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When discussing the nature and ambiguity of particles in curved space-times, one usually ends up at an expression for annihilation/creation operators that looks something like:

$$ a_{i}=\sum_{j}(\alpha_{ji}\hat{a}_{j}+\beta_{ji}^{*}\hat{a}_{j}^{*}), \quad \hat{a}_{j}=\sum_{i}(\alpha_{ji}^{*}a_{i}-\beta_{ji}^{*}a_{i}^{\dagger}). $$

The $\alpha_{ij}^{(*)}$ and $\beta_{ij}^{(*)}$ above representing Bogoliubov coefficients and the $a_{i}$/$\hat{a}_{i}$ representing those operators in two different bases. In a straightforward manner, one concludes that, for instance, when $\beta^{*}_{ij}\neq0$, the two sets of underlying modes are different and thus observers attributed to each of the modes will not observe the same particles.

My question is: what happens if that coefficient is equal to zero? Obviously, the two sets of modes i.e. the bases will be linearly dependent, but is there anything more?

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If the coefficient is zero, then both observers will count the same number of particles. An example of this is an unaccelerated observer in Minkowski space-time. But in a general curved space-time the coefficient is not equal to zero.

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  • $\begingroup$ Do you mean un-accelerated? The $\beta$'s are not zero when there is acceleration. $\endgroup$ – mike stone Sep 22 '19 at 13:12
  • $\begingroup$ I am also confused by this. The accelerated observer in flat spacetime sees the Unruh thermal state $\endgroup$ – Prof. Legolasov Sep 22 '19 at 16:16
  • $\begingroup$ You are right. I will correct. $\endgroup$ – Oбжорoв Sep 22 '19 at 16:26

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