samalkhaiat said:
SO where is the fasle part? If you change the partial derivitives to functional derivatives, then QFT can be written in terms of functional differential equations.
A unification would be a single differential equation. The condensed matter theorist's version of QFT is an effective field theory based on a single Schroedinger's equation. What I would like is a single differential equation underlying the standard model.
samalkhaiat said:
RUBBISH;the next logical step after QM was applying its principles to
i) relativistic particle;
ii) classical field theory;
Ooooops. What I should have written was that the concepts of renormalization and spontaneous symmetry breaking, which are so critical to the standard model's QFT, were first discovered by condensed matter theorists working with a field theory that was based on Schroedinger's equation. They weren't even doing relativistic work.
samalkhaiat said:
J.A. Wheeler called this thing "pregeometry" and thought that algebra is the mathematical system for describing such concept.
I agree that algebra is the key, but my guess is that it will be a continuous algebra. I'm going to the PANIC meeting in Santa Fe and will give a poster on how one can break the symmetry of the Geometric Algebra and get a result with the correct symmetry for the leptons. It's basically Clifford algebra stuff. Hope to see you.
samalkhaiat said:
The late David Bohm* called it "implicate order" which find a natural mathematical expression in terms of certain kinds of algebra. And there was M. sachs's theory of elementary matter which was described by Quaternions algebra. So, you see, the great men suggested algebra and excluded differential laws for simple reason; They wanted spacetime to come out (as well as particles) as a result of applying their algebric system on the "primeval matter".
I don't think we're at all in disagreement here. To me, the Dirac equation is a differential equation. It is written in terms of an algebra, the Dirac algebra, which is equivalent to a Clifford algebra \mathcal{CL}(3,1). In other words, the algebra defines a differential equation.
samalkhaiat said:
Because of the fact that the wave function F(x) is a number and not a column matrix, Schro.Eq. displays no SU(2) symmetry.
Any rep of SO(3) is also a rep of SU(2) and therefore possesses SU(2) symmetry as well as SO(3) symmetry. The concept you're thinking of is "faithful", and I agree that SO(3) reps are not faithful reps of SU(2). But they are, in fact, reps. But I didn't mean to distinguish between Schroedinger's, Pauli's and Dirac's equations.
There is an excellent introduction to the relationship between these equations by Hestenes. In it, he shows that Schroedinger's equation is not the equation for a spinless particle, but instead is the equation for a spin-1/2 particle with the spin stuck in one position. It's a fascinating read if you're interested in algebra:
Consistency in the formulation of the Dirac, Pauli, and Schroedinger theories
Consistency with the Dirac theory is shown to imply that the Schroedinger equation describes not a spinless particle as universally assumed, but a particle in a spin eigenstate. The bearing of spin on the interpretation of the Schroedinger theory discussed.
http://modelingnts.la.asu.edu/pdf/Consistency.pdf
Despite being weaned on spinors, I'm beginning to suspect that the density operator formalism is more fundamental. With density operators, the arbitrary phases in a spinor are canceled out so it seems like you end up collapsing an SU(2) rep back down to an SO(3) rep. However, it turns out that you don't actually lose complex phases as can be seen if you consider products of Stern-Gerlach type projection operators. For example:
(1+\sigma_z)(1+\sigma_x)(1+\sigma_y)(1+\sigma_z)
= \frac{1+i}{4}(1+\sigma_z)
Thus the effect of running a spin+1/2 beam of fermions through the above sequence of Stern-Gerlach filters would be to introduce a complex phase of 45 degrees (and a reduction in intensity).
I've only been thinking about this off and on for the last few weeks so please comment.
Some readers may be unfamiliar with density operators so:
This indistinguishability makes the density operator a perhaps more intuitive representation of a state than the state vector. For example, a state vector has an arbitrary overall phase factor, so two vectors differing by a phase factor represent the same state. Still, when the vector is put together with its dual vector to form a pure state density operator, the phase factor vanishes because of the complex conjugation. Hence, all state vectors representing the same state are represented by the same density operator.
http://arxiv.org/PS_cache/quant-ph/pdf/0408/0408094.pdf
samalkhaiat said:
Look Carl, I don't know, if you can show that j=1/2 REP can be realized as differential operator, then do it!
I'm not sure what you're asking for here. In geometric algebraic language, the massless Dirac equation is simply
\nabla \Psi = 0
which is about as simple a differential operator as one can hope for. The above differs from the usual Dirac equation in that it includes a set of four (independent) spinors in one equation. If you want to pick out just one of these, which is identical to the Dirac equation, then you multiply on the right by your favorite idempotent:
(\nabla \Psi) \iota = \nabla (\Psi \iota) = \nabla \psi = \gamma^\mu\partial_\mu \psi = 0[/itex]<br />
<br />
You can add mass back into these equations if you like. See equation 2.41 in this introductory article for details:<br />
http://arxiv.org/PS_cache/quant-ph/pdf/0509/0509178.pdf <br />
<br />
Oh, and "abd" means "all but dissertation". It means I left after I passed the PhD qualifying exams but without a dissertation. It seems to be a habit, as I'm also abd PhD in Math. at U. Washington.<br />
<br />
Carl
