How the EM energy-momentum tensor of vacuum state could be proportional to the metric?

Question

We read "everywhere" that, because of Lorentz invariance, the energy-momentum tensor of any field in the vacuum state should reduce to a constant multiplying the metric tensor (I'm using the metric signature $\eta = (1, -1, -1, -1)$ and units such that $c = 1$): $$\tag{1} T_{ab}^{(0)} = \rho_{\text{vac}} \, \eta_{ab}. $$ I agree with that. I'm considering the case of the electromagnetic field: $F_{ab} = \partial_a A_b - \partial_b A_a$, of energy-momentum (up to a constant factor) $$\tag{2} T_{ab} = F_a^{\; c} \, F_{cb} + \frac{1}{4} \, \eta_{ab} \, F_{cd} \, F^{cd}. $$ These components are such that their trace is 0, by construction: $\mathrm{Tr} (T_{ab}) \equiv T_a^{\; a} \equiv 0$. For an homogeneous and isotropic state, this tensor reduces to the one representing a perfect fluid: $$\tag{3} T_{ab} = (\rho + p) \, u_a \, u_b - \eta_{ab} \, p. $$ So the condition $\mathrm{Tr} (T_{ab}) = \rho - 3 p = 0$ imposes $p = \frac{1}{3}\, \rho$, as it should for random electromagnetic radiation. Whatever the energy density $\rho \ne 0$ (and $\rho < \infty$), this energy-momentum cannot reduces to (1). The classical solution to the vacuum state is to have $\rho_{\text{vac}} = 0$. My issue appears when I consider the "quantum mechanics vacuum" of the radiation field, according to which $\rho_{\text{vac}} \ne 0$ (the usual "naive" calculations actually give a divergent integral: $\rho_{\text{vac}} \rightarrow \infty$). In that case, how can we reconcile the 0 trace condition of (2)-(3) (which can't reduce to (1)) and the Lorentz invariance of the vacuum (i.e energy-momentum tensor (1)), which implies the vacuum pressure $p_{\text{vac}} = -\, \rho_{\text{vac}}$ instead of $p_{\text{vac}} = \frac{1}{3} \, \rho_{\text{vac}}$?

EDIT: Maybe my query isn't clear enough, so I'll try to be more precise in what I want to know. The vacuum is Lorentz invariant, so we need to get the tensor (1) with some quantum calculations. But then, (2) has a 0 trace, by construction. So it appears to be mathematically impossible to get (1) from (2), except in the special and trivial (non-quantum) case $\rho = 0$. The usual quantum calculation gives (this is the "naive" calculation) $$\tag{4} \rho_{\text{vac}} \propto \int_0^{\infty} \omega^3 \, d\omega = \infty. $$ Authors frequently use an "hard cutoff" at $\omega_{\text{max}}$ to get a finite integral, but this explicitly destroys the Lorentz invariance of the vacuum, so that $p = \frac{1}{3} \, \rho$, instead of the proper relation $p = -\, \rho$. So it is clear that the hard cutoff isn't right. If really $\rho_{\text{vac}}$ is finite and not 0, then the energy-momentum (2) can't be the right expression to be used in the quantum calculation.

Answer 1

You are correct, the classical EM field energy-momentum tensor has zero trace, thus cannot be, in any point of spacetime, equal to a non-zero non-infinite multiple of the Minkowski metric. The only way the equation (1) can hold is if the constant of multiplication $\rho_{vac} = 0$. Then, however, the energy-momentum tensor comes out as zero. Which is the best result for "vacuum state".

My issue appears when I consider the "quantum mechanics vacuum" of the radiation field, according to which $\rho_{\text{vac}} \ne 0$ (the usual "naive" calculations actually give a divergent integral: $\rho_{\text{vac}} \rightarrow \infty$).

QM vacuum of the radiation field is a general concept, the lowest possible state of the free quantum EM field. It does not imply any definite value for $\rho_{vac}$. This is because in quantum theory free of gravity, there is no unique definition of energy (and energy density); we can shift it by any constant. In this context, we usually base energy definition on a quantum Hamiltonian that really came from Poynting's formula for energy density of classical EM field, by applying some quantization procedure. Already this classical Poynting formula is not unique in classical EM theory.

When quantizing it, we get still more arbitrariness. There are different quantum Hamiltonians, leading to the same Maxwell equations. Canonical quantization leads to zero-point energy $\hbar\omega/2$ per oscillator which implies $\rho_{vac}=\infty$, which is cumbersome and there are good reasons to avoid it. However, it would "solve" your problem in a crazy way: $T = \rho_{vac}\eta$ for $\rho_{vac}=\infty$ means all diagonal components of the tensor are infinite, and trace cannot be calculated. What cannot be calculated cannot pose any problem. People will object to this "solution".

The so-called normal ordering quantization leads to $\rho_{vac} = 0$ which solves your problem nicely.

Personal opinion: I think the condition for a ground state of a quantum field to be Lorentz invariant is hard to motivate. There are states that aren't Lorentz-invariant, so why ground state? It's a quantum theory, and the Lorentz invariance should not be put on as high a pedestal automatically.

Answer 2

The formula $T_{\mu\nu}= \Lambda \eta_{\mu\nu}$ applies to a Lorentz vacuum state. The $p=\rho/3$ applies to a state that contains photons such a black-body radiation. Black body radiation has a preferred reference frame and is not Lorentz invariant: when you move with respect to the preferred frame the radiation appears hotter towards the direction you are moving in because of the Doppler Blue-shift.