Open quantum system approach to the Unruh-DeWitt detector in impulsive plane wave spacetimes

In the UD detector theory one is interested in obtaining the transition probabilities between various energy levels of $Q$ under the influence of the quantum field $\Phi$ in the corresponding background spacetime. To achieve that we study the reduced density matrix $\rho_{R}$ which describes the quantum state of $Q$ after integrating out the influence of the quantum field. The time evolution of $\rho_{R}$ is specified by

$\displaystyle\rho_{R}(Q_{+f},Q_{-f},\tau_{f})=\int_{-\infty}^{\infty}dQ_{+i}\int_{-\infty}^{\infty}dQ_{-i}\ {\cal J}_{R}(Q_{+f},Q_{-f},\tau_{f}|Q_{+i},Q_{-i},\tau_{i})\ \rho_{R}(Q_{+i},Q_{-i},\tau_{i})$

(5)

where ${\cal J}_{R}$ is the evolution operator of the reduced density matrix. In the open quantum system approach, in particular via the path-integral formalism, this evolution operator can be expressed as HPZ92; HM94

$\displaystyle{\cal J}_{R}(Q_{+f},Q_{-f},\tau_{f}|Q_{+i},Q_{-i},\tau_{i})=\int_{Q_{\pm i}}^{Q_{\pm f}}DQ_{\pm}\int_{CTP}D\Phi_{\pm}\,e^{iS[Q_{+},\Phi_{+}]-iS[Q_{-},\Phi_{-}]}$

(6)

where $CTP$ represents the so-called “closed-time-path” boundary condition for the path integrals over $\Phi_{\pm}$. Note that we have assumed that initially the detector and the field are uncorrelated to arrive at the expression in Eq. (6). Since the field $\Phi$ is at most quadratic in the action $S[Q,\Phi]$, the CTP path integrals can be evaluated exactly. The result is known as the influence action $S_{IF}$,

	$\displaystyle e^{iS_{IF}[\Delta,\Sigma]}$	$\displaystyle=$	$\displaystyle\int_{CTP}D\Phi_{\pm}\,e^{i(S_{F}[Q_{+}]+S_{I}[Q_{+},\Phi_{+}])-i(S_{F}[Q_{-}]+S_{I}[Q_{+},\Phi_{+}])}$		(7)
		$\displaystyle=$	$\displaystyle e^{i\lambda^{2}\int_{\tau_{i}}^{\tau_{f}}ds\int_{\tau_{i}}^{\tau_{f}}ds^{\prime}\,\Delta(s)D(s,s^{\prime})\Sigma(s^{\prime})-\frac{1}{2}\lambda^{2}\int_{\tau_{i}}^{\tau_{f}}ds\int_{\tau_{i}}^{\tau_{f}}ds^{\prime}\,\Delta(s)N(s,s^{\prime})\Delta(s^{\prime})}$

where $\Delta=Q_{+}-Q_{-}$ and $\Sigma=(Q_{+}+Q_{-})/2$. Here $D(s,s^{\prime})$ is called the dissipation kernel which is given by the retarded Green function of the field $\Phi$ in the corresponding spacetime,

$\displaystyle D(s,s^{\prime})=G_{ret}(z(s),z(s^{\prime}))=i\,\theta(z^{0}(s)-z^{0}(s^{\prime}))\left\langle[\Phi(z(s)),\Phi(z(s^{\prime}))]\right\rangle$

(8)

and $N(s,s^{\prime})$ is the noise kernel given by the Hadamard function of the field,

$\displaystyle N(s,s^{\prime})=G^{(1)}(z(s),z(s^{\prime}))=\frac{1}{2}\left\langle\{\Phi(z(s)),\Phi(z(s^{\prime}))\}\right\rangle$

(9)

Recall that $z(s)$ is the prescribed trajectory of the detector.

With the result of the influence action, the evolution operator can be written as

	$\displaystyle{\cal J}_{R}(Q_{+f},Q_{-f},\tau_{f}|Q_{+i},Q_{-i},\tau_{i})$	$\displaystyle=$	$\displaystyle\int_{Q_{\pm i}}^{Q_{\pm f}}DQ_{\pm}\,e^{i(S_{D}[Q_{+}]-S_{D}[Q_{-}]+S_{IF}[Q_{+},Q_{-}])}$		(10)
		$\displaystyle=$	$\displaystyle\int_{\Sigma_{i}}^{\Sigma_{f}}D\Sigma\int_{\Delta_{i}}^{\Delta_{f}}D\Delta\ e^{iS_{eff}[\Sigma,\Delta]}$

where $S_{eff}$ is the effective action

$\displaystyle S_{eff}[\Sigma,\Delta]=\int_{\tau_{i}}^{\tau_{f}}d\tau\,m\,[\partial_{\tau}\Sigma(\tau)\partial_{\tau}\Delta(\tau)-\omega_{0}^{2}\,\Sigma(\tau)\Delta(\tau)]+S_{IF}[\Delta,\Sigma]$

(11)

To evaluate the path integrals over $\Delta$ and $\Sigma$, we deal with their corresponding boundary conditions by splitting the functions into HM94

$\displaystyle\Sigma(s)=\Sigma_{cl}(s)+\sigma(s)\qquad;\qquad\Delta(s)=\Delta_{cl}(s)+\delta(s)$

(12)

where $\Sigma_{cl}$ and $\Delta_{cl}$ are solutions to the classical equations of motion,

	$\displaystyle\partial_{s}^{2}\Sigma_{cl}(s)+\omega_{0}^{2}\,\Sigma_{cl}(s)-\left(\frac{\lambda^{2}}{m}\right)\int_{\tau_{i}}^{\tau_{f}}ds^{\prime}\,D(z(s),z(s^{\prime}))\,\Sigma_{cl}(s^{\prime})=0$		(13)
	$\displaystyle\partial_{s}^{2}\Delta_{cl}(s)+\omega_{0}^{2}\,\Delta_{cl}(s)-\left(\frac{\lambda^{2}}{m}\right)\int_{\tau_{i}}^{\tau_{f}}ds^{\prime}\,\Delta_{cl}(s^{\prime})\,D(z(s),z(s^{\prime}))=0$		(14)

with the boundary conditions

	$\displaystyle\Sigma_{cl}(\tau_{i})=\Sigma_{i}\qquad$		$\displaystyle;\qquad\Delta_{cl}(\tau_{i})=\Delta_{i}$
	$\displaystyle\Sigma_{cl}(\tau_{f})=\Sigma_{f}\qquad$		$\displaystyle;\qquad\Delta_{cl}(\tau_{f})=\Delta_{f}$		(15)

On the other hand, $\sigma(s)$ and $\delta(s)$ are arbitrary functions with boundary conditions

$\displaystyle\sigma(\tau_{i})=\delta(\tau_{i})=0\qquad;\qquad\sigma(\tau_{f})=\delta(\tau_{f})=0$

(16)

With this splitting, the path integrals of the evolution operator in Eq. (10) can be further separated as

			$\displaystyle{\cal J}_{R}(Q_{+f},Q_{-f},\tau_{f}|Q_{+i},Q_{-i},\tau_{i})$		(17)
		$\displaystyle=$	$\displaystyle e^{iS_{eff}[\Sigma_{cl},\Delta_{cl}]}\int\,D\sigma\int\,D\delta\ e^{iS_{eff}[\sigma,\delta]-\lambda^{2}\int_{\tau_{i}}^{\tau_{f}}ds\int_{\tau_{i}}^{\tau_{f}}ds^{\prime}\,\Delta_{cl}(s)N(s,s^{\prime})\,\delta(s^{\prime})}$

where we have made use of the equations of motion in Eqs. (13) and (14).

First, we consider the prefactor involving $S_{eff}[\Sigma_{cl},\Delta_{cl}]$. Suppose that $u_{1}$ and $u_{2}$ are solutions to Eq. (13) with boundary conditions

$\displaystyle u_{1}(\tau_{i})=1=u_{2}(\tau_{f})\qquad;\qquad u_{1}(\tau_{f})=0=u_{2}(\tau_{i}),$

(18)

while $v_{1}$ and $v_{2}$ are solutions to Eq. (14) with boundary conditions

$\displaystyle v_{1}(\tau_{i})=1=v_{2}(\tau_{f})\qquad;\qquad v_{1}(\tau_{f})=0=v_{2}(\tau_{i})$

(19)

Using these two sets of fundamental solutions, we can write

	$\displaystyle\Sigma_{cl}(s)=\Sigma_{i}\,u_{1}(s)+\Sigma_{f}\,u_{2}(s)$		(20)
	$\displaystyle\Delta_{cl}(s)=\Delta_{i}v_{1}(s)+\Delta_{f}\,v_{2}(s)$		(21)

Putting these into $S_{eff}[\Sigma_{cl},\Delta_{cl}]$, the prefactor in Eq. (17) is simplified to

$\displaystyle e^{iS_{eff}[\Sigma_{cl},\Delta_{cl}]}=e^{ib_{1}\Sigma_{f}\Delta_{f}-ib_{2}\Sigma_{f}\Delta_{i}+ib_{3}\Sigma_{i}\Delta_{f}-ib_{4}\Sigma_{i}\Delta_{i}}\,e^{-a_{11}\Delta_{i}^{2}-a_{12}\Delta_{i}\Delta_{f}-a_{22}\Delta_{f}^{2}}$

(22)

where

	$\displaystyle b_{1}=m\left[\partial_{s}u_{2}\right]_{s=\tau_{f}}\qquad;\qquad b_{2}=m\left[\partial_{s}u_{2}\right]_{s=\tau_{i}}$
	$\displaystyle b_{3}=m\left[\partial_{s}u_{1}\right]_{s=\tau_{f}}\qquad;\qquad b_{4}=m\left[\partial_{s}u_{1}\right]_{s=\tau_{i}}$		(23)
	$\displaystyle a_{ij}=\frac{\lambda^{2}}{1+\delta_{ij}}\int_{\tau_{i}}^{\tau_{f}}ds\int_{\tau_{i}}^{\tau_{f}}ds^{\prime}\,v_{i}(s)N(s,s^{\prime})v_{j}(s^{\prime})$		(24)

In evaluating the path integrals in Eq. (17), we note that the function $\sigma(s)$ is linear in the effective action $S_{eff}[\sigma,\delta]$. Hence, the result of doing the path integral over $\sigma(s)$ gives a functional delta-function which basically enforces $\delta(s)$ to obey the equation of motion in Eq. (14). However, the boundary conditions for $\delta(s)$ are $\delta(\tau_{i})=\delta(\tau_{f})=0$, the only allowed solution would be $\delta(s)=0$. As a result, the path integrals over $\sigma(s)$ and $\delta(s)$ give a function of $\tau_{i}$ and $\tau_{f}$, ${\cal Z}[\tau_{t},\tau_{i}]$, which is independent of $\Delta_{cl}$.

One can obtain ${\cal Z}$ by imposing the normalization of the reduced density matrix, ${\rm Tr}(\rho_{R})=1$. From Eq. (5), we can see that this normalization condition requires the evolution operator to satisfy

	$\displaystyle\int_{-\infty}^{\infty}d\Sigma_{f}\,{\cal J}_{R}(\Sigma_{f},0,t_{f}|\Sigma_{i},\Delta_{i},t_{i})=\delta(\Delta_{i})$	$\displaystyle\Rightarrow$	$\displaystyle{\cal Z}\int_{\infty}^{\infty}d\Sigma_{f}\ e^{-ib_{2}\Sigma_{f}\Delta_{i}-ib_{4}\Sigma_{i}\Delta_{i}-a_{11}\Delta_{i}^{2}}=\delta(\Delta_{i})$		(25)
		$\displaystyle\Rightarrow$	$\displaystyle{\cal Z}=\frac{b_{2}}{2\pi}$

We therefore arrive at the final result in this derivation for the evolution operator

			$\displaystyle{\cal J}_{R}(Q_{+f},Q_{-f},\tau_{f}|Q_{+i},Q_{-i},\tau_{i})$		(26)
		$\displaystyle=$	$\displaystyle\left(\frac{b_{2}}{2\pi}\right)e^{ib_{1}\Sigma_{f}\Delta_{f}-ib_{2}\Sigma_{f}\Delta_{i}+ib_{3}\Sigma_{i}\Delta_{f}-ib_{4}\Sigma_{i}\Delta_{i}}\,e^{-a_{11}\Delta_{i}^{2}-a_{12}\Delta_{i}\Delta_{f}-a_{22}\Delta_{f}^{2}}$

Now we are in a position to evaluate the reduced density matrix at any time $\tau$. Suppose that initially $Q$ is in the ground state. Then its density matrix is given by

$\displaystyle\bar{\rho}_{R}^{(0)}(\Sigma_{i},\Delta_{i})=\sqrt{\frac{m\omega}{\pi}}\,e^{-\frac{m\omega}{4}\left(4\Sigma_{i}^{2}+\Delta_{i}^{2}\right)}$

(27)

where we have used an overbar to indicate the time-independent density matrix of an harmonic oscillator. The density matrix at time $\tau_{f}$ can be obtained from the evolution equation in Eq. (5) with the form of the evolution operator in Eq. (26) by integrating over $\Sigma_{i}$ and $\Delta_{i}$. Since it is at most quadratic in these variables up in the exponential, the integrals can be done exactly. Hence, the final density matrix can be expressed as

$\displaystyle\rho_{R}^{(0)}(\Sigma_{f},\Delta_{f},\tau_{f})={\cal N}\,e^{-{\cal A}\,\Delta_{f}^{2}-i\,2\,{\cal B}\,\Delta_{f}\Sigma_{f}-{\cal C}\,\Sigma_{f}^{2}}$

(28)

where

	$\displaystyle{\cal N}$	$\displaystyle=$	$\displaystyle\frac{b_{2}}{2\sqrt{\pi}}\left(\frac{b_{4}^{2}}{4m\omega}+\frac{m\omega}{4}+a_{11}\right)^{-1/2}$
	$\displaystyle{\cal A}$	$\displaystyle=$	$\displaystyle-\frac{1}{4}\left(\frac{b_{4}^{2}}{4m\omega}+\frac{m\omega}{4}+a_{11}\right)^{-1}\left(\frac{b_{3}b_{4}}{2m\omega}-a_{12}\right)^{2}+\frac{b_{3}^{2}}{4m\omega}+a_{22}$
	$\displaystyle{\cal B}$	$\displaystyle=$	$\displaystyle-\frac{b_{1}}{2}+\frac{b_{2}}{4}\left(\frac{b_{4}^{2}}{4m\omega}+\frac{m\omega}{4}+a_{11}\right)^{-1}\left(\frac{b_{3}b_{4}}{2m\omega}-a_{12}\right)$
	$\displaystyle{\cal C}$	$\displaystyle=$	$\displaystyle\frac{b_{2}^{2}}{4}\left(\frac{b_{4}^{2}}{4m\omega}+\frac{m\omega}{4}+a_{11}\right)^{-1}$		(29)

This density matrix will be crucial in our subsequent derivations of the expectation values of functions of $Q$ and its momentum $P$ with respect to time. Then, the corresponding transition probabilities from the ground state to the excited states will be expressed in terms of these expectation values.

From the density matrix $\rho_{R}$ in Eq. (28), one can evaluate the expectation value of an operator ${\cal O}$ in the quantum state which is initially in the ground state (Eq. (27)) by taking the trace ${\rm Tr}({\cal O}\rho_{R}^{(0)})$. Since the density matrix is gaussian, it is obvious that the expectation values $\langle Q\rangle$ and $\langle P\rangle$ vanish. On the other hand,

	$\displaystyle\langle Q^{2}\rangle_{f}^{(0)}$	$\displaystyle=$	$\displaystyle{\rm Tr}(Q^{2}\rho_{R}^{(0)}(\tau_{f}))$		(30)
		$\displaystyle=$	$\displaystyle\int_{-\infty}^{\infty}d\Sigma_{f}\,\Sigma_{f}^{2}\,\rho_{R}^{(0)}(\Sigma_{f},0,\tau_{f})$
		$\displaystyle=$	$\displaystyle\frac{1}{2{\cal C}}$
		$\displaystyle=$	$\displaystyle\frac{b_{4}^{2}}{2m\omega b_{2}^{2}}+\frac{m\omega}{2b_{2}^{2}}+\left(\frac{2}{b_{2}^{2}}\right)a_{11}$

The subscript $f$ shows that the expectation value is evaluated at time $\tau_{f}$ and the superscript $(0)$ indicates that the quantum state at the initial time $\tau_{i}$ is in the ground state.

In a similar fashion the expectation values $\langle P^{2}\rangle_{f}^{(0)}$ and $\langle\{Q,P\}\rangle_{f}^{(0)}$ are given by

			$\displaystyle\langle P^{2}\rangle_{f}^{(0)}$		(31)
		$\displaystyle=$	$\displaystyle{\rm Tr}(P^{2}\rho_{R}^{(0)}(\tau_{f}))$
		$\displaystyle=$	$\displaystyle\int_{-\infty}^{\infty}d\Sigma_{f}\int_{-\infty}^{\infty}d\Delta_{f}\left(\frac{\partial^{2}}{\partial\Delta^{2}_{f}}\delta(\Delta_{f})\right)\rho_{R}^{(0)}(\Sigma_{f},\Delta_{f},\tau_{f})$
		$\displaystyle=$	$\displaystyle 2\left({\cal A}+\frac{{\cal B}^{2}}{{\cal C}}\right)$
		$\displaystyle=$	$\displaystyle\left(\frac{1}{2m\omega b_{2}^{2}}\right)(b_{1}b_{4}-b_{2}b_{3})^{2}+\frac{m\omega b_{1}^{2}}{2b_{2}^{2}}+\left(\frac{2b_{1}^{2}}{b_{2}^{2}}\right)a_{11}+\left(\frac{2b_{1}}{b_{2}}\right)a_{12}+2a_{22}$

and

	$\displaystyle\langle\{Q,P\}\rangle_{f}^{(0)}$	$\displaystyle=$	$\displaystyle{\rm Tr}(\{Q,P\}\,\rho_{R}^{(0)}(\tau_{f}))$		(32)
		$\displaystyle=$	$\displaystyle(-i)\int_{-\infty}^{\infty}d\Sigma_{f}\int_{-\infty}^{\infty}d\Delta_{f}\left(-\frac{\partial}{\partial\Delta_{f}}\delta(\Delta_{f})\right)\left(\Sigma_{f}+\frac{\Delta_{f}}{2}\right)\rho_{R}^{(0)}(\Sigma_{f},\Delta_{f},\tau_{f})$
			$\displaystyle\ \ +(-i)\int_{-\infty}^{\infty}d\Sigma_{f}\int_{-\infty}^{\infty}d\Delta_{f}\left(-\frac{\partial}{\partial\Delta_{f}}\delta(\Delta_{f})\right)\left(\Sigma_{f}-\frac{\Delta_{f}}{2}\right)\rho_{R}^{(0)}(\Sigma_{f},\Delta_{f},\tau_{f})$
		$\displaystyle=$	$\displaystyle-\frac{2{\cal B}}{{\cal C}}$
		$\displaystyle=$	$\displaystyle\left(\frac{b_{4}}{m\omega b_{2}^{2}}\right)(b_{1}b_{4}-b_{2}b_{3})+\frac{m\omega b_{1}}{b_{2}^{2}}+\left(\frac{4b_{1}}{b_{2}^{2}}\right)a_{11}+\left(\frac{2}{b_{2}}\right)a_{12}$

The transition probabilities of the detector from one quantum state to the other can also be formulated using the density matrices. Suppose that the detector is in its ground state initially. It is described at the initial time $\tau_{i}$ by the density matrix $\bar{\rho}_{R}^{(0)}$ as given by Eq. (27). At time $\tau_{f}$, the quantum state of the detector evolves to $\rho_{R}^{(0)}(\tau_{f})$ in Eq. (28). The probability that the detector will be, for example, in the first excited state at time $\tau_{f}$ is given by ${\rm Tr}(\,\bar{\rho}_{R}^{(1)}\rho_{R}^{(0)}(\tau_{f}))$, where $\bar{\rho}_{R}^{(1)}$ is the density matrix of the first excited state of a harmonic oscillator as given by

$\displaystyle\bar{\rho}_{R}^{(1)}(\Sigma_{f},\Delta_{f})=\sqrt{\frac{m^{3}\omega^{3}}{4\pi}}\,(4\Sigma_{f}^{2}-\Delta_{f}^{2})\,e^{-\frac{m\omega}{4}(4\Sigma_{f}^{2}+\Delta_{f}^{2})}$

(33)

Therefore, this transition probability can be worked out exactly as

	$\displaystyle P_{0\rightarrow 1}(\tau_{f})$	$\displaystyle=$	$\displaystyle\int_{-\infty}^{\infty}d\Sigma_{f}\int_{-\infty}^{\infty}d\Delta_{f}\ \bar{\rho}_{R}^{(1)}(\Sigma_{f},\Delta_{f})\ \rho_{R}^{(0)}(\Sigma_{f},\Delta_{f},\tau_{f})$		(34)
		$\displaystyle=$	$\displaystyle\frac{2\,\sqrt{m^{3}\omega^{3}\,{\cal C}}\ (4{\cal A}-{\cal C})}{[(4{\cal A}+m\omega)({\cal C}+m\omega)+4{\cal B}^{2}]^{3/2}}$
		$\displaystyle=$	$\displaystyle\left[\langle Q^{2}\rangle_{f}^{(0)}\langle P^{2}\rangle_{f}^{(0)}-\frac{1}{4}[\langle\{Q,P\}\rangle_{f}^{(0)}]^{2}-\frac{1}{4}\right]\times$
			$\displaystyle\ \ \left\{\left[\langle P^{2}\rangle_{f}^{(0)}+\frac{m\omega}{2}\right]\left[\langle Q^{2}\rangle_{f}^{(0)}+\frac{1}{2m\omega}\right]-\frac{1}{4}\left[\langle\{Q,P\}\rangle_{f}^{(0)}\right]^{2}\right\}^{-3/2}$

This is the expression we shall use in our subsequent section in the evaluation of the transition probabilities of the UD detector from the ground state to the first excited state under the influence of the impulsive plane wave.

The quantum field theory in the impulsive plane wave spacetimes have been considered in Klimcik88; GV91. For a massless scalar field $\Phi$, the corresponding Klein-Gordon equation in this spacetime can be written as

$\displaystyle\left[-2\,\frac{\partial}{\partial u}\frac{\partial}{\partial v}+\sum_{a=1,2}\frac{\partial}{\partial x^{a}}\frac{\partial}{\partial x^{a}}-\sum_{a=1,2}\eta_{a}\,\delta(u)(x^{a})^{2}\frac{\partial^{2}}{\partial v^{2}}\right]\Phi=0$

(36)

The mode functions are defined as follows Klimcik88; GV91; Cho23. For the so-called in-modes, the mode function is that of a Minkowski mode function

$\displaystyle\Phi_{k_{-}\,\vec{k}}^{\rm in}(z)=N_{k_{-}}e^{-ik_{-}v}e^{-\frac{iu}{2k_{-}}}e^{i\vec{k}\cdot\vec{x}}$

(37)

for $u<0$. Here the normalization constant $N_{k_{-}}=[(2\pi)^{3}(2k_{-})]^{-1/2}$ and $z^{\mu}=(u,v,\vec{x})$ is a general spacetime point. For $u>0$, due to the delta function term in Eq. (36), we have the extra factor

$\displaystyle\Phi_{k_{-}\,\vec{k}}^{\rm in}(z)=N_{k_{-}}e^{-ik_{-}v}e^{i\vec{k}\cdot\vec{x}}\int\frac{d^{2}x^{\prime}\,d^{2}k^{\prime}}{(2\pi)^{2}}e^{-i(\vec{k}-\vec{k}^{\prime})\cdot(\vec{x}-\vec{x}^{\prime})}e^{-\frac{iu}{2k_{-}}}e^{\frac{i}{2}k_{-}\sum_{a=1,2}\eta_{a}(x^{a})^{2}}$

(38)

Similarly, one can also define the out-mode $\Phi_{k_{-}\,\vec{k}}^{\rm out}(z)$ as having the Minkowski form for $u>0$ and for $u<0$

$\displaystyle\Phi_{k_{-}\,\vec{k}}^{\rm out}(z)=N_{k_{-}}e^{-ik_{-}v}e^{i\vec{k}\cdot\vec{x}}\int\frac{d^{2}x^{\prime}\,d^{2}k^{\prime}}{(2\pi)^{2}}e^{-i(\vec{k}-\vec{k}^{\prime})\cdot(\vec{x}-\vec{x}^{\prime})}e^{-\frac{iu}{2k_{-}}}e^{-\frac{i}{2}k_{-}\sum_{a=1,2}\eta_{a}(x^{a})^{2}}$

(39)

Since we assume that the detector interacts with the quantum modes in the Minkowski vacuum before encountering the impulsive wave, we shall use the in-modes to construct the various two-point functions.

First, we look at the Wightman function from the in-modes,

$\displaystyle G_{+}(z,z^{\prime})=\int_{0}^{\infty}dk_{-}\int d^{2}k\ \Phi_{k_{-}\,\vec{k}}^{\rm in}(z)\Phi_{k_{-}\,\vec{k}}^{\rm in\,*}(z^{\prime})$

(40)

One can find a detailed derivation of the Wightman function in Cho23. We just quote the result here. For a massless scalar field,

$\displaystyle G_{+}(z,z^{\prime})=\frac{\sqrt{\Delta(u,u^{\prime})}}{8\pi^{2}}\left[\frac{1}{\sigma(z,z^{\prime})+i(u-u^{\prime})\epsilon}\right]$

(41)

where we have used the $i\epsilon$ prescription. Here we have introduced two biscalars $\Delta(u,u^{\prime})$ and $\sigma(z,z^{\prime})$. The world function $\sigma(z,z^{\prime})$ is related to the distance between two points $u$ and $u^{\prime}$ along a geodesic, while the van Vleck determinant $\Delta(u,u^{\prime})$ describes the focusing of geodesic flows. By analyzing these biscalars, one can have a better understanding of the properties of the spacetime.

For both $0>u>u^{\prime}$ and $u>u^{\prime}>0$, the world function is in the Minkowski form

$\displaystyle\sigma(z,z^{\prime})=-(u-u^{\prime})(v-v^{\prime})+\frac{1}{2}\sum_{a=1,2}(x^{a}-x^{\prime a})^{2}$

(42)

For $u>0$ and $u^{\prime}<0$, we have Cho23

			$\displaystyle\sigma(z,z^{\prime})$		(43)
		$\displaystyle=$	$\displaystyle\frac{1}{2}(u-u^{\prime})\Big\{-2(v-v^{\prime})$
			$\displaystyle\ \ +\sum_{a=1,2}(u-u^{\prime}-uu^{\prime}\eta_{a})^{-1}\left[(1-u^{\prime}\eta_{a})(x^{a})^{2}+(1+u\eta_{a})(x^{\prime a})^{2}-2x^{a}x^{\prime a}\right]\Big\}$

For the van Vleck determinant, we have for $0>u>u^{\prime}$ and $u>u^{\prime}>0$,

$\displaystyle\Delta(u,u)=1$

(44)

and for $u>0$ and $u^{\prime}<0$,

$\displaystyle\Delta(u,u^{\prime})=(u-u^{\prime})^{2}(u-u^{\prime}-uu^{\prime}\eta_{1})^{-1}(u-u^{\prime}-uu^{\prime}\eta_{2})^{-1}$

(45)

For the trajectory of the detector specified by $z^{\mu}=(u,u/2,\vec{0})$, the Wightman function in Eq. (41) can be simplified to

	$\displaystyle G_{+}(u,u^{\prime})$	$\displaystyle=$	$\displaystyle\frac{\sqrt{\Delta(u,u^{\prime})}}{8\pi^{2}}\left[\frac{1}{-(u-u^{\prime})^{2}/2+i(u-u^{\prime})\epsilon}\right]$		(46)
		$\displaystyle=$	$\displaystyle\sqrt{\Delta(u,u^{\prime})}\ G_{+}^{M}(u,u^{\prime})$

where $G_{+}^{M}(u,u^{\prime})$ is the Wightman function for the same trajectory in the Minkowski spacetime. To proceed we can choose the most convenient representation of the Minkowski Wightman function for our purpose. As we shall see, the best choice is

$\displaystyle G_{+\Lambda}^{M}(u,u^{\prime})=\frac{1}{4\pi^{2}}\int_{0}^{\Lambda}\,dk\,k\,e^{-ik(u-u^{\prime})}$

(47)

so that the resulting integrals will be in much simpler forms. Here we have introduced a momentum cutoff $\Lambda$ to regularize the $k$ integral, where $\Lambda$ will be treated as the largest scale in the problem.

In the influence action in Eq. (7), there are two more two-point functions we need to consider. The first one is the dissipation kernel defined by Eq. (8). This is necessary for the study of the classical equations of motion in Eqs. (13) and (14). Using the representation of the Wightman function above, we can write

	$\displaystyle D_{\Lambda}(u,u^{\prime})$	$\displaystyle=$	$\displaystyle i\,\theta(u-u^{\prime})\sqrt{\Delta(u,u^{\prime})}\left[G_{+\Lambda}^{M}(u,u^{\prime})-G_{+\Lambda}^{M}(u^{\prime},u)\right]$		(48)
		$\displaystyle=$	$\displaystyle\theta(u-u^{\prime})\frac{\sqrt{\Delta(u,u^{\prime})}}{2\pi^{2}}\int_{0}^{\Lambda}\,dk\,k\,\sin[k(u-u^{\prime})]$

The other two-point function is the noise kernel given by the Hadamard function

	$\displaystyle N_{\Lambda}(u,u^{\prime})$	$\displaystyle=$	$\displaystyle\frac{1}{2}\left[G_{+\Lambda}^{M}(u,u^{\prime})+G_{+\Lambda}^{M}(u^{\prime},u)\right]$		(49)
		$\displaystyle=$	$\displaystyle\frac{\sqrt{\Delta(u,u^{\prime})}}{4\pi^{2}}\int_{0}^{\Lambda}\,dk\,k\,\cos[k(u-u^{\prime})]$

As we have seen in Sec. II, the evaluation of the expectation values and transition probabilities relies crucially on the solutions $u_{1}$, $u_{2}$, $v_{1}$, and $v_{2}$ to the equations of motion in Eqs. (13) and (14). Hence, we need to study these equations more closely. We treat these equations as a initial value problem. Since the impulsive plane wave spacetime possesses no Cauchy surface, one can regard the constant $u$ surfaces as a substitute and use $u$ as the affine parameter for evolution.

Take $u_{i}$ as the initial parameter and $u$ as the final one. For $0>u>s>u_{i}$ and $u>s>u_{i}>0$, the van Vleck determinant $\Delta(s,s^{\prime})=1$. Using the form of the dissipation kernel in Eq. (48), the equation of motion in Eq. (13) can be expressed as

			$\displaystyle\partial_{s}^{2}f(s)+\omega_{0}^{2}\,f(s)-\left(\frac{\lambda^{2}}{m}\right)\int_{u_{i}}^{u}ds^{\prime}\,D_{\Lambda}(s,s^{\prime})\,f(s^{\prime})=0$		(50)
		$\displaystyle\Rightarrow$	$\displaystyle\partial_{s}^{2}f(s)+\omega_{0}^{2}\,f(s)-\left(\frac{\lambda^{2}}{2\pi^{2}m}\right)\int_{u_{i}}^{s}ds^{\prime}\,f(s^{\prime})\int_{0}^{\Lambda}dk\,k\,\sin[k(s-s^{\prime})]=0$

If we take the integration over $k$, the result will be

$\displaystyle\int_{0}^{\Lambda}dk\,k\,\sin[k(s-s^{\prime})]=(s-s^{\prime})^{-2}\,\Big(\sin[\Lambda(s-s^{\prime})]-\Lambda(s-s^{\prime})\cos[\Lambda(s-s^{\prime})]\Big)$

(51)

which is a function that concentrates around $s^{\prime}=s$ for large $\Lambda$. Therefore, to analyze Eq. (50), we can Taylor-expand the function $f(s^{\prime})$ around $s^{\prime}=s$.

$\displaystyle f(s^{\prime})=f(s)-(s-s^{\prime})\,f^{\prime}(s)+\frac{1}{2}(s-s^{\prime})^{2}f^{\prime\prime}(s)+\cdots$

(52)

and the last term in Eq. (50) becomes

			$\displaystyle-\left(\frac{\lambda^{2}}{2\pi^{2}m}\right)\int_{0}^{\Lambda}dk\,k\,\int_{u_{i}}^{s}ds^{\prime}\,\sin[k(s-s^{\prime})]\left[f(s)-(s-s^{\prime})\,f^{\prime}(s)+\frac{1}{2}(s-s^{\prime})^{2}f^{\prime\prime}(s)+\cdots\right]$		(53)
		$\displaystyle=$	$\displaystyle-\left(\frac{\lambda^{2}}{2\pi^{2}m}\right)\left[\Lambda\,f(s)-\frac{\pi}{2}f^{\prime}(s)+\cdots\right]$

where the ellipsis represent terms which are of the order of $1/\Lambda$ or higher. Putting this result into Eq. (50), we have

			$\displaystyle\partial_{s}^{2}f(s)+\omega_{0}^{2}\,f(s)-\left(\frac{\lambda^{2}}{2\pi^{2}m}\right)\left[\Lambda\,f(s)+\frac{\pi}{2}f^{\prime}(s)+\cdots\right]=0$		(54)
		$\displaystyle\Rightarrow$	$\displaystyle\partial_{s}^{2}f(s)+2\gamma\,f^{\prime}(s)+\omega^{2}f(s)=0$

where

$\displaystyle\omega^{2}=\omega_{0}^{2}-\frac{\lambda^{2}\Lambda}{2\pi^{2}m}\qquad;\qquad\gamma=\frac{\lambda^{2}}{8\pi m}$

(55)

Note that $\omega$ can be viewed as the renormalized frequency of the oscillator. This equation represents a damped harmonic oscillator with the damping constant $\gamma$.