# Dirac's field

1. Apr 30, 2007

### jostpuur

Why does everybody say that classical Dirac's field has no physical meaning? This doesn't fully make sense to me. Why should classical EM field be more real than Dirac's field? The idea of a classical field describing electrons sounds like something that could be used to describe large number of electrons as a continuous matter. I'm absolutely certain, that you could of course approximate large number of electrons at least somehow as a continuous matter, but if we do so, would it be precisly the Dirac's field that we need for it?

2. May 1, 2007

### Demystifier

In quantum field theory, Dirac field describes fermions rather than bosons, which means that it satisfies ANTIcommutation relations, rather than commutation relations. Consequently, in the classical limit, this field is decribed by anticommuting Grassmann numbers. Such numbers cannot not represent physically measurable quantities.

3. May 1, 2007

### quetzalcoatl9

Can you explain this a bit more?

My impression was that the Dirac field was not wrong per se, but that the explanation for the negative eigenvalue solutions was not attractive and, ultimately, unnecessary.

4. May 1, 2007

### jostpuur

I never though about anticommutation relations surviving in the classical limit, so I appreciate the answer. I haven't had time to wonder that matter much yet, but I could throw some other questions too to not make life too easy for those who try to answer. Do these strange properties of the field produce any real problems? Shouldn't it be enough if the energy, momentum and angular momentum are measurable?

5. May 1, 2007

### TriTertButoxy

This 'physical meaning' that people speak of is the ability to build up a field strength to measure at the classical level. To build up a macroscopic strength of any field you need to be able to have a large number of quanta (particles) with identical quantum numbers. If this is possible then you can get the field strength that is observed classically. These states are coherent states which are superpositions of multiparticle states:

$$|\alpha,\,\mathbf{p}\rangle=e^{|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}\,|n,\,\mathbf{p}\rangle\,,$$​

with $|n,\,\mathbf{p}\rangle=a^\dag_{\mathbf{p}}\ldots a^\dag_{\mathbf{p}}|0\rangle$ being defined as the $n$-particle state of momentum $\mathbf{p}$. The parameter $\alpha$ sets the amplitude (and phase) of the coherent state (plane wave).

Now to answer your question: we never speak of the electron field or the dirac field in undergraduate chemistry and physics courses because we can never build up macroscopic field strength. This is because as Demystifier mentioned, the Dirac (electron) field is quantized using anticommutation relations: $\{a^\dag_{\mathbf{p}},\,a^\dag_{\mathbf{p'}}\}\,=0$. What does this mean? It means we can create a 1-particle state by acting the creation operator on the vacuum state: $a^\dag_{\mathbf{p}}|0\rangle=|1,\,\mathbf{p}\rangle$, but the 2-particle state vanishes: $a^\dag_{\mathbf{p}}a^\dag_{\mathbf{p}}|0\rangle=-a^\dag_{\mathbf{p}}a^\dag_{\mathbf{p}}|0\rangle=0$, where in the first step, I've used the anticommutation relation (hard to tell since the $a^\dag_\mathbf{p}$'s are the same). All higher-particle states vanish for the same reason.

So, look up at the formula for the coherent state. The sum abruptly terminates after $n=1$ for the Dirac field, preventing us from building up that macroscopic strength we were looking for. This is why the Dirac field is said to be 'classically unphysical.'

Last edited: May 2, 2007
6. May 2, 2007

### Demystifier

TTB, good explanation!

7. May 2, 2007

### Jimmy Snyder

Thanks TriTertButoxy for a very lucid explaination. It explains more to me than I have been able to get out of reading QED and QFT books for months now.

Last edited: May 2, 2007
8. May 3, 2007

### Anonym

Pity, the statement “|n,p> being defined as the n-particle state of momentum p. The parameter alpha sets the amplitude (and phase) of the coherent state (plane wave)” is completely wrong.

Dany.

9. May 4, 2007

### Jimmy Snyder

Thanks. Can you correct the error in the completely wrong statement?

10. May 4, 2007

### Anonym

The error is due to confusion between the notion of the coherence and the coherent state. The coherent state is defined as superposition of eigenfunctions of the harmonic oscillator (Hermit polinoms) and not plane waves. The coherent state is the minimum uncertainty state delta x*delta p = h/2. Obviously, delta p not =0. You may define the coherent state as eigenstate of the annihilation operator. Therefore, the above answer has nothing to do with the OP original question (in addition, it describes the single particle state).

We treat the Dirac electron as carry the addition degree of freedom: spin. If that degree of freedom gets lost during the transition from Quantum world to the Classical world then the Dirac field can not be classical. The situation is very interesting. The spin of free moving electron was never measured and N.Bohr states that it is impossible in principle. However, if I consider the bounded solutions of the same Dirac equation (electron) in the presence of the external EM field the spin is measured easily in the different setups (the classical example is SG). Since the spin is conserved quantity only together with the orbital angular momentum it is reasonable to expect that it is unobservable in the absence of it. Certainly, it behaves here as the external degree of freedom. On the other side, the presence of spin rise the gauge symmetry to U(2), the electron participate in the weak interaction and therefore it is also the internal degree of freedom. I do not know the answer yet.

Regards, Dany.

11. May 4, 2007

### Jimmy Snyder

Thanks Dany, you ought to write a book on this stuff.

12. May 4, 2007

### jostpuur

I didn't understand the meaning of coefficents in the TriTertButoxy's equation, but I thought it's not important for the idea, so I ignored it. Is there problems with them?

13. May 5, 2007

### Anonym

It is not necessary:

1)E. Schrödinger, Die Naturwissenschaften, 14,664,(1926);
2)W. Heisenberg, Zs. Phys., 43, 172 (1927) (English tr. in W&Z);
3)W. Heisenberg, The Physical Principles of the Quantum Theory, Dover Pub. (1930)
4)R. Jackiw, JMP, 9, 339 (1968);
5)P. Carruthers, M. Nieto, Rev. Mod. Phys., 40, 411 (1968);
6)L. Mandel, E. Wolf, Optical coherence and quantum optics, Camb. 1995.

In that order and you will see that it is impossible to do better. I hope to write a short paper on the orthonormal and complete basis of coherent states.

Regards, Dany.

14. May 7, 2007

### strangerep

Orthonormal and complete? Not sure what you meant. - I was under the
impression that the coherent states are over-complete.

15. May 7, 2007

### Anonym

You are right. More precisely, complete but not orthonormal. That is the problem to be solved. It must be Hilbert space with well defined geometry.

Regards, Dany.

Last edited: May 7, 2007