FLRW Fields and Mode Quantization

A free scalar field on a spatially flat FLRW spacetime can be reduced to independent time-dependent oscillators. The reduction fixes the oscillator equation and its conserved symplectic normalization; it does not select a vacuum or an observer-independent particle number. Keeping those statements separate is the main conceptual task of cosmological mode quantization.

Required background. Covariant scalar fields fixes $P=\Box+m^2+\xi R$, and canonical quantization on curved backgrounds fixes the symplectic form. Helpful background. Adiabatic states and time-dependent particle creation develop later choices made on this normalized solution space.

Canonical oscillators from the scalar action

Use conformal time and the site’s $(+---)$ curvature convention,

$$ds^2=a^2(\eta)(d\eta^2-d\mathbf x^2), \qquad \mathcal H=\frac{a'}a, \qquad R=-6\frac{a''}{a^3}.$$

The action

$$S=\frac12\int d\eta\,d^3x\,a^2 \left[\phi'^2-(\boldsymbol\nabla\phi)^2 -a^2(m^2+\xi R)\phi^2\right]$$

gives $(\Box+m^2+\xi R)\phi=0$. Introduce the canonical field $\chi=a\phi$ and discard the resulting time-boundary term. Then

$$S=\frac12\int d\eta\,d^3x \left\{ \chi'^2-(\boldsymbol\nabla\chi)^2 -\left[a^2m^2-(1+6\xi)\frac{a''}{a}\right]\chi^2 \right\}.$$

Thus a Fourier mode $v_k$ of $\chi$ obeys

$$v_k''+\Omega_k^2v_k=0, \qquad \Omega_k^2=k^2+a^2m^2-(1+6\xi)\frac{a''}{a}.$$

This formula is a useful convention test. In four dimensions the site’s conformal value is $\xi=-1/6$; for $m=0$ the scale-factor potential then cancels. Sources that write $P=\Box+m^2-\xi_cR$ instead quote $\xi_c=+1/6$. The reduction and its sign bookkeeping are reviewed systematically in Parker and Toms 2009, ch. 2.

The Wronskian gives the commutator

Expand the real field as

$$\phi(\eta,\mathbf x)=\frac1{a(\eta)} \int\frac{d^3k}{(2\pi)^{3/2}} \left[ a_{\mathbf k}v_k(\eta)e^{i\mathbf k\cdot\mathbf x} +a_{\mathbf k}^{\dagger}v_k^*(\eta)e^{-i\mathbf k\cdot\mathbf x} \right],$$

where isotropy makes $v_k$ depend only on $k=|\mathbf k|$. The momentum conjugate to $\phi$ is

$$\pi=a^2\phi' =a\int\frac{d^3k}{(2\pi)^{3/2}} \left[ a_{\mathbf k}(v_k'-\mathcal Hv_k)e^{i\mathbf k\cdot\mathbf x} +a_{\mathbf k}^{\dagger}(v_k^{*'}-\mathcal Hv_k^*)e^{-i\mathbf k\cdot\mathbf x} \right].$$

With $[a_{\mathbf k},a_{\mathbf q}^{\dagger}]=\delta^3(\mathbf k-\mathbf q)$, the equal-time commutator is

$$\begin{aligned} [\phi(\eta,\mathbf x),\pi(\eta,\mathbf y)] &=\int\frac{d^3k}{(2\pi)^3} \left(v_kv_k^{*'}-v_k'v_k^*\right) e^{i\mathbf k\cdot(\mathbf x-\mathbf y)}. \end{aligned}$$

The two $\mathcal H|v_k|^2$ terms cancel. Therefore

$$v_kv_k^{*'}-v_k'v_k^*=i$$

is exactly the condition for $[\phi(\eta,\mathbf x),\pi(\eta,\mathbf y)]=i\delta^3(\mathbf x-\mathbf y)$. Differentiating the Wronskian and using the real-frequency mode equation gives zero, so normalization imposed on one Cauchy slice is preserved by exact evolution. Parker’s oscillator construction emphasizes that this canonical requirement precedes any particle interpretation Parker 1969, §§II–III.

One convenient normalized initial-data prescription is

$$v_k(\eta_0)=\frac1{\sqrt{2W_k}}, \qquad v_k'(\eta_0)= \left(-iW_k-\frac{W_k'}{2W_k}\right)v_k(\eta_0),$$

where $W_k(\eta_0)>0$ and $W_k'(\eta_0)$ are declared data. Substitution gives the Wronskian $i$ for any such positive $W_k$. Consequently, normalization alone cannot tell us which $W_k$ represents a physically preferred state.

For numerical evolution, monitor three logically independent residuals:

$$\epsilon_{\rm ode}=|v_k''+\Omega_k^2v_k|, \qquad \epsilon_W=|v_kv_k^{*'}-v_k'v_k^*-i|,$$

together with positivity of the Gaussian covariance. A small differential-equation residual does not guarantee the correct normalization, and neither check guarantees that the chosen covariance is positive.

Basis changes versus changes of state

Any second normalized complex solution can be written

$$\widetilde v_k=A_kv_k+B_kv_k^*, \qquad |A_k|^2-|B_k|^2=1.$$

There are two different operations hidden in this formula.

• If the creation and annihilation operators are transformed inversely, $\phi$ and all of its correlation functions are unchanged. This is a new complex basis for the same state.
• If the operators are held fixed and $\widetilde v_k$ is declared positive frequency, the two-point function changes. This is a different Gaussian state.

The same $B_k$ can therefore be a coordinate change or physical state data; the statement becomes meaningful only after saying what is held fixed. For a real field, reality relates Fourier amplitudes at $\mathbf k$ and $-\mathbf k$. Count modes once using either the displayed all-$\mathbf k$ annihilator convention or a half-space real-mode convention; imposing an extra identification between $a_{\mathbf k}$ and $a_{-\mathbf k}$ inside the all-$\mathbf k$ expansion produces a factor-of-two error.

Cosmic-time cross-check

Write $dt=a\,d\eta$, $H=\dot a/a$, and set $q_k=a^{3/2}\phi_k$. The original mode equation becomes

$$\ddot q_k+ \left[ \frac{k^2}{a^2}+m^2+\xi R -\frac32\dot H-\frac94H^2 \right]q_k=0,$$

while its canonical normalization is

$$q_k\dot q_k^*-\dot q_kq_k^*=i.$$

The effective frequency is visibly different from $\Omega_k$, but the field reconstructed from $q_k/a^{3/2}$ is the same field reconstructed from $v_k/a$. In particular,

$$a^3(\phi_k\dot\phi_k^*-\dot\phi_k\phi_k^*) =q_k\dot q_k^*-\dot q_kq_k^* =i.$$

This is the correct adversarial test: translate the momentum and symplectic form together with the field variable. Equal-time commutators and field correlators agree. An occupation obtained by diagonalizing either instantaneous oscillator Hamiltonian need not agree, because that extra particle basis is not invariant under a time-dependent canonical transformation.

Common pitfalls

Treating the Wronskian as a vacuum condition. The Wronskian fixes one real normalization condition. Infinitely many Bogoliubov-related normalized modes remain, so a state-selection criterion is still required.

Differentiating $v_k$ as though it were $\phi_k$. Because $\phi_k=v_k/a$, the momentum mode is $a(v_k'-\mathcal Hv_k)$, not $av_k'$. Dropping the $-\mathcal Hv_k$ term changes the covariance and can violate the uncertainty relation even when the Wronskian of $v_k$ is exact.

Exercises

1. Recover the commutator

Starting from the displayed mode expansion, compute $[\phi(\mathbf x),\pi(\mathbf y)]$ and show explicitly why the terms proportional to $\mathcal H$ cancel.

Solution

Only $[a_{\mathbf k},a_{\mathbf q}^{\dagger}]$ and its reversed ordering contribute. The first ordering gives $v_k(v_k^{*'}-\mathcal Hv_k^*)$, while the second gives $-(v_k'-\mathcal Hv_k)v_k^*$. Their $\mathcal H|v_k|^2$ pieces cancel, leaving $v_kv_k^{*'}-v_k'v_k^*=i$. Fourier inversion then gives

$$[\phi(\mathbf x),\pi(\mathbf y)] =i\int\frac{d^3k}{(2\pi)^3}e^{i\mathbf k\cdot(\mathbf x-\mathbf y)} =i\delta^3(\mathbf x-\mathbf y).$$

2. Translate the symplectic normalization

Show that $q=a^{3/2}\phi$ converts $a^3(\phi\dot\phi^*-\dot\phi\phi^*)=i$ into the ordinary oscillator Wronskian for $q$. Why does this not force the two instantaneous particle numbers to agree?

Solution

Since $\phi=a^{-3/2}q$,

$$\dot\phi=a^{-3/2}\left(\dot q-\frac32Hq\right).$$

The two terms proportional to $H|q|^2$ cancel in the antisymmetric product, so

$$a^3(\phi\dot\phi^*-\dot\phi\phi^*) =q\dot q^*-\dot q q^*.$$

The field algebra is therefore unchanged. Instantaneous particle number requires an additional split of phase space into positive and negative frequencies. A time-dependent rescaling changes the oscillator Hamiltonian and hence that split, so the intermediate occupations may differ even though all translated field observables agree.

References

• Parker, Leonard. “Quantized Fields and Particle Creation in Expanding Universes. I.” Physical Review 183 (1969): 1057–1068. DOI.
• Parker, Leonard, and David Toms. Quantum Field Theory in Curved Spacetime: Quantized Fields and Gravity. Cambridge University Press, 2009, ch. 2. Chapter DOI.