Regarding the Clifford algebra and spinors


by Kontilera
Tags: algebra, clifford, spinors
Kontilera
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#1
Feb15-13, 01:45 PM
P: 102
Hello! Im currently taking a course in RQM and have some questions for which I didnt get any satisfactory answers on the lecture. All comments are appricieted!

1. Is the gamma zero tensor some kind of metric in the space for spinors? When normalizing our solution to the Dirac equation it seems as we use them exactly as our metric in SR.

2. As I understood it the gamma matricies are Lorentz invariant tensors. However when taking the norm of a spinor describing a particle moving in lets say the z-direction we get a answer that depends on the energy (only one component of our 4-momentum). In other words it seems as if our normalization is not invariant under Lorentz transformations and therefore not gamma 0?

Thanks in advance!
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Bill_K
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#2
Feb15-13, 02:51 PM
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For a Lorentz 4-vector, xx is not invariant. To get an invariant you have to insert a metric, xηx where η = (-1, 1, 1, 1).

Likewise for spinors, ψψ is not an invariant, you must insert a 4x4 matrix that acts like a metric, and consider instead ψηψ. Now if a Dirac spinor transforms like ψ → Lψ under a Lorentz transformation, then ψηψ → ψLηLψ, and so the condition that this quantity is a Lorentz scalar is LηL = η. The solution is η = γ0.
Kontilera
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#3
Feb15-13, 07:33 PM
P: 102
Quote Quote by Bill_K View Post
For a Lorentz 4-vector, xx is not invariant. To get an invariant you have to insert a metric, xηx where η = (-1, 1, 1, 1).

Likewise for spinors, ψψ is not an invariant, you must insert a 4x4 matrix that acts like a metric, and consider instead ψηψ. Now if a Dirac spinor transforms like ψ → Lψ under a Lorentz transformation, then ψηψ → ψLηLψ, and so the condition that this quantity is a Lorentz scalar is LηL = η. The solution is η = γ0.
Yes! This is what I mean when I say that it seems as we use the gamma 0 tensor as a metric. However ψγ0ψ = 1 - (p_z/E+m)2 for a particle moving in the z direction. Is the RHS really a Lorentz scalar?

Is it mathematically correct to actually call γ0 our metric in the space of spinors? Or is this just a smiliarity of how we use an metric?

AdrianTheRock
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#4
Feb16-13, 05:48 AM
P: 136

Regarding the Clifford algebra and spinors


Surely, if the electron is moving along the z axis then [itex]p_z = p[/itex] and then
[tex]1 - (\frac{p}{(E + m)})^2 = \frac{E^2 + 2Em + m^2 - p^2}{(E + m)^2} = \frac{2m}{(E + m)}[/tex]
I'm no expert, but this doesn't look invariant to me.
Bill_K
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#5
Feb16-13, 06:22 AM
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The correctly normalized plane wave solution has an additional factor of ((E + m)/2m) in front. E.g. Bjorken and Drell vol I gives the plane wave solution as

ψ(pz) = ((E + m)/2m)(pz/(E + m), 0, 1, 0)

Multiply your answer of 2m/(E + m) by this additional factor squared, and you get 1.
andrien
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#6
Feb16-13, 07:26 AM
P: 987
for free particle spinors,the convention is
u-u=2mS[itex]\dagger[/itex]S,where S is two component spinor normalized by
S[itex]\dagger[/itex]S=1.
AdrianTheRock
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#7
Feb16-13, 10:01 AM
P: 136
Thanks Bill
Kontilera
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#8
Feb16-13, 12:39 PM
P: 102
Yeah I agree with you, but dont you see the problem Im pointing at? :)

1) Our metric (γ0) defines an inner product which gives us a norm, |ψ|2 = ψηψ.
2)When I calculate the norm of a spinor it is dependent of its energy.
3) Energy is not a Lorentz scalar so the norm is not invariant under Lorentz transformations.

So it seems to me that our metric is not Lorentz invariant. But I assume that there is some misunderstanding in my logic..
Bill_K
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#9
Feb16-13, 02:14 PM
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2)When I calculate the norm of a spinor it is dependent of its energy.
No sorry, if that's the case, you're using improperly normalized wavefunctions! See above:
The correctly normalized plane wave solution has an additional factor of ((E + m)/2m) in front. E.g. Bjorken and Drell vol I gives the plane wave solution as

ψ(pz) = ((E + m)/2m)(pz/(E + m), 0, 1, 0)
The norm of this wavefunction is ψψ = 1, as it should be. The plane wave solution is derived by taking a solution at rest and Lorentz transforming it, so its norm is guaranteed to be Lorentz invariant, by construction.

The full set of wavefunctions for given p consist of two positive energy solutions ur(p) and two negative energy solutions vr(p) where r = 1 is the spin coordinate. The normalization is:
ur(p)us(p) = δrs
vr(p)vs(p) = - δrs
ur(p)vs(p) = vr(p)us(p) = 0
Kontilera
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#10
Feb17-13, 06:39 AM
P: 102
Yeah, Im with you now. Dont know why but I regareded your additional normalization factor as a dimensionless constant which only normalizes for one inertial frame. :)


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