Vacuum energy density with respect to what reference frame? Wikipedia quotes the observed vacuum energy density to be around $10^{-9}\frac{J}{m^3}$. Whatever the real energy density is, my question is, with respect to what reference frame is this being quoted? 
 A: This has been answered elesewhere on this site, I think. Vacuum "energy" usually means that the energy-momentum tensor
$$
T_{\mu\nu}\equiv  \left[\matrix{E & 0&0&0\cr 0&P&0&0\cr 0&0&P&0\cr 0&0&0&P}\right]
$$
is of the form 
$$ 
T_{\mu\nu}=
 E g_{\mu\nu},
$$
where $g_{\mu\nu}={\rm diag}(1,-1,-1,-1)$ is the metric. As a consequence  the pressure  is given by $P=-E$ and   is  alo non-zero. When  $E$ is positive, $P$ is  negative (a tension). As the metric tensor is a Lorentz  invariant, the energy desity $E$  is independent of the reference frame. 
A: In any inertial frame. The vacuum state is locally Lorentz-invariant, which means that the vacuum expectation value of the energy-momentum-tensor is proportional to the Minkowski tensor. When you Lorentz-transform it, you get the same thing  in the new frame.
A: I will answer this time with question. With respect to what reference frame Gravitational Potential Energy is zero of object lying on the ground ? With respect to ground of course ! Or another similar question. With respect to what particle energy in a ground-state of infinite potential well is defined as : $$ {\frac {h^{2}}{8mL^{2}}} $$
By definition, it is lowest energy possible of particle in an infinite potential well.
Analogically, vacuum energy density is ground-state of the vacuum,- lowest internal energy of vacuum possible. Currently energy density value is not so clear. Estimates ranges from $10^{−9}$ to $10^{113} \left[\frac J {m^3}\right]$
However what IS clear is that ground-state energy of vacuum is NOT zero.
