4 ms·
Good question! You can definitely add a constant to the Hamiltonian without changing the local physics, but I was referring to the stress-energy tensor (i.e. en
by zachf 6y ago
Good question! You can definitely add a constant to the Hamiltonian without changing the local physics, but I was referring to the stress-energy tensor (i.e. energy density), not the Hamiltonian (total energy). The stress tensor can be locally negative but still integrate to a positive or zero value of total energy.
So let's talk about energy density, I'll call it T. I need two facts:
1. The Reeh-Schlieder theorem implies that any operator that the vacuum state of QFT is "separating", meaning that the only operator A that satisfies A|0> = 0 is the trivial solution A = 0. (Here |0> means vacuum.)
2. The energy density operator has zero expectation value in the vacuum state, <0|T|0> = 0. Why? Explicitly: it has to be constant because of translational symmetry, and it has to because 0 is the only value that is constant and integrates to 0 total energy.
From fact 1, we know that T|0> is not 0 in any nontrivial theory (otherwise T = 0 identically, and no state has energy density at all). From fact 2, it follows that T|0> is orthogonal to |0>. Therefore |0> and T|0> span a 2D subspace of the Hilbert space.
Now write down the matrix T in that space. It looks like
[0, b*]
[b, c ]
where b is nonzero because of the way the subspace is defined, and c is real because T is Hermitian. Such a matrix is never positive definite, therefore there exists a state with negative expectation value in this subspace. Explicitly, one of the eigenvalues of this matrix is c - sqrt(|b|^2 + c^2) which is negative because b is nonzero. The corresponding eigenvector therefore has negative energy density.
(Source for this argument: [0])
An example of such a state is the Casimir state, see [1].
[0] https://arxiv.org/abs/1803.04993 https://arxiv.org/abs/1803.04993
[1] https://en.wikipedia.org/wiki/Casimir_effect https://en.wikipedia.org/wiki/Casimir_effect
- ssivark 6y agoThanks for the nice explanation, and the Witten ref. So it seems that in QFT we must have at least one negative energy state (should we consider such a state “below” the vacuum? Maybe the message is that we should give up on a strict ordering once we have many local degrees of freedom). It’s interesting to interpret the Casimir effect in this language — is it that the location of negative energy density is specified by spontaneously broken translational symmetry (negative between plates and zero/positive outside). What property decides whether the system is sitting in the “vacuum” or the “Casimir” state? (If we’ve given up on “lowest energy”)
- zachf 6y agoWell the argument I presented shows that at least one state has a negative energy density at any given point (I suppressed the dependence on position, T = T (x) )— but the argument does not prove that this state has lower total energy than the vacuum because there could be lots of positive energy density somewhere else! In fact this has to be true, because as you correctly pointed out the hamiltonian has to be bounded below, and with some assumptions about the field theory you can show that the vacuum has 0 energy and every other state has higher total energy. You can even put constraints on how far away the positive energy density is from the negative energy density. These identities historically went by the strangely non-descriptive name of Quantum Inequalities, and there’s a nice modern proof of a similar result in conformal field theory due to Blanco and Casini [0]. Take the Casimir state as an example. In the usual setup there’s negative energy density between two perfectly conducting planes. Either the setup is unstable and the plates will attract, or else something is holding the plates apart which takes work (a source of positive energy density). It’s a bit unclear to me how to think about the former type of states, they’re unstable and so they can’t be energy eigenstates. Another instance of the Casimir effect is when you take a 2D CFT on a cylinder, in which case the normal vacuum state on flat space maps to a state of negative energy on the cylinder under the relevant conformal transformation. That’s a form of the Casimir effect, the energy has dropped to something negative whereas on flat space it was zero (from the algebraic perspective it arises from the conformal anomaly), but now that’s the new vacuum state and the new (negative) lower bound on the energy. [0] https://arxiv.org/abs/1309.1121 https://arxiv.org/abs/1309.1121