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?


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

  2. Have we ever experimentally found negative energy particles? Does negative energy particles mean in any way dark energy?

  • $\begingroup$ It will help to read up on the evolution/derivation of the Dirac equation as this is one of the many things that comes up. $\endgroup$ – Triatticus May 27 '19 at 4:25
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    $\begingroup$ The division of particles into particles and antiparticles is arbitrary. For example, is the electron particle and the positron antiparticle? Or is the electron antiparticle and positron particle? It doesn't matter. It is just a label. Both have a positive energy. $\endgroup$ – safesphere May 28 '19 at 11:30

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.

  • $\begingroup$ being negative energy means that the massless particle is on a trajectory that does not have enough energy to reach infinity that is wrong. Photon can have positive energy and still might not be able to reach infinity (provided its angular momentum is large enough). $\endgroup$ – A.V.S. May 27 '19 at 16:21
  • $\begingroup$ @A.V.S. Yes there is also the centrifugal potential barrier, that could prevent a positive energy massless particle from reaching infinity, but it would still in principle be able to tunnel through that barrier. $\endgroup$ – mmeent May 28 '19 at 6:37
  • $\begingroup$ "A negative energy particle is a particle whose binding energy is larger than its rest mass." Can you please do this with elementary particles? $\endgroup$ – Árpád Szendrei Jun 21 '19 at 12:59

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

$$\langle \hat{T}_{\mu\nu} \rangle_\Omega = \sum_n (2n |c_n|^2 T_{\mu\nu}[f_k, f_k^*] + n^{1/2} (n-1)^{1/2} c_n c_{n-2}^*T_{\mu\nu}[f_k, f_k] + n^{1/2} (n-1)^{1/2} c_n^* c_{n-2}T_{\mu\nu}[f^*_k, f^*_k])$$


$$T_{\mu\nu}[g, h] = (\partial_\mu g)(\partial_\nu h) - \frac{1}{2} \eta_{\mu\nu} (\partial_\sigma)(\partial^\sigma h)$$

and $|\Omega\rangle = \sum c_n |n\rangle$. Then take for instance the state $$\frac{1}{\sqrt{1 + \varepsilon^2}}(|0\rangle + \varepsilon |2\rangle)$$


$$\langle \hat{T}_{\mu\nu} \rangle_\Omega = (k_\mu k_\nu - \frac12 \eta_{\mu\nu} k_\sigma k^\sigma ) \frac{\varepsilon}{1 + \varepsilon^2} (2 \varepsilon - \sqrt{2} \cos(2 k_\rho x^\rho))$$

The sign of which depends on the last factor. For $\varepsilon$ small enough, there are spacetime regions for which the energy becomes negative.


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