It has been awhile since my last blog entry, but if you want to check out some my recent wanderings through physicsland, check out sci.physics.foundations, relativity, and research.

Here, I would like to show a rather simple calculation, which may cast a different light on how one needs to think about the canonical commutation relationship . I would very much like your comments in helping me sort this through. You may download this in pdf form at https://jayryablon.files.wordpress.com/2008/06/linear-mass-commutator-calculation.pdf.

**I. A Known Square Mass Commutation Calculation**

Consider a particle of mass *m* as a single particle system. Consider canonical coordinates , and that at least the space coordinates are operators. If we require that the mass *m *must commute with all operators, then we must have , and by easy extension, . It is well known that the commutation condition , taken together with the on-shell mass relationship and the single-particle canonical commutation relationship , where is the Minkowski tensor, leads inexorably to the commutation relationship:

(1.1)

where is the particle velocity (in c=1 units) along the *k*th coordinate. I leave the detailed calculation as an exercise for the reader not familiar with this calculation, and refer also to the sci.physics.research thread at http://www.physicsforums.com/archive/index.php/t-142092.html or http://groups.google.com/group/sci.physics.research/browse_frm/thread/d78cbfecf703ff6a.

I would ask for your comments on the following calculation, which is totally analogous to the calculation that leads to (1.1), but which is done using the linear mass *m* rather than the square mass , and using the Dirac equation written as , in lieu of what is, in essence, the Klein Gordon equation that leads to (1.1).

**2. Maybe New?? Linear Mass Commutation Calculation**

Start with Dirac’s equation written as:

. (2.1)

Require that:

(2.2)

Continue to use the canonical commutator . Multiply (2.1) from the left by noting that to write:

. (2.3)

This separates into:

. (2.4)

Now, use the canonical relation to commute the space (*k*) equation, thus:

. (2.5)

In the final line, we use Dirac’s equation written as , and specifically, the component equation .

If we require that , which is (2.2), then (2.5) reduces easily to:

, (2.6)

Finally, multiply from the left by , and employ and to write:

. (2.7)

If we contrast (2.7) to (1.1) written as , we see that the velocity has been replaced by the Dirac operator , that is, .

**3. Questions**

Here are my first set of questions:

1) Is the calculation leading to (2.7) correct, and is (2.7) a correct result, or have I missed something along the way?

2) If (2.7) is correct, has anyone seen this result before? If so where?

3) Now use the plane wave so that we can work with the Dirac spinors , and rewrite (2.7) as:

The upper member of (3.1) is an eigenvalue equation. Reading out this equation, I would say that the commutators are the eigenvalues of the Dirac matrices, which are:

and , (3.2)

in the respective Pauli/Dirac and Weyl representations, and that the *u *are the eigenvectors associated with these eigenvalues . Am I wrong? If not, how would one interpret this result? Maybe the commutators can be discussed in the abstract, but it seems to me that the commutators can only be discussed as the eigenvalues of the matrices with respect to the eigenstate vectors *u*. This, it seems, would put canonical commutation into a somewhat different perspective than is usual.

Just as Dirac’s equation reveals some features that cannot be seen strictly from the Klein Gordon equation, the calculation here seems to reveal some features about the canonical commutators that the usual calculation based on and cannot, by itself, reveal.

I’d appreciate your thoughts on this, before I proceed downstream from here.

Thanks,

Jay.

I haven’t checked it carefully for errors, but the conclusion seems natural.

For the Dirac equation, it’s well known that the velocity operator exists, and has eigenvalues of +-1 (that correspond to speeds +-c). I don’t deal with

\alpha_k, and I’m not sure about your choice of signature, but this should be what you get for a velocity operator.

No one uses the velocity operator because its eigenvalues are not states we can make. However, I think it is significant that the handed chiral states do have speed +-c but really aren’t so massless in that they contribute to the mass of the particle. The question is debated at length in the literature, but mostly quite some time ago. Google “Dirac” “velocity” “eigenstate”.

Comment by carlbrannen — June 22, 2008 @ 1:50 am |

Hi Carl,

The main result I derive from the foregoing, developed in detail at https://jayryablon.wordpress.com/files/2008/06/cannonical-commutators-2.pdf, is precisely what you state: that the fermions are constrained to have a velocity equal to the speed of light c, and so by any conventional analysis, the fermions must be massless. This means that for a fermion of any non-zero rest mass, there is something “wrong,” or at least “only approximate,” with the usual canonical commutation relationships. Would you concur?

As far as chirality goes, once there is any mass at all, one can non longer separate the chiral states as is done for what used to be thought of as “massless” neutrinos.

What might be, in your thinking, the best way to make sense of this “speed of light for fermions” result which I derived in https://jayryablon.wordpress.com/files/2008/06/cannonical-commutators-2.pdf not knowing if it was already known, and which you tell me is already known and debated in the literature. Especially, do you think that this might point toward gravitational theory and using Dirac’s equation with a vierbein? Because with mass comes gravitation, and the use of a Dirac equation which presupposes no gravitational field must therefore also, strictly speaking, presuppose no mass either, and thus, only luminous particles. (I do note that if all the particle are massless, one can still have traceless, luminous energy.)

Thanks,

Jay.

Comment by Jay R. Yablon — June 26, 2008 @ 8:20 pm |

Jay, the usual interpretation is that the electron is simply not an eigenstate of velocity. Instead, in a certain way, they are eigenstates of mass. My interpretation is that the fundamental objects are the chiral electrons and that these are massless, more or less. The “more or less” is because in my interpretation, mass is an interaction between the left and right particles; it’s a force between two particles like any other.

The left and right handed chiral halves are related by mass. That is, the Hamiltonian terms look like m \psi_L^\dag \psi_R and Hermitian conjugate, that is m \psi_R^\dag \psi_L.

As perturbations for Feynman diagrams, these amount to annihilation of a right(left) particle, and the creation of a left(right) particle, with a vertex amplitude of m.

My version of all this is to get the regular electron from the 3-d Feynman checkerboard. The left and right chiral particles get split into three preons each, call them -XL -YL, -ZL, and XR, YR, ZR. This would work for a particle whose spin is in the (1,1,1) direction, so the right handed particle moves in (1,1,1) and the left handed one moves back in (-1,-1,-1). But then the speed of the preons has to be c sqrt{3}, and the whole scheme is in violation of relativity. (However, it does get the masses right and Kea and I are working on the mixing matrices.)

In this scheme, gravity becomes a boson that influences the probability of an XL converting to a -YR or -ZR, but this is too long already.

Comment by carlbrannen — June 28, 2008 @ 2:58 am |