avoid confusing the single detector's inability to register arbitrarily close counts with real antibunching. First-order interference and second-order correlations contain different information: weak first-order coherence does not imply an absence of intensity bunching.^[ref]

The unit-normalized Fock vacuum is characterized byChoose the relativistic normalization of a one-particle state from the outset:Since any two creation operators commute, any permutation of their order gives exactly the same state vector. Thus these multiparticle quantum states are symmetric under exchange and describe bosons, without a minus sign or any additional exchange phase.

The Hilbert space is the bosonic Fock space, the direct sum of the symmetric $n$-particle Hilbert spaces, including the zero-particle sector. Sharp-momentum kets are distributionally normalized; square-integrable wave packets provide ordinary Hilbert space vectors. For repeated occupation of a discrete mode, a unit-normalized number state uses $(a^{†}{)}^n|0⟩/\sqrt{n!}$. That factorial does not belong in the above labelled continuum-ket convention.

The position operators and momentum operators are self-adjoint operators. Using conjugate linearity of the adjoint operator, $(cA+dB{)}^{\ast }=\overline{c}{A}^{\ast }+\overline{d}{B}^{\ast }$, givesWrite $D=\sqrt{2m\omega \hslash }$. Bilinearity and antisymmetry of the commutator, together with the canonical commutation relations, give $[P_j,X_k]=-i\hslash {\delta }_{jk}I$. HenceThe terms involving two position operators or two momentum operators vanish by the canonical commutation relations. Finally, reversing products under the adjoint operator, $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$, gives $[a_j,a_k{]}^{\ast }=-[a_j^{\ast },a_k^{\ast }]$. ThereforeHere $I$ is the identity operator. These are the relations for two independent pairs of creation and annihilation operators. The operator algebra in this question is understood on a common invariant operator domain, such as finite linear combinations of the number states constructed in part (c).

The common Fock vacuum obeys $N\psi ={\sum }_j{a}_j^{\ast }(a_j\psi )=0$. Put $B=(a_1^{\ast }{)}^{n_1}(a_2^{\ast }{)}^{n_2}$. By the commutator product rule and part (a),Moving $N$ through $B$ therefore givesDividing by $\sqrt{n_1!n_2!}$ provesThe constructed number state is nonzero: the relation $a_j(a_j^{\ast }{)}^r\psi =r(a_j^{\ast }{)}^{r-1}\psi$ follows from $[a_j,a_j^{\ast }]=I$ and $a_j\psi =0$. Commuting different modes and successively moving each creation operator across the inner product yieldsThus the given factorial denominator also gives unit norm, as expected for a number state.

Use the canonical commutation relationsWriting $c_j=\sqrt{m{\omega }_j/(2\hslash )}$, $d_j=1/\sqrt{2m\hslash {\omega }_j}$, direct expansion of the commutator givesThe Fock vacuum ${\psi }_{0,0}$ is annihilated by both $a_{j,-}$. Successive creation operators generateHence, in the order of part (a), the six energy eigenstates are proportional toIf the Fock vacuum has unit norm, dividing the general expression by $\sqrt{n_1!n_2!}$ gives normalized number states. Normalization is not required here, but using this convention makes the later change of basis unambiguous.

The three-point density correlation of an ideal quantum gas follows first for a number-conserving Gaussian quantum state. There are six pairings, one for each permutation of three annihilation operators. The identity gives the product of densities; each transposition has sign $ϵ$; both three-cycles have positive sign. ThusThe last two terms are complex conjugates. They equal $2\mathrm{Re}(g_{12}{g}_{23}{g}_{31})$, not an additional signed pair term. This is the permanent of a matrix $(g_{ij})$ for bosons and its determinant for fermions. Wick's theorem supplies the six terms only with the state assumption stated in part (a).

For completeness an exact bosonic Fock-state alternative can be given without that assumption. Let $\mathbf{\ell }$ be a mode-occupation decrement with ${\sum }_{\alpha }{\ell }_{\alpha }=3$. Define $A_{\mathbf{\ell }}(1,2,3)$ as the sum of ${\prod }_{j=1}^3{\phi }_{{\alpha }_j}({\mathbf{r}}_j)$ over all distinct ordered triples having these mode multiplicities. Acting three annihilation fields on the number state leaves mutually orthogonal states for different $\mathbf{\ell }$. ThereforeHere $N^{\underline{\ell }}=N!/(N-\ell )!$, with zero when $\ell >N$. This expression includes repeated-mode occupation corrections absent from the Gaussian permanent; for a single mode it reduces to $N(N-1)(N-2){\prod }_j|\phi ({\mathbf{r}}_j){|}^2$.

Choose orthonormal, flux-normalized incoming and outgoing channels and include all channels of this lossless scattering problem. Conservation of probability requires $\Vert \beta {\Vert }^2=\Vert \alpha {\Vert }^2$ for every input vector. With $\beta =S\alpha$,For a square channel matrix this also gives $S{S}^{†}=I$. A submatrix obtained by omitting loss channels need not be unitary, but that is not the stated two-channel model.

The given real Hadamard matrix is symmetric, unitary and its own inverse. HenceThis same transformation preserves the canonical commutation relations for bosons and anticommutation relations for fermions. To extract probabilities one must use normalized number states, including their factorial factors, even when the supplied input expression is unnormalized.

Displacing a one-dimensional quantum harmonic oscillator produces a linear force term and a scalar shift relative to the unshifted Hamiltonian. The scalar adds only a common phase. The linear term couples number states separated by one excitation. Arbitrary linear driving of the ground state creates a coherent state; a first-order expansion of that state contains only the vacuum and first excited state.^[ref]

For a number-conserving Gaussian pairing state, six permutations give this expression. Each transposition has sign $ϵ$, positive for bosons and negative for fermions; the two three-cycles have positive sign for either statistics. Equivalently it is the permanent or determinant of the three-by-three one-body density matrix. For bosonic number states, repeated-mode factorial corrections must be included instead of assuming that Wick theorem.

An internal qubit uses two distinguishable electronic or hyperfine structure quantum states, denoted $|g⟩,|e⟩$. The ion's centre-of-mass motion has a different Hilbert space. In one approximately harmonic direction, a number state $|n⟩$ has energy $E_n=\hslash \nu (n+1/2)$, where $\nu$ denotes a motional angular frequency, not ordinary frequency. The tensor product of quantum systems gives $|g,n⟩$. Cooling towards $n=0$ prepares motion; it does not automatically prepare the internal qubit. Nor does the ground state mean a stationary point particle: zero-point energy and position spread remain.

For several ions, Coulomb forces couple their displacements into collective normal modes. Each selected normal mode has its own quantum harmonic oscillator number states, while each ion retains internal qubit states. A pulse addressing one ion can change a shared mode. This supplies a route to correlations between distant internal quantum states; it also makes residual motional excitation and unwanted excitation of other normal modes relevant. A single-ion pulse calculation alone establishes no experimental multi-ion quantum gate performance.

A product of number states with equal fixed populations in two modes. Its total particle number is $2N$. It differs from a pair of coherent states, whose number distributions fluctuate.

For a two-dimensional anisotropic quantum harmonic oscillator in the indicated range, the six lowest number states, counted with multiplicity, have occupations $(0,0),(1,0),(0,1),(2,0),(1,1),(0,2)$. Their excitation energies, divided by $\hslash$, are $0,{\omega }_1,{\omega }_2,2{\omega }_1,{\omega }_1+{\omega }_2,2{\omega }_2$. These are ordered because ${\omega }_2<2{\omega }_1$ and $2{\omega }_2<3{\omega }_1$; every omitted state has at least three quanta and costs at least $3\hslash {\omega }_1$. At equal angular frequencies, the two-dimensional isotropic harmonic oscillator has energy eigenvalue $\hslash \omega (N+1)$ and quantum degeneracy $N+1$, since there are $N+1$ ordered nonnegative pairs with sum $N$.