Unruh effect derivation I was going through the derivation of the Unruh effect in chapter 5 of these lecture notes ("Lecture Series on
Relativistic Quantum Information" by Prof Ivette Fuentes). On p.26 the author introduces the "Unruh basis" and goes on to state that
$$A_k = \int C_{k'} a_{k'} dk'$$
and that
$$A_k = \cosh(r) a^I_k - \sinh(r) a^{II}_k{}^\dagger$$
I'm not entirely sure how these can be derived?
 A: For the computation you need to express the creation and annihilation operators of the Rindler observer in terms of the corresponding operators of an inertial observer. This can be achieved by a Bogolyubov transformation. In this case the Bogolyubov transformation uses the change of coordinates from the inertial to the accelerated observer.

I strongly recommend you these other notes. As you will see the key point is to use a Bogolyubov transformation for computing the creation $\hat{b}^+_\Omega$ and annihilation $\hat{b}^-_\Omega$ operators associated with the accelerated observer. The expected number of particles for this observer is given by:
$$\langle \hat{N}_\Omega \rangle = \langle 0_M|\hat{b}^+_\Omega\hat{b}^-_\Omega|0_M\rangle = \int_0^\infty \frac{\Omega}{\omega}|F(\omega,\Omega)|^2\text{d}\omega = \frac{1}{e^{2\pi c\Omega/a}-1}\delta(0), \tag{1}$$
where $\delta(0)$ is for the infinite volume of the space, and the function $F(\omega,\Omega)$ is given by:
$$F(\omega,\Omega) = \int_{-\infty}^{\infty} \frac{\text{d}u}{2\pi} \exp\left[i\Omega u + i\frac{c\omega}{a}e^{-au/c}\right], \tag{2}$$
which has the following property:
$$F(\omega,\Omega) = F(-\omega,\Omega) \exp\left(\frac{\pi c\Omega}{a}\right), \qquad a>0, \omega>0, \tag{3}$$
this property allows to compute $(1)$ easily. Then if you compare $(1)$ with the expected Bose-Einstein distribution for massless particles:
$$\frac{\langle \hat{N}_\Omega \rangle}{V} = \frac{1}{e^{E/k_BT}-1}, \tag{4}$$
being $E=\hbar \Omega$ Then you have:
$$\frac{E}{k_BT} = \frac{2\pi c \Omega}{a} \qquad \Rightarrow T = \frac{\hbar a}{2\pi c k_B}$$
