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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 Fock vacuum condition and the creation operator commutator giveEquality of the square-root product and $2E_{\mathbf{p}}$ holds on the support of the Dirac delta function.

The positive-energy mass shell has the Lorentz invariant measureHere both $p^2$ and the four-dimensional volume element are invariant, and positive energy is preserved by a proper orthochronous Lorentz transformation. Consequently the three-dimensional Jacobian determinant is $J=E_{\mathbf{\Lambda }p}/E_{\mathbf{p}}$. Transforming the Dirac delta function yieldsso its factor exactly cancels that of $2E_{\mathbf{\Lambda }p}$. Thus the displayed inner product is invariant under simultaneous transformation of both momenta. This is also why the one-particle completeness measure is $d^{3}p/((2\pi {)}^{3}2E_{\mathbf{p}})$.^[ref]

The canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, soDifferent species commute at equal times, so these products have no ordering ambiguity and $Q$ is Hermitian. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. Substituting the mode expansion of a free field and using the spatial Fourier transform identity givesSubtract the expression with species interchanged. The two-annihilator terms cancel after $\mathbf{p}\mapsto -\mathbf{p}$, as do the two-creator terms. Commuting different species in the remaining terms yieldsThere is no infinite vacuum contribution to the Noether charge to subtract: the contractions that would generate a species-diagonal constant are absent.

The bosonic canonical commutation relations giveDefine the creation operatorsTheir commutators with $Q$ are $[Q,c_{\pm }^{†}]=\pm c_{\pm }^{†}$. Since $Q|0⟩=0$, the one-particle states haveThe fixed-momentum states are delta-function normalized; a square-integrable wave packet of either species is a normalizable state with the same Noether charge eigenvalue. In this basis $Q={\int }_{\mathbf{p}}(c_{+}^{†}{c}_{+}-c_{-}^{†}{c}_{-})$, so it measures a difference of occupation numbers. The equal-time identity $i[Q,{\varphi }_1]={\varphi }_2$, $i[Q,{\varphi }_2]=-{\varphi }_1$ independently checks the sign against a.

For a Lagrangian density depending on $\varphi$, ${\partial }_t\varphi$, ${\partial }_i\varphi$ and ${\partial }_i{\partial }_j\varphi$, direct variation givesTake $\delta \varphi$ to have compact support, so every boundary term vanishes. One integration by parts for the first-derivative terms and two for the second-derivative term yield the Euler-Lagrange field equation with second derivativesThe plus sign on the last term is essential. For the action in a, it is particularly transparent to vary the Laplacian directly:Thus the classical equation of motion isHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. Substitution of the plane wave $\varphi =A{e}^{-i\omega t+i\mathbf{p}\cdot \mathbf{x}}$ givesA real classical solution combines complex-conjugate plane waves. Quantization assigns the positive branch to the excitation energy, giving the required dispersion relation. The negative-frequency branch supplies the creation operator part of the field, rather than negative excitation energy. Independently, the Hamiltonian in a gives $\dot{\varphi }=\pi$, $\dot{\pi }=-{\kappa }^2{\nabla }^4\varphi -m^2\varphi$, confirming the same sign.

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

Use the combinations $a_{\pm }=a\pm b$ at each momentum. Part a gives $[Q,a_{+}]=0$, $[Q,a_{-}]=-2a_{-}$, while their creation operators obey $[Q,a_{+}^{†}]=0$, $[Q,a_{-}^{†}]=2a_{-}^{†}$. Applying b givesThe creation operators transform by the Hermitian adjoint formulas with $e^{+2i\lambda }$. At $\lambda =\pi /2$, both exponentials are $-1$, soSubstituting into the fixed-time mode expansion of a free field of the complex scalar field givesThus $C$ exchanges particles and antiparticles, as required for charge conjugation. To see the charge reversal explicitly, the conserved charge $Q_{\mathrm{em}}={\int }_{\mathbf{p}}(a_{\mathbf{p}}^{†}{a}_{\mathbf{p}}-b_{\mathbf{p}}^{†}{b}_{\mathbf{p}})$ obeys $C{Q}_{\mathrm{em}}{C}^{†}=-Q_{\mathrm{em}}$. The generator $Q$ in the question is distinct from this charge.

The two occupation terms in the Hamiltonian are exchanged, with the same relativistic energy $E_{\mathbf{p}}$, and thereforeIn fact $[Q,H]=0$ because its mixing terms exchange equal-energy modes. With normal ordering, $:Q:=2{\int }_{\mathbf{p}}{c}_{-,\mathbf{p}}^{†}{c}_{-,\mathbf{p}}$ for $c_{-}=(a-b)/\sqrt{2}$. Then $C=e^{i\pi N_{-}}$ fixes the vacuum state and satisfies $C^2=1$. The construction is a bosonic species-swap generator.

Take parity symmetry in quantum field theory to preserve free time evolution, so the conjugate initial data reflect along with the field. Substitute the mode expansion of a free field into the defining parity operator relation and change the integration variable $\mathbf{p}\mapsto -\mathbf{p}$. Since $E_{-\mathbf{p}}=E_{\mathbf{p}}$, coefficient comparison, or comparison of both the field and its conjugate initial data, gives the intrinsic-even conventionHere parity symmetry in quantum field theory reverses spatial momentum and preserves time. The creation operator formulas follow by taking Hermitian adjoints. The given field relation has fixed the intrinsic phase to $+1$.

For one real scalar field, let $R$ act on one-particle wave functions by $(Rf)(\mathbf{p})=f(-\mathbf{p})$. It is a Hermitian operator with $R^2=1$, so $P_{-}=(1-R)/2$ is the orthogonal projector onto odd wave functions. Its second quantization is the odd-mode number operatorThe full momentum space is integrated in both expressions; the factors prevent double counting the pairs $\{\mathbf{p},-\mathbf{p}\}$. A suitable parity operator isFor verification, $[N_{\mathrm{odd}},a_{\mathbf{p}}]=-(a_{\mathbf{p}}-a_{-\mathbf{p}})/2$. Thus the even combination $a_{\mathbf{p}}+a_{-\mathbf{p}}$ is unchanged, and the odd combination $a_{\mathbf{p}}-a_{-\mathbf{p}}$ changes sign under conjugation by $e^{i\pi N_{\mathrm{odd}}}$, exactly reproducing $a_{\mathbf{p}}\mapsto a_{-\mathbf{p}}$. The same holds for creation operators. The odd-mode number operator has integer eigenvalues, so ${\mathcal{P}}^2=1$, and $\mathcal{P}|0⟩=|0⟩$. Multiplying by an overall constant phase would not change its conjugation action. This is parity as odd-mode number parity.

The quantized field expansion in the complete orthonormal basis of orbitals isThe complex conjugation in the creation operator expansion is essential for complex orbitals. Indeed, ${\hat{a}}_n^{†}=\int d^{3}r{\varphi }_n(\mathbf{r}){\hat{\psi }}^{†}(\mathbf{r})$ creates the wavefunction ${\varphi }_n$. The resolution of the identity in an orthonormal basis givesrecovering the canonical commutation relation and the spatial Dirac delta function.

The normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

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:

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.

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.

Put $C=\cos (\chi /2)$ and $S=\sin (\chi /2)$. Commutativity of the two creation operators and the binomial theorem giveThusThe state is a fixed-number two-mode condensate with binomial mode occupations; it is not a product of two independent Poisson-distributed Glauber coherent states.

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.

Bosonic creation operators commute, so the cross terms cancel:ThusThe square root of two in $(b^{†}{)}^2|0⟩=\sqrt{2}|2⟩$ is essential. Destructive interference removes the one-particle-per-output amplitude; the particles do not independently choose ports.

For photons this is the ideal Hong-Ou-Mandel effect. Its laboratory test combines one photon from each input at a balanced beam splitter and scans relative arrival delay while measuring output coincidences. Spatial, polarization and spectral indistinguishability must be controlled; distinguishable photons restore a nonzero coincidence baseline. The original experiment used down-converted photons, a translated beam splitter and coincidence counting, with an independently estimated accidental-coincidence background.^[ref]

The exponential input has norm squared $\exp (|{\alpha }_1{|}^2+|{\alpha }_2{|}^2)$. Its normalized form is a product of coherent states. Substitution of the inverse transformation givesSince the output creation operators commute, the normalized output factorizes as $|{\beta }_1⟩|{\beta }_2⟩$. For all nonnegative integers $n_1,n_2$,These are independent Poisson distributions, not a fixed-total-particle-number distribution. unitarity gives $|{\beta }_1{|}^2+|{\beta }_2{|}^2=|{\alpha }_1{|}^2+|{\alpha }_2{|}^2$. Equal complex inputs make ${\beta }_2=0$, so that port is certainly empty; opposite inputs empty port 1. This is amplitude interference of coherent inputs, distinct from the fixed-number two-particle interference in (c).

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.