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

It's often claimed that the Dirac sea is obsolete in quantum field theory.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

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?

• physics.stackexchange.com/a/19381/25851 – bolbteppa Nov 21 '19 at 17:55
• What is your question exactly? The point is renormalization (?) – lcv Nov 22 '19 at 11:21
• @lcv Are $|\Sigma\rangle$ and $|0\rangle$ related by renormalization? Or formulated differently, is $|0\rangle$ the ground state in the renormalized theory? If this is the case, I would love to read how this idea can be made precise – jak Nov 22 '19 at 11:52
• No what I meant is that the only problem lies in the renormalization. I'll write a short answer that should address your points. – lcv Nov 22 '19 at 13:29
• @lcv I'm looking forward to it! – jak Nov 22 '19 at 13:41

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.

• @BenCrowell you are absolutely right. I meant the (relativistic) electrons of course, see the edit. – lcv Nov 22 '19 at 15:08
• Thanks! Do you how we can renormalize the vacuum charge or energy in practice? Usually we regularize and then show that the cutoff drops out from observable quantities, e.g., by introducing renormalized charges. But I'm unsure how this really works for the vacuum charge or energy. How does the cutoff gets absorbed? I think it should be possible to introduce a bare vacuum charge and energy and then absorb the infinite contribution by defining that the renormalized vacuum charge and energy have the values that we measure in experiments (zero). However, I've never seen this spelled out explicitly – jak Nov 23 '19 at 7:03
• It's nothing fancy. It's what I just described. If what you need are energies of certain processes, simply subtracting the energy of the Dirac sea it's the way to go. Since this energy is infinite if you want to be a bit more formal, you first introduce a cutoff. Do the subtraction. At this point the end result (an energy difference) is independent of the cutoff as you said, and you can send the cutoff to infinity. Of course I only considered a free theory where the steps are simple... – lcv Nov 23 '19 at 14:22
• ...If your theory is interacting presumably you will get your answer in some form of weak coupling expansion. Then you will have to make sure that at every order of perturbation theory renormalization works. In fact we say a theory is renormalizable if you can get finite results by introducing a finite number of renormalization parameters. But as I said, this is somehow a mathematical complication or that's what people first thought when QFT was first 'cured'... – lcv Nov 23 '19 at 14:22
• ...Nowadays we thing there are some important physical questions and concepts behind the procedure of renormalization, for example the underlying nature of space (or time). But that's another question. – lcv Nov 23 '19 at 14:22

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.

• I challenge the downvoters. Ha. – my2cts Nov 21 '19 at 18:17
• It doesn't appear to me that you've taken much care in reading the question. In particular, Jackiw's Dirac Prize lecture directly and specifically contradicts your answer. – d_b Nov 21 '19 at 22:04
• @d_b I don't care if it was Dirac himself. The concept does not make any sense. Nonsense. – my2cts Nov 21 '19 at 22:10
• the bare mass and charge of individual particles are formally infinite too. Thus the same arguments regarding gravitational effects etc. apply. So would you say that QED is failed idea too? Secondly, what if there are an equal number of positively and negatively charged particles in the Dirac sea? see books.google.de/… – jak Nov 23 '19 at 6:45
• @jak QED is very accurate so in spite of the renormalisation difficulties it must be accepted. One day we may be able to fix these difficulties in a more satisfactory way. Dirac sea only makes ludicrously wrong predictions. – my2cts Nov 23 '19 at 8:12