Do virtual particles have negative mass? Do virtual particles (which are generated from quantum fluctuations) have negative mass? Because virtual particles have negative energy  which due to the Energy-mass equivalence causes mass to be negative too.
If this is wrong I think the reason would be "because probably $E=mc^2$ is not applicable to Quantum mechanics"
Note: Real particles have positive energy and virtual particles have a negative energy unless/until  it is near something with an extremely strong gravity such as black holes
 A: Virtual particles do not fulfill the dispersion relation for a single relativistic particle $p^2\not = m^2$ or if the rest mass of the concerned particle is zero (for instance a photon) $p^2\not =0$ ($p^2= p^\mu p_\mu$ is a short-cut for the 4-momentum $p^\mu$ squared).
Virtual particles are also often off-shell, because they are not on the hyperboloid defined by $m^2 = p_0^2 -\mathbf{p}^2\equiv p^2$ (note that an equation of $m^2 = y^2-x^2$ defines a hyperboloid).
A particle that fulfills $p^2= m^2$ (the dispersion relation) is called on-shell and is therefore not virtual.
Due to this terminology we don't assume a strange value for the mass in order to reinforce the dispersion relation. It is preferred to abandon the dispersion relation for virtual particles, but keep their mass equal to their rest mass.
In this post the convention $c=1$ is used.
Note that the use of virtual particles as intermediate states is a kind of trick, so they have very little in common with real particles (that fulfill the dispersion relation).
A: Feynman diagrams conserve energy and momentum at the vertices, so at tree level, the one virtual particle's $q_{\mu}$ is determined by the real external particles. This is where the Mandelstam variables become useful (see fig).

Consider a $s$-channel process, such as:
$$ e^+ + e^- \rightarrow \gamma^* \rightarrow X $$
where the $*$ indicated virtual and $X$ is a fermion/anti-fermion pair.
In the lab frame:
$$ p^{\mu}_1 = (E, \vec p) $$
$$ p^{\mu}_2 = (E, -\vec p) $$
so the virtual photon has:
$$ q^{\mu} = p^{\mu}_1 + p^{\mu}_2 = (2E, 0, 0, 0) $$
So it's a massive photon at rest. Clearly virtual.
If you consider $t$-channel 180 degree scattering in the above case, then the final states are:
$$ p^{\mu}_3 = (E, -\vec p) $$
$$ p^{\mu}_4 = (E, +\vec p) $$
so that the exchanged photon has:
$$ q_{\mu} = p^{\mu}_4
  - p^{\mu}_2 = (0, 2\vec p) $$
This is generally written as:
$$ Q^2 = -(q^{\mu}q_{\mu}) = 4||\vec p||^2 \approx 4E^2 $$
where the approximation is for $E \gg m_e$.
So here, the virtual photon has no energy, but large momentum (and negative mass-squared)...defintely virtual.
For any 2-particle EM scattering process, there is a frame (the Breit Frame) where the exchanged photon has no energy, and transfers 3 momentum.
Hence the interpretation is that the virtual photon probes structures with size $\hbar c/\sqrt{Q^2}$.
Finally, for identical particles you have $u$-channel in which the final state external lines are swapped. In 180 degree backscattering of indetical particles, you can work out that $q^{\mu} = (0,0,0,0)$, so no scattering occurs at all...that is, you can't distinguish forward and backward elastic scattering.
At higher order you have loops, and you have to integrate over all energy and momenta that are conserved at vertices, so you get way off-shell.
