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The Heisenberg picture evolution of the annihilation operator follows from $[H,a_{\mathbf{p}}]=-E_{\mathbf{p}}{a}_{\mathbf{p}}$. With $p^0=E_{\mathbf{p}}$, this gives the mode expansion of a free fieldDefine ${\Delta }^{+}(z)={\int }_{\mathbf{p}}{e}^{-ip\cdot z}/(2E_{\mathbf{p}})$. The time-ordered product then gives the Feynman propagatorThe last expression follows by closing the $p^0$ contour integral below for $z^0>0$ and above for $z^0<0$; the scalar Feynman propagator pole prescription supplies the required positive- and negative-frequency terms.

For $z^0\ne 0$ every term obeys the homogeneous Klein-Gordon equation. At $z^0=0$, the equal-time commutator of the fields vanishes, so the propagator is continuous and there is no derivative of a Dirac delta function. Its first time derivative has the jumpThe second time derivative therefore supplies precisely the source:Thus the defined Feynman propagator is a Green's function with source normalization $-i$; $i{\Delta }_F$ has source normalization $+1$. Keeping this factor avoids confusing the vacuum expectation value convention with conventions which define the propagator after dividing by $i$.^[ref]

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 $E_{\mathbf{p}}>0$.

Identify the nonrelativistic mass with the rest mass, $m=m_0$. Expanding the relativistic kinetic energy in $p^2/(m^2{c}^2)$ givesFirst-order perturbation theory requires $\Delta E_0=⟨0|H^{\prime }|0⟩$. Put $\alpha =\sqrt{m\hslash \omega /2}$, so $p=i\alpha (a^{†}-a)$. The annihilation relations are $a|0⟩=0$ and $⟨0|a^{†}=0$. In particular, $a^{†}|0⟩$ does not vanish. A short calculation retaining the operator ordering givesEquivalently, of the ordered words in $(a^{†}-a{)}^4$, only $a{a}^{†}a{a}^{†}$ and $aa{a}^{†}{a}^{†}$ have nonzero ground-state expectation values, respectively $1$ and $2$. Words with a creation operator on the far left or an annihilation operator on the far right vanish; the remaining unbalanced words also vanish. ThusThe correction is negative because the relativistic kinetic energy is smaller than $p^2/(2m)$ at fixed nonzero momentum. This relativistic oscillator ground-state kinetic correction requires $\hslash \omega \ll m{c}^2$.

First apply the commutator product rule in the form $[AB,C]=A[B,C]+[A,C]B$:Repeated use of the commutator product rule gives, for $n\ge 1$,This is the commutator with powers of an eigenoperator. For $n=0$, the commutator with the identity operator is zero, agreeing with the formula.

Each $a_k^{\ast }{a}_k$ is unchanged by taking the adjoint operator, so $N^{\ast }=N$ in this algebra. Taking the adjoint of a commutator yields $[N,(a_j^{\ast }{)}^n{]}^{\ast }=[a_j^n,N]=n{a}_j^n$. By antisymmetry,In particular, each creation operator raises the number operator eigenvalue by one, and each annihilation operator lowers it by one.

Inverting the definitions of the creation and annihilation operators givesSubstituting in the orbital angular momentum $L$ givesOperators belonging to distinct modes commute. Therefore the terms with two annihilation operators and those with two creation operators cancel, and $a_1^{\ast }{a}_2=a_2{a}_1^{\ast }$, $a_2^{\ast }{a}_1=a_1{a}_2^{\ast }$. The surviving terms give the orbital angular momentum in oscillator ladder operators:

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.

A bosonic coherent state obeys $a|\alpha ⟩=\alpha |\alpha ⟩$ and has expansion $e^{-|\alpha {|}^2/2}{\sum }_n{\alpha }^n|n⟩/\sqrt{n!}$. Its occupation is Poisson distributed, with mean and variance $|\alpha {|}^2$. Its nonzero field amplitude makes phase coherence natural, although physically observable phase comparisons require a reference.

For paired fermions, the corresponding Cooper-pair pseudospin coherent state isIt coherently superposes empty and occupied pair states and has an anomalous average $⟨c_{-\mathbf{k}\downarrow }{c}_{\mathbf{k}\uparrow }⟩\ne 0$. It is not an ordinary complex-eigenvalue coherent state of a single fermion annihilation operator. The broken-symmetry state has indefinite particle number; number projection can restore a fixed-number description without discarding the underlying pairing correlations.

Hartree-Fock-Bogoliubov theory approximates a two-body interaction using both the normal density ${\rho }_{ij}=⟨c_j^{†}{c}_i⟩$ and anomalous pairing tensor ${\kappa }_{ij}=⟨c_j{c}_i⟩$. Direct and exchange contractions determine a normal one-particle field, while anomalous contractions determine the pair potential. The resulting quadratic Hamiltonian is diagonalized by a fermionic Bogoliubov transformation. In a homogeneous singlet case its Nambu block isThe fields are determined self-consistently, rather than selected independently of the quasiparticle state. For example,together with the particle-number equation. The pairing field extends ordinary Hartree-Fock theory to states with pair correlations. This mean-field treatment omits fluctuations beyond its self-consistent quadratic description.

For bosons, write the field as a coherent condensate plus fluctuations, $\hat{\psi }=\varphi +\delta \hat{\psi }$. The same normal/anomalous strategy retains $\tilde{n}=⟨\delta {\hat{\psi }}^{†}\delta \hat{\psi }⟩$ and $\tilde{m}=⟨\delta \hat{\psi }\delta \hat{\psi }⟩$, with a bosonic, commutation-preserving Bogoliubov transformation. In the leading weakly depleted contact-gas limit, the condensate density $n_0$ and effective coupling $g>0$ give $E_k=\sqrt{{ϵ}_k({ϵ}_k+2g{n}_0)}$, with ${ϵ}_k={\hslash }^2{k}^2/(2m)$. This Bogoliubov theory connects low-momentum sound to high-momentum particle excitations. More elaborate self-consistent truncations must respect the symmetry constraints needed for a gapless condensate mode.

Condensate depletion means particles outside a macroscopically occupied bosonic single-particle mode. Interactions can deplete that mode even at zero temperature: the Bogoliubov transformation mixes creation and annihilation operators, so its quasiparticle vacuum contains particles at nonzero momentum. A weak homogeneous three-dimensional gas with repulsive scattering length $a_s$ has leading quantum depletion of a Bose condensateThermal quasiparticles add another contribution. Depletion and normal-fluid density are distinct: a Galilean-invariant fluid can have quantum depletion while retaining its full superfluid mass response at zero temperature.

For the fermionic pairing problem, mean-field decoupling of the reduced BCS pairing Hamiltonian introduces the anomalous amplitude $\Delta$. Each paired Nambu block has diagonal entries ${\xi }_{\mathbf{k}},-{\xi }_{\mathbf{k}}$ and off-diagonal pairing entries of magnitude $\Delta$. A phase convention with negative off-diagonal entries gives eigenvalues $\pm E_{\mathbf{k}}$, where $E_{\mathbf{k}}=\sqrt{{\xi }_{\mathbf{k}}^2+{\Delta }^2}$. Its canonical Bogoliubov rotation yields the BCS coherence factorsThe partner transformation has the complementary sign required by fermionic anticommutation. At zero temperature all positive-energy BCS quasiparticle modes are empty. Expanding the supplied electron operator therefore givesSince $E_{\mathbf{k}}^2-{\xi }_{\mathbf{k}}^2={\Delta }^2$,The electron occupation is close to one far below the normal-state Fermi surface, equals one half at $\xi =0$, and decays toward zero above it. Pairing smooths the Fermi step over an energy range of order $\Delta$. The positive-energy quasiparticle occupation is instead zero everywhere at this temperature; the electron sea and the quasiparticle vacuum are different descriptions of the same state. At finite temperature, $n_{\mathbf{k}\sigma }=[1-(\xi /E)\tanh (E/(2k_{B}T))]/2$ in the same mean-field model.

For a three-dimensional parabolic normal band, the total density of states including both spins is $D(\mu )=3N/(2\mu )$. In the weak-gap limit, replace it by this constant across the narrow pairing region and use $\mu \simeq E_F$. The number of electrons above the normal-state Fermi surface isThe integral is $[\xi -\sqrt{{\xi }^2+{\Delta }^2}{]}_0^{\infty }/2=\Delta /2$. Therefore the BCS population above the normal-state Fermi surface isThe constant-density-of-states approximation and a small gap relative to the chemical potential are essential. This is the question's explicitly defined fermionic population measure; it is not the fraction of electrons outside a single bosonic condensate orbital.

For an order-of-magnitude mercury estimate, use $T_c\simeq 4.2\mathrm{K}$ and the weak-coupling scale ${\Delta }_0\simeq 1.76k_B{T}_c\simeq 0.64\mathrm{meV}$, with a representative metallic $\mu \simeq 7\mathrm{eV}$. These input approximations give $\Delta N/N\sim 7\times 10^{-5}$, of order $10^{-4}$. The use of a weak-coupling gap is an estimate, not an assertion of mercury's exact measured gap.

For a copper oxide, an illustrative gap $20\mathrm{meV}$ and effective chemical-potential scale $0.5\mathrm{eV}$ give about three per cent in this formula. Choosing $0.2$–$1\mathrm{eV}$ instead gives roughly $7.5$–$1.5$ per cent. These are explicit example inputs: the paper specifies no compound, doping or band parameters from which a unique number follows. A real cuprate's anisotropic gap, quasi-two-dimensional band and strong correlations require a band-resolved occupation integral, so the reliable contrast is the much larger gap-to-electronic-energy scale, rather than a universal cuprate percentage.

Normalize the ring modes as ${\phi }_j(x)=L^{-1/2}{e}^{i{k}_{j}x}$, with distinct $k_j=2\pi {\ell }_j/L$ and integers ${\ell }_j$. Inserting $\hat{\psi }(x)={\phi }_1(x)a_1+{\phi }_2(x)a_2$ into the contact interaction givesOrthogonality makes $I_{ijkl}=1/L$ when $k_i+k_j=k_k+k_l$, and zero otherwise: this is momentum conservation in the projected Hamiltonian operator. There is one all-mode-1 term, one all-mode-2 term, and four mixed terms with one particle in each mode. A pair conversion between distinct ring modes would require $2k_1=2k_2$ and therefore vanishes.

Using $a_j^{†2}{a}_j^2=n_j(n_j-1)$ and $a_1^{†}{a}_2^{†}{a}_1{a}_2=n_1{n}_2$ yields the contact interaction between two ring modes,The factor four comes from two independent choices of ordering for the creation operators and annihilation operators. This is the two-mode projected model; a full ring contact interaction can also scatter particles into other modes while conserving total momentum.

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.

For independent bosonic modes, the canonical commutation relation isFor independent fermionic modes, the canonical anticommutation relations arewhere $[A,B]=AB-BA$ and $\{A,B\}=AB+BA$. The creation operator is the adjoint of the annihilation operator. In either case the mode's number operator iswith $a_i=b_i$ or $c_i$ as appropriate. It measures the mode's occupation number. Its possible eigenvalues areFor fermions, $c_i^2=0$ implies $n_i^2=c_i^{†}(1-c_i^{†}{c}_i)c_i=n_i$, so the number operator is a projection. This expresses the exclusion of double occupation of one fermionic mode. The total particle number is $N={\sum }_i{n}_i$.

Expand the anticommutators using the relations in part (a). The terms with two creation or two annihilation operators vanish, leavingEach Majorana fermion operator is Hermitian and squares to the identity. In particular, $({\gamma }_i^A{)}^{†}{\gamma }_i^A=1$, so this expression cannot distinguish an empty mode from an occupied one. A single Majorana operator is its own adjoint and does not supply distinct creation and annihilation operators for such an occupation.

An occupation can instead be assigned after pairing Majorana operators into a fermionic mode. For two distinct Majorana operators ${\gamma }_a,{\gamma }_b$,This is a nontrivial projection operator with eigenvalues zero and one. Thus occupation belongs to a chosen pair of Majorana operators, not to an individual Majorana operator.

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

The annihilation operator associated with the new coordinate $X=x-x_0$ isAs the initial state is the old vacuum,Equivalently, applying $a^{\prime }$ to the number-state expansion in part (c) and shifting its summation index returns $-\lambda$ times the same state. It is therefore a coherent state of the displaced oscillator.

The new Hamiltonian is $H=U_{\min }+\hslash \omega (a^{\prime †}{a}^{\prime }+1/2)$. In the Schrödinger picture its evolution givesConsequentlyThe displayed prefactor is only a global phase. Equivalently, in the Heisenberg picture $a^{\prime }(t)=a^{\prime }(0)e^{-i\omega t}$. The mean occupation $|\alpha (t){|}^2={\lambda }^2$ is constant, and $\hslash \omega ({\lambda }^2+1/2)$ reproduces the energy above the new minimum found in part (b). Its mean position and momentum likewise reproduce that part's oscillating wavepacket centre.

The Holstein–Primakoff transformation replaces each finite-dimensional spin algebra by a bosonic creation operator and annihilation operator, while restricting the local occupation number to $0\le n_j\le 2S$. The number $n_j=a_j^{†}{a}_j$ counts deviations from the fully polarized reference quantum state. Its zero-boson Fock vacuum has ${\tilde{S}}^z=S$.

The representation is exact on this physical subspace: the ordered square root gives ${\tilde{S}}^{+}|n⟩=\sqrt{n(2S-n+1)}|n-1⟩$ and ${\tilde{S}}^{-}|n⟩=\sqrt{(n+1)(2S-n)}|n+1⟩$, reproducing the spin commutation relations. For $n_j\ll S$,so the leading Hamiltonian is quadratic in bosons and describes spin waves.

The square root generates boson interactions beyond leading order and its operator ordering matters. Truncation also enlarges the Hilbert space to unphysical occupations unless the low-density approximation remains self-consistent. Large fluctuations, particularly near a gap closing or in low dimensions, can invalidate the linear spin-wave approximation even when its algebraic manipulation is straightforward.

Choose the current convention corresponding to $\delta \phi =i\alpha \phi$. Noether's theorem givesFor the requested terms with no $\sigma$ and $f=\sqrt{2}v$, the Lagrangian is $\frac{1}{2}(\partial \pi {)}^2$ and the current becomes $J^{\mu }=f{\partial }^{\mu }\pi$. The supplied covariant oscillator normalization cancels the invariant measure when the annihilation operator contracts with $a^{†}(p)$, soThe momentum is on the massless shell, $p^2=0$. Differentiating once more therefore givesThis Goldstone current matrix element in a phase-only scalar theory shows that a conserved current can have a nonzero matrix element proportional to a null momentum. Thus current conservation is fully consistent with the current creating a massless one-particle state from the vacuum; the nonzero coefficient $f$ is its Goldstone-current coupling. It is the divergence that vanishes, not the current matrix element. Reversing the current convention reverses the nonzero matrix element but leaves this zero divergence unchanged.

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]

For real angles write $u_p=\cos {\theta }_p$ and $v_p=\sin {\theta }_p$. For one independent pair $p>0$, the transformation and its conjugate readUsing the canonical anticommutation relations for the $\alpha$ modes givesThe remaining anticommutators vanish for independent momentum pairs. The displayed two-by-two matrix is unitary and its inverse is obtained by changing $i$ to $-i$.

When the same formula is indexed over both signs of momentum, its first row at $-p$ must agree with the conjugate of its second row at $p$. Thus a consistent global convention requiresThis is the odd-momentum fermionic Bogoliubov angle convention. If one checks only the first-row formulas, their cross anticommutator is $i\sin ({\theta }_p+{\theta }_{-p})$; its vanishing gives the condition modulo $\pi$, with possible signs absorbable in the $\alpha$ modes. Requiring both rows to describe the same operators fixes the convention above. Merely checking a separate unitary matrix at each momentum misses this redundancy: choosing ${\theta }_p={\theta }_{-p}=\pi /4$ in both first rows would give $\{c_p,c_{-p}\}=i$.

At $p=0$, the annihilation operator and its conjugate belong to one mode. The full displayed formula is consistent only with $v_0=0$; choose ${\theta }_0=0$ and ${\alpha }_0=c_0$. This mode is not a two-mode pair.

Since every quasiparticle energy is positive, the ground state is the vacuum of all the ${\alpha }_p$ annihilation operators, including ${\alpha }_0=c_0$. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation givesBoth annihilate the normalized pair state $(u_p+i{v}_p{c}_p^{†}{c}_{-p}^{†})|0⟩$. For example, applying ${\alpha }_p$ produces the two cancelling terms $i{u}_p{v}_p{c}_{-p}^{†}|0⟩$. ThusThe factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is $v_p^2$. ThereforeThere is no zero-mode contribution. For fixed $L$, the sum converges since $v_p^2=O(p^{-4})$.

A normalized simple excited state is $|k⟩={\alpha }_k^{†}|\Omega ⟩$ for $k>0$. Its excitation energy is $E_k$, and its total momentum is $k$: both terms in ${\alpha }_k^{†}=u_k{c}_k^{†}+i{v}_k{c}_{-k}$ add momentum $k$. Acting on the $k,-k$ pair givesThe affected pair now contains exactly one fermion instead of mean number $2v_k^2$; all other pairs are unchanged. ConsequentlyThis is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for $\mu >0$, with the value one at $\Delta =0$. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,so fermion number is not conserved when $\Delta \ne 0$, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added fermion.

Write $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2{\omega }_p]$, $p^{\mu }=({\omega }_p,\mathbf{p})$ and ${\omega }_p=\sqrt{{\mathbf{p}}^2+m^2}$. The Fourier expansion in the source is compatible with the oscillator canonical commutation relationsFor the free real scalar field, $\pi =\dot{\varphi }$ andSubstitution of the Fourier expansion produces spatial Dirac delta functions. Terms with two annihilation operators or two creation operators cancel: in $P_0$ their coefficient is $-{\omega }_p^2+{\mathbf{p}}^2+m^2=0$, and in $P_i$ their integral is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The two mixed terms combine, with normal ordering removing the vacuum constant, to giveIn particular, $P_i$ contains the lower component $p_i=-{\mathbf{p}}_i$ in this Minkowski metric convention.