In Hawking's famous paper "particle creation by black holes", he expands the real scalar field $\phi$ in two ways. I am confused why he makes the choice he does in the second way.

First in the distant past (before the star has collapsed), he expands the field in (2.3) as $$ \phi = \sum_{i} \big( f_i \mathbf{a}_{i} + f_i^{\ast} \mathbf{a}_{i}^{\dagger} \big) $$ for some mode functions $f_{i}$ and ladder operators $\mathbf{a}_{i}$.

In the second way, in the distant future (after the star has collapsed and so a black hole horizon has formed and so on), he expands the field in (2.4) as $$ \phi = \sum_{i} \big( p_i\mathbf{b}_{i}+p_i^{\ast}\mathbf{b}_{i}^{\dagger}+q_i\mathbf{c}_{i} + q_i^{\ast} \mathbf{c}_{i}^{\dagger} \big) $$ for some mode functions $p_{i}$, $q_{i}$ and ladder operators $\mathbf{b}_{i}$, $\mathbf{c}_{i}$.

My question boils down to: in the second expansion, why do we need two types of ladder operators $\mathbf{b}_{i}$ and $\mathbf{c}_{i}$? There seem to be two Fock spaces being referred to here?

In the text below (2.4) goes on to say that $p_i$ are mode functions which are purely outgoing (with zero Cauchy data on the event horizon), and $q_{i}$ are solutions which contain no outgoing component (with zero Cauchy data on $\mathscr{I}^{+}$, the distant future aka. future null infinity).

Why do the outgoing and incoming modes (at the horizon) seem to require separate Hilbert spaces? I do not understand this from Hawking's paper. Does the presence of the horizon split up your Hilbert space in a way analogous to how this happens in Rindler space (there, you need different ladder operators in the right and left Rindler wedges, and so this reminds me of this).


3 Answers 3


The answer is really quite simple: the field operator must be expanded in terms of a complete set of modes.

"Complete" in this case means "forming a complete basis for classical solutions on a Cauchy surface for the problem." Here ${f_i}$ is complete on Cauchy surface $J_-$ in Hawking's collapse spacetime, and ${p_i} \cup {q_i}$ is complete on Cauchy surface $J_+ \cup EH$. So the field operator can be expanded in terms of either set.

The Hilbert space is the same in both cases, just expressed in different bases. It's not so different from, say, relating $e^{ikx} \leftrightarrow {a^\dagger_k}$ for one complete set, while using $\sin(lx) \leftrightarrow {b^\dagger_l}$ and $\cos(lx) \leftrightarrow {c^\dagger_l}$ ($l>0$) for another in flat spacetime. (Note this is just illustrative, taken literally it would violate the positive frequency condition.)

So at a formal level, the answer has little to do with the existence of a horizon. It is merely convenient to split the late time modes into two subsets, each with a simple behavior at the horizon. You could just as well use a single letter to label all the modes, and split them up by (say) odd and even indices.

You might consult Dewitt 1975 for a nice review of the formalism.


All of the operators shown in the question act on the same Hilbert space. The operators $\mathbf{a}$ can be expressed in terms of $\mathbf{b}$ and $\mathbf{c}$, and the operators $\mathbf{b}$ and $\mathbf{c}$ can be expressed in terms of $\mathbf{a}$. This is implicit in the fact that the field operator $\phi$ can be expressed either in terms of $\mathbf{a}$ (Hawking's equation (2.3)) or in terms of $\mathbf{b}$ and $\mathbf{c}$ (Hawking's equation (2.4)), and Hawking shows it explicitly in equations (2.7)-(2.8).

The reason Hawking introduces these different ways of writing the same thing is explained on the same page as those equations:

  • In (2.3), the field operator is expressed using mode functions that have simple behavior on past null infinity, and he mentions that the fields are completely determined by specifying initial conditions there.

  • In (2.4), the field operator is expressed using mode functions that have simple behavior on future null infinity, but specifying data only one future null infinity is not sufficient, because the spacetime also has a future event horizon. Hawking chooses to use different symbols ($\mathbf{b}$ and $\mathbf{c}$) for the operators associated with data specified on future null infinity and the horizon, respectively. Together, these operators generate the full algebra, just like the original operators $\mathbf{a}$ do.


The reason you need two sets of modes, $b$ and $c$, when the black hole has formed is the same exact reason that you need two modes when describing the Unruh effect. They correspond to modes on either side of the horizon. The $b$ and $c$ modes end up entangled with one another. This means that the modes on one side of the horizon are entangled with the modes on the other side. Remember, quantum field states are given by specifying an occupation number to each mode. To say that the $b$ and $c$ modes are entangled just means that if you measure there are, say, $10$ particles in a particular $b$ mode coming out of the black hole, then there must also be $10$ partner particles in the $c$ mode which are on the other side of the event horizon. The stuff about the support on the horizons is just telling you where the modes "live." The upshot is that one set of the modes has support on one side of the event horizon and the other set has support on the other side. The union of both sets of modes is necessary to describe a state on the full Cauchy slice.

I've included a little picture for the Rindler case. (Hawking himself has very good pictures for the black hole case.) Here the particles are massless for simplicity. Also, because there are two horizons in the Rindler case, I had to draw four sets, to account for both right moving and left moving. The red lines are lines of constant phase. Notice the phase pile up of the modes on horizons. They go as $|u|^{i \nu}$ or $e^{i \nu \log |u|}$ (or $v$ for right moving) and have support only for $u>0$ or $u<0$ depending on the half it lives in. They are eigenmodes of the boost Killing vector field which is $u \partial_u - v \partial_v$. The massive modes aren't quite as easy to draw. However, close to the horizon, the general idea is the same regardless of mass. Once again, if a Rindler observer measures, say, $10$ particles in a particular Rindler mode, then there must be exactly $10$ particles in the partner mode on the other side of the horizon, because the modes are entangled.

enter image description here


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