In the context of BH evaporation, one can represent an initial vacuum as the final state containing particles, following Polchcinski (2016, page 9)

$$|0\rangle_{a}=\mathcal{N} \exp \left(\int_{0}^{\infty} \frac{d \omega}{2 \pi} e^{-\omega / 2 T_{\mathrm{H}}} b_{\omega}^{\dagger} \tilde{b}_{\omega}^{\dagger}\right) $$

where $\mathcal{N}$ is the normalization constant.

On the other hand, one can read in Lectures on quantum gravity edited by Gomberoff, Marolf (Springer) that the condition for normalizability is

the average total number of excitations must be finite. If it is not, the state $\left|0_{\rm in}\right\rangle$ does not lie in the Fock space built on the the state $\left|0_{\text {out }}\right\rangle$

We know that in the case of BH evaporation the number of excitations is infinite.

My question is: is that state actually normalizable, if yes, how to do this?


1 Answer 1


is that state actually normalizable...?

Yes. To see this, start with the state $U|0\rangle$ with $$ \newcommand{\ra}{\rangle} U = \exp\left( \int_0^\infty\frac{d\omega}{2\pi}\ f(\omega)\left(b_\omega\tilde b_\omega - b_\omega^\dagger\tilde b_\omega^\dagger\right) \right) \tag{1} $$ for some real-valued function $f(\omega)$. The operator $U$ is manifestly unitary, so the norm of $U|0\ra$ is the same as the norm of $|0\ra$. Use equation (3.7.50) in ref 1 to get the identity $$ U|0\ra = A|0\ra \tag{2} $$ with $$ A = \mathcal{N} \exp\left( -\int_0^\infty\frac{d\omega}{2\pi}\ \tanh\big(f(\omega)\big) b_\omega^\dagger\tilde b_\omega^\dagger \right) \tag{3} $$ and $$ \mathcal{N} = \exp\left( -\int_0^\infty\frac{d\omega}{2\pi}\ \log\Big(\cosh\big(f(\omega)\big)\Big) \right). \tag{4} $$ The function $f(\omega)$ can be chosen so that the state $A|0\ra$ is the state shown in the question, namely $$ -\tanh\big(f(\omega)\big) = e^{-\omega/2T}. \tag{5} $$ The only thing left to check is that (4) is finite. To see that it is, use the fact that for large $\omega$, equation (5) implies $$ -f(\omega)\approx e^{-\omega/2T}\ll 1, \tag{6} $$ which in turn implies $$ \log\Big(\cosh\big(f(\omega)\big)\Big) \sim e^{-\omega/T}, \tag{7} $$ so the integral (4) converges. If we're concerned that the manipulations leading to the identity (2) might not be well-defined, then we can use a regulator, like taking $\omega$ to be a discrete parameter and deferring the continuum limit until the end of the calculation, after all desired inner products have been calculated.


  1. Chapter 3 (Operators and States) in Barnett and Radmore (2002), Methods in Theoretical Quantum Optics (Oxford University Press)
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    $\begingroup$ The link defaults to the main page. $\endgroup$ Dec 25, 2023 at 3:41
  • $\begingroup$ @flippiefanus Thank you for letting me know! I removed the broken link. $\endgroup$ Dec 25, 2023 at 20:18

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