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On polynomials in $L^2$ of standard Gaussian measure, Gaussian integration by parts makes $D$ and $x-D$ adjoints, with commutator $[a^{-},a^{+}]=I$. The Probabilists' Hermite polynomials are obtained by applying $a^{+}$ repeatedly to $1$, and $a^{-}{\mathrm{He}}_n=n{\mathrm{He}}_{n-1}$. These give a Gaussian-space realization of creation and annihilation operators.

Promote the canonical field and momentum $\pi =f^{\prime }$ to operators satisfying the canonical commutation relation, $[\hat{f}(\tau ,\mathbf{x}),\hat{\pi }(\tau ,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$. A real field has Fourier reality condition ${\hat{f}}_{\mathbf{k}}^{†}={\hat{f}}_{-\mathbf{k}}$. Its oscillator expansion isThe creation and annihilation operators add or remove excitations in the selected mode basis, withCanonical normalization requires the Wronskian $u_k{u}_k^{\ast \prime }-u_k^{\ast }{u}_k^{\prime }=i$.

For the Bunch-Davies vacuum, choose the positive-frequency short-wavelength behavior $u_k\sim e^{-ik\tau }/\sqrt{2k}$ as $k|\tau |\to \infty$. For the proposed solution, direct differentiation givesSubstitution cancels every term in $u_k^{\prime \prime }+(k^2-2/{\tau }^2)u_k$. Its Wronskian is $i$, and its large-$|k\tau |$ limit is the required flat-spacetime positive-frequency mode. ThusA term with the conjugate frequency is another allowed classical solution, but would describe a different quantum state, through a Bogoliubov transformation. Its coefficient is set to zero by the early-time vacuum condition, not by absence of a second solution to the differential equation.

Write $f_s=(-1{)}^{1/2-s}$ for $s=\pm 1/2$, so $f_{-s}=-f_s$. Under the antiunitary operator $\hat{T}$, the coefficients and exponentials in the mode expansion of a Dirac field are conjugated as well as the creation and annihilation operators being transformed. ThusRelabel $q=p_T$ and $r=-s$. The stated Dirac spinor identities implyAlso $q_T\cdot x=-q\cdot x_T$. Substitution reconstructs the original Dirac field at the reflected time:The minus sign comes from reversing the spin label, not from anticommuting field operators. This is the relation between the time-reversal matrix and the charge-conjugation matrix in the supplied spin phases and intrinsic phase convention. Rephasing the charge-conjugation matrix or the intrinsic time-reversal phase can change its displayed form. It proves time reversal of a Dirac field without identifying an antiunitary operator with its finite-dimensional spinor matrix.

Use the Fourier normalization of the PDF. The creation and annihilation operators obeyThere is no $(2\pi {)}^3$ in this commutator because the expansion uses $(2\pi {)}^{-3/2}$. The Bunch-Davies vacuum satisfies ${\hat{a}}_{\mathbf{k}}|0⟩=0$. Write $u_k=f_k^{\ast }$ for the coefficient of the annihilation operator. Only the contraction $⟨0|{\hat{a}}_{\mathbf{k}}{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}|0⟩={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$ contributes, soThe given mode hasSince $a^{-2}=H^2{\tau }^2$, the equal-time two-point function of a de Sitter scalar therefore has dimensional power spectrumComparing with the dimensionless power convention in the hint givesEquivalently, angular integration expresses the correlation as ${\int }_0^{\infty }(dk/k){\Delta }_{\delta \varphi }^2\sin (kr)/(kr)$, where $r=|\mathbf{r}|$.

On superhorizon scales, $k\ll aH$ is $|k\tau |\ll 1$, andThis scale-invariant inflationary power spectrum gives equal variance per logarithmic wavenumber interval. It is exactly scale invariant in the massless constant-$H$ approximation here; slow evolution of $H$ and small field mass produce a small spectral tilt. Such quantum primordial perturbations supply the initial fluctuations that subsequently seed cosmological structure after conversion to curvature and density perturbations. The curvature conversion itself requires the metric perturbations excluded from this calculation.

The continuum correlation is understood with the usual infrared regulator, such as finite volume or finite inflationary duration. The strictly massless infinite-volume model has a logarithmic infrared divergence. Coincident products also require ultraviolet regularization or smearing; the finite mode spectrum is unaffected by these qualifications.

The Klein-Gordon inner product satisfies $(\overline{\alpha },\overline{\beta })=-(\beta ,\alpha )$, so $({\overline{\psi }}_i,{\overline{\psi }}_j)=-{\delta }_{ij}$. Orthogonality of positive and negative modes in the field expansion therefore gives $({\psi }_i,\Phi )=a_i$ and $({\overline{\psi }}_i,\Phi )=-a_i^{†}$.

Extracting the primed coefficient in the same way, and using antilinearity in the first entry,Consequently the Bogoliubov transformation of the creation and annihilation operators isThe minus sign comes from the negative norm of the conjugate modes, and the complex conjugates come from the first argument of the Hermitian pairing.

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.

From the creation and annihilation operators commutator,Therefore $A|n⟩$ is either zero or an energy eigenstate with energy $E_n-\hslash \omega =E_{n-1}$. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate, $A|n⟩=c_n|n-1⟩$. Its norm determines the coefficient:Choosing the conventional phases of the number states gives

Solving the given relation for the position operator givesPut $Q=A+A^{†}$. Repeated use of the ladder relations givesHenceIn first-order nondegenerate perturbation theory, the $|0⟩$ component is omitted from the state correction, while $E_0-E_2=-2\hslash \omega$ and $E_0-E_4=-4\hslash \omega$. It follows thatwhich is the first-order ground state of the quartic oscillator.

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

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 creation and annihilation operators obeyThe two-point correlation function in their vacuum has Fourier amplitude $|f_k{|}^2/a^2$. After cosmological horizon exit, $|k\tau |\ll 1$, and the stated Bunch-Davies vacuum mode and $a=-1/(H\tau )$ giveThe dimensionless power spectrum is consequently

At leading order in $|k\tau |$, the inflationary Fourier mode and its derivative areThus the leading parts of $\hat{f}$ and $\hat{\pi }$ contain the same quadrature of the creation and annihilation operators and satisfyTheir commutator consequently vanishes in this leading approximation. The exact nonzero result in part b is carried by the subleading decaying mode. Relative to the rapidly growing anticommutator, its effect is suppressed by powers of $|k\tau |$, which is the squeezing captured by the classicality parameter of a cosmological perturbation. The perturbation can therefore be treated as an effectively classical stochastic field on superhorizon scales even though its exact quantum commutator is unchanged.

Expanding the free field asand using the vacuum creation and annihilation operators algebra givesThus the dimensional power spectrum is $P_{\phi }(k,\tau )=H^2{\tau }^2/(2ck)$. Unlike a minimally coupled massless inflationary fluctuation, this conformally coupled field decays as $a^{-1}$ and does not freeze at late time.

Orthogonality of the energy eigenstates givesUsing the lowering-operator action from part (a) and shifting the summation index,Thus this coherent state is a normalized eigenstate of $A$ with eigenvalue $z$.