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Use the supplied oscillator expansion, with the boundary-term convention for momentum stated in part (b). Use a real polarization vector basis, as in the supplied unconjugated polarization completeness relation. Raise its second field index to obtainThe mixed commutator contains only the annihilation-creation and creation-annihilation terms. Their signs are both positive after combining the minus sign in the momentum expansion with the negative Minkowski metric oscillator commutator. Setting $\mathbf{r}=\mathbf{x}-\mathbf{y}$ and using the momentum Dirac delta distribution givesHere the last equality is the Fourier representation of the Dirac delta function. Similarly,because the integrand is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The momentum-momentum commutator is proportional to the same odd difference, now weighted by $|\mathbf{p}|/2$, and also vanishes. Therefore the equal-time canonical commutation relations areThese are identities of operator-valued distributions, understood after smearing. The extra indices in the TeX polarization relation are transcription defects; the PDF has the ordinary two-polarization completeness contraction used above. The direct momenta of part (b) also satisfy the canonical commutation relations after their boundary-generated shift, although they do not have the unmodified mode expansion printed here.

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.

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.

Write $X^i-y^i={\sum }_{r>0}{q}_r^i(\tau )\cos (r\sigma )$. The orthogonality relation ${\int }_0^{\pi }\cos (r\sigma )\cos (s\sigma )d\sigma =(\pi /2){\delta }_{rs}$ reduces the action to independent harmonic oscillators:The canonical commutation relations $[q_r^i,P_s^j]=i{\delta }^{ij}{\delta }_{rs}$ determine normalized annihilation operators at $\tau =0$Thus $q_r^i=\sqrt{2{\alpha }^{\prime }/r}(a_r^i{e}^{-ir\tau }+a_r^{i†}{e}^{ir\tau })$. These formulas derive the quantization from the action rather than import integer-moded open-string rules.

For the usual phased string oscillators, set ${\alpha }_r^i=-i\sqrt{r}{a}_r^i$ and ${\alpha }_{-r}^i=i\sqrt{r}{a}_r^{i†}$ for $r>0$. Then the half-integer open-string oscillator expansion and algebra areThe corresponding field momentum density is ${\Pi }_i={\dot{X}}^i/(2\pi {\alpha }^{\prime })$. Completeness of the mixed-boundary eigenfunctions gives $[X^i(\sigma ),{\Pi }_j({\sigma }^{\prime })]=i{\delta }_j^i{\delta }_{\mathrm{ND}}(\sigma ,{\sigma }^{\prime })$, where ${\delta }_{\mathrm{ND}}=(2/\pi ){\sum }_{r>0}\cos (r\sigma )\cos (r{\sigma }^{\prime })$. This is a distribution identity on the mixed-boundary function space, not an unrestricted value at a fixed endpoint with a Dirichlet boundary condition.

Classically, the transverse Virasoro algebra zero-mode generator isThere is no transverse momentum term. If the two light-cone directions are common directions with Neumann boundary conditions, the full zero-mode constraint adds ${\alpha }^{\prime }{p}_{\parallel }^2=-2{\alpha }^{\prime }{p}^{+}{p}^{-}$: $L_{0,\mathrm{cl}}={\alpha }^{\prime }{p}_{\parallel }^2+L_{0,\mathrm{cl}}^{\perp }$. This assumption about the longitudinal directions is needed to interpret oscillator levels as target-space masses.

Quantizing the symmetrically ordered transverse generator gives $L_0^{\perp }=N+E_0$, where the string level operator is $N={\sum }_{i,r>0}r{a}_r^{i†}{a}_r^i$. Each harmonic oscillator contributes $r/2$ to the vacuum energy. Use zeta function regularization and the Riemann zeta function with the actual half-integer spectrum:Consequently the Neumann-Dirichlet string zero-point energy isIn the common convention $L_0={\alpha }^{\prime }{p}_{\parallel }^2+N-a$, the normal-ordering constant of a string is $a=-\frac{1}{2}$. This positive shift differs from the $-1$ vacuum energy of 24 integer-moded transverse bosons. The difference between those two vacuum energies is $3/2$. To distinguish zero-mode conventions, the ND twist conformal weight is $1/16$ per transverse boson. With 24 bosons, the plane matter generator is $L_0^{\mathrm{plane}}={\alpha }^{\prime }{p}_{\parallel }^2+N+3/2$; its physical open-string condition $L_0^{\mathrm{plane}}-1=0$ is exactly the strip/light-cone constraint ${\alpha }^{\prime }{p}_{\parallel }^2+N+1/2=0$ used here. The transverse plane and strip constants differ by the central charge shift $c_{\perp }/24=24/24=1$. A common exponential frequency cutoff independently gives ${\sum }_{r>0}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, confirming the finite part and avoiding invalid termwise manipulation of divergent sums.

Let $a_r^i|0⟩=0$. The lowest levels of a fully transverse ND bosonic string areThe third level has two $r=1/2$ excitations; there is no $r=1$ oscillator. Its indices are symmetric because the creation operators commute. With $L_0=0$ and the common longitudinal momentum convention, the rest energies and masses obeyAt fixed positive $p^{+}$ the corresponding light-cone energies are $p^{-}=(N+1/2)/(2{\alpha }^{\prime }{p}^{+})$; the table lists excitation levels, not a spectrum that remains discrete if longitudinal momentum is varied continuously.

The surviving transverse rotations form $SO(24)$. The ground state is a scalar, the next level its vector, and the third level the symmetric square of the vector. Separating its trace givesThe trace state is proportional to ${\sum }_i(a_{1/2}^{i†}{)}^2|0⟩$; subtracting this trace gives the 299-dimensional symmetric traceless square.

There is a qualification to the printed “little group”. The little group with mixed string boundary conditions must preserve the endpoints as well as momentum. These boundary conditions break the full 26-dimensional Lorentz group, so the massive states above cannot be classified as representations of the unbroken 26-dimensional massive little group $SO(25)$. In a D1–D25 realization the common worldvolume has Lorentz group $SO(1,1)$ and trivial connected massive little group; $SO(24)$ acts on the ND coordinates as an internal rotation group. The scalar, vector and symmetric trace-free decomposition is under the surviving transverse SO(24), with this boundary-background qualification.

Invert the operator definitions to obtain $a=\sqrt{m\omega /(2\hslash )}\hat{x}+i\hat{p}/\sqrt{2\hslash m\omega }$ and the adjoint formula for $a^{†}$. The canonical commutation relation yields $[a,a^{†}]=1$, with $[a,a]=[a^{†},a^{†}]=0$. Substitution givesWrite $N=a^{†}a\ge 0$. Since $[N,a]=-a$ and $[N,a^{†}]=a^{†}$, these are lowering and raising operators. Applying $a$ to a ground state would lower its energy unless it vanishes, so $a|0⟩=0$. From the normalized unique ground state constructThe commutator gives the factorial normalization and $a|n⟩=\sqrt{n}|n-1⟩$. For any energy eigenstate, repeated lowering cannot pass below $N=0$; positivity of each lowered norm forces its $N$ eigenvalue to be a nonnegative integer and lowering eventually reaches the ground state. Its uniqueness then yields the entire eigenstate ladder. The usual oscillator on the real line has discrete complete spectrum because its potential is confining, so these states span its Hilbert space.

For the perturbation $\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$, nondegenerate time-independent perturbation theory gives $E_n^{\prime }=E_n+\lambda W_{nn}+{\lambda }^2{\sum }_{m\ne n}|W_{mn}{|}^2/(E_n-E_m)+O({\lambda }^3)$. The first-order term vanishes. Only $m=n-2,n+2$ contribute, with squared matrix elements $(\hslash \omega {)}^{2}n(n-1)$ and $(\hslash \omega {)}^2(n+1)(n+2)$ respectively. ThusThe absent downward transition for $n=0,1$ is automatically represented by $n(n-1)=0$. Direct substitution also givesFor $|\lambda |<1/2$ this is another oscillator with mass $m^{\prime }=m/(1-2\lambda )$ and frequency ${\omega }^{\prime }=\omega \sqrt{1-4{\lambda }^2}$. ThereforeIts expansion is $E_n[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with the perturbative result and sharpening its remainder. At the excluded endpoints confinement is lost; beyond them the quadratic Hamiltonian is not bounded below.

Put $N_a=a^{†}a$ and $N_b=b^{†}b$. The canonical commutation relations give $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$, and the analogous relations for $b$, with all cross-mode operators commuting. Consequently $[K_3,K_{+}]=K_{+}$ and $[K_3,K_{-}]=-K_{-}$. Alsoso $[K_{+},K_{-}]=N_a-N_b=2K_3$. This is the Schwinger boson representation of the angular momentum algebra, with $\hslash =1$ in the stated commutators.

The vacuum has $N_a=N_b=0$, and $(a^{†}{)}^2|0⟩$ has occupation numbers $(2,0)$, hence $K_3$ eigenvalue $1$. With $N=N_a+N_b$,Thus $\boxed{K^2(a^{†}{)}^2|0⟩=2(a^{†}{)}^2|0⟩}$. It is the highest-weight state of the spin-one irreducible representation. There are two other states in this multiplet, $a^{†}{b}^{†}|0⟩$ and $(b^{†}{)}^2|0⟩$, with $K_3$ eigenvalues $0,-1$. A normalized basis is $(a^{†}{)}^2|0⟩/\sqrt{2}$, $a^{†}{b}^{†}|0⟩$, $(b^{†}{)}^2|0⟩/\sqrt{2}$. This count assumes the usual two-mode bosonic Fock space; an extra unmentioned vacuum multiplicity could replicate the multiplet.

A Hermitian operator $A$ satisfies $⟨\phi ,A\psi ⟩=⟨A\phi ,\psi ⟩$ on its operator domain. HereTwo integration by parts operations, together with the boundary conditions in the Hamiltonian domain, show that the kinetic term is Hermitian; multiplication by the real function $V$ is Hermitian as well. Thus $H$ is Hermitian.

For a time-independent operator $A$, the Schrodinger equation and its adjoint give the Ehrenfest theoremUsing the canonical commutation relation $[x,p]=i\hslash$,Consequently

The canonical quantization of a real scalar field promotes the classical equal-time brackets to the canonical commutation relations, in units $\hslash =1$:The Dirac delta distribution expresses locality of the conjugate variables. These are identities of operator-valued distributions, interpreted after spatial smearing where necessary.

Use the free Hamiltonian operator and the canonical commutation relations. The kinetic momentum term givesThe Heisenberg picture equation therefore yields ${\dot{\varphi }}_1={\pi }_1$.

For the momentum commutator, the mass term gives $i{m}_1^2{\varphi }_1(\mathbf{x})$. The gradient term giveswhere integration by parts differentiates the field instead of the Dirac delta distribution. ThusCombining the two first-order operator equations gives the Klein-Gordon equation:Suitable decay or spatial smearing removes the boundary term in the integration by parts.

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 $D=26$.

Use the canonical commutation relation for the creation and annihilation operators and the transformation derived above. Directly,Requiring these to equal ${\delta }_{ik}$ and zero, respectively, and conjugating the equations yields the canonical identities for a bosonic Bogoliubov transformation:The commutator of the primed creation operators then vanishes by taking adjoints. These are exactly the corresponding positive-mode norm and positive-negative orthogonality conditions for the Bogoliubov transformation. They are sufficient for these algebraic commutators whenever the sums are well defined. Infinitely many modes introduce further questions of convergence and whether the two choices admit a common bosonic Fock space representation; the matrix identities alone do not settle unitary implementability.

With the symmetric Fourier normalization displayed in the question, the creation and annihilation operators obey the canonical commutation relationSince $\delta \hat{\varphi }=\hat{f}/a$, its vacuum two-point correlation function isComparison with the definition in the question givesUsing $|f_k{|}^2=(1+1/(k^2{\tau }^2))/(2k)$ and $a=-1/(H\tau )$,On superhorizon scales, $k\ll aH$ or $|k\tau |\ll 1$, so the scale-invariant inflationary power spectrum freezes to

For a real free scalar field on a globally hyperbolic spacetime, takeThe Klein-Gordon equation has a well-posed initial-value problem on every Cauchy hypersurface. Its complex solution space carries the conserved Klein-Gordon inner product. Choose modes withand expandThe canonical commutation relations produce the bosonic Fock space, and $a_i|0⟩=0$ defines the chosen vacuum.

The choice of positive-frequency solutions is extra structure. In a time-dependent geometry there is no preferred time translation, so different choices mix $u_i$ with $u_i^{\ast }$ and give the vacuum ambiguity in a nonstationary spacetime. In a stationary spacetime, a timelike Killing vector field defines positive frequency by $i{\mathcal{L}}_K{u}_i={\omega }_i{u}_i$, ${\omega }_i>0$, and hence the preferred vacuum state in a stationary spacetime.

If the spacetime is stationary in the remote past and future, define normalized in- and out-modes there. Their Bogoliubov transformation isThe expected out-particle number in the in-vacuum isThus nonzero beta coefficients measure particle creation by the intervening time dependence.

The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

For part (a), the adjoint operator isUsing the canonical commutation relation $[X,P]=i\hslash I$,Thus $A$ and $A^{†}$ are the annihilation and creation operators of the quantum harmonic oscillator.

For part (b), writeSince $[A,K]=A^{†}$ and $[A^{†},K]=A$, differentiation of the unitary conjugation givesThe initial conditions are $A(0)=A$ and $A^{†}(0)=A^{†}$. Solving this pair of linear ordinary differential equations gives the Bogoliubov transformationThis is the transformation implemented by the single-mode squeeze operator.

For part (c), invert the definition of $A$:The transformation above impliesThe ground state has zero position and momentum means and variances $\hslash /(2m\omega )$ and $m\hslash \omega /2$. Hence the squeezed vacuum state $|\gamma ⟩=S(\gamma )|0⟩$ hasThe reciprocal squeezing factors cancel, soThe state therefore continues to saturate the Heisenberg uncertainty relation.

The exact mode functions satisfy the Wronskian normalizationUsing the Dirac delta function generated by the ladder-operator commutator therefore gives the exact canonical commutation relationThis remains true on superhorizon scales, so the exact operators do not literally commute. A claim of classical behavior must instead compare this fixed commutator with the growing statistical fluctuations, or appeal to quantum decoherence.

The constant scalar-field shift symmetry has momentum-space actionFor the canonical commutation relation $[\varphi (\mathbf{x}),\Pi (\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$, its Noether charge can be writtenIndeed, $i[Q,\varphi (\mathbf{x})]=c$.

The creation and annihilation operators obey the canonical commutation relationsSince $\delta \hat{\varphi }=\hat{f}/a$, the vacuum two-point function identifiesOn superhorizon scales, $|k\tau |=k/(aH)\ll 1$, so