# 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?

• Do you mean un-accelerated? The $\beta$'s are not zero when there is acceleration. – mike stone Sep 22 '19 at 13:12