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Use the resonant interaction picture and define $g>0$ by the one-atom Jaynes-Cummings model in part (a). Equal coupling to both atoms gives the Tavis-Cummings modelThe initial state has two excitations and is symmetric under atom exchange. Excitation conservation and this symmetry confine the evolution toThe antisymmetric atomic spin singlet state is uncoupled and unpopulated. Applying the annihilation operator and creation operator givesThe first off-diagonal entry combines two indistinguishable atomic emission paths; the second also includes the $\sqrt{2}$ enhancement for creating the second photon. Its eigenvalues are $0,\pm \sqrt{6}g$. Put $\Omega =\sqrt{6}g$ and $K=H_I/\hslash$, so $K^3={\Omega }^{2}K$. Its exponential isApplying this to $|A⟩$ yieldsAll three states have the same bare energy on resonance, so restoring the free Schrödinger-picture evolution adds only a common global phase. The expression starts at $|ee;0⟩$, has initial derivative $-i\sqrt{2}g|B⟩$, and its three squared probability amplitudes sum to one.

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.

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.

Treat $\varphi ,{\varphi }^{†}$ as independent variables when taking the Legendre transform. Their canonical momenta areThe canonical quantization of a complex scalar field imposes, at equal times,with $[\varphi ,{\varphi }^{†}]=[\varphi ,\varphi ]=[\pi ,{\pi }^{†}]=[\pi ,\pi ]=0$, together with their adjoints, and $[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$. These are canonical commutation relations for two real scalar degrees of freedom, written in a complex basis. Local fields are operator-valued distributions, so these identities are understood after smearing or with a regulator.

The Hamiltonian density is $\pi \dot{\varphi }+{\pi }^{†}{\dot{\varphi }}^{†}-\mathcal{L}$, givingThe Heisenberg equation of motion is $\dot{O}=i[H,O]$. For example, $i[\int {\pi }^{†}\pi ,\varphi ]={\pi }^{†}$, while commuting $\pi$ with the gradient term and integrating its derivative of a Dirac delta distribution gives ${\nabla }^2{\varphi }^{†}$. ThusBoth fields satisfy the Klein-Gordon equation. Their spatial Fourier modes therefore have frequencies $\pm E_p$, $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. A complex scalar field has independent coefficients for these two frequency sectors, so, defining the invariant measure $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, its mode expansion isOne can recover the mode operators at any fixed time using complex scalar mode inversion:Substitution of the expansion verifies these inverses: the spatial integral selects the appropriate momentum, while the two frequency sectors acquire factors $E_p\pm E_q$. The unwanted sector vanishes when the delta function sets $E_p=E_q$. The canonical commutation relations then give, for example,Likewise $[b(p),b^{†}(q)]$ has the same value. The cross commutators vanish; their coefficients contain $E_p-E_q$ on the corresponding momentum delta function. Conversely, these oscillator brackets reproduce the equal-time field brackets, confirming the normalization of the measure.

Insert the mode expansion into $H$ and integrate over space. The terms $a^{†}{b}^{†}$ and $ba$ enforce opposite spatial momenta; their coefficient is $-E_p^2+{\mathbf{p}}^2+m^2=0$. For the diagonal terms the coefficient is $E_p^2+{\mathbf{p}}^2+m^2=2E_p^2$. Combining this with the two invariant measures leavesIn a box, $E_0={\sum }_{\mathbf{p}}{E}_p$, one half-quantum for each of the two real oscillator species. It diverges as the cutoff is removed. Normal ordering removes this vacuum energy, yielding the normal-ordered Hamiltonian of a free complex scalar field:In this nongravitational free theory an additive constant in $H$ changes only an overall state phase, not Heisenberg equations of motion, energy differences or scattering probabilities. It is therefore consistent to choose the Fock vacuum energy as zero. Both creation operators raise the energy: $[H,a^{†}(p)]=E_p{a}^{†}(p)$ and $[H,b^{†}(p)]=E_p{b}^{†}(p)$. Repeated creation builds a bosonic Fock space of spin-zero particles and antiparticles, with the same positive mass and energy. The negative-frequency part of the field does not describe negative-energy physical particles.

The global phase symmetry of a complex scalar field, with orientation $\varphi \mapsto e^{-i\alpha }\varphi$, has the indicated Noether current. Taking its divergence, the two first-derivative terms cancel, givingThe same calculation holds for its normal-ordered version. Integrating its temporal component gives a conserved charge when spatial boundary flux vanishes.

Here the ordering convention is important. Direct substitution into the literally ordered product $i\int ({\varphi }^{†}\dot{\varphi }-{\dot{\varphi }}^{†}\varphi )d^{3}x$ yieldsThe pair terms again vanish, now because their coefficient is a difference of equal energies. Reordering the $b$ term produces a divergent vacuum constant: with a finite-mode regulator $Q_{\mathrm{bare}}=N_a-N_b-N_{\mathrm{modes}}$. Thus the requested finite charge formula uses zero vacuum charge, or equivalently the vacuum subtraction of a complex scalar charge. With that conventional subtraction,The subtraction changes neither current conservation nor the action of $Q$ on fields; without it, the unrenormalized product does not literally equal the displayed normal-ordered operator.

Finally the oscillator commutators giveor $\boxed{Q{a}^{†}=a^{†}(Q+1),Q{b}^{†}=b^{†}(Q-1)}$. Acting on a charge eigenstate, the first creator raises its charge by one and the second lowers it by one. Starting from the neutral Fock vacuum, their one-particle states therefore have charges $+1$ and $-1$ respectively. This is the complex scalar charge operator $N_a-N_b$, whereas the energy counts the sum of the two occupations. A multiparticle state has additive charge equal to the number of particles minus the number of antiparticles.

Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

For the complex scalar field, the canonical momenta are $\pi ={\dot{\varphi }}^{†}$ and ${\pi }^{†}=\dot{\varphi }$. Split $H=H_0+H_I$. The interaction picture is obtained from the Schrödinger picture by $|{\Psi }_I(t)⟩=e^{i{H}_{0}t}|{\Psi }_S(t)⟩$ and $O_I(t)=e^{i{H}_{0}t}{O}_S{e}^{-i{H}_{0}t}$, choosing a common time origin. The fields therefore obey the free Klein-Gordon equation, while the states evolve through $H_I(t)$. The canonical quantization of a complex scalar field givesThe adjoint expansion contains $a^{†}$ and $b$. Particle and antiparticle modes are independent; their nonzero commutators areAll cross-species commutators, and all pairs of creation operators or of annihilation operators, vanish. These formulas reproduce $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=[{\varphi }^{†}(\mathbf{x}),{\pi }^{†}(\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$ and the remaining vanishing equal-time commutators.

There are no derivative interactions, so the interaction Hamiltonian is $H_I(t)=-\int d^{3}x{\mathcal{L}}_I(x)$. Differentiating the interaction picture state gives $i{\partial }_t|{\Psi }_I⟩=H_I(t)|{\Psi }_I⟩$. Its time-evolution operator satisfiesIteration gives ordered integration regions $t_1>\cdots >t_n$. Each is $1/n!$ of the full region with a time-ordered product, yielding the Dyson seriesWith the usual asymptotic scattering prescription, taking $t_0\to -\infty$ and $t\to +\infty$ provesTime ordering is essential because different interaction Hamiltonians need not commute.

At first order, use the connected tree scattering amplitude, equivalently the normal-ordered quartic vertex with the vacuum and one-particle mass contributions removed. This qualification matters if the local product is interpreted literally without normal ordering: single-vertex tadpole diagrams and vacuum diagrams are not the requested connected two-particle amplitude. The complex scalar quartic contact vertex is obtained fromIn a fully connected matrix element, an incoming particle and outgoing antiparticle attach to the two $\varphi$ factors; an outgoing particle and incoming antiparticle attach to the two ${\varphi }^{†}$ factors. Each pair has two assignments, giving $2!2!=4$. The external relativistic normalization of a one-particle state cancels each mode measure: for example $[\varphi (x),a^{†}(p_1)]=e^{-i{p}_{1}x}$ and $[b(q_2),\varphi (x)]=e^{i{q}_{2}x}$. Consequently the connected integrand is $4e^{-i(p_1+q_1-p_2-q_2)\cdot x}$, andThus the invariant connected amplitude isThe Dirac delta function expresses four-momentum conservation.

For the identity contribution, commute $a(p_2)$ through $a^{†}(p_1)$ and independently $b(q_2)$ through $b^{†}(q_1)$; all terms leaving a creation operator against the left Fock vacuum or an annihilation operator against the right Fock vacuum vanish. HenceThere is no exchange term between the two distinct particle and antiparticle species.

For $d\ge 2$, let $\omega$ be a Conformal Killing vector field, put $\sigma =\partial \cdot \omega /d$, and assign the massless scalar field scaling dimension $\Delta =(d-2)/2$. In a finite conformal transformation with metric scale ${\Omega }^2$, the transformation law is ${\varphi }^{\prime }(x^{\prime })=\Omega (x{)}^{-\Delta }\varphi (x)$. Its infinitesimal fixed-coordinate form isFor $\mathcal{L}=(\partial \varphi {)}^2/2$, contract the Conformal Killing equation with ${\partial }_{\mu }\varphi {\partial }_{\nu }\varphi$. Direct differentiation then givesHere $2(1+\Delta )=d$. Flat conformal Killing integrability gives $\square \sigma =0$, with the Lorentzian version obtained by replacing the flat metric consistently. Thus the action variation is a boundary integral, vanishing for suitable boundary conditions. This proves conformal invariance with field dimension $(d-2)/2$, without using the field equation. The connected finite symmetry follows by integrating the infinitesimal generators. Equivalently its conformal currents follow from the improved stress-energy tensor of a free massless scalar.

In one dimension the metric conformality condition is vacuous. The same action-variation calculation contains $-{\varphi }^2{\omega }^{\prime \prime \prime }/4$, so the free action is invariant under the projective subgroup ${\omega }^{\prime \prime \prime }=0$, rather than arbitrary reparametrizations. The usual field-theory conformal-invariance claim above concerns $d\ge 2$.

In $d=2$, $\Delta =0$, so the field transforms simply as a scalar. There is no improvement term and no derivative of the conformal scale in the variation. All local independent reparametrizations of the two null coordinates preserve the action; after Euclidean continuation they are holomorphic and antiholomorphic maps. The conformal algebra is therefore infinite-dimensional, in contrast to $d>2$.

Now use Lorentzian coordinates with spatial period $\ell$. The canonical momentum is $\pi =\dot{\varphi }$, and canonical quantization imposes $[\varphi (t,x),\pi (t,y)]=i{\delta }_{\ell }(x-y)$, with the periodic delta distribution ${\delta }_{\ell }(s)={\ell }^{-1}{\sum }_{n\in \mathbb{Z}}{e}^{2\pi ins/\ell }$. Solving the wave equation and imposing reality gives a convenient oscillator normalization:Strict spatial periodicity excludes a winding term. The canonical commutation relation is equivalent towith ${\alpha }_n^{†}={\alpha }_{-n}$ and the corresponding tilded relation. Both oscillator families commute with $q,p$. To verify the normalization, the zero-mode commutator supplies $i/\ell$, while each oscillator family supplies $i(2\ell {)}^{-1}{\sum }_{n\ne 0}{e}^{2\pi in(x-y)/\ell }$. Their sum is exactly $i{\delta }_{\ell }(x-y)$. The positive-index modes are annihilation operators and the negative-index modes are creation operators, giving two independent bosonic Fock spaces together with the free-particle zero mode.

Define ${\alpha }_0={\tilde{\alpha }}_0=p/\sqrt{4\pi }$ and the normal ordering quadratic generatorsThese are the Fourier modes of the two chiral components of the stress-energy tensor. The oscillator operator commutators immediately give $[L_m,{\alpha }_n]=-n{\alpha }_{m+n}$. Applying this to both factors of $L_n$ gives $(m-n)L_{m+n}$, except for a scalar term when $m+n=0$, produced by reordering creators and annihilators. Its value is independent of the momentum sector. Evaluate it in the formal zero-momentum oscillator vacuum. For $m>0$,The two contractions of each pair explain the factor $1/2$. Hence the free-boson Virasoro central term yieldsThe tilded modes give a second commuting Virasoro algebra with $\bar{c}=1$. The adjoint condition $L_m^{†}=L_{-m}$ and the positive oscillator norms make this a unitary highest-weight Virasoro representation in each momentum sector. In a fixed momentum sector, $L_0=p^2/(8\pi )+{\sum }_{n>0}{\alpha }_{-n}{\alpha }_n$, so the oscillator ground state is a highest-weight state of weight $p^2/(8\pi )$ in either sector. For a noncompact field the zero mode is a free particle: definite momentum ground states are generalized states, rather than a normalizable zero-mode vacuum. The oscillator representation and central term remain well defined. With the standard cylinder vacuum-energy subtraction, the Hamiltonian isThe constant is the Casimir energy of a two-dimensional conformal field theory; omitting it amounts to a different additive Hamiltonian convention, not a change of the plane Virasoro algebra.

In the two-body centre-of-mass frame, spatial reflection reverses the relative coordinate of a quark-antiquark pair. Its orbital state of angular momentum $L$ has orbital parity $(-1{)}^L$, as follows from $Y_{Lm}(-\hat{\mathbf{r}})=(-1{)}^L{Y}_{Lm}(\hat{\mathbf{r}})$. Opposite constituent intrinsic parities multiply to minus one. Thus the meson parity and charge conjugation rules begin withTo obtain charge conjugation, consider a self-conjugate flavor state and write its spin-orbital wavefunction using ordered creation operators $b_{\sigma }^{†}(\mathbf{p})d_{\tau }^{†}(-\mathbf{p})$. Charge conjugation interchanges the quark and antiquark operators. Reordering them back to quark-first order produces a fermionic sign minus one. Interchanging the two spin-$1/2$ labels has eigenvalue $-1$ in the spin singlet state ($S=0$) and $+1$ in the spin triplet ($S=1$), namely $(-1{)}^{S+1}$. Reversing the relative momentum contributes $(-1{)}^L$. ThereforeReciprocal phases for the two conjugate creation operators cancel, so the result is independent of that convention. This applies to a state that is mapped to itself by charge conjugation, not to every electrically neutral flavor state. For example a neutral meson with unequal quark and antiquark flavors can be exchanged with a different antimeson. Charged states have no individual $C$ eigenvalue.

Restrict flavor to the $u,d$ isospin doublet. Its product with the conjugate doublet decomposes as $2\otimes \overline{2}=3\oplus 1$. Convenient normalized flavor states, with irrelevant overall phase choices, areOrbital excitation normally costs energy, so the simplest low-lying quark model states have $L=0$. Coupling the two constituent spins gives $S=0$ or $S=1$, and then $J=S$. The isotriplet with $S=0$ is the pion triplet, while the isotriplet with $S=1$ is the rho meson triplet. The isosinglet with $S=1$ is the nonstrange component of the omega meson. Their quantum numbers areThe charged pions have $J^P=0^{-}$ and the charged rho mesons have $J^P=1^{-}$; charge conjugation exchanges the positive and negative members. The same tensor product also allows an $S=0$, $I=0$ pseudoscalar state, so symmetry alone does not assert that only the listed mesons exist or predict their masses. The physical isosinglet spectrum involves further dynamics and, when strange flavor is included, mixing.

For a vector-meson decay to two spinless pions, conservation of angular momentum forces their relative orbital value to be $L=1$. Their combined parity is $(-1{)}^2(-1{)}^L=-1$, and the neutral charge combination has $C=(-1{)}^L=-1$, matching both ${\rho }^0$ and $\omega$. Thus $C$ and $P$ alone do not distinguish these two decays. The decisive restriction is isospin together with Bose statistics.

Two isotriplet pions have $3\otimes 3=1\oplus 3\oplus 5$, or total $I=0,1,2$. Under exchange the isospin-$I$ state has sign $(-1{)}^{2-I}$: the $I=2$ highest state and its lowered partners are symmetric, the $I=1$ highest combination $(|+0⟩-|0+⟩)/\sqrt{2}$ is antisymmetric, and the singlet $(|+-⟩+|-+⟩-|00⟩)/\sqrt{3}$ is symmetric. The $L=1$ orbital state is antisymmetric. Since pions are spin-zero bosons, their total state must be symmetric, requiring the antisymmetric isospin sector $I=1$. This allows ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ but excludes an isospin-conserving $\omega \to {\pi }^{+}{\pi }^{-}$ amplitude.

There is no analogous obstruction for three pions. In a Cartesian isospin basis $a,b,c$, their totally antisymmetric combination ${ϵ}_{abc}|{\pi }^a{\pi }^b{\pi }^c⟩$ is an $I=0$ state. It can multiply a totally antisymmetric momentum amplitude proportional in the rest frame toMomentum conservation ${\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3=0$ makes this amplitude change sign under every exchange of two pion labels. The product of the flavor and momentum factors therefore obeys Bose statistics. The cross product supplies $J=1$ and an even orbital parity; multiplying by the three negative pion intrinsic parities gives total $P=-1$. In the charged component, charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$ and changes the momentum factor's sign, giving $C=-1$. Thus $\omega \to {\pi }^{+}{\pi }^{-}{\pi }^0$ is compatible with $I=0$ and $J^{PC}=1^{--}$.

The same distinction is encoded in G parity: $G=C(-1{)}^I$ for the neutral member gives $G_{\pi }=-1$, $G_{\rho }=+1$, $G_{\omega }=-1$. An $n$-pion state has $G=(-1{)}^n$, so the G-parity selection rule for pion multiplicities allows two pions for the rho and three for the omega. The printed prohibition of the omega two-pion mode is an exact-isospin strong-interaction selection rule. In the physical theory, isospin-breaking effects allow a small two-pion mode; it is not an absolute prohibition. The scope of C parity is also stated in the Particle Data Group quark-model review, and the physical two-pion mode is recorded in its omega listing.

Use a mostly-plus target metric tensor and the standard open-string mode expansion convention ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$. The integer string oscillator level is $N={\sum }_{n\ge 1}{\alpha }_{-n}\cdot {\alpha }_n$, whose eigenvalues are nonnegative integers. Then $L_0=N+{\alpha }^{\prime }{p}^2$, so the string mass-shell condition givesThe ground state is a tachyon, and the level-one state is massless. The physical string state conditions in old covariant string quantization areOnly the positive-mode Virasoro constraints annihilate a ket; their negative-mode partners annihilate the corresponding bra. Imposing both signs on every ket would contradict the nonzero Virasoro central extension.

The central charge comes from normal ordering the quadratic oscillator expressions. Classically their Poisson brackets have no central term. Quantum mechanically, moving annihilation operators through creation operators produces double contractions. From the oscillator commutator one obtains $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$, which gives the noncentral part $(m-n)L_{m+n}$.

The coefficient is fixed by the free-boson Virasoro central term. On a formal zero-momentum vacuum, for $m>0$,The two contraction pairings giveHere ${\eta }_{\mu \nu }{\eta }^{\mu \nu }=D$; the time coordinate contributes one, just as each spatial coordinate does. Comparing with the Virasoro algebra provesThis counts covariant target coordinates. It is distinct from the $D-2$ physical transverse polarizations. Including the central charge of reparameterization ghosts gives $c_{\mathrm{total}}=D-26$; cancellation of the worldsheet Weyl anomaly gives the usual critical dimension of the bosonic string $D=26$.

A general level-one state isThe zero-mode constraint requires $k^2=0$. Since $[L_1,{\alpha }_{-1}^{\mu }]={\alpha }_0^{\mu }$, the remaining nontrivial positive-mode constraint is $k\cdot ϵ=0$; all $L_m$ with $m\ge 2$ act trivially at this level. Thus the polarization vector is transverse to a nonzero null momentum.

There is also a null string state $L_{-1}|0;k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|0;k⟩$. It is physical when $k^2=0$ and orthogonal to every physical polarization. The null-state quotient of a string therefore identifiesChoose $k=(E,E,0,\dots ,0)$. Transversality sets ${ϵ}^1={ϵ}^0$, and the null-state identification removes this common component. The remaining $D-2$ spatial components have positive norm. Hence the physical level-one states form a massless vector with $D-2$ polarizations, or 24 at the critical dimension. The removed component is gauge redundancy. States violating the constraints, including independent timelike negative-norm polarizations, are unphysical; physical null states are removed by the quotient rather than counted as additional particles.

Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and the free real scalar field Lagrangian density $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. Its conjugate canonical momentum is $\pi =\dot{\varphi }$. Spatial translation symmetry and Noether's theorem give the canonical stress-energy tensorVanishing boundary flux then makes the four-momentum of a free real scalar field constant. In particular the contravariant spatial momentum isThe minus sign is from ${\partial }^i=-{\partial }_i$; it is essential for the following mode calculation.

In the quantized theory, insert the mode expansions into this integral, using a Hermitian symmetric ordering or normal ordering. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The spatial integral is $(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}\pm \mathbf{k})$, leavingEach last term integrates to zero: replace $\mathbf{p}$ by $-\mathbf{p}$ and commute the two annihilation operators or the two creation operators. The mixed terms reorder by the canonical commutation relation. Their vacuum contribution is proportional to the integral of the odd vector $\mathbf{p}$ and vanishes with an inversion-symmetric regulator; normal ordering removes it directly. Thus the normal-ordered free scalar four-momentum has spatial partThese manipulations can be made first with box modes and a symmetric momentum cutoff, then used on finite-particle wave packets.

The commutator derivation identity givesFor $U(\mathbf{y})=e^{-i\mathbf{P}\cdot \mathbf{y}}$, the commutator expansion for exponential conjugation therefore gives the translation of a scalar creation operatorThe vacuum has zero momentum, so $U|0⟩=|0⟩$. Hence the one-particle momentum eigenstate obeys $U|\mathbf{q}⟩=e^{-i\mathbf{q}\cdot \mathbf{y}}|\mathbf{q}⟩$: translation preserves its momentum and multiplies it by the associated phase.