= Solution
Quantize the real massless <scalar field> using the normalized in-modes:
$$
\widehat\phi=\sum_j\left(a_jf_j+a_j^\dagger\bar f_j\right),\qquad[a_j,a_k^\dagger]=\delta_{jk},\qquad a_j|0_{\rm in}\rangle=0.
$$
Use the <Klein-Gordon inner product> convention that is antilinear in its first argument. Mode normalization gives
$$
(f_j,f_k)=\delta_{jk},\qquad(\bar f_j,\bar f_k)=-\delta_{jk},\qquad(f_j,\bar f_k)=0.
$$
The out annihilation operator is $b_i=(p_i,\widehat\phi)$. Applying the <Bogoliubov transformation> and the negative norm of conjugate modes gives
$$
\boxed{b_i=\sum_j\left(\bar A_{ij}a_j-\bar B_{ij}a_j^\dagger\right).}
$$
Both the complex conjugation and the minus sign follow from the inner-product convention. These operators annihilate the <out-vacuum>, but need not annihilate the <in-vacuum>. The <particle number operator> in out-mode $i$ is $N_i^{\rm out}=b_i^\dagger b_i$. In the <in-vacuum>, the only nonzero contraction is $\langle0_{\rm in}|a_ja_k^\dagger|0_{\rm in}\rangle=\delta_{jk}$, obtained from the <canonical commutation relation>. Thus
$$
\boxed{\langle0_{\rm in}|N_i^{\rm out}|0_{\rm in}\rangle=\sum_{j,k}B_{ij}\bar B_{ik}\delta_{jk}=\sum_j|B_{ij}|^2=(BB^\dagger)_{ii}.}
$$
This is <particle number from Bogoliubov coefficients>: mixing with negative-frequency in-modes produces out-particles even though no in-particles were present. For continuous mode labels the sum becomes an integral; normalized wave packets make an individual occupation number well-defined. No explicit collapse geometry or thermal spectrum is needed for this conclusion.
Back to article page
