Why do we need these two sets of modes in the gravitational collapse?

Consider the gravitational collapse spacetime:

Hawking argues in his paper$^{[1]}$ about black hole radiation that the massless scalar field $\phi$ can be decomposed as

$$\phi = \sum_i \{p_i b_i+p_i^\ast b_i^\dagger+q_i c_i+q_i^\ast c_i^\dagger\} \tag{2.4}$$

where $\{p_i\}$ is a complete orthonormal family of solutions of the Klein-Gordon equation containing only positive frequencies with respect to $\mathscr{I}^+$ and have zero Cauchy data on the horizon $\mathfrak{H}$.

On the other hand $\{q_i\}$ is a complete orthonormal family of solutions of the Klein-Gordon equation with no outgoing component (by the way, I don't know how he rigorously define no outgoing component, would it mean that $q_i(x) = 0$ for $x\in \mathscr{I}^+$?)

Now I don't get what is being done here. If $\{p_i\}$ is a complete set, it seems to me that by definition we can write

$$\phi = \sum_i\{ p_i b_i+p_i^\ast b_i^\dagger\},$$

and similarly for $\{q_i\}$ if it's a complete set.

Why do we need to use two complete sets at once?

I believe this has something to do with the initial value formulation with initial data on a Cauchy surface. But I'm failing to get the precise connection here which explain rigorously why we need this decomposition to analyze the problem of Hawking radiation.

$[1]$ Hawking, S., "Particle creation by black holes", https://link.springer.com/article/10.1007/BF02345020

• You're correct. $p$ is the outgoing radiation while $q$ is the one engulfed by the event horizon. The horizon alone is not Cauchy surface, but $\mathfrak{H} \cup \mathscr{I}^+$ is - any future causal curve ends up on $\mathfrak{H}$ or $\mathscr{I}^+$, so you need Cauchy data on both the surfaces to get a complete set of mode solutions. – Avantgarde Aug 28 '18 at 18:01
• @Avantgarde, but a timelike curve can end at $i^+$. So to get a Cauchy surface don't we need to set $\Sigma = \mathfrak{H}\cup \mathscr{I}^+\cup i^+$? And by the way, on a Cauchy surface we should specify initial data isn't it? This doesn't seem like initial data (it's on the far future by the way). – user1620696 Aug 28 '18 at 19:58
• Ah yeah, I missed the point $i^+$. Having data defined on a Cauchy surface lets us determine the future and past. – Avantgarde Aug 28 '18 at 20:45
• There is no "the" gravitational collapse spacetime. There are many different spacetimes that demonstrate some type of gravitational collapse. Do you mean the Schwarzschild spacetime? The Schwarzschild spacetime doesn't even describe gravitational collapse -- it describes an eternal black hole, which was not formed by gravitational collapse. – Ben Crowell Aug 29 '18 at 0:41
• @BenCrowell, indeed I do agree I was imprecise. I added the diagram of the gravitational collapse spacetime I have in mind. As you can see, I'm not considering Schwarzschild spacetime. – user1620696 Aug 29 '18 at 1:46

1. At first one might suspect that Hawking is considering a complex scalar field (corresponding to 2 real scalar fields), but actually Hawking explicitly mentions above eq. (2.2) that $\phi$ is just 1 real scalar field. The real scalar field $\phi$ is moreover assumed massless.
2. Instead the doubling in eq. (2.4) [as compared to eq. (2.3)] is caused by the fact that the final Cauchy surface consists of 2 parts: The event-horizon ${\scr H}^+$ and ${\scr J}^+$, cf. above comment by user Avantgarde. This is similar to a 1D scattering experiment: The incoming wave from ${\scr J}^-$ is partly transmitted to the event-horizon ${\scr H}^+$ and partly reflected to ${\scr J}^+$, cf. OP's Figure (or Hawking's Fig. 3).
• Thanks for the answer. So picking a basis with modes $\{p_i\}$ vanishing at the horizon and of positive frequency with respect to the time parameter of $\mathscr{I}^+$ and modes $\{q_i\}$ vanishing at $\mathscr{I}^+$ is a choice physically motivated by the "scattering picture" imagining the horizon as a potential barrier and trying to make sense of "the part of the wave transmitted and reflected"? – user1620696 Sep 4 '18 at 1:54