# Is the QFT Hamiltonian on an eternal Schwarzschild black hole background unbounded below?

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)?

• 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. Feb 8, 2023 at 3:41
• 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. Feb 9, 2023 at 15:20
• 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). Feb 13, 2023 at 18:37
• 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. Feb 14, 2023 at 3:54
• 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. Feb 14, 2023 at 5:36

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