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Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

For the real scalar field, the mostly-plus Minkowski metric gives a positive kinetic energy. Its canonical momentum and Hamiltonian operator areThe Euler-Lagrange field equation is $(\square -m^2)\varphi =0$, or $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. Canonical quantization of a real scalar field promotes the fields to Hermitian operators and replaces the equal-time Poisson brackets by canonical commutation relations:The Heisenberg equation of motion then gives $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, so the classical field equation remains an operator identity. Products at coincident points need a regulator; equivalently, the field is an operator-valued distribution.

Resolve the free field into positive- and negative-frequency plane waves. With $E_{\mathbf{k}}=\sqrt{|\mathbf{k}{|}^2+m^2}$ and $k\cdot x=-E_{\mathbf{k}}t+\mathbf{k}\cdot \mathbf{x}$,Reality pairs the two terms by Hermitian conjugation. The equal-time canonical commutation relations are equivalent toFor example, the two mixed terms in $[\varphi ,\pi ]$ each supply half the Dirac delta function. Thus the normalization $1/\sqrt{2E_{\mathbf{k}}}$ is fixed, not optional. Each independent mode is a quantum harmonic oscillator, with $a$ and $a^{†}$ its annihilation operator and creation operator.

Choose the Fock vacuum by $a(\mathbf{k})|0⟩=0$. Repeated creation operators construct the bosonic Fock space. Substitution into the Hamiltonian operator yieldswhere a finite volume $V$ can be used as a regulator. Normal ordering sets the flat-space vacuum reference to zero. The spatial momentum operator isIn particular, $[H,a^{†}(\mathbf{k})]=E_{\mathbf{k}}{a}^{†}(\mathbf{k})$ and $[\mathbf{P},a^{†}(\mathbf{k})]=\mathbf{k}{a}^{†}(\mathbf{k})$. A one-particle state therefore has positive energy, momentum $\mathbf{k}$, and invariant mass $m$. A scalar field transforms in the trivial spin representation, so its particles have spin zero. Because the field is real there is only one set of oscillators: the particle is its own antiparticle, with no distinct conserved particle-minus-antiparticle charge. The free theory does conserve its occupation-number sum, though generic scalar interactions need not.

A momentum eigenstate extends throughout space. A localized particle is described by a superposition such asIts wave packet evolves through the phase $e^{-i{E}_{\mathbf{k}}t}$; a narrow packet has group velocity $\mathbf{v}=\mathbf{k}/E_{\mathbf{k}}$, whose magnitude is at most one. Localization and dispersion concern these superpositions, rather than classical trajectories attached to individual field modes.

Relativistic spacetime behavior is encoded by microcausality. Directly from the oscillator commutators,This is a Lorentz-invariant distribution and vanishes at equal time. Any spacelike separation can be transformed to an equal-time separation, so $[\varphi (x),\varphi (y)]=0$ for spacelike $x-y$. Thus local operations at spacelike-separated points commute. The vacuum Wightman function need not vanish outside the light cone, but that correlation does not transmit a controllable signal. The Feynman propagator iswhich propagates positive energy forward and negative energy backward in the time-ordered product. For causal response one uses the retarded Green function, whose support lies in or on the future light cone.

Finally, commuting creation operators implyThe multiparticle state is symmetric under the particle exchange operator, which is Bose statistics. For a normalized single mode, the occupation states are $(a^{†}{)}^n|0⟩/\sqrt{n!}$ with $n=0,1,2,\dots$: any number of identical bosons can occupy it. There is no Pauli exclusion principle for these particles. This construction agrees with the Spin-statistics theorem relating integer spin to bosonic exchange symmetry under relativistic locality and positive-energy assumptions. It demonstrates the required scalar case without assuming that theorem as the quantization prescription.

For the relativistic particle phase-space action, the first-class constraint $\phi =(p^2+m^2)/2$ generatesIndeed the integrand varies by $\frac{1}{2}d[ϵ(p^2-m^2)]/dt$. The canonical gauge transformation is an invariance when the gauge parameter vanishes at fixed temporal endpoints, or when all fields and the parameter are periodic. The boundary restriction matters for the proper-time modulus.

Normalize the worldline interval to $[0,1]$. Thenis invariant because $\delta s=ϵ(1)-ϵ(0)=0$. Every allowed $e$ in its orbit can be written $e(t)=s+\dot{ϵ}(t)$: set $ϵ(t)={\int }_0^t[e(u)-s]du$. Thus $s$ remains a gauge-invariant integration variable, not another removable nonconstant mode. For a worldline circle the constant gauge parameter is a residual zero mode. Without the endpoint restriction, the assertion that $s$ is invariant would not hold.

The worldline gauge-orbit determinant is the Jacobian from gauge-orbit coordinates $ϵ$ to the nonconstant part of $e$ is the Faddeev-Popov determinant of ${\partial }_t$. Equivalently, the gauge-fixing identity has the formThe determinant is taken between the appropriate boundary-condition spaces, with the modulus removed; on a circle the prime also removes the constant parameter. Gauge fixing therefore leaves a factor $\det {}^{\prime }{\partial }_t$ and a modulus measure, after division by any residual gauge volume. Even though this determinant is field independent in the present Abelian example, it is the required change-of-variables Jacobian. A Grassmann integral over the Faddeev-Popov ghosts exponentiates it:The overall determinant phase depends on the integration convention and can be absorbed into normalization. Zero modes and the same endpoint restrictions must be treated separately rather than included in an invertible determinant.

For the free-ended open string, take $0\le \sigma \le \pi$. A canonical cosine expansion at a fixed time isIt implements the Neumann boundary conditions and has ${\alpha }_n^{†}={\alpha }_{-n}$. With ${\alpha }_0^m=p^m/\sqrt{\pi T}$, its Nambu-Goto phase-space action, up to a total time derivative, isReality requires ${\lambda }_{-n}={\lambda }_n^{\ast }$. Numerical factors can be absorbed into these Lagrange multipliers. In this covariant quantization of the bosonic string the oscillators retain all $D$ spacetime components, in contrast to the transverse oscillators in the preceding solution. The canonical commutation relations areThe oscillator vacuum $|0;p⟩$ is annihilated by ${\alpha }_n^m$ for $n>0$. Its momentum label will sometimes be suppressed.

Define the matter Virasoro algebra generators using normal ordering:No additive intercept is included in this definition of $L_0$. For $n>0$ the indices of the two factors in each term add to $n$; they cannot both be negative. After normal ordering there is a positive-mode annihilation operator on the right, possibly accompanied by the zero mode. Hence $L_n|0;p⟩=0$ for every $n>0$. In $L_0$, commuting positive modes past negative modes formally adds $\frac{D}{2}{\sum }_{k>0}k$. This divergent constant needs a prescription, and a finite shift is an ordering ambiguity. Our convention instead puts the physical string intercept into the constraint $L_0-a$.

With this convention the matter Virasoro algebra isA different additive constant in $L_0$ would change the linear-in-$m$ central term, so stating the convention is essential.

For the worldsheet ghost fields, useand choose a ghost oscillator vacuum withThen $c_0|0{⟩}_{\mathrm{gh}}\ne 0$ since $\{b_0,c_0\}=1$. This choice specifies the ghost zero-mode doublet; it is not a claim that both zero modes annihilate one state. With the printed ghost Virasoro zero-mode convention, $L_0^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ and $L_m^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ for $m>0$. The latter follows by putting positive ghost modes on the right; a possible contraction requires $m=0$ and is absent here.

Apply the supplied BRST charge to the matter state times this ghost oscillator vacuum. Terms with a rightmost $c_m$, $m>0$, vanish, as do the positive-mode ghost generators. ThusThese one-ghost states are independent, as can also be seen by applying $b_0,b_m$. Therefore the BRST physical-state constraints of an open string areFor the matter oscillator vacuum, $L_0={\alpha }^{\prime }{p}^2$, so $p^2=1/{\alpha }^{\prime }$ and $M^2=-1/{\alpha }^{\prime }=-2\pi T$: the physical ground state is a tachyon. The momentum must satisfy this equation; the zero-momentum oscillator vacuum by itself would not be BRST-closed.

Matter and ghost generators commute with one another. Add their two algebras and write ${\mathcal{L}}_m=L_m+L_m^{\mathrm{gh}}$. The given ghost constant must be retained:Define the shifted generators ${\hat{\mathcal{L}}}_m={\mathcal{L}}_m-{\delta }_{m,0}$. ThenBRST nilpotence requires cancellation of the anomalous central term in this shifted constraint algebra, together with the intercept one already present in the charge. At $m=2,n=-2$ the remaining anomalous coefficient is $(D-26)/2$, so $D=26$. At this value the shifted total generators obey the Witt algebra. The unshifted ${\mathcal{L}}_m$ still have the displayed linear zero-mode shift; it must not be silently discarded.

Fix a classical globally hyperbolic spacetime with metric signature $(-+++)$; in this solution take $\hslash =1$. A free real Klein-Gordon field can be specified bywhere $\mu$ is its mass, $\xi$ its curvature coupling and $\mathcal{R}$ the scalar curvature. Global hyperbolicity ensures a well-posed initial-value problem on a Cauchy hypersurface $\Sigma$ and the existence of retarded and advanced propagators. Thus compactly supported field and normal-derivative data determine a classical solution. This fixes the dynamics, but not a Fock vacuum.

For real solutions with suitable support or falloff, the symplectic form on scalar-field solutions iswith $n$ the future unit normal. The field equation makes the current conserved, so this symplectic form is independent of $\Sigma$ when boundary flux vanishes. Quantize the initial data by the canonical commutation relation: with $\pi =n^a{\nabla }_a\varphi$ and the delta function defined relative to $d\Sigma$, $[\hat{\varphi }(x),\hat{\pi }(y)]=i{\delta }_{\Sigma }(x,y)$ and the two equal-field commutators vanish. Equivalently, construct the field algebra using the causal propagator. A state on that algebra is additional input.

To construct a particle representation, complexify the classical solution space. Its conserved Klein-Gordon inner product isIt is indefinite on all complex solutions. Choose a complete positive-norm subspace and an orthonormal mode basis $u_i$ with $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$. Such a choice is encoded by a compatible complex structure on the Klein-Gordon solution space. Its positive subspace gives the one-particle Hilbert space, and the associated bosonic Fock space contains symmetrized many-particle states. The field expansion isThe creation operator $a_i^{†}$ adds a particle in mode $i$, the annihilation operator removes one, and the number operator is $N_i=a_i^{†}{a}_i$. For continuous mode labels the sums and Kronecker deltas become integrals and delta functions, or one can work with normalized wave packets.

The ambiguity is precisely that the field equation and global hyperbolicity do not select that positive subspace. A different normalized basis may mix $u_j$ and $u_j^{\ast }$ by a Bogoliubov transformation, and then its annihilation operators mix $a_j$ and $a_j^{†}$. Its Fock vacuum and number operators differ. The Hadamard condition constrains physically acceptable short-distance singularities and allows local renormalization, but it still leaves many states. Hence there is generally no observer-independent particle count on an arbitrary dynamical geometry.

In a stable strictly stationary spacetime, a globally future timelike Killing vector field $K$ gives a preferred time translation. Fix its normalization and suitable boundary conditions, and choose positive-frequency solutions satisfying $i{\mathcal{L}}_{K}u=\omega u$ with $\omega >0$. The corresponding positive spectral subspace gives the preferred vacuum state in a stationary spacetime. Unitary changes of basis within it leave the vacuum and the particle notion unchanged. This construction assumes a well-defined positive stationary generator; stationarity by itself is insufficient if $K$ becomes spacelike, as in a Kerr ergoregion, or if unstable or zero modes obstruct the ground-state construction. It selects a preferred ground state under the stated assumptions, not a unique state among all thermal and excited states.

If the geometry is suitably stationary in the asymptotic past and future, choose those preferred mode spaces separately, giving the in-vacuum and out-vacuum. Propagate the past modes through the intervening region using the field equation and compare them to the future modes using the conserved Klein-Gordon inner product. Adopt the conventionThe canonical identities for a bosonic Bogoliubov transformation read $\alpha {\alpha }^{†}-\beta {\beta }^{†}=I$ and $\alpha {\beta }^T=\beta {\alpha }^T$. Extracting the future annihilation operator with the same inner product givesIn the in-vacuum, only $⟨a_j{a}_k^{†}⟩={\delta }_{jk}$ contributes to $⟨b_i^{†}{b}_i⟩$. Therefore the particle number from Bogoliubov coefficients isNonzero $\beta$ is the production of future particles from the past vacuum. Summing over future modes gives the total expected particle number when that sum is finite. For infinitely many modes, a Hilbert-Schmidt operator $\beta$ is the condition for unitary implementability between these pure bosonic Fock space representations; finite-volume or wave-packet calculations must respect the relevant measures and convergence. Particle production is determined by the negative-frequency mixing, rather than by identifying a single instantaneous vacuum throughout the time-dependent region.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, the corresponding coefficient is a difference of those square roots and multiplies ${\delta }^3(\mathbf{p}+\mathbf{q})$; it vanishes because $E_{\mathbf{p}}=E_{-\mathbf{p}}$. Taking the adjoint gives $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ is the vacuum expectation value of the time-ordered product of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,Here $\theta$ is the Heaviside step function, and changing $\mathbf{p}$ to $-\mathbf{p}$ in the second term gives the last line.

For the Fourier transform convention ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for $t>0$, and above for $t<0$, reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, soIt is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.

The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.

Evaluate the field variance in the zero-occupation Fock vacuum, ${\hat{a}}_{\mathbf{k}}|0⟩=0$, with $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}]={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The canonical commutation relation alone does not specify a quantum state; this vacuum assumption is needed for the requested expression without occupation factors. In the Heisenberg picture, use the position-space mode expansion of a free fieldThis is the position-space form of the shorthand oscillator expansion in the question; equivalently its Fourier operator uses the creation operator with opposite wavevector. The annihilation operators give $⟨{\hat{a}}_{\mathbf{k}}{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}⟩={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$, while all other vacuum pairings vanish. The mean field fluctuation is zero, so the two-point function is its coincident-point vacuum field variance. Assuming spatial isotropy, angular integration gives the variance per logarithmic wavenumber interval:Here the dimensionless power spectrum is denoted ${\Delta }_v$, as in the question, rather than the alternative notation ${\Delta }_v^2$. A coincident vacuum field product needs smearing or ultraviolet regulation: the unregulated full vacuum variance is generally divergent. With a specified finite wavenumber band, the same expression holds with those integration limits. This does not affect the mode-by-mode dimensionless power spectrum.

For the leading nonzero slower branch, put $x=-k\tau =k/(aH)>0$ and $v_k=c{x}^{1/2-\nu }/\sqrt{2k}$. Since the physical fluctuation is $\delta \chi =v/a$, its dimensionless power spectrum isFor a $k$-independent coefficient $c$, this is a blue spectral tilt, because $3-2\nu >0$ in the specified mass range. At fixed $k$ it decays as $a^{-(3-2\nu )}$, consistently with the field-amplitude decay found above.

To include the other branch explicitly, the general superhorizon expression isIf $c_{-}\ne 0$, it approaches the displayed target spectrum with $c=c_{-}$. A pure faster classical branch would instead have power proportional to $x^{3+2\nu }$. For a normalized quantum mode, however, the Wronskian normalization $v_k{v}_k^{\ast \prime }-v_k^{\ast }{v}_k^{\prime }=i$ requires both independent real solutions; its slower coefficient cannot vanish. Thus the target is the leading superhorizon quantum spectrum, while the faster classical solution remains part of the complete mode analysis.

Distinguish the slow-roll approximation parameter $ϵ$ from the infinitesimal positive contour regulator $\epsilon$. In the interaction picture, the ordered unequal-time Wick contraction isA differentiated vertex field instead supplies $u_k^{\ast \prime }(\tau )$. These conjugates are fixed by the annihilation operators and creation operators and cannot be dropped while retaining the same contour.

Convert the interaction time integral to conformal time. Since $dt=ad\tau$ and each cosmic-time derivative is $a^{-1}{\partial }_{\tau }$, the integrated cubic vertex isThe in-in formalism supplies $-2i$ multiplying the expectation of the Hamiltonian, so its negative sign gives $+2i$ multiplying this vertex. At nonzero external momenta, the connected contractions join each of the three external fields to one vertex field. There are six bijections, not three. Choosing which external field meets the undifferentiated vertex gives three cyclic choices; swapping the two differentiated fields gives a further factor of two. This is the Wick-pairing multiplicity for a zeta zeta-prime-squared vertex.

For example, before the internal momenta are integrated, one cyclic contribution includesThe factors of $2\pi$ cancel to leave one overall $(2\pi {)}^3$ and the external momentum delta. Thus the properly normalized reduced expression isThe vacuum prescription for inflationary in-in integrals damps the early-time oscillations and selects the Bunch-Davies vacuum. Internal self-contractions correspond to disconnected zero-momentum tadpole terms and are excluded from this connected three-point function.

The printed intermediate expression has unconjugated modes and only three literal cyclic terms with coefficient $-2i$. The conjugate form of the result above would use $-4i$, together with the conjugated contour. A literal $-2i$ with only three terms gives half the final printed answer. The six-contraction expression above is the consistent reduction of the supplied Hamiltonian and leads to that final answer.

Set $\hslash =1$. Work on the strip $0\le \sigma \le \pi$ in conformal gauge, with worldsheet signature $(-,+)$ and target signature $(-,+,\dots ,+)$. In light-cone gauge in string theory, the independent fields are the $d=D-2=24$ transverse coordinates. Put the Neumann boundary condition at $\sigma =0$ and the Dirichlet boundary condition at $\sigma =\pi$; reversing the endpoints exchanges cosine and sine modes without changing the spectrum. The fixed endpoint is $X^i(\tau ,\pi )=y^i$.

Varying the transverse Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^i=0$ and the spatial boundary contribution $-(2\pi {\alpha }^{\prime }{)}^{-1}\int d\tau [X^{\prime i}\delta X^i{]}_0^{\pi }$. At the free endpoint this vanishes precisely when $X^{\prime i}=0$; at the fixed endpoint $\delta X^i=0$. Separation of variables then gives $\cos (r\sigma )$ with $\cos (r\pi )=0$, hence $r=n+\frac{1}{2}$, $n=0,1,\dots$. These are Neumann-Dirichlet open-string boundary conditions. No dynamical transverse worldsheet zero mode survives: a constant solution must equal the prescribed $y^i$, and a linear-in-$\sigma$ solution violates the free-end condition.