# Is Dirac's equation still useful after QED is developed?

1. Jan 13, 2013

### wdlang

Dirac's equation is just a low energy limit of QED

it is not exact

2. Jan 13, 2013

### andrien

it is exact.

3. Jan 13, 2013

### Staff: Mentor

It predicts a g-factor of 2, which is not correct (but a good approximation) - not surprising, as Dirac's equation is not QED.

4. Jan 13, 2013

### andrien

that g factor of 2 which is not exact can also be derived from dirac's qed. No additional pauli type interaction is required.it is exact.

5. Jan 13, 2013

### wdlang

so is it worthwhile to study Dirac equation?

it seems to be between the schrodinger equation and QED

6. Jan 13, 2013

### andrien

dirac eqn,quantized electromagnetic field and with renormalization scheme you get qed.

7. Jan 13, 2013

### tom.stoer

It depends what you mean.

Dirac eq. in QED is exakt as an Operator eq. when taking into account renormalization.

Dirac eq. as eq. in rel. QM is an approx. only.

Afaik rel. QM is used in several applications like (numerical) quantum chemistry, atomic and molecular physics etc.

8. Jan 13, 2013

### vanhees71

I think it is important to note clearly that, as tom.stoer wrote in the previous posting, that a one-particle interpretation of the Dirac equation in analogy to the one-particle interpretation of the nonrelativistic Schrödinger equation is inconsistent.

Dirac solved this problem by invoking a genius argument to extend the one-particle interpretation to a many-particle interpretation by filling up his fictitious sea and interpreting holes in the sea as positrons (after some quibbles, because in an earlier attempt in interpreting the holes he tried to interpret them as protons, but Oppenheimer pointed out that the reinterpreted holes (in the non-interacing case!) must have the same mass as an electron but of course must carry the opposite charge). As far as I know, Diracs hole theoretical reinterpretation of the unquantized Dirac equation is, with some difficulties, quite equivalent to modern QED, but it's unnecessarily complicated to deal with.

Thus, in my opinion it is way better to study relativistic QT right away from the modern point of view as a local microcausal QFT than to struggle with the clumsy old hole theory. It is also clear, in which context the Dirac equation of an electron in an electrostatic field can serve as an approximative solution of the bound-state problem in atomic physics. This approximation naturally appears in the soft-photon resummation techniques for electron-nucleus scattering, leading to a Bethe-Salpeter ladder in leading order, which is equivalent to solving the Dirac equation for a electron in the electrostatic field of the nucleus. It also provides the systematic way to calculate the radiative corrections to this picture, which have played soch an important role in the development of modern QED, because this leads to a full explanation of the Lamb shift. Also the anomalous magnetic moment of the electron is just easily explained by the radiative corrections to the electron-positron-photon vertex. That's way easier in QED (with Feynman diagrams at hand to organize the calculation) than with hole theory, which I wouldn't know how to use for this purpose.

S. Schweber, QED and the Men who Made It.

9. Jan 13, 2013

### DrDu

Actually, the value 2 for the g factor is the non-relativistic result. So you don't need the Dirac equation at all to derive it.

10. Jan 13, 2013

### tom.stoer

No, it isn't. You can't derive g=2 w/o using SR.

11. Jan 13, 2013

### dextercioby

Sure you can. Check out the work by Levy-Leblond in the '60s. And particularly his article with the reference:

Comm. math. Phys. 6, 286--311 (1967)

which is freely available on the internet through <Projecteuclid>.

Last edited: Jan 13, 2013
12. Jan 13, 2013

### dextercioby

The study of Dirac's equation as a 1-particle equation is useful for at least 3 reasons:

* purely historical (physics is generally taught from a historical perspective for a good reason).

* specially-relativistic QED doesn't come out of thin air. You can't just write down the Lagrangian without explaining where it comes from.

* if a study of quantum mechanics is made with emphasis on the Schrödinger equation, which is not in agreement with special relativity, why not try to write down an equation for 1 particle in agreement with special relativity and try to apply it to the H atom for example...

13. Jan 13, 2013

### tom.stoer

Interesting!

Last edited by a moderator: Jan 13, 2013
14. Jan 14, 2013

### vanhees71

The only trouble is that this is not a unique prescription. The point is to "gauge" the Pauli equation. The analysis by Levy-Leblond can be shortened a bit. The basic point, why he gets the correct gyro-factor of 2 is that he takes a first-order differential equation representation of the wave function's dynamics as Dirac did for his equation. This derivation can be much shortened by just taking the analysis of the unitary ray representations of the Galilei group.

Non-relativistically, i.e., from the analysis of the Galilei group you get
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}$$
as the free-particle Hamiltionian of particles of any spin. If you "gauge" this, using the principle of minimal coulpling, you don't get the correct coupling of the spin to the em. field.

Now you can use the following trick. You just write
$$\hat{H}=\frac{(\vec{\sigma} \cdot \hat{\vec{p}})\cdot (\vec{\sigma} \cdot \hat{\vec{p}})}{2m},$$
which is the same as above, because the momentum components commute, and the Pauli matrices generate the Clifford algebra of 3D Euclidean space, i.e., fullfil the anticommutation relations
$$\{\sigma_j,\sigma_k \}=2 \delta_{jk}.$$
If you now do the "gauging" via minimal coupling, you find
$$\displaystyle{\hat{H}_{\text{em}}=\frac{[\vec{\sigma} \cdot (\hat{\vec{p}}-q \vec{A}(t,\hat{\vec{x}}))] \cdot [\vec{\sigma} \cdot (\hat{\vec{p}}-q \vec{A}(t,\hat{\vec{x}}))]}{2m} + q \Phi(t,\hat{\vec{x}})},$$
and this gives a different result than the naive "gauging", because the momenta and the vector potential components don't commute. Of course $\Phi$ and $\vec{A}$ are the scalar and vector potential of the electromagnetic field.

Multiplying out the above Hamiltonian you get the correct Pauli Hamiltonian for spin-1/2-particles:
$$\displaystyle{\hat{H}_{\text{em}}=\frac{[\hat{\vec{p}}-q \vec{A}(t,\hat{\vec{x}})]^2}{2m} -\frac{q}{2m} g_s \hat{\vec{S}} \cdot \vec{B}(t,\hat{\vec{x}})+q \Phi(t,\hat{\vec{x}}) \quad \text{with} \quad g_s=2.}$$
Without the trick to introduce the vector potential into the non-naive Hamiltonian you wouldn't have gotten the coupling of the spin with the magnetic field $\propto \vec{B} \cdot \hat{\vec{S}}$ with $\vec{B}=\vec{\nabla} \times \vec{A}$ at all.

That this is the "correct" equation for an elementary particle can only be derived by taking the non-relativistic limit of the Dirac equation.

What's of course right is that spin is not a relativistic feature per se but it emerges as "naturally" from the analysis of the unitary ray representations of the Galilei group as it arises from the analogous analysis for the Poincare group.

15. Jan 14, 2013

### DrDu

Yes, but this holds also true in case of the Dirac equation. Gauging the Klein Gordon equation evidently also does not yield a g factor as there is no spin.

16. Jan 14, 2013

### andrien

the point emphasized here by vanhees is treated in sakurai advanced quantum mechanics.Also it is somewhat artificial(it was pointed out by feynman first).Also the paper relies on linear factorization which is at heart of dirac formalism so it is not a new idea.

17. Jan 14, 2013

### vanhees71

The difference is that the Dirac equation for a spin-1/2 particle naturally emerges from the analysis of the proper-orthochronous Poincare group augmented with parity invariance. The most natural Lagrangian for a spin-1/2 particle turns out to be the Dirac Lagrangian, leading to the first-order equation (in space and time). Then minimal coupling of an Abelian gauge field leads to QED with a (tree-level) value $g=2$.

In the case of the non-relativistic Pauli equation, it's more of an artificial trick. Of course, finally the correct dynamics only emerges from observations. One has simply to check, how particles behave in nature, i.e., whether the one or the other way of "gauging" leads to the correct coupling of the em. field to the particle's degrees of freedom.

18. Jan 14, 2013

### dextercioby

Group theory aside, it is no more artificial to linearize the Schrödinger equation than to linearize the Klein-Gordon equation. Both attempts lead to better equations, the Dirac one in special relativity and Pauli one in Galilean relativity.

19. Jan 15, 2013

### vanhees71

Well, the KG equation (or better its quantized form) is a nice application, when applicable, e.g., if you like to describe pions.