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Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.

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 $-{\delta }^{(3)}$.

Take a periodic comoving box of volume $L^3$, with $\mathbf{k}=2\pi \mathbf{n}/L$. A real field has the expansionwhere $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{\mathbf{q}}^{†}]={\delta }_{\mathbf{kq}}$ and the other ladder-operator commutators vanish. The coefficient of $e^{i\mathbf{k}\cdot \mathbf{x}}$ is therefore $w_k{\hat{a}}_{\mathbf{k}}+w_k^{\ast }{\hat{a}}_{-\mathbf{k}}^{†}$. This opposite momentum is required by reality; the single-oscillator shorthand with the same label on both operators must not be read literally as every traveling-wave Fourier coefficient.

The quadratic action has canonical momentum $\hat{\pi }=a^3\delta \dot{\hat{\varphi }}$. Its equal-time field commutator requiresChoose the vacuum annihilated by all $a_{\mathbf{k}}$ and the subhorizon positive-frequency condition. For constant $H$ and negligible effective mass, put $x=k/(aH)$, $C_k=L^{-3/2}H/\sqrt{2k^3}$. The candidate is $w_k=C_k(i+x)e^{ix}$. Since $\dot{x}=-Hx$,Direct substitution makes ${\ddot{w}}_k+3H{\dot{w}}_k+(k^2/a^2)w_k$ identically zero. Its Wronskian is $2iH{C}_k^2{x}^3=i/(L^3{a}^3)$, which has the required normalization. ThusThis is the existing Bunch-Davies mode normalization in a finite comoving volume. It is exact for a massless minimally coupled test field in de Sitter and the leading approximation for a light slowly evolving field. The homogeneous zero mode needs separate treatment.

Use units $\hslash =c=1$ and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, so $p\cdot x=E_{p}t-\mathbf{p}\cdot \mathbf{x}$. The canonical momentum of the real scalar field is $\pi =\dot{\varphi }$, and the Hamiltonian operator isCanonical quantization of a real scalar field promotes $\varphi$ and $\pi$ to operator-valued distributions satisfying the equal-time canonical commutation relationProducts and operator commutators can first be interpreted after spatial smearing. An additive vacuum constant in $H$ does not affect the following equations. In the Heisenberg picture, $\dot{O}=i[H,O]$. Applying the canonical commutation relation to the quadratic Hamiltonian operator, and integrating the gradient term by parts, givesThus the Klein-Gordon equation is an operator equation. The spatial Fourier transform reduces it to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, with $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches, together with the Hermitian adjoint condition ${\varphi }^{†}=\varphi$, give the mode expansion of a free fieldThe measure $d\mu (p)$ is Lorentz invariant; its normalization fixes that of the creation and annihilation operators rather than imposing any extra physical assumption.

For invariant-normalized scalar mode extraction, take the Klein-Gordon inner product of $f_p$ with the field. The positive-frequency contribution has coefficient $E_p+E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$. The negative-frequency contribution has coefficient $E_p-E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}+\mathbf{q})$, so it vanishes because $E_{-p}=E_p$. ConsequentlyTaking the Hermitian adjoint, with the order of the scalar factors immaterial, yieldsEquivalently $a(\mathbf{p})=\int d^{3}x{f}_p^{\ast }(E_p\varphi +i\pi )$ and $a^{†}(\mathbf{q})=\int d^{3}y{f}_q(E_q\varphi -i\pi )$. At a common time their operator commutator therefore isThe same calculation gives $[a(\mathbf{p}),a(\mathbf{q})]=[a^{†}(\mathbf{p}),a^{†}(\mathbf{q})]=0$: the surviving factor is an energy difference supported at opposite spatial momenta. These are the bosonic canonical commutation relations in relativistic normalization of a one-particle state.

Choose the Fock vacuum by $a(\mathbf{p})|0⟩=0$ and $⟨0|0⟩=1$. Only the annihilator-creator product contributes to the Wightman function, givingUsing time ordering and the reverse product for $y^0>x^0$, the Feynman propagator, with the convention $D_F=i{\Delta }_F$ in this calculation, isIn particular the denominator in the spatial measure is $2E_p$ in both terms.

To verify the four-dimensional Fourier transform representation, let $\tau =x^0-y^0$ and integrate the energy variable first:The Feynman i-epsilon prescription puts the positive-energy pole at $E_p-i0$ and the negative-energy pole at $-E_p+i0$. For $\tau >0$, the contour integral closes in the lower half-plane with clockwise orientation. The residue theorem multiplies the residue $i{e}^{-i{E}_p\tau }/(2E_p)$ by $-i$, giving $e^{-i{E}_p\tau }/(2E_p)$. For $\tau <0$, it closes counterclockwise above: multiplying the residue $-i{e}^{i{E}_p\tau }/(2E_p)$ by $i$ gives $e^{i{E}_p\tau }/(2E_p)$. Thus $I(\tau )=e^{-i{E}_p|\tau |}/(2E_p)$. After inserting the spatial exponential, reversing $\mathbf{p}$ in the negative-time term gives exactly the time ordering expression above. HenceThis equality is understood as a distribution limit; the same prescription gives $(\square +m^2)D_F(x-y)=-i{\delta }^4(x-y)$.

Use $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Vary the Maxwell action with respect to the electromagnetic four-potential. Antisymmetry of the electromagnetic field tensor gives $\delta \mathcal{L}=-F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. After integration by parts, the Euler-Lagrange field equation isA gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. It therefore leaves the Maxwell action and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for canonical quantization: before gauge fixing, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.

Choose the Lorenz gauge $D={\partial }_{\mu }{A}^{\mu }=0$, attainable by solving $\square \chi =-D$ with suitable boundary data. The Euler-Lagrange field equation then reduces toResidual Lorenz gauge symmetry consists of transformations with $\square \chi =0$. The gauge condition and this residual redundancy ultimately leave two transverse polarizations of a free photon.

For the specified kinetic density ${\mathcal{L}}_C=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$, one has $\partial {\mathcal{L}}_C/\partial ({\partial }_{\mu }{A}_{\nu })=-{\partial }^{\mu }{A}^{\nu }$. Its Euler-Lagrange field equation is therefore $-\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction, sinceThe divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a Feynman gauge action; the additional physical-state condition will implement the Lorenz gauge. One must not impose $D=0$ inside the original action before varying it.

Take the covariant components $A_{\nu }$ as canonical coordinates. The canonical momentum computed from ${\mathcal{L}}_C$ isThe equal-time canonical Poisson brackets are $\{A_{\mu }(t,\mathbf{x}),{\Pi }^{\nu }(t,\mathbf{y})\}={\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$, with coordinate-coordinate and momentum-momentum brackets zero. Applying canonical quantization givesIn particular $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The negative temporal sign is the source of the indefinite state-space norm in covariant photon quantization.

The spatial Fourier transform of the wave equation gives an oscillator of frequency $\omega =|\mathbf{k}|$ for each component. The Hermitian adjoint condition on the electromagnetic four-potential then yields the mode expansion of a free fieldThese are massless on shell modes. Consistently with the preceding equal-time canonical commutation relation,The normalization is the same as for the scalar modes, with $-g_{\mu \nu }$ replacing the species identity matrix.

Gupta-Bleuler quantization starts with the Fock vacuum annihilated by every $a_{\mu }$ and imposes only the annihilation, or positive-frequency part of a quantum field, of the Lorenz gauge condition:Imposing the entire Hermitian field divergence as a strong operator identity would be incompatible with the four-component canonical commutation relation. The weaker condition suffices: between two physical states its positive-frequency part vanishes on the ket, and its negative-frequency part vanishes on the bra. Hence $⟨{\mathrm{phys}}^{\prime }|{\partial }_{\mu }{A}^{\mu }|\mathrm{phys}⟩=0$.

For a one-photon state $|\mathbf{k},ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(\mathbf{k})|0⟩$, commuting the constraint through the creator givesThus the allowed photon polarization vectors satisfy $\boxed{k\cdot ϵ=0}$. Their state inner product, with the common momentum-normalization factor suppressed, is $⟨{ϵ}^{\prime }|ϵ⟩=-{ϵ}^{\prime \ast }\cdot ϵ$. Momentum eigenstates are delta-normalized; the same statements hold with finite norms for normalizable wave packets.

Rotate the spatial axes so that $k=\omega (1,0,0,1)$. The constraint is ${ϵ}^0={ϵ}^3$, so the complete allowed family isThe first vector is the allowed longitudinal gauge polarization, proportional to the null four-momentum $k$. Its state has zero norm because $k^2=0$; it is also orthogonal to every constrained state because $k\cdot ϵ=0$. A purely spatial vector $(0,0,0,1)$ alone is not an allowed physical polarization: its time component must accompany it to satisfy the constraint. The other two vectors are transverse polarizations, perpendicular to $\mathbf{k}$, with positive unit norms. For the displayed family,The timelike negative-norm excitation has been excluded by the constraint, and the remaining form is positive semidefinite. The Gupta-Bleuler null-state quotient identifies $ϵ$ and $ϵ+\lambda k$, removing the null direction. It leaves exactly two transverse physical photon polarizations with a positive inner product; their bosonic Fock space is the physical photon state space.

Finally use Feynman slash notation, $/a={\gamma }^{\mu }{a}_{\mu }$, and impose four-momentum conservation $p+q=k+k^{\prime }$. The external Dirac spinors obeySet $r=p-k$ and $r^{\prime }=p-k^{\prime }$. Replacing $ϵ$ by $k$ in the first term of the electron-positron annihilation into two photons amplitude, the rightmost contracted vertex obeysSince the Clifford algebra implies $(/r+m)(/r-m)=(r^2-m^2)I_4$, that term becomesFor the second term, four-momentum conservation gives $k=q+r^{\prime }$. The contracted vertex is now adjacent to the outgoing positron spinor on the left, soConsequentlyThe two terms cancel, proving the two-photon fermion Ward identityExchanging the two photon labels proves the same result for ${ϵ}^{\prime }=k^{\prime }$. By linearity the amplitude is invariant under $ϵ\mapsto ϵ+\lambda k$ on either external photon leg. Therefore the null longitudinal gauge polarization has zero production amplitude: physical scattering produces only the two transverse photon polarizations. This cancellation also shows why both photon orderings in the tree-level Feynman diagrams are essential; either term alone fails the Ward identity.

Take $0\le \sigma <2\pi$. Variation of the conformal-gauge action, with periodic variations, gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Hence $X^{\mu }=X_R^{\mu }(\tau -\sigma )+X_L^{\mu }(\tau +\sigma )$. In an uncompactified direction, periodicity allows equal linear slopes and independent periodic parts. Expanding the periodic parts in a Fourier series yields the closed-string mode expansionHere ${\alpha }_m$ labels the right-moving sector and ${\tilde{\alpha }}_m$ the left-moving sector; exchanging the names is harmless if done consistently. Integration of the canonical momentum density $(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}^{\mu }$ over the string gives $p^{\mu }$, fixing the zero-mode coefficient. Reality gives ${\alpha }_m^{†}={\alpha }_{-m}$ and the analogous relation for tildes, with ${\alpha }_0={\tilde{\alpha }}_0=\sqrt{{\alpha }^{\prime }/2}p$.

The closed-string physical-state Virasoro conditions arewhere $L_0={\alpha }^{\prime }{p}^2/4+N$ and ${\tilde{L}}_0={\alpha }^{\prime }{p}^2/4+\tilde{N}$. Thus closed-string level matching gives $N=\tilde{N}$, and $M^2=-p^2=4(N-1)/{\alpha }^{\prime }$. The massless level is $N=\tilde{N}=1$. Its general state isThe string oscillator commutators give $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$: the two terms obtained by commuting through the quadratic generator coincide and cancel its factor $1/2$. Therefore $L_1$ replaces ${\alpha }_{-1}^{\mu }$ by ${\alpha }_0^{\mu }$, while $L_{m\ge 2}$ produces positive string oscillators that annihilate this level. The same holds independently for tildes. The complete polarization conditions areNull string states identify ${\zeta }_{\mu \nu }$ with ${\zeta }_{\mu \nu }+p_{\mu }{\lambda }_{\nu }+{\tilde{\lambda }}_{\mu }{p}_{\nu }$, with transverse $\lambda ,\tilde{\lambda }$. At nonzero momentum this leaves $24\cdot 24=576$ physical polarizations. Their symmetric traceless, antisymmetric and transverse-trace parts are the graviton, Kalb–Ramond field and dilaton.

The original PDF's condition ${\alpha }^{\prime }{p}^2=4$ is inconsistent with calling the displayed excited states massless. It is the on-shell condition of the string oscillator vacuum as a physical tachyon. For the level-one state, that value would give $L_0={\tilde{L}}_0=2$, violating the physical conditions. The ket $|0;p⟩$ used as the basis for a massless excitation must instead carry the excitation's momentum $p^2=0$; by itself it is then not an on-shell physical vacuum state.

For the compact coordinate, $X^1(\tau ,\sigma +2\pi )=X^1(\tau ,\sigma )+2\pi wR$, $w\in \mathbb{Z}$. Its momentum is $n/R$, $n\in \mathbb{Z}$, since the centre-of-mass wavefunction is single-valued on the circle. Thus its zero modes becomewithThese are momentum and winding modes. Let $k$ be the 25-dimensional noncompact momentum and $M^2=-k^2$. The two zero-mode constraints now giveTheir difference and average give, respectively,This also displays T-duality $R\leftrightarrow {\alpha }^{\prime }/R$, $n\leftrightarrow w$, with one chiral momentum reversed.

At a generic radius, the massless solutions are only $n=w=0$, $N=\tilde{N}=1$. Their fields in 25 dimensions are a graviton (275 polarizations), antisymmetric two-form (253), dilaton (one), two abelian vectors from $G_{a1}$ and $B_{a1}$ (23 each), and the radius scalar (one). Hence the generic massless spectrum has $275+253+1+46+1=576$ polarizations, with gauge group $U(1{)}_L\times U(1{)}_R$. Here generic excludes isolated radii where exceptional massless ground states of a bosonic circle occur. Indeed $N=\tilde{N}=0$ forces $nw=0$, and becomes massless at $R=|n|\sqrt{{\alpha }^{\prime }}/2$ or $R=2\sqrt{{\alpha }^{\prime }}/|w|$. These additional thresholds matter in the bosonic theory.

At the self-dual circle $R=\sqrt{{\alpha }^{\prime }}$, masslessness and level matching readSince $N,\tilde{N}\ge 0$, their sum is at most two. For sum two, $n=w=0$ and level matching gives the ordinary $(1,1)$ family. For sum one the four possibilities areThey have one right-moving ${\alpha }_{-1}$ or one left-moving ${\tilde{\alpha }}_{-1}$ excitation, as their levels specify. Each family has 24 physical polarizations: a 25-dimensional vector with 23 polarizations and one compact-oscillator scalar. Thus these families add 96 states.