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About degrees of freedom of fermions 
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#1
Dec513, 09:08 PM

P: 86

There are something I don't get about the degrees of freedom(dof).
For massive dirac spinor, there are four complex components or 8 dofs. But for electron/position, there are only 4 dofs in total ( electron up &down, position up&down). Does it mean the equation of motion eliminate the other four dofs? I don't think so if KG equation doesn't eliminate any dof. Actually, if we write the EOM of massive dirac spinor in terms of left and righthanded weyl spinor, righthanded spinor can be expressed in terms of the derivative of lefthanded one and vice versa. Does it mean the two helicity spinors weyl spinors are not independent? I hope not, since they represent distinct spins. Then it comes to majorana spinor. We all know they can be described by only lefthand or righthand weyl spinor. Two complex components amount to 4 dofs. But obviously there are only two dofs (spin up and down). 


#2
Dec513, 10:50 PM

P: 865

##(p_\mu \gamma^\mu + m)u(p) = 0##. This equation has two independent solutions for ##u(p)##. The other two complex degrees of freedom are forced to be zero. ##(p^2  m^2)u(p) = 0## which is the KleinGordon equation in momentum space. So the Dirac equation is a much stronger constraint than the KleinGordon equation: when you impose the Dirac equation, you automatically impose the KleinGordon equation PLUS a constraint on the spinor structure. 


#3
Dec513, 11:35 PM

P: 86




#4
Dec613, 12:27 PM

P: 865

About degrees of freedom of fermions
##(p_\mu \gamma^\mu + m)u(p) = 0## To see what sort of constraint this is, look at the case ##p = (m, 0, 0, 0)## (the case of a particle at rest). Then this equation looks like ##(m\gamma^0 + m) u(p) = 0## I want to rewrite this as ##2m\frac{1}{2}(1 + \gamma^0)u(p) = 0## because the matrix ##\frac{1}{2}(1 + \gamma^0)## is a projection matrix that projects onto a twodimensional subspace of the fourdimensional vector space in which ##u(p)## lives. For example in one possible basis, [tex]\gamma^0 = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)[/tex] so that [tex]\frac{1}{2}(1 + \gamma^0) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)[/tex] Therefore in this basis the equation ##2m\frac{1}{2}(1 + \gamma^0)u(p) = 0## has the effect of setting the first two complex components of ##u(p)## to zero. This is one example of the fact that for any momentum ##p## the matrix ##(p_\mu \gamma^\mu + m)## is essentially a projection matrix onto a twodimensional subspace of the fourdimensional spinor space in which ##u(p)## lives, and so the Dirac equation has the effect of projecting out two of the four complex degrees of freedom of ##u(p)##. 


#5
Dec613, 12:52 PM

P: 86




#6
Dec613, 02:14 PM

P: 906

what do yo mean by vector fields?
The vector fields are introduced for gauge invariance.... 


#7
Dec613, 03:06 PM

P: 86




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