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About degrees of freedom of fermions

by karlzr
Tags: degrees, fermions, freedom
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karlzr
#1
Dec5-13, 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 right-handed weyl spinor, right-handed spinor can be expressed in terms of the derivative of left-handed 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 left-hand or right-hand weyl spinor. Two complex components amount to 4 dofs. But obviously there are only two dofs (spin up and down).
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The_Duck
#2
Dec5-13, 10:50 PM
P: 842
Quote Quote by karlzr View Post
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?
Yes. The Dirac equation in momentum space is

##(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.

Quote Quote by karlzr View Post
I don't think so if KG equation doesn't eliminate any dof.
Note that multiplying the above equation by ##(p_\mu \gamma^\mu - m)## gives

##(p^2 - m^2)u(p) = 0##

which is the Klein-Gordon equation in momentum space. So the Dirac equation is a much stronger constraint than the Klein-Gordon equation: when you impose the Dirac equation, you automatically impose the Klein-Gordon equation PLUS a constraint on the spinor structure.
karlzr
#3
Dec5-13, 11:35 PM
P: 86
Quote Quote by The_Duck View Post
Yes. The Dirac equation in momentum space is

##(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.
Is it possible to come to this conclusion from pure algebra. I don't understand which two complex dofs are set to zero. Is it because the left-hand and right-hand components couple in Dirac equation so they are not independent of each other?

Quote Quote by The_Duck View Post
Note that multiplying the above equation by ##(p_\mu \gamma^\mu - m)## gives

##(p^2 - m^2)u(p) = 0##

which is the Klein-Gordon equation in momentum space. So the Dirac equation is a much stronger constraint than the Klein-Gordon equation: when you impose the Dirac equation, you automatically impose the Klein-Gordon equation PLUS a constraint on the spinor structure.
That makes sense.

The_Duck
#4
Dec6-13, 12:27 PM
P: 842
About degrees of freedom of fermions

Quote Quote by karlzr View Post
Is it possible to come to this conclusion from pure algebra. I don't understand which two complex dofs are set to zero.
Consider ##u(p)##, the Fourier component of the Dirac field with momentum ##p##. This is a complex four-component spinor, so we might think that it has four independent complex degrees of freedom. But in fact ##u(p)## must obey

##(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 two-dimensional subspace of the four-dimensional 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 two-dimensional subspace of the four-dimensional 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)##.
karlzr
#5
Dec6-13, 12:52 PM
P: 86
Quote Quote by The_Duck View Post
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 two-dimensional subspace of the four-dimensional 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)##.
That clears my doubt about fermions. Can this argument be applied to vector field, like photon? It's said the equation of motion projects out one dof. So there are only 3 dofs for massive vector fields (no gauge invariance for massive vector fields).
ChrisVer
#6
Dec6-13, 02:14 PM
P: 751
what do yo mean by vector fields?
The vector fields are introduced for gauge invariance....
karlzr
#7
Dec6-13, 03:06 PM
P: 86
Quote Quote by ChrisVer View Post
what do yo mean by vector fields?
The vector fields are introduced for gauge invariance....
But if the vector field is massive, then the longitudinal mode becomes physical and there is no gauge invariance. Mass term violates gauge invariance.


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