What is really a negative energy particle (and why is it different from an anti-particle)? I have read this question:
What is negative Energy/Exotic Energy?
This does not really give an answer.
Why does a particle (charged) change sign passing the event horizon?
Black holes and positive/negative-energy particles
Where John Rennie says:


NB "positive" and "negative" doesn't mean "particle" and "anti-particle" (for what it does mean see below), and the black hole will radiate equal numbers of particles and anti-particles.


and then he says:


When you quantise a field you get positive frequency and negative frequency parts. You can sort of think of these as representing particles and anti-particles.


Now I am a little confused. Are negative energy particles anti-paricles (I understand they are not), but then what is the real difference? Is a negative energy particle the same as dark energy?
Question:


*

*What is the real difference between negative energy particles and anti-particles?

*Have we ever experimentally found negative energy particles? Does negative energy particles mean in any way dark energy?
 A: Not to put too fine a point on it but the big difference between negative energy particles and antiparticles is that negative energy particles have negative energy, that is, for some state $\Omega$ of negative energy,
$$\langle \hat{H} \rangle_\Omega < 0$$
Or, to be more specific since we're doing general relativity, consider the operator of the stress-energy tensor $T_{\mu\nu}$, then, for a null vector $k$, 
$$\langle \hat{T}_{\mu\nu}(x)\rangle_\Omega k^\mu k^\nu < 0$$
which has the benefit of being Lorentz invariant. 
On the flipside, for reasonable quantum fields, antiparticles have a positive energy. Consider for instance the usual case of a Dirac field. The Hamiltonian (density) operator for it is (in momentum space)
$$\hat{H} = \sum_s \vec{p} (\hat{a}^{s\dagger}_{\vec{p}} \hat{a}^s_{\vec{p}} + \hat{b}^{s\dagger}_{\vec{p}} \hat{b}^s_{\vec{p}})$$
$a^\dagger$ the creation operator for fermions and $b^\dagger$ for antifermions. You can observe that the role of particles and antiparticles is symmetric in the Hamiltonian : any particle will have the same energy as an antiparticle. 
On the other hand, consider the usual scalar field, with field operator defined as
$$\phi(x) = \sum_k f_k(x) \hat{a}_k + f^*_k(x) \hat{a}^\dagger$$
with $f_k$ the usual modes $f_k \propto e^{ik_\mu x^\mu}$. The (renormalized) stress-energy tensor, adapted from the classical theory, is
$$\langle \hat{T}_{\mu\nu} \rangle_\Omega = \sum_n (2n |c_n|^2 T_{\mu\nu}[f_k, f_k^*] + n^{1/2} (n-1)^{1/2} c_n c_{n-2}^*T_{\mu\nu}[f_k, f_k] + n^{1/2} (n-1)^{1/2} c_n^* c_{n-2}T_{\mu\nu}[f^*_k, f^*_k])$$
with 
$$T_{\mu\nu}[g, h] = (\partial_\mu g)(\partial_\nu h) - \frac{1}{2} \eta_{\mu\nu} (\partial_\sigma)(\partial^\sigma h)$$
and $|\Omega\rangle = \sum c_n |n\rangle$. Then take for instance the state 
$$\frac{1}{\sqrt{1 + \varepsilon^2}}(|0\rangle + \varepsilon |2\rangle)$$
Then
$$\langle \hat{T}_{\mu\nu} \rangle_\Omega = (k_\mu k_\nu - \frac12 \eta_{\mu\nu} k_\sigma k^\sigma ) \frac{\varepsilon}{1 + \varepsilon^2} (2 \varepsilon - \sqrt{2} \cos(2 k_\rho x^\rho))$$
The sign of which depends on the last factor. For $\varepsilon$ small enough, there are spacetime regions for which the energy becomes negative. 
A: Anti particles can be thought of as negative energy particles traveling backwards in time. In effect their reversed direction of time means they can really be understood as positive energy particles with all their charges reversed. Actual negative energy particles which would be traveling forward in time  ( or positive energy particles traveling backward in time ) can’t exist as “real” particles in QFT because this would mean that QFT isn’t bound from below. However, there have been a significant number of papers written  to attempt to incorporate these negative energy particles to solve the cosmological constant problem in QFT by credible physicists including one noble laureate.These negative energy particles are in fact valid solutions of the relativistic equations. The utilization of these negative energy particles do require a modification of QFT because they they generate unphysical negative probabilities
A: The simple (simplistic) answer is:

A negative energy particle is a particle whose binding energy is larger than its rest mass.

This is easiest to understand for massive particles. The potential well of a black hole is (in some sense) infinitly deep. Hence, if you put a particle deep enough into the well its binding energy will become bigger than its rest mass.
It becomes slightly harder to visualise for massless particles (since their rest mass is by definition zero). In this case, being negative energy means that the massless particle is on a trajectory that does not have enough energy to reach infinity. (Even if it would tunnel through all potential barriers on the way.)
By contrast an anti-particle is simply a particle that has the opposite conserved charges to some other particle. (It is a relative term) Both particles and anti-particles can be both positive and negative energy.
For a non-rotating (Schwarzschild) black hole, all negative energy trajectories lie inside the event horizon. Rotating (Kerr) black holes on the other hand have negative energy trajectories outside of the event horizon. These are what allow energy to be extracted from a rotating black hole through the Penrose process.
