How to rule out a classical interpretation for negative-energy states?

1. Jul 17, 2014

bcrowell

Staff Emeritus
The relativistic mass-energy-momentum relation $m^2=E^2-p^2$ predates quantum mechanics by a couple of decades. It allows a particle such as an electron to have a negative mass-energy. If it's 1906, and you're shown this equation, do you have any way to show that the negative-energy solutions can't be of interest based on the known (classical) laws of physics?

If an electron initially has an energy-momentum vector inside the future light cone, then a boost can never bring it into the spacelike region (tachyonic) or the past-timelikelight cones (what we would now call antimatter). This shows that no continuous process of acceleration can change a positive-energy solution to a negative-energy one, and that means that if the negative-energy solutions exist, they're qualitatively different particles with different characteristics.

Normally we imagine that knowledge of a particle's world-line and mass suffice to tell us its energy-momentum vector. If we were going to believe in the negative-energy solutions, we'd have to believe that the world-line had an additional property, a sort of "arrowhead" that told us which direction its tangent vector pointed. This doesn't seem like a huge problem, though. Particles do have various properties such as charge that exist in addition to their world-lines.

It would be possible for a positive-energy particle and a negative-energy particle to act on each other without violating conservation of energy or Newton's third law, in such a way that each would accelerate indefinitely. I guess this could have been considered either scary or exciting in 1906, before anyone could have conceived of the Dirac sea.

Last edited: Jul 17, 2014
2. Jul 17, 2014

atyy

3. Jul 17, 2014

atyy

If I write the Lorenz force law and Newton's second law for the dynamics of a charged particle in an electric field, it doesn't seem obvious to me that it would behave differently from a positive mass particle with opposite charge.

Would gravitation help? There is a "positive mass" theorem there. http://en.m.wikipedia.org/wiki/Positive_energy_theorem

Last edited: Jul 17, 2014
4. Jul 17, 2014

Staff: Mentor

The idea of the Dirac sea was gradually forgotten about when it was realised Quantum Field Theory naturally incorporates antiparticles with the introduction of this concept.

That said some people still muck around with it eg:
http://www.ma.utexas.edu/mp_arc/c/10/10-196.pdf
'This is a multi-particle theory and solutions of the Dirac equation are promoted to quantum field operators. In this framework it is convenient to abandon the hole theory and introduce the anti-particles as separate entities. Today hole theory is mostly regarded as inessential and possibly misleading - see the introduction in Weinberg. Still the idea retains a certain raw appeal and it seems like a good idea to keep our options open. A difficulty with taking hole theory seriously is that a satisfactory mathematical framework has apparently not been developed in detail. The purpose of this paper is to fill this gap by giving a construction of the Dirac field operator based on hole theory. The basic idea is that if an n-fermion state is modelled by an n-fold wedge product of Hilbert spaces, then the Dirac sea should be described as an infinite wedge product of Hilbert spaces.'

Thanks
Bill

5. Jul 17, 2014

atyy

Last edited: Jul 17, 2014
6. Jul 17, 2014

bcrowell

Staff Emeritus
That's about tachyons, which would have imaginary mass and spacelike energy-momentum vectors. I'm talking about particles with real mass and timelike energy-momentum vectors in the past light-cone.

I'm talking about particles with negative mass-energy; there's no reason for them to have negative mass.

7. Jul 17, 2014

strangerep

Well, let's do a scattering "experiment"...

Suppose we arrange a head-on collision between a particle #1 of mass $m$, energy $+E$, and momentum $+P$ (approaching from $-\infty$ on the x-axis), and a particle #2 of mass $m$, energy $-E$, and momentum $-P$ (approaching from $+\infty$ on the x-axis).

(Both particles are spinless and chargeless.)

What happens?

Hint: it's rooly, rooly cool! Just what everyone has always wanted.

8. Jul 18, 2014

DrDu

I always considered QFT to exactly equivalent to the introduction of the Dirac sea. You also redefine anihilation operators of particles with negative energy as creation operators of positrons and the normal ordering is equivalent to brushing infinite self energy terms under the carpet.

9. Jul 18, 2014

vanhees71

Indeed, at least for QED, one can show that Dirac's "hole theory" (referring to Dirac's original interpretation of the positrons as holes in the Dirac sea filled with electrons) is equivalent to the modern QFT formulation of QED.

However, the Dirac sea approach to relativistic QT is a bit unlogical. You start with a single-particle approach, namely with the attempt to build a relativistic theory analogous to Schroedinger's non-relativistic wave mechanics, describing a single particle's state with a square integrable wave function. This formalism assumes that you have a situation where you always deals with one particle, free or moving in an external field.

As it turns out for the relativistic case already the motion of a particle in an external field (potential) cannot be described when restricting the energy spectrum to positive values (or any values with a lower boundary, i.e., the assumption that you have a stable ground state, which is mandatory to have a "stable universe" at all. That's why Dirac introduced his trick, working for fermions: He considered the negative-energy states filled with electrons and came to the conclusion that in fact he deals with a many-body theory with holes equivalent to positively charged antielectrons (nowadays called positrons).

The QFT approach is a bit more consistent, because it starts from the fact that at relativistic energies a single-particle description doesn't work and that you need to consider a many-body approach, for which QFT is the most convenient formulation. Then the entire trick is to write a creation operator with the opposite momentum in front of the negative-frequency modes, and you get a antiparticle with positive energy and the opposite charge than the particles (but with the same mass as the particles). With this Feynman-Stueckelberg trick you get a theory with an energy spectrum bounded from below, i.e., with a stable ground state as wanted. So that's the more convincing approach to relativistic QT than the old Dirac approach.

10. Jul 18, 2014

Staff: Mentor

I take your word that its exactly equivalent but QFT seemed to handle it more naturally.

I always found it a bit illogical - sea of what - positrons or electrons - seems to work with either.

Same here. I always found QFT quite natural. Standard QM treats position as an observable and time a parameter - treating both as a parameter ie using a field - just seemed natural. Everything is a field.

Thanks
Bill

Last edited: Jul 18, 2014
11. Jul 18, 2014

DrDu

I have a solid states physics background and the dirac picture makes much sense there. E.g. in a semiconductor, the electrons near the top of the valence band have a negative effective mass, so that the holes have a positive one. There is in deed a positively charged background - of the atomic cores - which compensates for the negative charge of the electrons in the valence band. So there, the Dirac sea describes actual physics.

12. Jul 18, 2014

bcrowell

Staff Emeritus
Maybe I shouldn't have posted this in the quantum physics subforum, but my question isn't how we should interpret the negative-energy states quantum mechanically. My question is what rules out such states classically.

13. Jul 18, 2014

bcrowell

Staff Emeritus
I would say that if they have no charge, then they don't interact. They pass through each other, and their states are unchanged. Was that your point?

14. Jul 18, 2014

atyy

I don't think such states can be ruled out classically, at least not from the links in post #5. However, those don't address nonlinear stability, so that may change considerations.

15. Jul 18, 2014

Staff: Mentor

$E=\gamma m c^2$, $p=\gamma m v$
So negative mass also has negative energy, and negative momentum. I think classically, there is nothing that prevents them from existing (you can just plug in negative values in all Newtonian and relativistic equations), but they would give consequences never observed. They would also give an easy way to get limitless energy if they can be produced.

16. Jul 18, 2014

bcrowell

Staff Emeritus
I'm not asking about negative mass. I'm asking about negative values of E that are solutions of the equation $m^2=E^2-p^2$. The equations $E=m\gamma$ and $p=m\gamma v$ are not true in general; they only hold for particles with energy-momentum vectors inside the future-directed light cone. They fail for massless particles, and also for everything else outside the future light cone.

One could argue that negative-energy states *are* observed, in the Dirac "hole" interpretation.

Right. This is what I was alluding to in the final paragraph of #1.

17. Jul 18, 2014

Staff: Mentor

Where, if not in the future light cone, would those particles be?

But then you are beyond 1906.

18. Jul 18, 2014

Jilang

19. Jul 18, 2014

Renormalized

Negative frequencies are good

First, in science nothing is proved wrong. It is simply less credible than other possibilities. In this case, the other possibility is Quantum Field Theory. QFT interprets the “negative energies” as negative frequencies. Negative frequencies are required to preserve causality. Without them, causal relationships can occur outside the light cone. (i.e. faster than light communication) Feynman explains this in the 1986 Dirac Memorial Lecture: The lecture is also available from Amazon as a pamphlet.

The Feynman Lectures on Physics, Volume 2, chapter 28 also points out a problem with the quantum mechanical interpretation of an electron. It will have infinite energy. That problem is solved with the renormalization procedure of QFT.

Last edited by a moderator: Sep 25, 2014
20. Jul 18, 2014

bcrowell

Staff Emeritus
In the past light cone, because they have negative energy.

Right.