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An annihilation operator lowers the occupation number operator eigenvalue by one and annihilates the Fock vacuum.

The covariant free-photon oscillator construction uses four polarization vectors and $[a^{\lambda },a^{{\lambda }^{\prime }†}]=-{\eta }^{\lambda {\lambda }^{\prime }}$. Starting from a positive Fock vacuum, it induces an indefinite Hermitian form on the multiparticle state space: the temporal oscillator has negative norm while the three spatial oscillators have positive norm. It is therefore not the physical positive Hilbert space. The Gupta-Bleuler null-state quotient selects a positive physical space with two transverse photon polarizations. Continuum momentum oscillators and their states are understood after smearing or finite-volume regularization.

With the mode normalization given, the oscillator canonical commutation relations are $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$, with the other two commutators zero. Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p^0=E_{\mathbf{p}}$. The convention compatible with the requested numerator iswithout an additional factor of $i$ outside the vacuum expectation value. Put $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. The Fock vacuum is annihilated by $a_{\mathbf{p}}$, so only $a{a}^{†}$ contributes to the two Wightman functions. Consequently,In the second term we changed $\mathbf{p}\mapsto -\mathbf{p}$ to give the same spatial exponential.

Now perform the energy contour integralHere the residue step uses $E>0$; the massless zero-momentum point is interpreted through the smeared distribution limit, not as an isolated normalized oscillator. The Feynman i-epsilon prescription puts the positive-energy pole below the real axis and the negative-energy pole above it: $p^0=E-i0$ and $p^0=-E+i0$. For $\tau >0$, close in the lower half-plane, clockwise; the residue theorem gives $-i$ times the residue $i{e}^{-iE\tau }/(2E)$, hence $e^{-iE\tau }/(2E)$. For $\tau <0$, close in the upper half-plane, counterclockwise; the negative-energy residue is $-i{e}^{iE\tau }/(2E)$, again giving a positive $e^{iE\tau }/(2E)$. These are exactly the two time-ordered terms. Restoring the spatial integral proves the scalar Feynman propagator pole prescription:The prescription in this formula supplies the pole convention left unspecified in the printed display; an unprescribed ordinary real-axis integral would not be well-defined. It is a distribution limit after smearing, not an absolutely convergent four-dimensional integral. With this normalization the derivative jump of a free scalar time-ordered two-point function gives $(\square +m^2){\Delta }_F=-i{\delta }^4(x-y)$, an independent check of both the numerator and the sign.

The pole displacement is exaggerated in this original schematic. The contour orientation and selected pole reproduce the time ordering of the Feynman propagator.

For covariant quantization of the bosonic string,All other commutators between independent canonical variables vanish. The covariant Fock vacuum obeys ${\alpha }_k^m|0;p⟩=0$ for every $k>0$ and every $m$. Normal ordering moves these annihilating string oscillators to the right and givesThen ${\hat{N}}_{\mathrm{cov}}|0;p⟩=0$. Its commutator $[{\hat{N}}_{\mathrm{cov}},{\alpha }_{-k}^m]=k{\alpha }_{-k}^m$ grades a Fock state basis by $N=\sum k{n}_{km}$, although its inner product is indefinite because of the timelike oscillator.

The quantum Virasoro algebra has central charge $D$:The Virasoro central extension comes from commutators needed to order infinite oscillator sums; it is a quantum effect absent from the classical Poisson brackets. Replacing a classical Poisson bracket by a commutator cannot recover that term without a regularized ordering calculation.

If a state were annihilated by every nonzero ${\hat{L}}_j$, the $j=1,k=-1$ commutator would imply ${\hat{L}}_0|\psi ⟩=0$. The $j=2,k=-2$ commutator would then imply $(D/2)|\psi ⟩=0$. For $D>0$, only the zero vector is annihilated by all nonzero Virasoro constraints. This explains why only positive modes annihilate a physical string state.

The vacuum is a physical string state when $p^2=2\pi Ta$. At level one all states have the form $|A;p⟩=A_m{\alpha }_{-1}^m|0;p⟩$. Since $[{\hat{L}}_j,{\alpha }_k^m]=-k{\alpha }_{j+k}^m$, the positive-mode Virasoro constraints giveThus the complete level-one conditions and norm arewith the common vacuum normalization suppressed.

For $a>1$, choose spacelike momentum $p^m=(0,k,0,\dots )$, $k^2=2\pi T(a-1)$. A purely timelike polarization has $A\cdot p=0$ and norm $-1$. Hence a negative-norm physical string state exists when $a>1$.

For $a<1$, the mass-shell condition gives $M^2=2\pi T(1-a)>0$. In a rest frame, $A\cdot p=0$ forces $A_0=0$, leaving $D-1$ positive-norm vector-particle polarizations, those of a massive vector.

For $a=1$ and nonzero null momentum, $A\cdot p=0$ leaves a null direction $A_m\propto p_m$. The corresponding null string state is $p\cdot {\alpha }_{-1}|0;p⟩=\sqrt{\pi T}{\hat{L}}_{-1}|0;p⟩$. It is orthogonal to every physical string state because ${\hat{L}}_1$ annihilates them. Quotienting by this gauge redundancy, $A_m\sim A_m+\zeta p_m$, leaves $D-2$ positive vector-particle polarizations. Thus the level-one spectrum agrees with light-cone gauge in string theory at $a=1$. This level-one argument alone does not establish consistency or absence of negative norms at all higher levels.

In the displayed version of light-cone gauge in string theory, the oscillators have $D-2$ transverse components andThe center-of-mass Poisson brackets remain $\{x^m,p_n\}={\delta }_n^m$, and the transverse oscillator Poisson brackets are $\{{\alpha }_j^a,{\alpha }_k^b\}=-ij{\delta }^{ab}{\delta }_{j+k,0}$. Other independent brackets vanish. With $\hslash =1$, canonical quantization gives the canonical commutation relationsThe oscillator Fock vacuum at momentum $p$ is defined by ${\alpha }_k^a|0;p⟩=0$ for $k>0$. It is the ground state of one string, rather than the empty spacetime vacuum. Set $a_k^a={\alpha }_k^a/\sqrt{k}$ for $k>0$. The normal ordering prescription gives the string level operatorA Fock state basis is obtained by applying ${\prod }_{k,a}((a_k^a{)}^{†}{)}^{n_{ka}}/\sqrt{n_{ka}!}$ to $|0;p⟩$, with string level operator eigenvalue $N={\sum }_{k,a}k{n}_{ka}$. In particular, its level-one states areThey transform as the transverse vector of the little group rotation subgroup $SO(D-2)$, precisely the vector-particle polarizations of a massless vector. A massive vector would instead require $D-1$ vector-particle polarizations. This conclusion uses a quantization compatible with the Lorentz group of the bosonic string theory.

The classical mass constraint alone has no quantum zero-point energy shift. Its quantum version includes the normal-ordering constant of a string $a$:Masslessness at level one fixes $a=1$, soThus the ground state is a tachyon. Without the quantum ordering shift, the displayed classical constraint would give $M_1^2=2\pi T$ and would not support the stated massless interpretation. In the usual transverse vacuum regularization, $a=(D-2)/24$; consistency with $a=1$ also gives the critical dimension of string theory $D=26$.