Information Retrieval from Black Holes

Abstract

It is generally believed that, a black hole, originating from collapse of matter, erases all the information about the initial state. In other words, the initial configuration of matter forming a black hole cannot be retrieved by future asymptotic observers, through local measurements. This is in sharp contrast with the expectation from a unitary evolution in quantum theory and leads to (a version of) the black hole information paradox. Classically, no-hair theorems guarantee that, apart from mass, charge and angular momentum, nothing is expected to be revealed to such asymptotic observers after the formation of a black hole. On the other hand, semi-classically, black holes evaporate after their formation through the emission of Hawking radiation. The dominant part of the radiation is expected to be thermal and hence one cannot have any knowledge about the initial data. However, there can be distortions in the Hawking radiation from thermality, which even though not strong enough to make the evolution unitary, do carry some part of information regarding the in-state. In this chapter, we show how one may decipher the information about the in-state of the field from such distortions. In particular, distortions of a particular kind can be used to reconstruct the initial data completely.

Appendices

Appendix A

A.1 Spectrum Operator

Using the expression for the correction term over the vacuum thermal spectrum, we can obtain the distortion from thermal Hawking radiation for one particle initial state of the field which is undergoing the collapse as,

$$\begin{aligned} N_{\Omega } = \left[ \left| \int _0^{\infty } \frac{d \tilde{\omega }'}{\sqrt{4\pi \tilde{\omega }'}}\alpha ^*_{\Omega \tilde{\omega }'}f(\tilde{\omega }')\right| ^2+\left| \int _0^{\infty } \frac{d \tilde{\omega }}{\sqrt{4\pi \tilde{\omega }}}\beta _{\Omega \tilde{\omega }}f(\tilde{\omega })\right| ^2\right] . \end{aligned}$$

(10.54)

It must be noted that the expression in Eq. (10.54) is general enough to include cases when the Bogoliubov coefficients as in Eq. (10.9) are modified by back-reaction, angular momentum, quantum gravity etc. In any case, the non-vacuum part of the radiation spectra provides a constraint for $f(\omega )$ in form of Eq. (10.54). Using Eq. (10.14), we can rewrite Eq. (10.54) as

$$\begin{aligned} N_{\Omega } = \frac{1}{4 \pi }\frac{1}{4 \pi \kappa } \frac{1}{\sinh {\frac{\pi \Omega }{\kappa }}}\left[ \left| \tilde{F}\left( \frac{\Omega }{\kappa }\right) \right| ^2+\left| \tilde{F}\left( -\frac{\Omega }{\kappa }\right) \right| ^2\right] . \end{aligned}$$

(10.55)

We can decompose $|\tilde{F}\left( \Omega /\kappa \right) |^2$ into symmetric $\tilde{S}\left( \Omega /\kappa \right)$ and an anti-symmetric part $\tilde{A}\left( \Omega /\kappa \right)$

$$\begin{aligned} \left| \tilde{F}\left( \frac{\Omega }{\kappa }\right) \right| ^2 = \tilde{S}\left( \frac{\Omega }{\kappa }\right) + \tilde{A}\left( \frac{\Omega }{\kappa }\right) . \end{aligned}$$

(10.56)

With this decomposition, we realize from Eq. (10.55) that the symmetric part of $|\tilde{F}\left( \Omega /\kappa \right) |^2$ is entirely characterized by the distribution function $N_{\Omega }$ of the radiation,

$$\begin{aligned} \tilde{S}\left( \frac{\Omega }{\kappa }\right) = 8 \pi ^2 \kappa N_{\Omega } \sinh {\frac{\pi \Omega }{\kappa }}. \end{aligned}$$

(10.57)

Further, if the in-state is normalized to unity, we have

$$\begin{aligned} \int _{-\infty }^{\infty }d \left( \frac{\Omega }{\kappa } \right) e^{-\frac{\pi \Omega }{\kappa }} \left[ \tilde{S}\left( \frac{\Omega }{\kappa }\right) + \tilde{A}\left( \frac{\Omega }{\kappa }\right) \right] = 8 \pi ^2, \end{aligned}$$

(10.58)

which together with Eq. (10.57) regulates the integral (and hence the asymptotic behavior) of $\tilde{A}\left( \kappa \right)$. Apart from this constraint, $\tilde{A}\left( \kappa \right)$ is a completely arbitrary anti-symmetric function. Therefore, the radiation spectra fixes the symmetric part of the probability density in the Fourier space corresponding to z. However, the anti-symmetric part of this probability density remains largely unspecified.

In terms of the function g(z) defined in Eq. (10.15), the symmetric part $\tilde{S}\left( \frac{\Omega }{\kappa }\right)$ can be written as

$$\begin{aligned} e^{\pi \frac{\Omega }{\kappa }}&\times \int _{-\infty }^{\infty } d y \int _{-\infty }^{\infty } d z g(z-y/2)g^*(z+y/2) e^{-i\frac{\Omega }{\kappa } y} \nonumber \\&+e^{-\pi \frac{\Omega }{\kappa }}\int _{-\infty }^{\infty } d y \int _{-\infty }^{\infty } d z g(z-y/2)g^*(z+y/2) e^{i\frac{\Omega }{\kappa } y } = 2 \tilde{S}\left( \frac{\Omega }{\kappa }\right) . \end{aligned}$$

(10.59)

As, we see that $F(\kappa )$ is “momentum space representation” conjugate to g(z), the above expression can be written in terms of the Wigner function corresponding to the phase space of $(z, \Omega /\kappa )$,

$$\begin{aligned} e^{\pi \frac{\Omega }{\kappa }}\int _{-\infty }^{\infty } d z \mathcal{W}_{g}\left( z,\frac{\Omega }{\kappa }\right) + e^{-\pi \frac{\Omega }{\kappa }}\int _{-\infty }^{\infty } d z \mathcal{W}_{g}\left( z,-\frac{\Omega }{\kappa }\right) = 2 \tilde{S}\left( \frac{\Omega }{\kappa }\right) , \end{aligned}$$

(10.60)

where the Wigner function is defined as,

$$\begin{aligned} \mathcal{W}_{g}\left( z,\frac{\Omega }{\kappa }\right) =\int _{-\infty }^{\infty } d y g(z-y/2)g^*(z+y/2) e^{-i\frac{\Omega }{\kappa } y}. \end{aligned}$$

(10.61)

Also, with the relation

$$\begin{aligned} \left| F\left( \frac{\Omega }{\kappa }\right) \right| ^2 = \int _{-\infty }^{\infty } d z \mathcal{W}_{g}\left( z,\frac{\Omega }{\kappa }\right) , \end{aligned}$$

(10.62)

we obtain,

$$\begin{aligned} e^{\pi \frac{\Omega }{\kappa }}\left| F\left( \frac{\Omega }{\kappa }\right) \right| ^2 + e^{-\pi \frac{\Omega }{\kappa }}\left| F\left( -\frac{\Omega }{\kappa }\right) \right| ^2 = 2 \tilde{S}\left( \frac{\Omega }{\kappa }\right) , \end{aligned}$$

(10.63)

which is an obvious illustration of Eq. (10.57). Therefore, integrating the relation Eq. (10.63) over the frequency range at $\mathcal{J}^{+}$ we obtain the relation

$$\begin{aligned} 2 \int _0^{\infty } d\left( \frac{\Omega }{\kappa }\right) \tilde{S}\left( \frac{\Omega }{\kappa }\right)= & {} \int _{0}^{\infty }d \left( \frac{\Omega }{\kappa }\right) e^{\frac{\pi \Omega }{\kappa }} \int _{-\infty }^{\infty } d z \mathcal{W}_{g}\left( z,\frac{\Omega }{\kappa }\right) \nonumber \\+ & {} \int _{0}^{\infty }d \left( \frac{\Omega }{\kappa }\right) e^{-\frac{\pi \Omega }{\kappa }} \int _{-\infty }^{\infty } d z \mathcal{W}_{g}\left( z,-\frac{\Omega }{\kappa }\right) \nonumber \\= & {} \int _{-\infty }^{\infty }d y e^{\pi y}\left| F\left( y\right) \right| ^2. \end{aligned}$$

(10.64)

Although, the state which would be completely specified, if we know F(y), remains arbitrary apart form this constraint, the symmetric part $\tilde{S}\left( \Omega /\kappa \right)$ which is completely specified through the non-vacuum distortion, fixes the expectation of the exponentiated momenta conjugate to $z (=\log {\omega /C})$.

For a n-particle state

$$\begin{aligned} |\Psi \rangle = \int _0^{\infty }\prod _{i=1}^n \frac{d \omega _i}{\sqrt{2\pi \omega _i}} f(\omega _1,...\omega _n) \hat{a}^{\dagger }(\omega _i)|0 \rangle _M, \end{aligned}$$

(10.65)

the radiation profile over the thermal component fixes the expectation of a single particle exponentiated momentum, i.e.,

$$\begin{aligned} 16 \int _0^{\infty } d\left( \frac{\Omega }{\kappa }\right) \pi ^2 \kappa \left( \sinh {\frac{\pi \Omega }{\kappa }}\right) N_{\Omega } = \frac{1}{n}\langle \Psi | e^{\frac{\pi \hat{\Omega }}{\kappa }}|\Psi \rangle . \end{aligned}$$

(10.66)

Additional information about the initial state can only be obtained from the spectrum if the initial state has some symmetries. We will discuss few interesting cases below.

• If F(y) is a real and symmetric function, then we see from Eq. (10.63) that it gets completely specified in terms of $\tilde{S}(\Omega /\kappa )$. As, a result the initial state also gets completely specified as g(z) can be obtained by the inverse Fourier transform. However, by virtue of the properties of Fourier transform, g(z) also happens to be real and symmetric. This symmetry corresponds to a duality in the frequency space distribution about the surface gravity parameter $\kappa$. These states are very special class of initial states whose information get coded entirely in the radiation from the black hole within the framework of standard unitary quantum mechanics.

• For a slightly more general case, the reality condition on g(z) can be traded for by imposing relation between $F(\Omega /\kappa )$ and $F(-\Omega /\kappa )$, which is to specify the symmetry of F(y) in the positive and negative half planes. Such a specification of symmetry constrains the distribution F(y) to remain arbitrary in one of the half planes and amounts to reducing the degrees of freedom by half. Let us assume $F(\Omega /\kappa )$ is real, that means

  $$\begin{aligned} g(z)=g^*(-z). \end{aligned}$$

  (10.67)

  Now additionally if we impose,

  $$\begin{aligned} F\left( -y\right) = K\left( y\right) F\left( y\right) , \end{aligned}$$

  (10.68)

  for a specified function K(y), then

  $$\begin{aligned} \int _{-\infty }^{\infty }d z g(z) e^{-i y z } = \int _{-\infty }^{\infty }d z g(z) e^{i y z} K\left( y\right) . \end{aligned}$$

  (10.69)

  Therefore, using the condition Eq. (10.67), we can obtain from Eq. (10.69)

  $$\begin{aligned} g^*(z)= & {} \int _{-\infty }^{\infty }d z' g(z') \int _{-\infty }^{\infty }d y K\left( y\right) e^{iy (z'-z)} \nonumber \\= & {} \int _{-\infty }^{\infty }d z' g(z') \tilde{K} (z'-z), \end{aligned}$$

  (10.70)

  where $\tilde{K}(q)$ is the inverse Fourier transform of K(y). Therefore, for such a symmetry in the probability amplitude, the state can be recovered from

  $$\begin{aligned} F^2\left( \frac{\Omega }{\kappa } \right) = 16 \pi ^2 \kappa \frac{\left( \sinh {\frac{\pi \Omega }{\kappa }}\right) }{e^{\frac{\pi \Omega }{\kappa }}+\left( K\left( \frac{\Omega }{\kappa }\right) \right) ^2e^{-\frac{\pi \Omega }{\kappa }} } N_{\Omega }. \end{aligned}$$

  (10.71)

Therefore, we see that the symmetry of the prescribed class for one particle sate encodes the entire information of the in-state in the resulting radiation from the black hole. If the initial condition of the collapse demands symmetry of such kinds, the resulting mixed state has enough information in the spectra to completely specify the state. We will further consider some other classes of symmetries in the initial data for spherically symmetric collapse models and their imprints in the non-vacuum distortions.

A.2 Real Initial Distribution

For real distributions, the Fourier transform will satisfy

$$\begin{aligned} |F(y)|^2 = |F(-y)|^2 \end{aligned}$$

(10.72)

Therefore, $|K(y)|=1$ and we have the relation

$$\begin{aligned} \left| F\left( \frac{\Omega }{\kappa } \right) \right| ^2+\left| F\left( -\frac{\Omega }{\kappa } \right) \right| ^2 = 8 \pi ^2 \kappa \tanh {\frac{\pi \Omega }{\kappa }}N_{\Omega }. \end{aligned}$$

(10.73)

For a symmetric algebraic operator of y

$$\begin{aligned} \hat{\mathcal{O}}_{\text {even}}(y)=\hat{\mathcal{O}}_{\text {even}}(-y), \end{aligned}$$

(10.74)

the expression

$$\begin{aligned} 8 \pi ^2 \kappa\times & {} \tanh {\frac{\pi \Omega }{\kappa }}\mathcal{O}_{\text {even}}\left( \frac{\Omega }{\kappa } \right) N_{\Omega } \nonumber \\= & {} \mathcal{O}_{\text {even}}\left( \frac{\Omega }{\kappa } \right) \left| F\left( \frac{\Omega }{\kappa } \right) \right| ^2 + \mathcal{O}_{\text {even}}\left( -\frac{\Omega }{\kappa } \right) \left| F\left( -\frac{\Omega }{\kappa } \right) \right| ^2 \end{aligned}$$

(10.75)

which on integration over the whole frequency range gives the expectation value of the operator

$$\begin{aligned} \int _{0}^{\infty } d\left( \frac{\Omega }{\kappa } \right) 8 \pi ^2 \kappa \tanh {\frac{\pi \Omega }{\kappa }}\mathcal{O}_{\text {even}}\left( \frac{\Omega }{\kappa } \right) N_{\Omega }= & {} \int _{0}^{\infty } d\left( \frac{\Omega }{\kappa } \right) \Bigg [\mathcal{O}_{\text {even}}\left( \frac{\Omega }{\kappa } \right) \left| F\left( \frac{\Omega }{\kappa } \right) \right| ^2 \nonumber \\+ & {} \mathcal{O}_{\text {even}}\left( -\frac{\Omega }{\kappa } \right) \left| F\left( -\frac{\Omega }{\kappa } \right) \right| ^2\Bigg ] \nonumber \\= & {} \int _{-\infty }^{\infty }\mathcal{O}_{\text {even}}(y)|F(y)|^2. \end{aligned}$$

(10.76)

By similar logic, one can argue that expectation of all odd algebraic operators vanish in this case, i.e., with a symmetric $|F(y)|^2$,the expectation value for an odd observable

$$\begin{aligned} \langle \hat{\mathcal{O}}_{\text {odd}}(y) \rangle = \int _{-\infty }^{\infty } dy \mathcal{O}_{\text {odd}}(y)|F(y)|^2 =0. \end{aligned}$$

(10.77)

Thus in this scenario, expectation of all algebraic operators in y will be given in terms of spectral distortion. Any general operator $\hat{\mathcal{O}}(y)$ can be decomposed in terms of its even and odd parts

$$\begin{aligned} \hat{\mathcal{O}}(y) =\hat{\mathcal{O}}_{\text {even}}(y) + \hat{\mathcal{O}}_{\text {odd}}(y). \end{aligned}$$

(10.78)

Therefore, to obtain $\langle \hat{\mathcal{O}}(y) \rangle$ one only requires $\langle \hat{\mathcal{O}}_{\text {even}}(y) \rangle$, which can be easily obtained from Eq. (10.76). Similarly for the generalized symmetry class

$$\begin{aligned} \langle \hat{\mathcal{O}}(y) \rangle = \int _{0}^{\infty } dy\left[ \mathcal{O}\left( y \right) \left| F\left( y \right) \right| ^2 + \mathcal{O}\left( -y \right) \left| F\left( -y \right) \right| ^2\right] \nonumber \\ = \int _{0}^{\infty } dy(\mathcal{O}\left( y \right) +|K(y)|^2\mathcal{O}\left( -y \right) )\left| F\left( y \right) \right| ^2\nonumber \\ =16 \pi ^2 \kappa \int _{0}^{\infty } d\bar{\Omega } \frac{[\mathcal{O}\left( \bar{\Omega } \right) +|K(\bar{\Omega })|^2\mathcal{O}\left( -\bar{\Omega } \right) ]\sinh {\bar{\Omega }}}{e^{\frac{\pi \Omega }{\kappa }}+\left| K\left( \bar{\Omega }\right) \right| ^2e^{-\frac{\pi \Omega }{\kappa }} } N_{\Omega },\nonumber \\ \end{aligned}$$

(10.79)

where we have used the expression of $\left| F\left( y \right) \right| ^2$ in the range $y \in (0,\infty )$ from Eq. (10.71), in the third equality. Thus even with the specified symmetry class K(y), all the algebraic operators on the momentum space become fixed.

A.3 State for Step Function Support

Let us excite some right-moving modes beyond $x_i^+$ (for simplicity we work with single particle states), such that the normal ordered operator $\hat{T}_{++}(x^+)$ has support only in the region inside the horizon, i.e.,

$$\begin{aligned} \langle \hat{T}_{++}(x^+) \rangle _{\text {Regularized}}= h(x^+)\Theta (x^+ - x_i^+), \end{aligned}$$

(10.80)

for some well behaved function $h(x^+)$ and the step function $\Theta (x^+)$.

If the single particle state is taken to be in the frame of observers which would have described the linear dilaton vacuum, then

$$\begin{aligned} |\Psi \rangle = \int _\omega f(\omega ) \hat{a}_{\omega }^{\dagger }|0\rangle , \end{aligned}$$

(10.81)

where $\int _{\omega }$ stands for $\int d\omega /\sqrt{4\pi \omega }$ and the right-moving quantum field is given on $\mathcal {J}_L^{+}$ as

$$\begin{aligned} \hat{f}_{+}(y^+) = \int _\omega (\hat{a}_{\omega } u_\omega (y^+) + \hat{a}_{\omega }^{\dagger } u^*_\omega (y^+) ), \end{aligned}$$

(10.82)

with mode functions $u_\omega (y^+)$. Then the equation Eq. (10.80) can be re-written as

$$\begin{aligned} \left| \int _\omega f(\omega )u'_\omega (y^+) \right| ^2 = h_1(y^+)\Theta (y^+ - y_i^+), \end{aligned}$$

(10.83)

where$'$ denotes a derivative with respect to $y^+$ and $y_i^+$ marking the location corresponding to $x_i^+$. The function $h_1(y^+)$ absorbs the Jacobian of transformation from $x^{\pm }$ basis to $y^{\pm }$ basis,

$$\begin{aligned} T_{++}(x^+)=\frac{\partial y^{\mu }}{\partial x^+}\frac{\partial y^{\nu }}{\partial x^+}T_{\mu \nu }(y^+)= \frac{\partial y^{+}}{\partial x^+}\frac{\partial y^{+}}{\partial x^+}T_{++}(y^+) \end{aligned}$$

(10.84)

The condition Eq. (10.83) can be realized by

$$\begin{aligned} \int _\omega f(\omega )u'_\omega (y^+) = \tilde{h}(y^+)\Theta (y^+ - y_i^+), \end{aligned}$$

(10.85)

with some other well behaved function $\tilde{h}(y^+)$. Owing to the conformal flatness of the two dimensional spacetime and the conformal nature of minimally coupled massless scalar field, the mode functions can be written as

$$\begin{aligned} u'_\omega (y^+)=-i\omega u_\omega (y^+). \end{aligned}$$

(10.86)

Therefore, we only require to have

$$\begin{aligned} \int _\omega f(\omega )\omega u_\omega (x^+) = \tilde{h}(x^+)\Theta (x^+ - x_i^+) =\zeta (x^+), \end{aligned}$$

(10.87)

where we have absorbed the factor i in the redefinition of $\tilde{h}(x^+)$. Taking the inner product of the Eq. (10.87) with itself and using the orthonormal properties of the mode functions, we write

$$\begin{aligned} \int _\omega \omega ^2 |f(\omega )|^2 = (\zeta (x^+),\zeta (x^+)). \end{aligned}$$

(10.88)

Fo r the states satisfying Eq. (10.88), the expression Eq. (10.87) can be inverted for given $\zeta (x^+)$, using the completeness of mode functions to obtain a consistent state. Therefore, the one particle states respecting Eq. (10.88) will have mode excitations beyond $x_i^+$.

A.4 Information Retrieval for the CGHS Black Hole

Using a particular representation of the Gamma function

$$\begin{aligned} \Gamma [z] = i^z \int _{-\infty }^{\infty } d q e^{z q} e^{- i e^q}, \end{aligned}$$

(10.89)

we can write down a product formula

$$\begin{aligned} \Gamma [i(\bar{\omega }-\bar{\omega }')]\Gamma [-i(\bar{\omega }-\bar{\omega }'')]&=e^{-\pi \bar{\omega }} e^{\frac{\pi }{2}(\bar{\omega }'+\bar{\omega }'')} \nonumber \\&\times \int _{-\infty }^{\infty }d q_1 d q_2 e^{i\bar{\omega }(q_{1}-q_{2})} e^{i(\bar{\omega }'q_1-\bar{\omega }''q_2)}e^{- i e^{q_1} + i e^{q_2}}, \end{aligned}$$

(10.90)

which appears in the spectrum operator expression. The correction in the vacuum thermal radiation as received by asymptotic left moving observer can be expressed as

$$\begin{aligned} 2\pi \lambda \sinh {\pi \bar{\omega }} N_{\bar{\omega }}&=\int _0^{\infty }\int _0^{\infty } \frac{d\bar{\omega }'}{\bar{\omega }'}\frac{d\bar{\omega }''}{\bar{\omega }''}\text {Sym}_{\bar{\omega }}\left[ \frac{\Gamma [i(\bar{\omega }-\bar{\omega }')]\Gamma [-i(\bar{\omega }-\bar{\omega }'')]}{\Gamma [-i\bar{\omega }']\Gamma [i\bar{\omega }'']} \right] \nonumber \\&\times \left| y_i^{+}\right| ^{i(\bar{\omega }'-\bar{\omega }'')}f(\bar{\omega }')f^{*}(\bar{\omega }''), \end{aligned}$$

(10.91)

where $\text {Sym}_{x}[f(x)] =(f(x)+f(-x))/2$ with $\bar{\omega '} =\omega '/\lambda$. Using Eq. (10.53), the spectral distortion can be re-written as,

$$\begin{aligned} 2\pi \lambda \sinh {\pi \bar{\omega }} N_{\bar{\omega }}&=\int _0^{\infty } d\bar{\omega }'d\bar{\omega }''\int _{-\infty }^{\infty }d q_1 d q_2\text {Sym}_{\bar{\omega }}[e^{-\pi \bar{\omega }} e^{i\bar{\omega }(q_{1}-q_{2})}] \nonumber \\&\times e^{i(\bar{\omega }'q_1-\bar{\omega }''q_2)}e^{- i e^{q_1} + i e^{q_2}}g(\omega ')g^{*}(\omega ''). \end{aligned}$$

(10.92)

Using, $g(\bar{\omega })$, introduce yet another function

$$\begin{aligned} \chi (q)=e^{-i e^q}\int _0^{\infty } d\bar{\omega } e^{-i \bar{\omega } q} g(\bar{\omega }), \end{aligned}$$

(10.93)

to express the spectral distortion as

$$\begin{aligned} 2\pi \lambda \sinh {\pi \bar{\omega }} N_{\bar{\omega }}&= \int _{-\infty }^{\infty }d q_1 d q_2\text {Sym}_{\bar{\omega }}[e^{-\pi \bar{\omega }} e^{i\bar{\omega }(q_{1}-q_{2})}] \chi (q_1)\chi ^*(q_2).\end{aligned}$$

(10.94)

$$\begin{aligned} 4\pi \lambda \sinh {\pi \bar{\omega }} N_{\bar{\omega }}&=\int _{-\infty }^{\infty }d q_1 d q_2[e^{-\pi \bar{\omega }} e^{i\bar{\omega }(q_{1}-q_{2})} + e^{\pi \bar{\omega }} e^{-i\bar{\omega }(q_{1}-q_{2})}] \chi (q_1)\chi ^*(q_2), \end{aligned}$$

(10.95)

which simply gives

$$\begin{aligned} 4\pi \lambda \sinh {\pi \bar{\omega }} N_{\bar{\omega }}=e^{\pi \bar{\omega }} |\mathcal{F}_{\chi }(\bar{\omega })|^2 + e^{-\pi \bar{\omega }} |\mathcal{F}_{\chi }(-\bar{\omega })|^2, \end{aligned}$$

(10.96)

with $\mathcal{F}_{\chi }(\bar{\omega })$ being the Fourier transform of $\chi (q)$ w.r.t. $\bar{\omega }$

$$\begin{aligned} \mathcal{F}_{\chi }(\bar{\omega }) = \int _{-\infty }^{\infty } d q e^{-i \bar{\omega } q} \chi (q), \end{aligned}$$

(10.97)

which gives Eq. (10.96) as the analogue of the Eq. (10.63) for the spherical symmetric collapse. Therefore, we can follow the same steps as outlined in Sect. 10.3 and Sect. 10.4 to recover informations regarding $\mathcal{F}_{\chi }(\bar{\omega })$. Using the inverse transformations Eqs. (10.97), (10.93) and (10.53) we can recover the information regarding the field state $f(\bar{\omega })$ using the moments of $\mathcal{F}_{\chi }(\bar{\omega })$.