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The stated annihilation conditions can be solved algebraically, even though (b) shows they are not ground-state conditions for the full printed Hamiltonian. Invert the canonical transformation:For $|{\psi }_{\lambda }⟩=\mathcal{N}{e}^{\lambda a^{†}{b}^{†}}|0,0⟩$, the commutators giveThus both $A$ and $B$ annihilate it precisely whenThe normalized two-mode squeezed vacuum isThe factorial from the exponential cancels those generated by the two creation operators. Normalization follows from a geometric series; the problem does not require its value, but it makes the subsequent partial trace explicit.

For comparison, the actual ground wave function of the printed stable Hamiltonian isIt follows directly from the two normal-mode Gaussian vacua. The proposed $|{\psi }_{\lambda }⟩$ is instead the exact ground state of the distinct pair-coupling Hamiltonian $\hslash \omega (a^{†}a+b^{†}b+1)+g(ab+a^{†}{b}^{†})$ under its stability condition. Omitting the remaining terms of $kxy$ silently would change the problem.

Use ${ϵ}_{xyz}=+1$ and the supplied commutators. Antisymmetry also gives $[K_a,L_b]={ϵ}_{abc}{K}_c$. Expanding the combinations,The mixed commutator cancels:ThusThere are two mutually commuting three-generator rotation algebras. As real compact Lie algebras,For example, the anti-Hermitian basis $-i{\sigma }_a/2$ obeys the same commutation convention in each $\mathfrak{su}(2)$ factor. This is an algebra statement. Globally, the simply connected group is $\mathrm{Spin}(4)\cong SU(2)\times SU(2)$, andDropping that central quotient would incorrectly identify the global groups.

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.

Put $X={\sum }_s{a}_s^{†}{b}_s$ and $Y=X^{†}$, and take real hopping amplitude $t$ and real $\mathcal{E}$. Write the anti-Hermitian ansatz with its ordering visible:Using $Yf(\Delta )=f(\Delta +2)Y$, its commutator with $\mathcal{E}$ isBoth terms equal their required hopping coefficients $-t$ ifThe shifted expression is then $f(\Delta +2)=t/[\mathcal{E}(\Delta +2)-\mathcal{E}(\Delta )]$. For a transition from difference $d$ to $d+2$, the left-ordered factor evaluates as $f(d+2)=t/[\mathcal{E}(d+2)-\mathcal{E}(d)]$, the final-minus-initial energy denominator.

The formula describes occupation-dependent virtual hopping on actual transitions with nonzero energy gaps. Perturbation theory also requires the hopping matrix element, including its bosonic occupation factors, to be small compared with that gap. The resonance obstruction to perturbative elimination at a zero denominator on a nonzero hop requires degenerate perturbation theory. Values on unreachable states whose hopping amplitude vanishes need not determine the generator.

The second-order terms aresoHere $S$ is the first-order generator. The full second-order commutator includes terms changing the number difference by four. To retain only fixed site occupancies, keep the products $XY$ and $YX$.

With $f=f(\Delta )$ and $f_{+}=f(\Delta +2)$, the mixed commutators giveThus occupancy-preserving second-order bosonic hopping givesFor bosons, normal ordering givesConsequently an explicit form in terms of the supplied function isThe operator $P$ exchanges spin labels between sites while preserving their total occupancies. Defining dimensionless Schwinger-boson spins ${\mathbf{s}}_a=\frac{1}{2}{a}_s^{†}{\mathbf{\sigma }}_{s{s}^{\prime }}{a}_{s^{\prime }}$ and similarly for $B$,No factor of ${\hslash }^2$ is needed for these dimensionless spins; for physical spin operators ${\mathbf{S}}_a=\hslash {\mathbf{s}}_a$ it becomes $2{\mathbf{S}}_a\cdot {\mathbf{S}}_b/{\hslash }^2$. The spin exchange belongs in the occupancy-preserving part, whereas two hops in the same direction do not.

A linear number-difference energy gives only a one-body level renormalization. Let $\mathcal{E}(\Delta )=E_c+\gamma \Delta$, with $\gamma \ne 0$. Then $f=t/(2\gamma )$ is constant, so the exchange coefficient cancels. Since $[X,Y]=\Delta$,The same-direction commutators vanish as well, so this is the whole second-order term. This agrees with expanding the two-site single-particle splitting $\sqrt{{\gamma }^2+t^2}$ with the appropriate assignment of levels. A collection of noninteracting bosons remains noninteracting after a change of single-particle basis. The case $\gamma =0$ is resonant and cannot be treated by this elimination.

A nonlinear energy generally makes the virtual denominators occupancy-dependent. Then $f(\Delta )$ and $f(\Delta +2)$ need not agree, and the occupancy-preserving effective Hamiltonian contains density-dependent shifts and a spin-exchange coefficientmultiplying ${\mathbf{s}}_a\cdot {\mathbf{s}}_b$. Its sign depends on the function and sector; nonlinearity alone does not fix a universal exchange sign. The full second-order result can additionally contain pair hopping between different occupation sectors.

For a concrete repulsive example, take $\mathcal{E}=U{\Delta }^2/4$, $U>0$, and restrict to two bosons with one on each site. Here $f(0)=-t/U$, $f(2)=t/U$, $N_a=N_b=1$, givingThe triplet energy shift is $-4t^2/U$ and the singlet shift is zero. This is ferromagnetic bosonic superexchange in this example: the symmetric spin sector can lower its energy through virtual double occupancy. The sign must not be imported from the fermionic Hubbard problem. The example specifies an additional energy function solely to illustrate the general expression; it is not an asserted numerical input.

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.

Using $[x,p]=i\hslash$, the Heisenberg picture equations for the quantum harmonic oscillator areSolving these coupled linear equations givesThe oscillator annihilation operator and creation operator areThey obey $[a,a^{†}]=1$ and $H_0=\hslash \omega (a^{†}a+1/2)$. Substitution of the position and momentum solutions, or the corresponding commutators with $H_0$, yields

Define the signed dimensionless displacementThe spatial translation operator is $T(x_0)=e^{-ip{x}_0/\hslash }$, and the phases of the new energy eigenstates are fixed by $|n^{\prime }⟩=T(x_0)|n⟩$. Their expansion coefficients are thereforeSince $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$,Set $A=-\lambda a^{†}$ and $B=\lambda a$. Their commutator is $[A,B]={\lambda }^{2}I$, which commutes with both operators. Thus the stated special case of the Baker--Campbell--Hausdorff formula is applicable and givesBecause $a|0⟩=0$,Taking the overlaps with $⟨n|$ therefore provesThis is the coherent state after shifting an oscillator equilibrium. Its normalization follows from $e^{-{\lambda }^2}{\sum }_n{\lambda }^{2n}/n!=1$.

There is a sign qualification in the printed expansion: $(m\omega x_0^2/(2\hslash ){)}^{n/2}=|\lambda {|}^n$ agrees with $(-\lambda {)}^n$ when $x_0\le 0$, equivalently $\gamma \ge 0$. For a general signed displacement $x_0>0$, the odd coefficients have the opposite sign in the specified translated basis. The squared displacement alone fixes the occupation probabilities, not these coherent relative phases. The two-exponential identity also requires the central-commutator condition used above; it is not valid in that simple form for arbitrary operators.

For a local spin quantized along its positive $\tilde{z}$ axis, the exact Holstein–Primakoff transformation isThe bosonic operators obey $[a_j,a_k^{†}]={\delta }_{jk}$. The physical local subspace is $0\le n_j\le 2S$; the larger unconstrained boson Fock space is not itself the finite spin Hilbert space. The ordering of the square roots matters, since $n_j$ does not commute with $a_j$.

This represents spin lowering using a bosonic creation operator, making a fully polarized spin state the boson vacuum. On the physical subspace it reproduces the spin commutators and the Casimir ${\tilde{\mathbf{S}}}_j^2=S(S+1)$. Expanding the square roots for small occupation relative to $S$ organizes a controlled large-$S$ linear spin-wave approximation; it is not an assumption that a spin has an infinite number of physical levels.

Separate the Klein-Gordon equation as $\varphi (t,x)=e^{\pm i\omega t}X(x)$. The spatial equation is $X^{\prime \prime }+k^{2}X=0$, with ${\omega }^2=k^2+m^2$. The first Dirichlet boundary condition removes the cosine solution, and the second requires $\sin (kL)=0$. ThusThere is no nonzero $n=0$ Dirichlet mode. The real field is built from both complex-conjugate time dependences, with mode functionsThe sine functions satisfy ${\int }_0^L\sin (k_{n}x)\sin (k_{m}x)dx=(L/2){\delta }_{nm}$. Consequently the factor $1/\sqrt{{\omega }_{n}L}$ gives unit Klein-Gordon inner product for each positive-frequency mode and the canonical equal-time field commutator after quantization. A general solution is a superposition of these modes, rather than a claim that a single mode exhausts all solutions.

The Holstein–Primakoff transformation realizes a spin algebra using bosonic creation operators and bosonic annihilation operators. A boson counts a unit of spin deviation from the local positive quantization axis: ${\tilde{S}}^z=S-n$ with $n=a^{†}a$. The square root in ${\tilde{S}}^{+}$ supplies the correct spin-ladder matrix elements. In particular,Their commutator is $2(S-n)$ on every physical number state, reproducing the Spin commutation relations. Although the oscillator Fock space is infinite, the physical subspace is onlyThe order of the square root and oscillator operator matters in obtaining these coefficients.

When the spin deviations are dilute, expand the square root in $n/(2S)$. The leading transverse operators are linear in the bosonic creation operators and bosonic annihilation operators and turn exchange interactions into a quadratic Hamiltonian, which can be diagonalized. Higher powers describe interactions between spin waves. The method therefore organizes corrections by inverse powers of $S$ about an ordered classical configuration. Large $S$ alone is insufficient: the average depletion must remain small compared with $S$. Possible problems include the occupation constraint after truncation, noncollinear order requiring site-dependent axis rotations, zero modes, and infrared fluctuations that destroy the assumed dilute expansion.

Promote the mode expansion of a free field coefficients to annihilation operators and creation operators, replacing $a^{\ast }$ by $a^{†}$. The canonical commutation relations areTogether with the normalization in part a, they give $[\varphi (t,\mathbf{x}),\pi (t,\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$ and vanishing equal-time commutators between two fields or two conjugate momenta.

Insert the mode expansion of a free field in the Hamiltonian. The spatial integral produces $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. The $aa$ and $a^{†}{a}^{†}$ coefficients cancel because ${\omega }_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The remaining terms giveThis is a continuum of quantum harmonic oscillators. In a finite periodic volume $V$, $(2\pi {)}^3{\delta }^3(0)=V$. The Fock vacuum satisfies $a_{\mathbf{p}}|0⟩=0$, so its vacuum energy density isNormal ordering would subtract this constant; here it must be retained because it is the quantity being calculated.^[ref]

One standard five-operation version of the Tietze transformations is: invert a relator; conjugate a relator by a word; multiply a relator on either side by another relator; insert or delete the empty relator; and introduce or eliminate a generator together with a defining relation $y=w$, where $w$ uses the other generators and an eliminated generator occurs nowhere else. The elementary relator operations are reversible. Renaming and reordering the generators and relators is harmless. The Tietze theorem states that two finite presentations define isomorphic groups exactly when they are related by a finite sequence of these transformations.

For a connected cell complex with one vertex, take one generator for each oriented one-cell and one relator from the attaching loop of each two-cell. Higher-dimensional cells do not change the fundamental group. This is the cellular presentation theorem, obtained by applying van Kampen to the two-skeleton. Conversely a finite presentation is realized by a bouquet of its generator circles with one disk attached along each relator word, giving a finite connected complex with the required fundamental group.

Triangulate each relator polygon by adding diagonals from one boundary vertex. Each diagonal is a new one-cell, and every triangular two-cell has an attaching word of length at most three. Algebraically, for a word $s_1\cdots s_{\ell }$ with $\ell \ge 4$, introduce successive partial-product generators $y_2,\dots ,y_{\ell -2}$. Use relatorsEliminating the new generators recovers the original word; words already of length at most three need no change. This proves the triangular-presentation claim geometrically and algebraically.

For the free abelian group of rank two, useIts two relations say $c=ab=ba$. For the requested transformation sequence, replace the second relator by the product of the first and second, freely reducing it to $ab{a}^{-1}{b}^{-1}$. Invert the first relator, giving $c{b}^{-1}{a}^{-1}$, the definition $c=ab$. Eliminate $c$ together with this defining relation. The remaining presentation is exactly the stated two-generator commutator presentation.

The construction is a triangular group presentation.

On a hemisphere of SU(2) as the three-sphere, write $A=u_{0}I+i{u}_i{\sigma }_i$, with real coefficients satisfying $u_0^2+|\mathbf{u}{|}^2=1$. Right multiplication by $I+i{v}_j{\sigma }_j$ gives $d{u}_i=(u_0{\delta }_{ij}+u_k{ϵ}_{jki})v_j$. This right multiplication flow defines left-invariant tangent fields. With the displayed factor $i$, they obey $T_{j}A=-A{\sigma }_j$, since ${\partial }_{u_i}{u}_0=-u_i/u_0$. Applying the commutator to $A$ and using the Pauli matrix commutator identity gives $[T_i,T_j]=-2i{ϵ}_{ijk}{T}_k$. The entries of $A$ contain every coordinate, so this determines the vector-field identity, not just its action in one isolated representation.

With real complete polarization vectors, the covariant photon Fock space oscillator commutators and ${\sum }_{\lambda ,{\lambda }^{\prime }}{ϵ}_{\mu }^{\lambda }{ϵ}^{\nu {\lambda }^{\prime }}{\eta }_{\lambda {\lambda }^{\prime }}={\delta }_{\mu }{}^{\nu }$ imply the equal-time canonical commutation relations $[A_{\mu }(\mathbf{x}),-{\dot{A}}^{\nu }(\mathbf{y})]=i{\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$. The two mixed creation-annihilation terms add; the two coordinate-coordinate or momentum-momentum terms cancel after their polarization sums. The sign of the time-component oscillator is essential to the identity. Boundary-induced canonical transformation in Feynman gauge explains the momenta for the unsubtracted density.

Collapsing an essential meridian circle of a torus to one point gives the pinched torus. In the torus's CW structure, the two-cell attaches to two circles by the commutator. Collapsing one circle leaves one one-cell and a two-cell attached by a null-homotopic word. Thus the quotient is homotopy equivalent to $S^1\vee S^2$. Its cellular chain groups are $\mathbb{Z}$ in degrees zero, one and two, with zero differentials, so its integral homology groups are $H_0=H_1=H_2=\mathbb{Z}$ and vanish in all higher degrees. This illustrates that collapsing a circle can preserve a top-dimensional homology class even while reducing the first homology rank.

For a uniform pro-p group with $U_i=G^{p^i}$, the map $x\mapsto x^{p^a}$ transports congruences byCommutator collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$, and the power map is an isomorphism $U_b/U_{b+1}\to U_{b+1}/U_{b+2}$. Thus a nontrivial first difference moves down exactly one layer when taking a $p$th power. Iterating gives the equivalence; completeness supplies unique roots when the congruence is read in reverse.

An element $x$ of a Hopf algebra $H$ is primitive whenNo grading is required. Applying the counit to either tensor factor shows $\epsilon (x)=0$. If $x,y$ are primitive, the algebra-homomorphism property of comultiplication givesbecause elements in the two separate tensor factors commute. Thus the primitive elements form a Lie algebra under the commutator.

In a universal enveloping algebra over $\mathbb{Q}$, the primitive elements are exactly the original Lie algebra. The Poincaré-Birkhoff-Witt theorem identifies the associated graded algebra with a symmetric algebra; in characteristic zero, a nonzero homogeneous polynomial of degree greater than one cannot be additive under the diagonal. The highest-degree part of a primitive element must therefore have degree one. Positive characteristic of a field changes this conclusion: if $x$ is primitive in characteristic $p$, then $x^p$ is primitive too, since the intermediate binomial coefficients vanish modulo $p$.

For a Lie algebra representation with generators referred to an orthonormal invariant form, the sum ${\sum }_a{t}_a{t}_a$ commutes with every generator. Indeed, the commutator is proportional to a contraction of antisymmetric structure constants of a Lie algebra with the symmetric product $t_a{t}_b+t_b{t}_a$. The Schur lemma makes it scalar in an irreducible complex representation; that scalar is its quadratic Casimir eigenvalue. With trace normalization $\mathrm{tr}(t_a{t}_b)=C(R){\delta }_{ab}$, taking the trace gives $C(R)\dim \mathfrak{g}=C_2(R)\dim R$.