Annihilation and creation operators time-dependence in curved spacetime (Again)

Question

Let me ask this question again to hopefully get an answer.

Consider a free scalar field $\phi$ on a curved spacetime. The way we define the vacuum is by decomposing the field in terms of mode functions and associate to them creation and annihilation operators $a_k$, $a_k^{\dagger}$ via the inner product defined on the space of solutions of the KG equation (namely, we define $a_k:=\langle f_k, \phi \rangle$, similarly for $a^{\dagger}$) and this definition ensures that the $a_k$, $a_k^{\dagger}$ operators in the field expansion are time-independent (follows from the definition of the product). Then the vacuum is that state $|0\rangle$ s.t. $\forall k$, $a_k|0\rangle=0$.

In a dynamical background (i.e. a non-static spacetime) the Hamiltonian of the system is time dependent, thus we can let $a_k$ and $a_k^{\dagger}$ evolve (non trivially) in the Heisenberg picture, $a_k(t)$.

My question then becomes: Consider an asymptotically flat (in the past and in the future) spacetime with a field as above, see Birrell & Davis for an example, and two sets of mode functions $\{f_k\}$ and $\{g_k\}$ s.t. the first one in the past and the second one in the future, are asymptotically Minkowskian. Call $a_k$ and $b_k$ the relative creation operators.

In light of the Heisenberg evolution of these operators $a_k(t)$,$b_k(t)$, can I interpret $a_k$ as the creation operator of the $\{f_k\}$ mode functions in the infinite past and $b_k$ as the creation operator of the $\{g_k\}$ mode functions in the infinite future? A friend of mine even suggested that $a_k(+\infty)=b_k$ and $b_k(-\infty)=a_k$, is it right?

For context: I was trying to understand operationally the expectation value $${}_{in}\langle 0|N_{out}|0\rangle_{in},$$ where $N_{out}$ is the number operator for the $b_k$ operators and $|0\rangle_{in}$ is the $a_k$ vacuum.

My reasoning is, assuming that the answer to the above question is positive: the inertial observer in the past (i.e. mode functions $\{f_k\}$) prepares the state of the field in $|0\rangle_{in}$ (that is, $a_k(-\infty)|0\rangle_{in}=a_k|0\rangle_{in}=0$ for all $k$). The inertial observer in the future $\{g_k\}$ measures (Heisenberg picture again) ${}_{in}\langle 0|N_{out}(+\infty)|0\rangle_{in}$ particles, being $b_k(+\infty)=b_k$ we can then apply the well known Bogoliubov transformations and conclude the known result $${}_{in}\langle 0|N_{out}(+\infty)|0\rangle_{in}={}_{in}\langle 0|N_{out}|0\rangle_{in}=\Sigma|\beta_k|^2.$$

I hope I've been clear (enough).

Answer 1

I'll first give some general information that I think is useful to this question, and then I'll proceed to give you a more direct answer.

There is no general objective way to define creation and annihilation operators in a non-stationary spacetime.

The reason can be boiled down to the fact that the vacuum you described, which is annihilated by all annihilation operators, minimizes the Hamiltonian. However, the Hamiltonian is the quantity conserved by time translation symmetry. In the absence of time translation symmetry, you won't really have an objective definition of a Hamiltonian. More specifically, if your spacetime evolves on time, you are choosing a preferred notion of time. A different choice of what "time" means could lead to different results.

What happens in the scenario you described is that you can assign ladder operators at past and future infinities. These operators can indeed be evolved throughout the spacetime, but nothing ensures you will get the relations $a_k(+\infty) = b_k$ and $b_k(-\infty) = a_k$. In fact, you often won't. This is what lies behind the Hawking effect, for example.

Direct answer starts here

You can indeed interpret the $a_k$ as being associated to the $f_k$ mode functions and similarly for the $b_k$. In that region, you are effectively working in a stationary spacetime, so there are no weird tricks. Since the mode functions in the past and future are chosen arbitrarily, nothing ensures $a_k(+\infty) = b_k$ or anything of the sort. However, you will typically get an expression like $b_k = \alpha_k a_k + \beta_k a_k^\dagger$, for example. Using these Bogoliubov transformations you can then perform the calculations you mentioned and get to the desired result.

I am particularly fond of the discussions given by Wald about this. I recommend Chap. 14 of his General Relativity Book, Chap. 5 of his QFTCS book (this chap is about the Unruh effect, but the calculations are similar), and his paper "On particle creation by black holes", Commun. Math. Phys. 45, 9–34 (1975).