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  What's the role of the Dirac vacuum sea in quantum field theory?

+ 2 like - 0 dislike
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It's often claimed that the Dirac sea is obsolete in quantum field theory. On the other hand, for example, Roman Jackiw argues in [this paper][1] that

Once again we must assign physical reality to Dirac’s negative energy
sea, because it produces the chiral anomaly, whose effects are experimentally observed, principally in the decay of the neutral pion to two photons, but there are other physical consequences as well.

Moreover, Roger Penrose argues in his book "Road to Reality" (Section 26.5) that there are two "proposals" for the fermionic vacuum state:

 - the  state $|0 \rangle$ which is "totally devoid of particles", and
 - the the Dirac sea vacuum state $|\Sigma\rangle$, "which is completely full of all negative energy electron states but nothing else".

If we use $|0 \rangle$, we have the field expansion $\psi \sim a + b^\dagger$ where $a$ removes a particle and $b$ creates an antiparticle. But if we use $|\Sigma\rangle$, we write the field expansion as $\psi \sim a + b$ where now $b$ removes a field from the Dirac sea which is equivalent to the creation of an antiparticle. 

He later concludes (Section 26.5)

 the two vacuua that we have been considering namely $|0 \rangle$ (containing no particles and antiparticles) and $|\Sigma\rangle$ (in which all the negative-energy particle states are fulled) can  be considered as being, in a sense, effectively equivalent despite the fact that $|0 \rangle$ and $|\Sigma\rangle$ give us different Hilbert spaces. We can regard the difference between the $|\Sigma\rangle$ vacuum and the $|0 \rangle$ vacum as being just a matter of where we draw a line defining the "zero of charge".

This seems closely related to the issue that we find infinity for the ground state energy and the total ground state charge as a result of the commutator relations which is often handled by proposing normal ordering. To quote again [Roman Jackiw][2]

 Recall that to define a quantum field theory of fermions, it is necessary to fill the negative-energy sea and to renormalize the infinite mass and charge of the filled states to zero. In modern formulations this is achieved by “normal ordering” but for our purposes it is better to remain with the more explicit procedure of subtracting the infinities, i.e. renormalizing them.

---

So is it indeed valid to use the Dirac sea vacuum in quantum field theory? And if yes, can anyone provide more details or compare the two approaches in more detail?


  [1]: https://arxiv.org/abs/hep-th/9903255
  [2]: https://arxiv.org/abs/hep-th/9602122

asked Nov 22 in Theoretical Physics by JakobS (105 points) [ revision history ]

1 Answer

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The Dirac equation for hydrogen is a single particle equation for an electron in an external Coulomb field. There is no Lorentz invariant Dirac equation for 2 or more interacting particles producing results conforming to experiment. Thus the Dirac equation (and the Dirac sea) is not more than a very limited motivation for a quantum field theory of electrons. (But augmented by correction terms from form factors and self-energy, its positive energy part plays a role as an effective equation for single particles.)

The version of quantum field theory (QFT) relevant for practical use is the renormalized version with dressed (renormalized) physical particles and fields There the Dirac sea (which is a concept related to bare, unphysical particles) plays no role at all.

Whether it is used somewhere on the road to motivating the formulas for renormalized QFT is a matter of personal taste; it is certainly not needed. 

For a treatment in which no Dirac sea ever appears but anomalies are properly accounted for see, e.g., Weinberg's QFT volumes.

answered Nov 29 by Arnold Neumaier (14,069 points) [ revision history ]
edited 12 hours ago by Arnold Neumaier
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Yes,  the Dirac equation plays no role at all in quantum field theory. But, augmented by correction terms from form factors and self-energy, its positive energy part plays a role as an effective equation for single particles. The spin of a dressed particle is the same as that of the bare particle; the dressing transform is the limit of a unitary transformation which preserves the single-particle representations of the Poincare group.

Do you understand, Arnold, what you are saying? The Dirac equation plays no role, but augmented with correction terms (expressed via the Dirac equation solutions)  one obtains something. Something from nothing. Not too convincing to me.

The Dirac equation plays no role at all in quantum field theory. This is a simple fact.

But from the latter one can calculate form factors and self-energy. Then one can use this to get an effective 1-particle theory (not another QFT) which contains meaningful information

The right statement is the following: The Dirac equation plays a fundamental role as a decent approximation to the real electron, see, for example Hydrogen levels obtained within the Dirac theory. The rest is small radiative corrections. The Dirac sea is a meaningful idea borrowed from solid (or condensed) state physics; that's why it works. In addition, it inevitably introduces many-particle (particle-antiparticle) picture unlike the Schrödinger equation.

Instead of making empty claims point to a paper demonstrating that a Lorentz invariant Dirac equation for 2 or more interacting particles produced results conforming to experiment! The Dirac equation for hydrogen is a single particle equation for an electron in an external Coulomb field, not a 2-particle equation!
 

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...and their interaction is not described by Dirac's equation!

It really depends on how we introduce the interaction. Our originally (blind) guess furnished with the corresponding counter-terms is some sort of physical interaction.

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