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The field operator in curved space-time can be expanded as
$$\hat{\phi}(x)=\sum_{n}\left(\hat{a}_n f_n(x)+\hat{a}^\dagger _n f^* _n (x)\right)$$ where $f(x)$ is the positive frequency mode and $f^*(x)$ is the negative frequency mode. We define the vacuum state as $\hat{a}_n \left|0\right>_a=0$ But we can choose a different set of positive and negative frequency modes, and an expansion $$\hat{\phi}(x)=\sum_{n}\left(\hat{b}_n h_n(x)+\hat{b}^\dagger _n h^* _n (x)\right)$$ Now the vacuum would be define as $\hat{b}_n \left|0\right>_b=0$. The mode functions $f_n$ are normalized $(f_m,f_n)=\delta_{m n}$, $(f_m,f^*_n)=0$ and $(f^*_m,f^*_n)=-\delta_{m n}$. From the two different expansions of $\hat{\phi}$ we have $$\sum_{n}\left(\hat{a}_n f_n(x)+\hat{a}^\dagger _n f^* _n (x)\right)=\sum_{n}\left(\hat{b}_n h_n(x)+\hat{b}^\dagger _n h^* _n (x)\right)$$ Taking the inner product with $f_n$ on each side, we get $$\hat{a}_m=\sum_n (h_n,f_n)\hat{b}_n+\sum_n (h^*_n,f^*_n)\hat{b}^\dagger_n \equiv \sum_n \alpha_{m n}\hat{b}_n + \sum_n \beta_{m n}\hat{b}^\dagger_n$$ Thus the vacuum $\left|0\right>_a$ satisfies $$0=\hat{a}_m \left|0\right>_a=\left(\sum_n \alpha_{m n}\hat{b}_n + \sum_n \beta_{m n}\hat{b}^\dagger_n\right)\left|0\right>_a$$ If we suppose we had just one mode, with a relation $$\left(b+\gamma b^\dagger\right)\left|0\right>_a=0$$ The solution to this equation is of the form: $$\left|0\right>_a=Ce^{\mu \hat{b}^\dagger \hat{b}^\dagger}\left|0\right>_b$$ Where $C$ is a normalization constant and $\mu$ is a number.
My question is how you derive the solution of the equation : $$\left(b+\gamma b^\dagger\right)\left|0\right>_a=0$$

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    As far as I know, there is no way to derive the solution for vacuum of a different set of modes. It is an ansatz, and it is motivated from other principles. In the case of black hole evaporation, one important motivation is translational invariance (see page 9 of https://arxiv.org/abs/1609.04036) – Avantgarde Jul 27 '17 at 22:37

1 Answers1

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In a truly general spacetime, picking out "the" correct vacuum state is basically not possible (because the question doesn't make sense). If someone hands you a state you can tell them how that state will evolve in time, but there is no special vacuum state to favour. Note this is roughly also true of classical fields in curved spacetime; in a general spacetime there is no unique rest frame for a fluid.

When we specialise a bit, however, we can often pick out preferred vacua using a mix of mathematical and physical considerations. The most obvious example is Minkowski spacetime. There, we have good reason to treat the usual inertial-frame decomposition into Fourier modes as special, because

  1. we observe the "inertial" particle detectors don't click;
  2. these modes have a special relationship with one of the timelike Killing vectors (see e.g. Birrell and Davies)

Most compellingly, however, when the Hamiltonian is written in terms of these modes, the vacuum state (the state which is annihilated by the annihilation operator) relates to the spectrum of the Hamiltonian in the way we are familiar with from our studies of quantum theory:

  • The vacuum is also the lowest-energy eigenstate of the Hamiltonian.
  • When the creation operator acts on the vacuum, the result is another eigenstate of the Hamiltonian.

In a general spacetime, these properties cannot be met by any vacuum choice. In a time-evolving spacetime, for example, the vacuum is not a stationary state of the Hamiltonian, which is one way of viewing the resulting "particle production".

We can extend this specialness somewhat to spacetimes which are at least asymptotically flat, by comparing with the vacua which have the above properties in the asymptotic regions.

AGML
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  • My question is how you derive the solution of the equation : $$\left(b+\gamma b^\dagger\right)\left|0\right>_a=0$$ I know and i prove that if i take this solution : $$\left|0\right>_a=Ce^{\mu \hat{b}^\dagger \hat{b}^\dagger}\left|0\right>_b$$ the equation is satisfied. – christos leonidou Jul 27 '17 at 16:45
  • There is no unique solution in a general spacetime. You need to impose extra conditions like those on the Hamiltonian above to pick one out. – AGML Jul 27 '17 at 16:48