Does the vacuum energy of the Casimir effect gravitate?

Question

Since there is negative energy density between the plates, would this lead to anti gravity or perhaps gravity, since the vacuum energy between the plates is technically not less than 0, but less than surrounding vacuum.

Answer 1

Whether there is non-zero energy between the plates depends on the definition of energy density between the plates we adopt.

If we define it as expectation value of the standard Hamiltonian $H$ with zero-point terms $\hbar \omega_\mathbf{k}$, in ground state it comes out infinite. If we define it using the normally ordered Hamiltonian $H_{normal}$ without the $\hbar \omega_\mathbf{k}$ terms, it comes out zero.

There is no unique Hamiltonian to use to define energy - one can shift it by any finite constant and get equally good Hamiltonian.

For a simpler example, free particle in classical mechanics can be described using Hamiltonians $H=\frac{p^2}{2m}+H_0$ where $H_0$ is any real constant. But despite this we prefer the definition of energy $E=\frac{1}{2}mv^2$ to $E = \frac{1}{2}mv^2 + H_0$, because $H_0$ is just a useless symbol to carry around that changes nothing about the behaviour of the system or our predictions about it.

Similarly in EM field, we have good reasons to prefer one single definition of EM field energy, even if there is infinity of different possible Hamiltonians. Some people prefer the one based on $H$ with the zero point terms, other prefer the one based on $H_{normal}$ without them (because they cause energy to be infinite).

Of course, one can try to disambiguate this energy density definition by pointing to the idea that non-zero EM energy should gravitate. So maybe there is a way to find the "correct" energy density, via measuring the actual gravity effects in between the plates.

Needless to say, such gravity effects were never detected, so energy density equal to zero is the best estimate we have, and points to the definition based on the normally ordered Hamiltonian as the most appropriate one (if we care about EM energy formula being consistent with both general relativity and observations).

Answer 2

As of today, this is unclear. More generally, it is unclear how the quantum zero point energy of the vacuum couples to gravity, resulting in the famous Cosmological Constant Problem.

A relevant reference to check is How Does Casimir Energy Fall?

Doubt continues to linger over the reality of quantum vacuum energy. There is some question whether fluctuating fields gravitate at all, or do so anomalously. Here we show that for the simple case of parallel conducting plates, the associated Casimir energy gravitates just as required by the equivalence principle, and that therefore the inertial and gravitational masses of a system possessing Casimir energy E_c are both E_c/c 2 . This simple result disproves recent claims in the literature. We clarify some pitfalls in the calculation that can lead to spurious dependences on coordinate system.