- #1
Chopin
- 368
- 13
I've been watching Sidney Coleman's QFT lectures (http://www.physics.harvard.edu/about/Phys253.html, with notes at http://arxiv.org/pdf/1110.5013.pdf), and I'm now on to the spin 1/2 part of the course. We've gone through all the mechanics of constructing irreducible representations [itex]D^{(s1,s2)}[/itex] of the Lorentz group, and have begun to build Lagrangians out of [itex]D^{(1/2,0)}[/itex] and [itex]D^{(0,1/2)}[/itex], the right- and left-handed Weyl spinors.
One of the requirements of a Lagrangian density is that it transform like a scalar, so he turns to the task of finding objects that can be built out of spinors that transform like a scalar. He notes that if you have a spinor field [itex]u(x)[/itex], then you can construct a field which transforms like a vector by doing [itex]V^\mu(x) = u^*(x)\sigma^\mu u(x)[/itex], where [itex]\sigma^\mu = (1, \stackrel{\rightarrow}{\sigma})[/itex].
Seeing that, I would expect us to go on to construct a scalar field by taking [itex]\partial_\mu\cdot V^\mu(x) = \partial_\mu\cdot(u^*(x)\sigma^\mu u(x))[/itex], since taking the four-divergence is normally how one constructs a scalar field out of a vector. However, he instead constructs the object [itex]u^*(x)\sigma^\mu\cdot\partial_\mu u(x)[/itex], and uses that to build a Lagrangian, which when minimized produces the Weyl Equation. I'm having trouble understanding how this works--is that equivalent to putting the derivative outside like I did above, or is it something different? And if it's different, why would one want to do it that way, and how does one show that it transforms like a scalar?
One of the requirements of a Lagrangian density is that it transform like a scalar, so he turns to the task of finding objects that can be built out of spinors that transform like a scalar. He notes that if you have a spinor field [itex]u(x)[/itex], then you can construct a field which transforms like a vector by doing [itex]V^\mu(x) = u^*(x)\sigma^\mu u(x)[/itex], where [itex]\sigma^\mu = (1, \stackrel{\rightarrow}{\sigma})[/itex].
Seeing that, I would expect us to go on to construct a scalar field by taking [itex]\partial_\mu\cdot V^\mu(x) = \partial_\mu\cdot(u^*(x)\sigma^\mu u(x))[/itex], since taking the four-divergence is normally how one constructs a scalar field out of a vector. However, he instead constructs the object [itex]u^*(x)\sigma^\mu\cdot\partial_\mu u(x)[/itex], and uses that to build a Lagrangian, which when minimized produces the Weyl Equation. I'm having trouble understanding how this works--is that equivalent to putting the derivative outside like I did above, or is it something different? And if it's different, why would one want to do it that way, and how does one show that it transforms like a scalar?