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

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It's often claimed that the Dirac sea is obsolete in quantum field theory.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

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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?

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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, 2019 by (15,488 points)
edited Dec 8, 2019

On the same footing one might conclude that the very Dirac equation plays no role at all, does it?

One more question to Arnold: what spin has a dressed particle?

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.

The point is that the (positive energy part of the) Dirac equation can handle only a single electron, and hence has no place in quantum field theory. There is no consistent Lorentz invariant multiparticle Dirac equation.

Rubbish! In QFT all particles have "positive energies".

...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.

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!

As far as I remember, when we introduce the occupation number formalism (= "secondary quantization" = QFT), we use the corresponding one-particle (or one-quasiparticle) equations; thus the operators satisfy formally the same equations supplied with many-particle boundary (asymptotic) conditions.

No. As Weinberg shows, one may use for a free field theory any irreducible representation of the Poincare group. In the massive spin 1/2 case it happens to be equivalent to the representation on the positive frequency solutions of the free Dirac equation, but nowhere is made use of this fact.

The positive frequency solutions of the multiparticle Dirac equation contain mostly unphysical solutions, even in the free case. Thus one cannot second-quantize the Dirac equation itself, only its positive frequency part!

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I would propose to look at this all in the following way.

Consider states of QED containing a single electron with energy below the positron creation threshold.

The sea itself, as non observable thing, is meaningless as any non observable thing. But in case of Dirac equation, it is not that non observable; rather, the holes in it are positive-energy and positive-charge particles - positrons. So, the Dirac sea is not detached from us.

But let us consider energies smaller than the hole creation threshold. Do "negative energy" states participate in the real (or dressed) electron dynamics? The answer is yes - they are involved in the radiative corrections to zeroth-order Dirac equation solutions. Often they are called "virtual particles".

In order to simplify things, let us consider a spinless non relativistic particle in a 1D box: $0\le x\le L$ with some poitential in it, say a linear potential $U(x)=k\cdot x$. The boundary conditions for the solutions are reflecting: $\Phi(0)=0,\; \Phi(L)=0$. And the initial condition is such that only the ground state may exist - because of lack of energy to be in the excited states. In that case the exact solution is well known - it is a combination of Airy functions $\Phi_0 (x)\cdot \text{e}^{-iE_0t}$ with a certain ground energy $E_0(L,k)$. No excited state can be found experimenatlly in the ground state.

On the other hand, we may solve this well posed problem by the perturbation theory with considering the potential $U(x)$ a perturbation to the pure "box solutions" $\varphi(x)^{(0)}=\psi_n(x)\sqrt{\frac{2}{L}}\sin\left(\frac{\pi\cdot n}{L}\right), \; E_n^{(0)}\propto \left(\frac {\pi \cdot n}{L}\right)^2$, $$\Phi_0 (x,t) = \text{e}^{-i\cdot t\sum_{n\ge 0}^{\infty} E_0^{(n) }} \cdot\sum_{m\ge 0}^{\infty} C_m\varphi^{(0)}_m(x)\;\;\; (1).$$ The zeroth-order solutions $\varphi^{(0)}_n(x)$ are some sort physical as they may be decent approximations to the exact solutions in the solutions like this one: $$\Phi(x,t)=\sum_{n\ge 0}^{E_{\text{max}}} A_n \varphi_n(x)\text{e}^{-iE_n t}\approx \sum_{n\ge 0}^{E_{\text{max}}}A_n^{(0)}\varphi_n(x)^{(0)}\text{e}^{-iE_n^{(0)} t}\;\;\; (2).$$ They may be "found" experimantally.

However, if we express, for example, the exact ground state solution via zeroth-order ones (1), they may not be considered physical since they are just numerical correction to the initially inexact $\varphi_0(x)$. One can call such numerical corrections as due to "virtual states", but nevertheless they may not be found experimentally (except for the zeroth-order one, of course).

Here the excited states $E_{n\gt 0}$ may be considered as "hole states" - those with a positive and higher energies than the ground state. Now it is up to you to decide if they "play some role".

answered Dec 9, 2019 by (102 points)
edited Dec 9, 2019

The question was about quantum field theory whereas you write about a single particle....

As a QFT I just considered the QED with a single electron with the enrgy below the positron creation. Maybe I was not clear about it, but I meant exactly this. One can "feel" the presence of other excited states ("hole states") in the right way.

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