= Solution
Use a free real <scalar field> obeying a linear <Klein-Gordon equation>, for example $(\Box-\mu^2-\xi R)\phi=0$. This linearity is the assumption that permits mode evolution by a <Bogoliubov transformation>. Work with a complete set of normalized modes, or normalizable wave packets when the spectrum is continuous. The stationary asymptotic regions supply time translations relative to which <positive-frequency solutions> and their complex conjugates can be distinguished.
For solutions define the conserved <Klein-Gordon inner product>
$$
(u,v)=i\int_\Sigma d\Sigma^a\left(u^*\nabla_av-v\nabla_au^*\right).
$$
Here $\Sigma$ is a <Cauchy surface> with future-directed normal. Its conservation follows by applying the wave equation to the divergence of its current, provided boundary flux is included or vanishes. Let $u_j$ be a positive-frequency basis in the past and $v_s$ one in the future, each propagated as a solution through the intermediate geometry. Normalize them by $(u_i,u_j)=\delta_{ij}$, $(u_i^*,u_j^*)=-\delta_{ij}$ and $(u_i,u_j^*)=0$, with the corresponding identities for the future basis. The two field expansions are
$$
\widehat\phi=\sum_j(a_ju_j+a_j^\dagger u_j^*)
=\sum_s(b_sv_s+b_s^\dagger v_s^*).
$$
The <canonical commutation relations> give $[a_i,a_j^\dagger]=\delta_{ij}$ and $[b_s,b_t^\dagger]=\delta_{st}$. The <in-vacuum> satisfies $a_j|0_-\rangle=0$, while the <out-vacuum> satisfies $b_s|0_+\rangle=0$. They are generally different vacua.
Fix the coefficient convention by expanding future modes in past modes:
$$
v_s=\sum_j(\alpha_{sj}u_j+\beta_{sj}u_j^*),\qquad
\alpha_{sj}=(u_j,v_s),\qquad \beta_{sj}=-(u_j^*,v_s).
$$
The minus sign is due to the negative norm of the conjugate modes. Taking the <Klein-Gordon inner product> with the field, which is antilinear in its first entry, gives
$$
\boxed{b_s=\sum_j(\alpha_{sj}^*a_j-\beta_{sj}^*a_j^\dagger).}
$$
Both frequency sectors are needed: if $\beta\ne0$, a future annihilation operator contains a past creation operator. This is the mechanism of <particle number from Bogoliubov coefficients>, not a classical nonzero mean field.
Computing $[b_s,b_t^\dagger]$ and $[b_s,b_t]$ explicitly gives the <canonical identities for a bosonic Bogoliubov transformation>:
$$
\sum_j(\alpha_{sj}\alpha_{tj}^*-\beta_{sj}\beta_{tj}^*)=\delta_{st},
\qquad
\sum_j(\alpha_{sj}\beta_{tj}-\beta_{sj}\alpha_{tj})=0.
$$
Completeness also gives the inverse $a_j=\sum_s(\alpha_{sj}b_s+\beta_{sj}^*b_s^\dagger)$. Thus the evolved past vacuum, viewed in the future <Fock space>, obeys
$$
\left[\alpha^Tb+\beta^\dagger b^\dagger\right]|\Psi\rangle=0.
$$
For a finite-mode treatment with invertible $\alpha$, solve this condition by a <multimode squeezed vacuum>:
$$
|\Psi\rangle=C\exp\left(\frac12\sum_{s,t}Z_{st}b_s^\dagger b_t^\dagger\right)|0_+\rangle,
\qquad Z=-(\alpha^T)^{-1}\beta^\dagger.
$$
The canonical identities make $Z$ symmetric. Commuting $b_s$ through the exponential gives $b_s|\Psi\rangle=\sum_tZ_{st}b_t^\dagger|\Psi\rangle$, which directly verifies the vacuum condition. Its normalization is $|C|=\det(I-ZZ^\dagger)^{1/4}$. A general initial state is obtained by applying its initial creation-operator polynomial and expressing those operators in the future basis. In infinitely many modes, a common unitary Fock representation requires the creation-mixing coefficients to be Hilbert-Schmidt; finite wave-packet counts avoid treating delta-normalized modes as individual normalizable states. These are the <bosonic mode mixing implementability> conditions, rather than an assumption of arbitrary finite total creation over an infinite time interval.
For the requested <number operator> of one future mode, set $N_s=b_s^\dagger b_s$. Expectations below are in the evolved past vacuum $|\Psi\rangle$, equivalently in $|0_-\rangle$ using the Heisenberg-mode operators above. Acting on the past vacuum gives $b_s|0_-\rangle=-\sum_j\beta_{sj}^*a_j^\dagger|0_-\rangle$, whose squared norm is
$$
\boxed{n_s\equiv\langle N_s\rangle=\sum_j|\beta_{sj}|^2.}
$$
The mean particle number vanishes exactly when that mode has no negative-frequency mixing.
The second moment needs the anomalous contraction
$$
c_s\equiv\langle b_sb_s\rangle=-\sum_j\alpha_{sj}^*\beta_{sj}^*.
$$
The <canonical commutation relations> give $N_s^2=b_s^{\dagger2}b_s^2+N_s$. Moreover,
$$
b_s^2|0_-\rangle=c_s|0_-\rangle
+\sum_{j,k}\beta_{sj}^*\beta_{sk}^*a_j^\dagger a_k^\dagger|0_-\rangle.
$$
Its vacuum and two-particle pieces are orthogonal. Applying the commutation relations twice shows
$$
\langle0_-|a_k a_j a_l^\dagger a_m^\dagger|0_-\rangle
=\delta_{jl}\delta_{km}+\delta_{jm}\delta_{kl}.
$$
The two pairings each contribute $n_s^2$ to the squared norm. Hence the full answer is
$$
\boxed{\langle N_s^2\rangle=2n_s^2+n_s+|c_s|^2
=2\left(\sum_j|\beta_{sj}|^2\right)^2+\sum_j|\beta_{sj}|^2
+\left|\sum_j\alpha_{sj}\beta_{sj}\right|^2.}
$$
Thus $\operatorname{Var}(N_s)=n_s(n_s+1)+|c_s|^2$. These are the <out-mode number fluctuations in the in-vacuum>. Omitting $c_s$ would assume a property not given in the general question. A thermal single mode has $c_s=0$, giving $\langle N_s^2\rangle=2n_s^2+n_s$. A single-mode <squeezed vacuum state> instead has $|c_s|^2=n_s(n_s+1)$, giving $\langle N_s^2\rangle=3n_s^2+2n_s$.
If the counted observable is the total number over a finite set of future modes, its moments follow by retaining cross-mode correlations as well. Define $C_{st}=\langle b_s^\dagger b_t\rangle$ and $M_{st}=\langle b_sb_t\rangle$. The same vacuum pairings give $\langle N_sN_t\rangle=C_{ss}C_{tt}+|C_{st}|^2+|M_{st}|^2+\delta_{st}C_{ss}$. Thus $\langle N_{\rm tot}\rangle=\operatorname{tr}C$ and $\langle N_{\rm tot}^2\rangle=(\operatorname{tr}C)^2+\operatorname{tr}(C^2)+\operatorname{tr}(M^\dagger M)+\operatorname{tr}C$, with $C_{st}=\sum_j\beta_{sj}\beta_{tj}^*$ and $M_{st}=-\sum_j\alpha_{sj}^*\beta_{tj}^*$. Independent thermal counts cannot be assumed for a general correlated squeezed state.
For <Hawking radiation>, choose the incoming vacuum at <past null infinity> of a collapse spacetime and propagate late outgoing modes backwards. Near a nonextremal forming <event horizon>, the logarithmic tortoise-coordinate divergence leads to the <Hawking exponential ray map>
$$
U_H-U=Ae^{-\kappa u},
$$
where $u$ is late retarded time, $U$ is an early regular affine null coordinate and $\kappa>0$ is the final <surface gravity>. A late positive-frequency mode $e^{-i\omega u}$ therefore becomes proportional to $(U_H-U)^{i\omega/\kappa}$ on rays escaping just before horizon formation. This is not positive-frequency with respect to $U$: it has both frequency signs and hence nonzero $\beta$.
The thermal factor can be seen directly. With $x=U_H-U>0$, $a=\omega/\kappa$ and a regulator $\varepsilon>0$, the relevant Fourier integrals are
$$
\int_0^\infty x^{ia}e^{-(\varepsilon\pm i\omega')x}\,dx
=\Gamma(1+ia)(\varepsilon\pm i\omega')^{-1-ia}.
$$
As $\varepsilon\downarrow0$, the arguments of the two complex factors approach $\pm\pi/2$. Their squared-modulus ratio for the suppressed and enhanced frequency branches is $e^{-2\pi a}$. Thus the <thermal ratio of Hawking Bogoliubov coefficients>, combined with the canonical normalization, yields the bosonic occupation
$$
\boxed{n_\omega=\frac1{e^{2\pi\omega/\kappa}-1},\qquad
T_H=\frac\kappa{2\pi}}
$$
in units $\hbar=c=k_B=1$. Scattering through the exterior potential multiplies the asymptotic occupation by the appropriate <greybody factor>.
A complete future mode description includes modes crossing the horizon as well as modes reaching <future null infinity>. The global incoming vacuum develops correlations between these sectors; restricting to the exterior gives approximately thermal outgoing occupation rather than a claim that the complete pure state became a thermal mixed state. This is the particle-production picture of a collapsing <black hole>, not an incoming thermal bath on an eternal geometry. Including backreaction turns the emitted positive energy into black-hole mass loss. The renormalized quantum stress tensor is not subject to the pointwise classical positivity used in the <black-hole area theorem>, so this does not contradict its classical proof.
Back to article page
