# Gordon decomposition an right current

1. Sep 21, 2007

### OOO

Gordon decomposition an "right" current

I've got a question about the Dirac equation and the Gordon decomposition of the dirac current. The Gordon decomposition is

$$\bar \psi \gamma^\mu \psi = - \frac{1}{m}\Im (\bar \psi \partial^\mu \psi) - \frac{1}{2m}\partial_\mu ( \bar \psi \sigma^{\mu\nu} \psi)$$

where h=c=1. As is well known the divergence of the left hand side vanishes if we simply use the dirac equation and its adjoint. But the divergence of the second term on the right hand side (the so called spin current) vanishes as well because it is antisymmetric in $$(\mu,\nu)$$. As a consequence the first current on the rhs (the so called convective) must also be conserved.

So in total we have three conserved currents, namely

$$j^\mu = \bar \psi \gamma^\mu \psi$$

$$j^\mu_{conv} = - \frac{1}{m}\Im (\bar \psi \partial^\mu \psi)$$

$$j^\mu_{spin} = -\frac{1}{2m} \partial_\mu (\bar \psi \sigma^{\mu\nu} \psi)$$

Firstly: that seems a bit much to me. Especially considering the fact that charge conservation results from gauge invariance I would expect that there is exactly one conserved current as there is exactly one gauge generator for electromagnetism.

Secondly: how do I know that it is really the first (i.e. dirac) current which enters the Maxwell equations as its sources ? Of course we could look at the well known Lagrangian of Dirac+Maxwell and say "oh it works, why bother". But couldn't we assume that it is actually the convective current or the spin current or even an unspecified linear combination of the three which acts as the source of the electromagnetic field ?

2. Sep 21, 2007

### reilly

The Gordon Decomposition(GD) splits the canonical current into that due to a moving point charge -- like e v, e charge, v speed -- and due to a point magnetic moment -- the sigma mu nu term --.Why is the gamma mu term the right one? -- assume no anomalous moments. History, NR limit, what else could it be? gauge invariance all are important, but Dirac's genius is the main reason gamma mu is the right current for Maxwell....

A sure fire way to see what's right and what's not is: compute the electron scattering crosssection from a fixed positive charge with an arbitrary mixture of convective and magnetic currents. Check against some data -- switch to proton and alpha particles a la Rutherford. Back scattering is the big deal -- how is back scattering related to the convective/magnetic ratio? Also think about the anomalous magnetic moment of, say, the electron --what's with anomalous.
Good questions

Regards,
Reilly Atkinson

3. Sep 21, 2007

### OOO

Thank you Reilly,

I'm not sure that I have understood what you are trying to tell me. It sounds a bit shaky to me that you quote Dirac's genius as a reason. Of course one needs genius to find something new that also works. But 80 years later I suppose there should be some sound explanation "why" it worked so well. The phrase "no anomalous moments" urges one to think like: ah, okay, everything is normal - so I will believe it too.

Checking against the data is certainly what we should always do, yes. But I am asking myself if there is some deeper reason why the ratio between both currents is precisely 1 for the electron. Maybe it's just that I haven't grasped what you were saying about back scattering and so on.

4. Sep 21, 2007

### nrqed

I am not sure but at first sight it seems to me that maybe parity is a key issue. If you want parity to be a valid symmetry, the currents must be combined in just the right way.
But I may be wrong, I haven't looked at this in a very long time.

5. Sep 21, 2007

### Hans de Vries

There are two conserved currents, so the sum is also conserved.....

Both are sources of the electromagnetic field. Long distance fields are dominated by
the first because the charge related fields fall of by 1/r^2 while the spin related fields
fall of by 1/r^3.

In QED the spin current plays the most important role. The spin current, which is
the curl of the spin density, also causes, according to Stokes' law, effective global
electric currents. This is just the same as what happens in magnetic materials:

If the spin density stays equal then the "little circular currents" cancel each other,
but there is an effective global current circling at the edge of the magnetic material
where there is a gradient. The same happens with the electron's wave-function.

See for instance: Sakurai, chapter 3.5: "Gordon decomposition of the vector current"

Regards, Hans

6. Sep 21, 2007

### OOO

...and so does any linear combination...

Thank you for pointing this out to me. Indeed the spin current looks like the kind of trick one does when deriving the Maxwell equations in matter. I haven't seen it from this angle yet. But it sounds perfectly natural.

But then I am wondering if there is probably some way to reexpress the Dirac equation by absorbing magnetization and polarization into the electromagnetic field tensor in the usual way, i.e.: F+S=G where F=(E,B); S=(P,M); G=(D,H). Does this make sense to you ?

I will take a look at Sakurai. I haven't touched this book many times because it is mainly about nonrelativistic QM, at least as far as I remember.

7. Sep 21, 2007

### OOO

Now I have looked it up. But I'm afraid I haven't found anything about the Gordon decomposition in Sakurai 3.5. It is just the usual angular momentum algebra. We are talking about the chapter named "Eigenvalues and Eigenstates of Angular Momentum", right ?

Edit: Sorry, I see, there is another Book by Sakurai. I didn't know. So you are referring to "Advanced Quantum Mechanics". This will probably take me some time to get a copy of.

Last edited: Sep 21, 2007
8. Sep 22, 2007

### Hans de Vries

I was indeed referring to "Advanced Quantum Mechanics". This is a classic. There should
be books like these which handle the Dirac equation in the modern chiral representation.
The chiral form is assumed to be a better representation of the underlying physics since
the successes of the electroweak theory. It also maps perfectly to the Poincaré group of
rotations, boosts and translations.

Regards, Hans

9. Sep 24, 2007

### OOO

I have taken a look on Sakurai's second volume and I can see a bit more clearly now that the Dirac current is enforced by gyromagnetic ratio of the electron. I would have been happier if I had found some statement about "why" the gyromagnetic ratio is 2 (neglecting the higher order terms) for the electron. But maybe I have to swallow that as an experimental fact.

As to my first question, I'll repeat it once more: how can it be that there is only a gauge group with one generator, i.e. U(1), but there are three separately conserved currents: dirac current, convective current, spin current ?

10. Sep 24, 2007

### Demystifier

The Noether current is derived from the symmetry of the action, given the equations of motion. It is unique. The conservation of other currents is a pure consequence of the equation of motion, so they are not unique.

BTW, for some additional currents see also Eq. (11) in
http://xxx.lanl.gov/abs/hep-th/0702060

11. Sep 24, 2007

### OOO

Thank you for the answer, Demystifier
I don't understand this. If the equations of motion (the Dirac equation) are unique, why should a current that is a consequence of these equations not be unique ?

12. Sep 24, 2007

### OOO

For those who are interested: I have found a paper that deals with exactly these matters

http://arxiv.org/pdf/quant-ph/0305175

Unfortunately the author admits coming to no definite conclusion as well. At least he is able to show that there is an infinite number of conserved currents, each being constructed by yet another application of the Dirac operator.

Demystifier, I think I understand now what you meant by saying "not unique as a consequence of the equation of motion". From the above paper it looks like this could also be rephrased in terms of the non-uniqueness of the Lagrangian (I guess similar to the case of the energy-momentum tensor of the gauge field, which is also not unique).

13. Sep 24, 2007

### Hans de Vries

Well, the probability density of the Dirac equation can equally well be interpreted as charge density AND as a spin density. These three have always the same ratio. There are no area's in where there is more charge and less spin for example. So, I think you can say there is only one conserved current with multiple consequences.

Regards, Hans

14. Sep 24, 2007

### OOO

You mean that $$\psi^\dagger \psi$$ can be interpreted as a spin density ? Where can I read about this ? I'm not that much an attentive reader of textbooks, but I remember Ryder (and I think also Bjorken-Drell) talking about the difficulties with several definitions of "spin". So what kind of spin density is $$\psi^\dagger \psi$$ ?

I assume you mean ratio in the sense of an unspecified ratio resulting from arbitrary normalization, not a fixed definite ratio, don't you ?

15. Sep 24, 2007

### Hans de Vries

That would be $$\bar{\psi}\psi$$ or rather $$\bar{\psi}\gamma^0\psi$$. They are the same in the particle's restframe and otherwise relate as m and E. There is a special operation which returns you the spin density by its spin components.

If some area represents 1% of the probability then it also represents 1% of the charge and 1% of the spin/magnetic moment. For molecular modeling you need to assign such a charge and magnetic moment to the area (and all other area) to compute the fields which are seen by other electrons and the protons in the nuclei.

Regards, Hans

Last edited: Sep 24, 2007
16. Sep 24, 2007

### OOO

I'm afraid I haven't understood where this comes from. But anyway, some things are clearer for me now.

Thanks a lot for all your answers.

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