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

Question

It's often claimed that the Dirac sea is obsolete in quantum field theory.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 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

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.

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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?

Answer 1

I think the only problem lies in the renormalization, but it's not really a conceptual problem, rather a mathematical one (oh well..). Let me try to explain.

Consider a simple tight binding Hamiltonian in one dimensions. The dispersion (one-particle energy) is $-t\cos(k)$. For $-\pi/2 \le k\le \pi/2$ (and zero chemical potential) the dispersion is negative. To minimize the energy the particles will fill those states (one per state according to Pauli principle). There is your Dirac sea. Actually in this case it's called Fermi sea. Excitation above it are particles while removal of particles from the sea are holes (antiparticles).

In case of the relativistic electrons the dispersion is $\epsilon_p = \pm \sqrt{c^2 p^2+m^2c^4} $ where $c$ is the speed of light $p$ the momentum and $m$ the electron's mass. These are two hyperbolae in the energy-momentum plane and clearly there are negative energies. Now fill all the states with negative energy and you'll get the Dirac sea. The state that corresponds to this situation is what you call $|\Sigma\rangle$. The only problem with respect to the previous situation is that the total energy of the Dirac sea is formally minus infinity (the integral of $-\sqrt{c^2 p^2+m^2c^4}$ in $dp/(2\pi)$ from minus to plus infinity). Call $E_0$ such energy (formally infinity). An electron with momentum $p$ will have energy $|\epsilon_p| + E_0$ but in experiments we will always measure energy differences with respect to the Dirac sea which is $|\epsilon_p|$.

The problem remains how to properly define the state $|\Sigma\rangle$ given that its energy $E_0$ is infinite. The way to resolve this (and related) issues is the subject of 'renormalization'.

For example you can take a (large) cutoff in momentum space. At this point the energy of the Dirac sea is finite. Do all your calculations and send the cutoff to infinity at the end.

Answer 2

The Dirac sea at the time may have seemed an idea worth entertaining. In modern days one has to immediately conclude that it is fundamentally flawed and indefensible. Unfortunately some courses still present it as sensible.

The Dirac sea implies uniform infinite charge density throughout the universe. Fluctuations would give large effects. The repulsion would represent an energy do unimaginably huge that it will dwarf the infamous zero point energy. It s gravitational effect would require the universe to be much smaller the football size implied by ZPE. An unoccupied orbital will not behave like a positron. There will be electron correlation effects that imply that imply a different mass and magnetic moment (g-factor) from an electron. Also there should be Auger like decay mechanisms.

The Dirac sea is an instructive idea but not in any way an acceptable explanation of the positron.

Deleting this answer nor renormalisation can save this ludicrously flawed concept.