Regarding the Clifford algebra and spinors

Kontilera

Hello! I´m 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!

Bill_K

For a Lorentz 4-vector, x·x is not invariant. To get an invariant you have to insert a metric, x·η·x where η = (-1, 1, 1, 1).

Likewise for spinors, ψ^†ψ is not an invariant, you must insert a 4x4 matrix that acts like a metric, and consider instead ψ†ηψ. Now if a Dirac spinor transforms like ψ → Lψ under a Lorentz transformation, then ψ^†ηψ → ψ^†L^†ηLψ, and so the condition that this quantity is a Lorentz scalar is L^†ηL = η. The solution is η = γ⁰.

Kontilera

Quote by Bill_K

For a Lorentz 4-vector, x·x is not invariant. To get an invariant you have to insert a metric, x·η·x where η = (-1, 1, 1, 1).

Likewise for spinors, ψ^†ψ is not an invariant, you must insert a 4x4 matrix that acts like a metric, and consider instead ψ†ηψ. Now if a Dirac spinor transforms like ψ → Lψ under a Lorentz transformation, then ψ^†ηψ → ψ^†L^†ηLψ, and so the condition that this quantity is a Lorentz scalar is L^†ηL = 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 ψ^†γ⁰ψ = 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

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

The correctly normalized plane wave solution has an additional factor of ((E + m)/2m)^½ in front. E.g. Bjorken and Drell vol I gives the plane wave solution as

ψ(p_z) = ((E + m)/2m)^½(p_z/(E + m), 0, 1, 0)

Multiply your answer of 2m/(E + m) by this additional factor squared, and you get 1.

andrien

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

Thanks Bill

Kontilera

Yeah I agree with you, but dont you see the problem I´m pointing at? :)

1) Our metric (γ⁰) defines an inner product which gives us a norm, |ψ|² = ψ^†ηψ.
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

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)^½ in front. E.g. Bjorken and Drell vol I gives the plane wave solution as

ψ(p_z) = ((E + m)/2m)^½(p_z/(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 u_r(p) and two negative energy solutions v_r(p) where r = ±1 is the spin coordinate. The normalization is:
u_r(p)u_s(p) = δ_rs
v_r(p)v_s(p) = - δ_rs
u_r(p)v_s(p) = v_r(p)u_s(p) = 0

Kontilera

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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AdrianTheRock	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 Blog Entries: 1 Recognitions: Science Advisor	The correctly normalized plane wave solution has an additional factor of ((E + m)/2m)^½ in front. E.g. Bjorken and Drell vol I gives the plane wave solution as ψ(p_z) = ((E + m)/2m)^½(p_z/(E + m), 0, 1, 0) Multiply your answer of 2m/(E + m) by this additional factor squared, and you get 1.

andrien	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.

Kontilera	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. :)