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Let $b_k=\cosh {\theta }_k{a}_k+\sinh {\theta }_k{a}_{-k}^{†}$ for real even ${\theta }_k$, and let every $b_k$ annihilate a normalized vacuum state. Inverting the Bogoliubov transformation gives $a_k=\cosh {\theta }_k{b}_k-\sinh {\theta }_k{b}_{-k}^{†}$. Only $⟨b_k{b}_q^{†}⟩={\delta }_{kq}$ survives, proving the displayed vacuum expectation values. Taking the Hermitian adjoint gives the same real negative pair amplitude for two creation operators. The normal occupation number is quadratic at small ${\theta }_k$, whereas the pair amplitude is linear. For one opposite-momentum pair the two-mode squeezed vacuum is $\mathrm{sech}\theta {\sum }_{n\ge 0}(-\tanh \theta {)}^n|n,n⟩$, fixing the sign convention directly.

The four-momentum operator collects the generators of spacetime translations in quantum field theory. With physical speed of light $c$,For the mostly-minus Minkowski metric,All components of $P^{\mu }$ therefore have units of momentum; $P\cdot a$ has units of action. The natural units convention $c=1$ gives the common abbreviation $P^{\mu }=(H,\mathbf{P})$.

A simultaneous momentum eigenstate satisfies $P^{\mu }|p⟩=p^{\mu }|p⟩$, where $p^{\mu }=(E/c,\mathbf{p})$ is its four-momentum. On a one-particle state of rest mass $m$, the mass-shell relation is $p_{\mu }{p}^{\mu }=m^2{c}^2$.

In a translation-invariant theory the four translation generators commute. A translation-invariant vacuum state has zero spatial momentum; after subtracting the additive vacuum energy, it also has zero Hamiltonian eigenvalue. This fixes the vacuum origin of the four-momentum spectrum.

For any fixed $r\ge 0$, the source contribution obeys $c{\varphi }_1\ge cr$, because $c<0$ and ${\varphi }_1\le r$. Equality occurs only at $({\varphi }_1,{\varphi }_2,{\varphi }_3)=(r,0,0)$ when $r>0$. Thus the global ground state lies on the positive first axis. Minimize there:For $0\le r\le v=\mu /\sqrt{\lambda }$, $f^{\prime }(r)<0$. For $r>v$, $f^{\prime \prime }(r)=3\lambda r^2-{\mu }^2>0$, so $f^{\prime }$ increases strictly from $c$ to $+\infty$. There is exactly one root $\tilde{v}>v$, and it is the unique global minimum. ThereforeThis argument selects the true ground state even if the full cubic has additional negative stationary roots.

Expand ${\varphi }_1=\tilde{v}+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The source is linear, so it shifts the expansion point but has zero Hessian matrix. The scalar mass matrix is diagonal at the new vacuum expectation value:Eliminating $\lambda$ with the stationary equation givesAll three squared masses are positive. The masses themselves are the positive square roots of these expressions.

The source explicitly reduces the stated SO(3) symmetry to SO(2); the selected vacuum state preserves that remaining SO(2). There is consequently no spontaneously broken exact continuous global symmetry in field theory to which the Goldstone theorem applies. For small $|c|$, the two angular excitations are pseudo-Goldstone bosons. Their squared masses vanish as $c\to 0^{-}$, while $\tilde{v}\to \mu /\sqrt{\lambda }$ and $m_h^2\to 2{\mu }^2$.

The Goldstone theorem states that spontaneous breaking of a continuous internal global symmetry in field theory in a local relativistic quantum field theory produces a zero-mass Goldstone boson for each independent broken generator, under the usual positive-state-space and conserved-current hypotheses. For $G\to H$ there are $\dim G-\dim H$ symmetry-required Goldstone bosons. This count assumes a Lorentz-invariant vacuum state; nonrelativistic counting can differ.

At the classical level, invariance of the scalar potential gives ${\partial }_{i}V(T^a\varphi {)}_i=0$. Differentiating at a stationary vacuum expectation value ${\varphi }_0$ yieldsThese are the Goldstone directions in the scalar mass matrix. Broken generators have $T^a{\varphi }_0\ne 0$, so their independent directions have zero quadratic mass with a canonical kinetic term. The theorem guarantees these modes; it does not exclude further accidental zero-mass modes. A local gauge symmetry is a redundancy and does not meet the physical global-symmetry hypothesis: the Higgs mechanism can instead supply longitudinal particle polarizations of massive gauge bosons.

Since the hypercharge is zero, electric neutrality means $T^{3}v=0$. Choose the real-triplet vacuum $v_i=v{\delta }_{i3}$, with $v^2=m^2/(4\lambda )$. With canonical real-scalar kinetic density $-\frac{1}{2}(D_{\mu }\Phi {)}^T{D}^{\mu }\Phi$, insert this vacuum into the covariant derivative. The hypercharge field $B_{\mu }$ has no coupling to this scalar, while only $T^{1}v,T^{2}v$ are nonzero. ThusThe charged combinations $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$ haveBoth $T^3$ and the independent hypercharge generator annihilate the vacuum. The connected unbroken subgroup is thereforeThe electromagnetic generator $Q=T^3+Y/2$ belongs to it, but it is only one linear combination of two surviving neutral generators. This model has two massless neutral gauge fields, rather than the single massless photon and massive neutral vector of the usual scalar-doublet electroweak model. A neutral-field rotation does not create a massive eigenvalue from this zero block.

The two transverse triplet components are eaten, leaving one real, electrically neutral radial scalar $H={\varphi }_3-v$. The radial eigenvalue of the Hessian found in part i isIndeed the radial potential is a constant plus $4\lambda v^2{H}^2+4\lambda v{H}^3+\lambda H^4$, fixing the factor of two between its quadratic coefficient and its mass squared. The physical degree-of-freedom count is preserved: three real scalars plus four massless vectors give $3+4\times 2=11$, while one scalar, two massive vectors and two massless vectors give $1+2\times 3+2\times 2=11$.

Gauge invariance of the scalar potential implies the infinitesimal identityThe irrelevant common factor $i$ from the real representation's infinitesimal rotation has been divided out. Differentiate with respect to ${\varphi }_k$ and evaluate at the stationary vacuum $v$:The second term vanishes by stationarity, leavingWhenever $T^{a}v\ne 0$, this is a nonzero vector in the kernel of the scalar mass-squared matrix, so it is a zero-eigenvalue direction. It is tangent to the degenerate vacuum orbit. In a global theory it is a Goldstone direction; in the gauged theory the corresponding would-be Goldstone field supplies a longitudinal polarization of a massive vector.

For the specific potential, stationarity at a nonzero vacuum gives $v^2=m^2/(4\lambda )$, assuming $m^2>0$ and $\lambda >0$. Its Hessian isThe antisymmetry of the triplet generators gives $v_i(T^{a}v{)}_i=0$, verifying the claimed kernel directly.

Goldstone's theorem states that in a relativistic quantum field theory, spontaneous breaking of an exact continuous global internal symmetry produces a massless Goldstone boson for each independent broken generator. More precisely, a conserved global current whose charge acts nontrivially on the vacuum expectation value has a massless pole in an appropriate current-field correlation function. The usual counting assumes a Lorentz invariant vacuum state and the standard locality and positivity hypotheses.

A local gauge symmetry is a gauge redundancy, so it does not by itself imply massless physical Goldstone bosons. In the Higgs mechanism, the corresponding scalar directions provide the longitudinal polarizations of massive gauge bosons.

Use the standard canonical anticommutation relations for the two distinct fermionic orbitals. In particular $\{a,b^{†}\}=\{a^{†},b\}=0$, in addition to the relations explicitly listed in the paper; these mixed relations are part of the usual two-orbital fermionic Fock space interpretation. Normalize the original vacuum state by $⟨0|0⟩=1$ and fix the orderingThe canonical anticommutation relations give $a^2=(a^{†}{)}^2=0$ and $a{a}^{†}=1-a^{†}a$. Consequently the number operator is an orthogonal projector:The same holds for $N_b$. Hence each eigenvalue obeys $n^2=n$ and is either $0$ or $1$, expressing the Pauli exclusion principle. Also $[N_a,N_b]=0$, so the two number operators have a simultaneous orthonormal basis of Fock states:Each number operator has eigenvalues $0,1$, each with state degeneracy two. For $N_a$, the zero eigenspace is spanned by $|00⟩,|01⟩$ and the unit eigenspace by $|10⟩,|11⟩$; for $N_b$, these are respectively $|00⟩,|10⟩$ and $|01⟩,|11⟩$. Any normalized linear combination within one such eigenspace is also an eigenstate of that number operator.

The ordering convention will matter below: $b^{†}{a}^{†}|0⟩=-|11⟩$ and

Let $N_c=c^{†}c$, $N_d=d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Substitution of the Bogoliubov transformation, with normal ordering using $d{d}^{†}=1-d^{†}d$ and $c{c}^{†}=1-c^{†}c$, givesFor example, the $v^{2}d{d}^{†}$ term in $a^{†}a$ contributes $v^2-v^2{N}_d$; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequentlyTo eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfyingWith $u=\cos \theta$, $v=\sin \theta$, take $0<\theta <\pi /4$, so $\tan 2\theta =\Delta /\epsilon$. Setting $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$ givesThis positive choice fixes a convenient convention for the vacuum state in part (d). It yieldsBoth number operators independently have eigenvalues $0,1$. Hence the energy eigenvalues and state degeneracies areThere is one fermionic Bogoliubov quasiparticle of energy $E$ per occupied transformed orbital. The two Nambu spinor matrix eigenvalues $\pm E$ should not be mistaken for two independently occupiable levels of opposite energy; the normal ordering constant and the two positive quasiparticle excitation costs give the four physical energy eigenvalues above.

An independent diagonalization of a matrix checks both the pairing sign and the additive constant. In the ordered orthonormal basis $|0⟩,|11⟩$, the even-fermion parity block isbecause $ab|11⟩=-|0⟩$ and $b^{†}{a}^{†}|0⟩=-|11⟩$. Its characteristic polynomial is $x^2-2\epsilon x-{\Delta }^2$, giving $x=\epsilon \pm E$. The odd-fermion parity Fock states $a^{†}|0⟩,b^{†}|0⟩$ are unaffected by pairing and both have energy $\epsilon$.

Equivalently, with the Nambu spinor $\Psi =(a,b^{†}{)}^{\mathsf{T}}$,Replacing $\Psi$ by $R(c,d^{†}{)}^{\mathsf{T}}$ gives $E{c}^{†}c-Ed{d}^{†}+\epsilon =E(N_c+N_d)+\epsilon -E$, reproducing the result by normal ordering.

The canonical anticommutation relations make $(a^{†}{b}^{†}{)}^2=0$, so the proposed exponential function terminates exactly:Use the inverse Bogoliubov transformation $c=ua-v{b}^{†}$, $d=v{a}^{†}+ub$, and the actions $a|11⟩=b^{†}|0⟩$ and $b|11⟩=-a^{†}|0⟩$. The two vacuum state conditions becomeBoth therefore require $B=v/u$. Normalizing this quantum state gives $1=A^2(1+B^2)$. Choosing positive $A$, which fixes an irrelevant overall global phase, yields $A=u$ because $u^2+v^2=1$. Thus the two-mode fermionic paired vacuum isThe other sign for $A$ describes the same physical quantum state. The positive relative coefficient is fixed by the paper's pairing order and the convention $|11⟩=a^{†}{b}^{†}|0⟩$.

As a direct eigenvector check, $\Delta v=(E-\epsilon )u$ and $\Delta u=(E+\epsilon )v$ implyIt is therefore the unique ground state, as also follows from positive quasiparticle excitation energy $E$. It has definite even fermion parity but is not an eigenstate of the original total number operator: the empty-pair and occupied-pair probabilities are $u^2$ and $v^2$. In particular $⟨N_a⟩=⟨N_b⟩=v^2$. The pair amplitude is $⟨ab⟩=-uv$, consistently with $ab|11⟩=-|0⟩$.

For the limiting check $\Delta \to 0^{+}$ at fixed positive $\epsilon$, $A\to 1$, $B\to 0$, so $|\tilde{0}⟩\to |0⟩$. For $\epsilon \to 0^{+}$ at fixed positive $\Delta$, $A\to 1/\sqrt{2}$, $B\to 1$, giving equal amplitudes for the empty and occupied pair. These checks also fix which branch of the Bogoliubov transformation produces the ground state.

The canonical anticommutation relations imply $a^2=(a^{†}{)}^2=0$. Thus the number operator satisfiesIt is a orthogonal projection, so its eigenvalues obey $n_a^2=n_a$ and are $\boxed{n_a\in \{0,1\}}$. The same argument gives $\boxed{n_b\in \{0,1\}}$. This is the Pauli exclusion principle for each orbital. The two number operators commute, so they have simultaneous eigenstates.

Take the vacuum state to have unit norm and fix the double-occupation convention $|11⟩=a^{†}{b}^{†}|0⟩$. An orthonormal basis of the four-dimensional fermionic Fock space isEach individual number operator has two-dimensional eigenspaces: for $N_a$ these are spanned by $\{|00⟩,|01⟩\}$ and $\{|10⟩,|11⟩\}$; for $N_b$ they are spanned by $\{|00⟩,|10⟩\}$ and $\{|01⟩,|11⟩\}$. Higher occupation numbers vanish because the square of a fermionic creation operator is zero. Reversing the order of the two fermionic creation operators changes only the sign convention for $|11⟩$.

Write $Q=c^{†}c+d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Expanding the number operators and restoring the displayed order with the canonical anticommutation relations givesThe minus sign in the first line follows from $c^{†}{d}^{†}+dc=-(cd+d^{†}{c}^{†})$. Substitution into the Hamiltonian operator yieldsLet $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$. A convenient real Bogoliubov transformation hasThese obey $u^2+v^2=1$, $u^2-v^2=\epsilon /E$ and $2uv=\Delta /E$. Equivalently $u=\cos \theta$, $v=\sin \theta$ with $0<\theta <\pi /4$ and $\tan (2\theta )=\Delta /\epsilon$. The coefficient of $P$ vanishes, while the coefficient of $Q$ is $E$. The constant isThus the diagonalized two-mode fermion pairing Hamiltonian isThe two transformed fermionic occupation numbers independently take values zero or one. The energy eigenvalues and their eigenspace dimensions are thereforeThe positive quasiparticle energy $E$ makes the transformed vacuum state the unique ground state. The constant $\mu$ is essential for the correct ground-state energy.

As a separate sign and spectrum check, use the original fermionic Fock space basis. Both singly occupied states are energy eigenstates with energy $\epsilon$, since the pairing terms annihilate them. In the ordered even-fermion parity basis $(|0⟩,a^{†}{b}^{†}|0⟩)$,The off-diagonal signs follow from $b^{†}{a}^{†}=-a^{†}{b}^{†}$ and $ab{a}^{†}{b}^{†}|0⟩=-|0⟩$. Its characteristic polynomial is $z^2-2\epsilon z-{\Delta }^2$, giving the remaining two eigenvalues $\epsilon \pm E$, in agreement with the Bogoliubov transformation.

The inverse Bogoliubov transformation gives $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$. Because $(a^{†}{b}^{†}{)}^2=0$ by the canonical anticommutation relations, the exponential truncates:The fermionic signs are fixed byIt follows thatBoth annihilation conditions hold precisely when $B=v/u$. Normalizing the paired fermionic vacuum, using the orthogonality of its two Fock states, gives $A^2(1+B^2)=1$. Choose the irrelevant overall phase so that $A>0$. The result isEquivalently,Its positive pair coefficient follows from the specified order $ab+b^{†}{a}^{†}$ in the Hamiltonian operator. The state has even fermion parity and is a quantum superposition of zero and two original particles. In the limit $\Delta \to 0^{+}$, $A\to 1$ and $B\to 0$, so it reduces to the original vacuum state, as expected for $\epsilon >0$.

In spontaneous symmetry breaking, the action respects a symmetry while a selected vacuum state is invariant only under a proper subgroup. A nonzero vacuum expectation value can identify that choice. For a local gauge theory, gauge transformations are redundancies rather than transformations between physically different states; a gauge-fixed scalar expectation value describes the Higgs mechanism and its massive excitations. It should not be interpreted as breaking an observable local redundancy.

Write the adjoint scalar field as $\Phi ={\Phi }_0+\phi$, with Hermitian gauge generators normalized by $\mathrm{tr}(t^a{t}^b)={\delta }^{ab}/2$. Since ${\Phi }_0$ is constant, the part of the covariant derivative linear in the gauge field is $D_{\mu }{\Phi }_0=ig{A}_{\mu }^a[t^a,{\Phi }_0]$. Both $t^a$ and ${\Phi }_0$ are Hermitian, so their commutator is anti-Hermitian. The scalar kinetic term therefore contributesMatching coefficients gives the adjoint scalar gauge-boson mass matrixIt is positive semidefinite: for real coefficients $x_a$, its quadratic form is $2g^2\mathrm{tr}(C^{†}C)\ge 0$, where $C=[x_a{t}^a,{\Phi }_0]$. Its null space consists of generators commuting with the vacuum, the Lie algebra of its centralizer.

For a diagonal vacuum ${\Phi }_0=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$, the matrix unit $E_{ij}$ obeys $[E_{ij},{\Phi }_0]=({\lambda }_j-{\lambda }_i)E_{ij}$. The two normalized Hermitian generators built from $E_{ij},E_{ji}$ each acquireThe diagonal generators $t^3,t^8$ commute with every such vacuum and remain massless. Orthogonality of these root generators also shows that the mass matrix is diagonal in the supplied Gell-Mann matrices basis. Take $g>0$ and $v\ne 0$ below; $v=0$ restores the full SU(3) group.

The beta barium borate crystal supplies spontaneous parametric down-conversion: a weak ultraviolet pump can produce a pair of lower-frequency photons, with their frequencies and wavevectors selected by conservation of energy and phase matching. A suitably aligned and compensated source makes the two polarization alternatives coherent, giving the stated Bell state. The crystal alone does not guarantee that every collected pair has that state.

The two polarizing beam splitters route horizontal polarization and vertical polarization to different ports. Take $H$ from arm $a$ to the trigger $T$ and $V$ to arm $c$. The ordinary beam splitter in arm $b$ is a lossless, polarization-independent 50:50 splitter, with the unused input in the vacuum state. It splits probability amplitudes, rather than measuring which output a photon takes. The half-wave plate changes the relative phase of its principal polarizations by $\pi$; orient it so that a $V$ photon in $c$ becomes $(H+V)/\sqrt{2}$ in $c^{\prime }$, up to a common phase. Finally, photodetectors absorb arriving photons and supply electrical clicks. They need not resolve photon number for the ideal four-photon, fourfold-coincidence calculation: a click at every detector already requires one photon at each.

Choose the relative wave phases so that the final polarizing beam splitter sends $H_{c^{\prime }}$ to $H_2$, $V_{c^{\prime }}$ to $V_1$, $H_d$ to $H_1$ and $V_d$ to $V_2$. Such phase conventions must be used consistently: an uncompensated relative phase would give a phase-shifted GHZ state.

There is at most one field photon in b. Projection onto its one-photon sector givesBefore detector absorption, the normalized conditional joint state is $|g,1⟩$ up to an overall phase. A destructive photon count changes the field to its vacuum but leaves the atom in $|g⟩$. This conditional state is meaningful whenever the one-photon outcome has nonzero probability.

For one electromagnetic normal mode, $a$ and $a^{†}$ are the annihilation operator and creation operator, with commutator $[a,a^{†}]=I$. Their action on normalized number states isThey remove or add one photon, whose energy is $\hslash \omega$. The vacuum state $|0⟩$ is normalized and obeys $a|0⟩=0$. It has no photons, but retains zero-point energy $\hslash \omega /2$ for the quantum harmonic oscillator Hamiltonian $H=\hslash \omega (a^{†}a+1/2)$. The coherent state $|\alpha ⟩=D(\alpha )|0⟩$ is a displaced vacuum state.

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.

Zero-point fields are the irreducible electromagnetic field fluctuations in the vacuum state. Each normal mode has ground energy $\hslash \omega /2$. Although $⟨\mathcal{E}⟩=⟨\mathcal{B}⟩=0$ in the vacuum state, d gives $⟨{\mathcal{E}}^2⟩=⟨{\mathcal{B}}^2⟩=1$. These are quantum fluctuations, not a classical random coherent field with a known instantaneous value.

Two consequences are the Lamb shift, where coupling to the quantized electromagnetic field changes atomic energy levels, and spontaneous emission, where an excited chemical atom can emit a photon into an initially empty normal mode. In the latter process the creation operator has a nonzero matrix element $⟨1|a^{†}|0⟩=1$; stimulated-plus-spontaneous factors have the form $n+1$, which does not vanish at $n=0$. Descriptions splitting spontaneous emission into vacuum fluctuations and radiation reaction depend on operator-ordering conventions; the measurable total rate does not. Zero-point fields do not mean that the vacuum state contains a nonzero mean photon number.

Define $E_{\mathbf{p}}=\sqrt{{\kappa }^2|\mathbf{p}{|}^4+m^2}$ and ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The Hermitian quantum field has the mode expansion of a free fieldwhere the annihilation operators and creation operators obey the bosonic canonical commutation relationsBecause $\pi =\dot{\varphi }$, its mode expansion of a free field isThe two nonzero cross commutators then giveand the equal-time field-field and momentum-momentum commutators vanish. This verifies the factor $1/\sqrt{2E_{\mathbf{p}}}$ without requiring Lorentz invariance.

Substitution into the Hamiltonian, followed by normal ordering, yieldsConsequently $[:H:,a_{\mathbf{p}}^{†}]=E_{\mathbf{p}}{a}_{\mathbf{p}}^{†}$: one application of a creation operator adds precisely the required quasiparticle energy. Before normal ordering there is the usual constant zero-point vacuum energy. If $m=0$, the exactly zero-momentum mode has $E_0=0$ and is a free canonical coordinate, not an oscillator with a normalizable oscillator vacuum state; in a finite box handle that mode separately or introduce a small positive mass. The formula applies directly to all modes with $E_{\mathbf{p}}>0$.