# Bogoliubov coefficient (not) being equal to zero

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?

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.

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