Potential energy in the Dirac equation

[snoopies622]

Why does the Dirac equation not have a potential energy term? The Schrödinger equation does, and the Dirac equation is supposed to be the special relativity version of the Schrödinger equation, no?

 

[DarMM]

It can actually, it's just an artefact of the way physics is commonly taught. When
you see the Dirac Equation you are usually on the way to Quantum Field Theory where
the function obeying the Dirac equation is understood as a field and not a wavefunction. In this case a potential term wouldn't be physically sensible.

However in relaitivistic chemistry, where the Dirac equation is used as a relativisitc model
for the electron, potential terms are added.

The only problem with adding potential terms is that the Dirac equation with a potential is often
fraught with technical issues related to operator theory. For example, if you're familiar with
Helium treated with the Schrödinger equation (perturbation of Hydrogen, various expansions
can be used to compute spectra, e.t.c.), you know that Helium is perfectly well behaved with
a stableground state.

However it isn't known if Dirac Helium has any stable state, that is if the electrons are bound to
the nucleus at all.

In addition, many potentials cause the Dirac equation to have non-unitary time evolution.

 

[dextercioby]

You can add a Coulomb term to the 1 particle Dirac eqn. and get a good approximation of the H-atom spectrum, up to quantum field effects, such as the Lamb shift.

 

[vanhees71]

That's all true, but one should emphasize that the Dirac equation is not so easily interpreted simply as the relativistic version of the Schrödinger (or more accurately the Pauli equation) of an elementary spin-1/2 particle.

The reason is that the naive interpretation of the Dirac equation as an equation of motion for a single-particle wave function doesn't work out, because even for the free Dirac particle the energy is not bounded from below, i.e., there is no stable ground state. For free particles you could say, I just only consider the positive-energy states as physical, but as soon as you introduce a potential, there is always a finite probability to also find negative-energy states in the time evolved wave function.

The way out for Dirac was the idea to interpret the Dirac equation as a many-body description, where all states with negative energy are filled with particles and renormalizing the infinite energy of this state to be 0 and this as the vacuum state. With the Dirac sea occupied, no particles could be scattered into it, because of the Pauli exclusion principle (i.e., the fermionic nature of spin-1/2 particles).

Coupling the Dirac field to the electromagnetic field (not only a potential but in a Poincare-covariant way), via the "minimal coupling" lead to the socalled "hole-theoretical formulation" of QED. This is, however, an awfully complicated way to look at the problem. After all hole theory is a many-body interpretation of the single-particle Dirac equation, and it's way more convenient to start in a QFT formulation right from the beginning. Then there is no more Dirac sea (which is unobservable anyway) and everything is more consistent to begin with. Another great advantage of QFT is that it can also be easily applied to bosons (integer-spin particles).

The reason, why a description of the hydrogen atom in the "naive" single-particle way works to a certain extent is the smallness of the electromagnetic coupling constant (Sommerfeld fine-structure constant). One can show this by typical soft-photon-resummation techniques also from QFT. A very good explanation of this systematic approach to the hydrogen atom in relativistic QED can be found in Weinberg, Quantum Theory of Fields, Vol. I.

 

[snoopies622]

Thanks all, much to learn here. One question:

...the function obeying the Dirac equation is understood as a field and not a wavefunction.

Isn't a wavefunction a kind of field?

 

[Bill_K]

the function obeying the Dirac equation is understood as a (multiparticle) field and not a (single particle) wavefunction.

 

[PhotonCurve]

Why does the Dirac equation not have a potential energy term? The Schrödinger equation does, and the Dirac equation is supposed to be the special relativity version of the Schrödinger equation, no?

You can have a potential term for the Dirac Equation. I'll tell you a more exotic case, a Lippmann-Schwinger approach for instance. You'd write out the Dirac Hamiltonian

$$H_D \phi = (c \alpha \cdot P + \beta mc^2)\phi$$

That's just

$$H_D| \phi > = E| \phi >$$

A potential term is just an ''interaction term'' for scattering experiments so we can add this in

$$(H_D + V)| \psi > = E|\psi>$$

in the case of the continuity of eignevalues $$\phi \rightarrow \psi$$ and $$V \rightarrow 0$$ [a] solution can be found using the Lippmann-Schwinger equation, just as an example. The potential, since we are dealing with electrically-charged particles, might be

$$V = e \phi$$

which is known as the electrostatic potential.

 

[mpv_plate]

Isn't a wavefunction a kind of field?

Wavefunction is something else than a field.

Wavefunction is a function of the possible configurations of a system. The variables can be position, momentum, spin and other quantities if needed. The wavefunction returns a complex number (probability amplitude) as a value.

Quantum field is a function of space and time only (no momentum, no spin). The quantum field returns an operator (not a probability amplitude) as a value.

 

[Hyper_ilyas]

reply

Why does the Dirac equation not have a potential energy term? The Schrödinger equation does, and the Dirac equation is supposed to be the special relativity version of the Schrödinger equation, no?

as in all equations we can add a potentiel in the Dirac's equation, and all what we do is adding the potentiel to the Dirac's Hamiltonian like the coulamb potential v=e/r , or the step potential or young milles potential..

 

[PhotonCurve]

Wavefunction is something else than a field.

No, the wave function is a field of probabilities.

 

[mpv_plate]

No, the wave function is a field of probabilities.

Yes, but in general it is a field in a different "space" than a quantum field of QFT.

 

[snoopies622]

$$(H_D + V)| \psi > = E|\psi>$$

Ah, there we go. :)

The wavefunction returns a complex number (probability amplitude)... the quantum field returns an operator.

Thanks mpv_plate. I've never been clear on what was meant by , "in quantum field theory, the fields become operators." This helps.