The argument is often given that the early attempts of constructing a relativistic theory of quantum mechanics must not have gotten everything right because they led to the necessity of negative energy states. What's so wrong with that? Why can't we have negative energy states?

As I understand it, we know now that these "negative energy states" correspond to antiparticles. So then, what's the difference between a particle with negative energy and an antiparticle with positive energy? It seems to me that there really is no difference, and that the viewpoint you take is simply a matter of taste. Am I missing something here?


5 Answers 5


The usual argument is that negative energy states are inherently unstable; if energy states are not bounded from below, a negative energy state can always become more negative, emitting positive energy radiation continuously. It turns out, this is more or less what it is believed that happened in the inflationary era:

1) an accelerated expanding cosmos

2) all the positive energy matter we see today.

So negative energy states are only "bad" (or let say just wildly inconvenient) in our currently asymptotically flat space-time, but they probably existed at the very beginning in vast quantities. They probably marginally exist still today in the form of dark energy.

However, i am confused why people extrapolate the idea that states will always try to decay to lower energy states (even if already negative): What happens at a more fundamental level is that systems try to achieve equilibrium by spreading energy evenly across degrees of freedom of all fields. Entropy is nothing but a logarithm in the number of available states reachable for a degree of freedom at a given, well defined energy. This entropy has a minima at zero energy, not at $- \infty$, as would be implied by the common lore. So it is not unreasonable to expect that, negative energy systems would decay to higher energy states, toward the zero energy states that we associate with the vacuum.

  • $\begingroup$ While I recognize that you might have the right idea about entropy, you're contradicting yourself here: the first part of your answer says that negative energy states will decay to states of more negative energy, but the second part says that they don't. Which is it? $\endgroup$
    – David Z
    Commented Nov 9, 2011 at 17:35
  • $\begingroup$ yes, the first part is trying to answer according to the current lore. The last part is just me trying to stir up the notions as we believe to understand them. In any case, it might be well possible that there are systems of both kinds (unstable negative energy and stable negative energy states) or even that the stable systems are just adiabatic approximations of the unstable ones (small time scales) $\endgroup$
    – lurscher
    Commented Nov 9, 2011 at 17:41
  • $\begingroup$ The point of energy is that it is distributed with an additive constraint--- the total is fixed. If you remove the additive constraint, it isn't that the energy runs to -infinity, it runs to -infinity while dumping more energy in modes that simultaneously run to +infinity. So the distribution of energies broadens until it is arbitrarily broad--- there is no additive conserved quantity anymore limiting growth, and you have instability. This is not conjecture, you can see it explicitly in phi^3 model as the field runs away to infinity, producing positive energy domain wall to balance energy. $\endgroup$
    – Ron Maimon
    Commented Jan 25, 2012 at 5:41
  • $\begingroup$ @lurscher Your argument that entropy should be minimum at most stable state is misleading. It is Helmotz free energy that is minimum which is A=U-TS, which suggest that U should be minimum. $\endgroup$
    – Aman pawar
    Commented Jan 21, 2019 at 16:21

The problem is that interacting systems, like particles, tend to transition into states of lower energy. (Technically, the universe transitions into states of higher entropy, but in the context of a particle that usually means lower energy.) So in order for particles to be stable, the energy spectrum has to have a lower bound. Otherwise, a particle could just keep dropping to lower and lower energy states, emitting photons at every step.

Now, there is a sense in which a positive energy antiparticle state can just as well be considered a negative energy particle state. The solution to the Dirac equation looks the same in either case. In the early days of relativistic QM, it never occurred to anyone that there was any interpretation of these solutions other than being negative-energy states, which led to the invention of ideas like the Dirac sea, and the identification of holes in the sea with antiparticles. But by the time quantum field theory came along, people realized that it just made more sense to include antiparticles as proper objects in the theory, rather than trying to explain them as holes, because then there was no need to bother with negative energy states at all.

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    $\begingroup$ You wrote: "Otherwise, particle could just keep dropping to lower and lower energy states, emitting photons at every step." But the negative energy in question is about energy spectrum of a free particle, no interaction with photons is involved. If the particle is in interaction with the photon filed, it is not a free and cannot have certain energy. Besides, how about 4-momentum conservation law for emission/absorption of photons by a free particle? $\endgroup$ Commented Nov 9, 2011 at 11:00
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    $\begingroup$ the fact that we see systems to transition to lower energy states is based on our experience from positive energy systems: more fundamentally, what happens is that systems tend to split the energy evenly across degrees of freedom. It makes complete sense for negative energy systems to thermodinamically evolve into states with higher energy (i.e: toward zero energy per degree of freedom) $\endgroup$
    – lurscher
    Commented Nov 9, 2011 at 15:18
  • $\begingroup$ @lurscher: that thought did occur to me, but I have heard from other sources that particles in negative-energy states still tend to lose energy. I wanted to do an explicit calculation of the entropy to resolve this one way or the other but I didn't have time. $\endgroup$
    – David Z
    Commented Nov 9, 2011 at 17:32
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    $\begingroup$ @lurscher:You have said in your answer, that the energy of the particles should tend to go to the zero energy states, to achieve 'minimum' entropy. I can't understand, how is this consistent with the fact as david as outlined, that the ' the universe transitions into states of higher entropy'. Also, I don't really understand how you can use statistical concepts such as entropy for a single particle. $\endgroup$
    – user7757
    Commented Dec 30, 2012 at 4:57
  • $\begingroup$ @DavidZ: Why can't a system be bounded from below by a negative energy? $\endgroup$
    – BMS
    Commented Jan 6, 2015 at 5:49

For very simple cases only, free quantum fields, we can certainly map negative frequencies (not energies, but the two things are conflated by most authors) to positive frequencies and vice versa in various ways. The details of this for the Klein-Gordon field are published as EPL 87 (2009) 31002, http://arxiv.org/abs/0905.1263v2; for the electromagnetic field, there is http://arxiv.org/abs/0908.2439v2 (which I recently almost completely rewrote). Up to a point, these papers put Vladimir Kalitvianski comment into one mathematical form (but other mathematical forms for his comment are certainly possible). FWIW, the presence of measurement incompatibility is tied in with whether one allows negative frequency modes.

HOWEVER, I have no idea what the construction in those papers looks like if one uses similar mathematical transformations for the whole of the standard model of particle physics. In fact, over a number of years I have failed to get such an approach to work. It's necessary to get it right for the whole of a system that comes close to reproducing the phenomenology of the standard model (or something slightly different in an experimentally useful way or in a way that is useful for engineering) before many Physicists are likely to take the idea very seriously.

The stability argument given by David Zaslavsky is completely right by conventional wisdom, but it assumes, for starters, that energy and action are viable concepts in a QFT context. In the algebraic context I currently work in, energy and action are not viable concepts. There is also no "axiom of stability" in quantum field theory, so there is no proof of a no-go theorem that there is no way to ensure stability except by having only positive frequencies; there is, instead, an "axiom of positive frequency" in the Wightman axioms. Note that a well-formulated axiom of stability would be far less theoretical and more natural than an axiom of positive frequency.

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    $\begingroup$ as i commented on David's answer, if one wants to extrapolate thermodynamics arguments about stability to negative energy systems, one might want to consider the fact that positive energy systems go to lower energy states because that is what happens if you split their energy evenly across degrees of freedom. With negative energy systems the same principle would make then decay into higher energy levels (toward zero-energy states) $\endgroup$
    – lurscher
    Commented Nov 9, 2011 at 15:32
  • $\begingroup$ @lurscher Interesting comment, but I'm not yet sure what to make of it in detail. You've clearly thought about this, so are there any references that you think are relevant? $\endgroup$ Commented Nov 9, 2011 at 16:15
  • $\begingroup$ sorry, i don't have any concrete reference that i'm aware. However i think the main problematic assumption here is that the Boltzmann factor $e^{ - \beta E_{i} }$ is unchanged when a system allows negative energy states. In this case, there is a certain $E_0$ for which entropy is exactly 0 (the Lorentz-invariant vacuum), so the Boltzmann factor needs to change near the vacuum energy (or any entropy local minima, for what its worth) $\endgroup$
    – lurscher
    Commented Nov 9, 2011 at 17:36

A negative kinetic energy is not physical. It is supposed to be observable, as well as the particle velocity and mass. So it is just a non physical solution. On the other hand, for completeness of Fourier transformation, those negative frequencies must be present in the solution. They were made present as "antiparticle" solutions in a multi-particle construction. It means, the Dirac's equation solutions came in handy in QED and are not really physical in one-particle relativistic QM.


I complete analogy with classical mechanics:

We define the proper velocity: $$ \eta ^\mu :=\frac{dx^\mu}{d\tau}, $$ where $\tau$ is proper time. We likewise define (relativistic) momentum: $$ p^\mu :=m\eta ^\mu . $$ And finally we define the (relativistic) energy (up to multiples of $c$) as the time-component of $p^\mu$. This happens to be $$ \frac{mc^2}{\sqrt{1-(v/c)^2}}, $$ which obviously must be positive. Thus, in order to be consistent with our relativistic definition of energy, we can't have particles with negative energy. This almost makes it tautological, but it is straightforward and precise.

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    $\begingroup$ This is total nonsense. The definitions proposed here are not even viable definitions, because they don't handle the zero-mass case. The relativistic relation that works for all masses, including zero mass, is $m^2=E^2-p^2$, which is perfectly compatible with $E<0$. If this elementary classical argument had been correct, then it would have been a schoolboy mistake for Dirac to propose his picture of the Dirac sea. $\endgroup$
    – user4552
    Commented May 21, 2015 at 15:39

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