Lab Notes for a Scientific Revolution (Physics)

June 30, 2008

Foldy-Wouthuysen, continued

Just for the heck of it, I did a calculation of what happens to the mass matrix M\equiv \beta m during the transformation from the Dirac-Pauli representation to the Newton-Wigner representation via Foldy-Wouthuysen.  This is shown in:

https://jayryablon.files.wordpress.com/2008/06/foldy-wouthuysen.pdf

Not sure where to go from there, but I’ll be away the rest of the week on vacation, so I’ll take another look when I return.

Interested in any further thoughts anyone may have.

Jay

June 29, 2008

Might Foldy-Wouthuysen Transformations Contain a Hidden Fermion Mass Generation Mechanism?

I have been looking over the following three links for the Foldy-Wouthuysen transformation from the Dirac-Pauli to the Newton-Wigner representation of Dirac’s equation:

The first shows the calculation itself of this transformation:

I: http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf.

The second, an excellent and lucid exposition of the physics (why this is of interest), is to be found at:

II: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.3209&rep=rep1&type=pdf.

The third, dealing with Zitterbewegung motion and the velocity operator in the Dirac-Pauli representation, is at:

III: http://en.wikipedia.org/wiki/Zitterbewegung.

What I would like to discuss, for the purpose of getting your reactions as to whether I am on a sensible track, is the possibility that a mechanism for generating fermion mass may be hidden in all of this.

I say this in particular because in the Dirac-Pauli representation, the velocity operator is given by:

v^{k} =\alpha ^{k} (1)

where \alpha ^{k} = \gamma ^{0} \gamma^{k} , see reference III.  Further, the eigenvalues of this velocity operator constrain the velocity of the Fermion of be the speed of light, see reference II in the middle of page 3.  This means that the fermion must be massless and luminous, in the Dirac-Pauli representation.  Why this is so, has long been a mystery, and is thought not to make any sense, for obvious reasons.

Now, transform into the Newton-Wigner representation via Foldy-Wouthuysen.  The velocity operator in Newton-Wigner now takes the classical form:

v^{k} =dx^{k} /dt   (2)

where x^{k}  is the position operator.  But even more importantly, Newton-Wigner permits a range of eigenvalues less than the speed of light, and so, the fermions permitted by Newton-Wigner are massless and sub-luminous.

Following this to its logical conclusion, this seems to suggest that somewhere hidden in the Foldy-Wouthuysen transformation, we have gone from a fermion which is massless and luminous, to one which has a finite, non-zero rest mass and travels at sub-luminous velocity.  It seems, then, that it would be important to specifically trace how the velocity operator (1) of the Dirac-Pauli representation with \pm c eigenvalues transforms into the velocity operator (2) of Newton-Wigner which allows a continuous, sub-luminous velocity spectrum, and at the same time, to trace through how the rest mass goes from necessarily zero (with decoupled chiral components), to non-zero with chiral couplings.

By doing so, perhaps one would find a mechanism for generating fermion masses.

One contrast to make here: think about how vector boson masses are generated.  One starts with a Lagrangian in which the boson mass term is omitted entirely.  Then, via a well-knows technique, one breaks the symmetry and reveals a boson mass.  Perhaps the mystery of luminous velocity eigenvalues in the Dirac-Pauli representation is telling us a similar thing: Start out with a Dirac-Pauli Lagrangian in which the mass of the fermion is zero, i.e., without a mass term.  Then, the +/- c velocity eigenvalues make sense.  Transform that into the Newton-Wigner representation.  Somewhere along the line, a mass must appear, because a subliminous velocity appears.

I will, of course, try to pinpoint how this all happens, if it does indeed happen.  But I would for now like some reactions as to the tree up which I am barking.

Thanks,

Jay.

June 19, 2008

A New Lab Note: Commutation of Linear Rest Mass with Canonical Position

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 \left[x_{j} ,p_{k} \right]=i\eta _{jk} ;\; j,k=1,2,3.  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 x_{\mu } , and that at least the space coordinates x_{j} ;\; j=1,2,3 are operators.  If we require that the mass m must commute with all operators, then we must have \left[x_{\mu } ,m\right]=0, and by easy extension, \left[x_{\mu } ,m^{2} \right]=0.  It is well known that the commutation condition \left[x_{\mu } ,m^{2} \right]=0, taken together with the on-shell mass relationshipm^{2} =p^{\sigma } p_{\sigma } and the single-particle canonical commutation relationship \left[x_{j} ,p_{k} \right]=i\eta _{jk} ;\; j,k=1,2,3, where diag\left(\eta _{\mu \nu } \right)=\left(-1,+1,+1,+1\right) is the Minkowski tensor, leads inexorably to the commutation relationship:

\left[x_{k} ,p_{0} \right]=-ip_{k} /p^{0} =-iv_{k}    (1.1)

where v_{k} is the particle velocity (in c=1 units) along the kth 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 m^{2} , and using the Dirac equation written as m\psi =\gamma ^{\nu } p_{\nu } \psi , in lieu of what is, in essence, the Klein Gordon equation m^{2} \phi =p^{\sigma } p_{\sigma } \phi that leads to (1.1).

2.  Maybe New?? Linear Mass Commutation Calculation

 Start with Dirac’s equation written as:

m\psi =\gamma ^{\nu } p_{\nu } \psi .  (2.1) 

Require that:

\left[x_{\mu } ,m\right]=0   (2.2)

 Continue to use the canonical commutator \left[x_{j} ,p_{k} \right]=ig_{jk} .  Multiply (2.1) from the left by x_{\mu } noting that \left[\gamma ^{\nu } ,x_{\mu } \right]=0 to write:

x_{\mu } m\psi =\gamma ^{\nu } x_{\mu } p_{\nu } \psi =\gamma ^{0} x_{\mu } p_{0} \psi +\gamma ^{j} x_{\mu } p_{j} \psi .  (2.3) 

This separates into:

 \left\{\begin{array}{c} {x_{0} m\psi =\gamma ^{0} x_{0} p_{0} \psi +\gamma ^{j} x_{0} p_{j} \psi } \\ {x_{k} m\psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} x_{k} p_{j} \psi } \end{array}\right. .  (2.4)

  Now, use the canonical relation \left[x_{j} ,p_{k} \right]=i\eta _{jk} to commute the space (k) equation, thus:

 \begin{array}{l} {x_{k} m\psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} x_{k} p_{j} \psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} \left(p_{j} x_{k} +i\eta _{jk} \right)\, \psi } \\ {=\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} p_{j} x_{k} \psi +i\gamma _{k} \psi } \\ {=\gamma ^{0} x_{k} p_{0} \psi +mx_{k} \psi -\gamma ^{0} p_{0} x_{k} \psi +i\gamma _{k} \psi } \end{array} .  (2.5)

In the final line, we use Dirac’s equation written as mx_{\mu } \psi =\gamma ^{\nu } p_{\nu } x_{\mu } \psi =\gamma ^{0} p_{0} x_{\mu } \psi +\gamma ^{j} p_{j} x_{\mu } \psi , and specifically, the \mu =k component equation \gamma ^{j} p_{j} x_{k} \psi =mx_{k} \psi -\gamma ^{0} p_{0} x_{k} \psi .

 If we require that \left[x_{\mu } ,m\right]=0, which is (2.2), then (2.5) reduces easily to:

 \gamma ^{0} \left[x_{k} ,p_{0} \right]\psi =-i\gamma _{k} \psi ,  (2.6)

Finally, multiply from the left by \gamma ^{0} , and employ \gamma ^{0} \gamma _{k} \equiv \alpha _{k} and \gamma ^{0} \gamma ^{0} =1 to write:

\left[x_{k} ,p_{0} \right]\, \psi =-i\alpha _{k} \psi .  (2.7) 

If we contrast (2.7) to (1.1) written as \left[x_{k} ,p_{0} \right]\phi =-iv_{k} \phi , we see that the velocity p_{k} /p^{0} =v_{k} has been replaced by the Dirac operator \alpha _{k} , that is, v_{k} \to \alpha _{k} .

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 \psi =ue^{ip^{\sigma } x_{\sigma } } so that we can work with the Dirac spinors u\left(p^{\mu } \right), and rewrite (2.7) as:

\left\{\begin{array}{c} {\left(\alpha _{k} -\lambda \right)\, u=0} \\ {\lambda =i\left[x_{k} ,p_{0} \right]} \end{array}\right.  

The upper member of (3.1) is an eigenvalue equation.  Reading out this equation, I would say that the commutators \lambda =i\left[x_{k} ,p_{0} \right] are the eigenvalues of the Dirac \alpha _{k} matrices, which are:

{\bf \alpha }=\left(\begin{array}{cc} {0} & {{\bf \sigma }} \\ {{\bf \sigma }} & {0} \end{array}\right) and {\bf \alpha }=\left(\begin{array}{cc} {-{\bf \sigma }} & {0} \\ {0} & {{\bf \sigma }} \end{array}\right) ,  (3.2)

in the respective Pauli/Dirac and Weyl representations, and that the u are the eigenvectors associated with these eigenvalues \lambda =i\left[x_{k} ,p_{0} \right].  Am I wrong?  If not, how would one interpret this result?  Maybe the commutators \left[x_{j} ,p_{k} \right]=i\eta _{jk} can be discussed in the abstract, but it seems to me that the commutators \lambda =i\left[x_{k} ,p_{0} \right] can only be discussed as the eigenvalues of the matrices \alpha _{k} 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 \left[x_{\mu } ,m^{2} \right]=0 and m^{2} =p^{\sigma } p_{\sigma } cannot, by itself, reveal.

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

Thanks,

Jay.

Blog at WordPress.com.