Consider the $t=0$ Cauchy slice of the maximally extended Schwarzschild black hole. Let the parts of the slice to the left and right of the bifurcation surface have Hilbert spaces $\mathcal{H}_L $ and $\mathcal{H}_R$ respectively. Roughly speaking, the overall Hilbert space is $\mathcal{H}=\mathcal{H}_L\otimes\mathcal{H}_R$. I won't worry about technical details here.

The Hamiltonian $H$ is the conserved charge associated to the Killing vector field $\partial_t$. It generates translations of the $t$ coordinate. Naively, I'd expect to be able to write: $$\tag{1} H = H_L +H_R$$ where $H_L$ and $H_R$ act on the left and right Hilbert spaces, and have the same, positive spectrum. But I'm not so sure about the relative sign, since "time runs backwards" on the left portion of the spacetime. That is, perhaps we instead have: $$\tag{2} H = -H_L + H_R.$$

I can't firmly convince myself which sign is correct, but it seems like an important sign to get right. If (1) is true then $H$ is bounded below, with a unique ground state $|\psi\rangle$ whose energy we can set to zero so that $H|\psi\rangle =0$. If (2) is true then $H$ is unbounded below, and $H$ has many zero-eigenstates.

Which is correct, (1) or (2)?

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    $\begingroup$ Interesting... user @devCharaf posted a fairly in-depth answer, and then deleted it immediately after I asked whether he had used ChatGPT to help write it. $\endgroup$ Feb 8, 2023 at 3:41
  • $\begingroup$ Without taking the time to write write an actual answer, you may wish to look at arxiv.org/abs/1804.01081 for how this works in JT gravity (which is technically simpler). The latter of your two Hamiltonians turns out to vanish, and hence time "running backwards" on one side corresponds to a pure gauge transformation just like any bulk diffeomorphism. $\endgroup$ Feb 9, 2023 at 15:20
  • $\begingroup$ Thanks, I'll give it a read. Do you know if this is special to JT gravity, or do you know the answer in general? (Note my question isn't actually about quantum gravity: I'm fixing the background). $\endgroup$ Feb 13, 2023 at 18:37
  • $\begingroup$ Oh, well if you're just fixing a background my comment does not apply. In your case I would suggest working out the total H from the field theory. You will find that the sign depends on how you chose the future oriented vector field. $\endgroup$ Feb 14, 2023 at 3:54
  • $\begingroup$ Yes, the sign appears to depend on some choices/conventions. But the question of whether $H$ is bounded below or has a spectrum symmetric about $0$ is well-defined, I think. $\endgroup$ Feb 14, 2023 at 5:36

1 Answer 1


The correct expression is (2), i.e. $H = -H_L + H_R$.

The intuition I have behind this is that, in the left part of the spacetime, time runs "backwards" with respect to the right part of the spacetime. This means that we need to change the sign of the Hamiltonian in order to take this into account. Indeed, generally speaking, the Hamiltonian generates the time evolution of states. If time runs "backwards" in one part of the spacetime, then we need to change the sign of the Hamiltonian in that region in order to get the correct time evolution. In other words, instead of evolving states forward in time, the Hamiltonian would evolve them backwards in time.

As you mentionned, it implies that the Hamiltonian is unbounded below, and there are many zero-eigenstates, which correspond to the left- and right-moving excitations near the horizon of the black hole (known as "soft hair").

Edit to clarify

In general relativity, time is a coordinate, and the direction of time is determined by the choice of time orientation, which is a choice of a future-pointing timelike vector field. In the case of the maximally extended Schwarzschild black hole, the future-pointing timelike vector field changes direction across the bifurcation surface, so that on one side it points towards the future and on the other side it points towards the past. When we say that "time runs backwards" in one part of the spacetime, what we mean is that the future-pointing timelike vector field in that part of the spacetime points towards the past, rather than the future. In order to take this into account, we need to change the sign of the Hamiltonian in that part of the spacetime, because the Hamiltonian generates time evolution along the direction of the future-pointing timelike vector field. If the future-pointing timelike vector field points towards the past, then the Hamiltonian will generate time evolution that goes "backwards" in time, rather than forwards, so we need to change the sign of the Hamiltonian to reflect this.

  • $\begingroup$ Sorry but this is just not precise enough for me. What does "time runs backwards mean"? With respect to what? Time always runs forwards with respect to time (duh) so it can't be that... $\endgroup$ Feb 13, 2023 at 18:35
  • $\begingroup$ @nodumbquestions I edited my answer to provide precision. Let me know if it clarifies what I meant ! $\endgroup$
    – Baloo
    Feb 14, 2023 at 9:36

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