What happens to electron spin under Lorentz transformation?

[alemsalem]

When you start in the rest frame of the electron, the Spinor w(p = 0) = (1 0 0 0 ) represents a positive energy state with spin up in the Z direction u = (0, 0 0 1),, that is the spinor is an eigenspinor of the operator S . u, where S is a 4 dimensional operator (S0, S)

after a Lorentz transformation (A) to a frame where the electron has momentum p (w(p) = S w(0)).

now w(p) is an eigenstate of the operator S . u1 where u1 = A u and its magnitude |w(p)|^2 has changed by a factor of gamma which is fine if you want to relate the probability density in both frames ( the volume element changed by a corresponding factor),, but if you want to calculate the total probability that the electron is in a spin state (Sz up or down) you don't get 1.

So how do we understand the spin of a moving electron, and how does it transform relativistically? shouldn't we be able to say that the electron is is in some eigenstate of an operator of the form S . v where both of them are three dimensional?

Thanks!

[A. Neumaier]

When you start in the rest frame of the electron, the Spinor w(p = 0) = (1 0 0 0 ) represents a positive energy state with spin up in the Z direction u = (0, 0 0 1),, that is the spinor is an eigenspinor of the operator S . u, where S is a 4 dimensional operator (S0, S)
No. w represents a positive energy state with spin up in the direction of 4-momentum p =(m,0,0,0). After a Lorentz transformation, Sw represents a positive energy state with spin determined by an appropriate projection to the positive energy subspace in the direction of 4-momentum Lp, where S is the spinor representation and L the vector representation of the Lorentz transformation.

[alemsalem]

No. w represents a positive energy state with spin up in the direction of 4-momentum p =(m,0,0,0). After a Lorentz transformation, Sw represents a positive energy state with spin determined by an appropriate projection to the positive energy subspace in the direction of 4-momentum Lp, where S is the spinor representation and L the vector representation of the Lorentz transformation.

I'm having trouble understanding it physically, if there is a magnetic field in the z direction, what's the energy of w1(0) = (1 0 0 0 ) and w2 = (0 1 0 0 ).

Thanks.

[Bill_K]

the spinor is an eigenspinor of the operator S . u, where S is a 4 dimensional operator (S0, S)


Alemsalem, the four-dimensional analog of the spin vector is not a 4-vector, as you're apparently thinking. Spin is, by definition, the quantity that's conjugate (in the sense of Hamiltonian mechanics) to infinitesimal rotations. Three-dimensional rotations form a 3-vector, and so S is a 3-vector. But four-dimensional rotations (Lorentz tranformations) form an antisymmetric rank 2 tensor (note there are six of them) and so the angular momentum of a system must also be represented by an antisymmetric tensor. For a Dirac particle this is σμν

[alemsalem]

Alemsalem, the four-dimensional analog of the spin vector is not a 4-vector, as you're apparently thinking. Spin is, by definition, the quantity that's conjugate (in the sense of Hamiltonian mechanics) to infinitesimal rotations. Three-dimensional rotations form a 3-vector, and so S is a 3-vector. But four-dimensional rotations (Lorentz tranformations) form an antisymmetric rank 2 tensor (note there are six of them) and so the angular momentum of a system must also be represented by an antisymmetric tensor. For a Dirac particle this is σμν

I think I get that Ji = σjk, but my broblem is with the fact that S is not unitary, and that the spinor magnitude changes by a factor of gamma, which when you want to relate the probability densities it works just fine because it compensates the change in volume,, what then is the total probability for the spinor w(p) to be found in either of the eigenstates of J3 = σ12, is it gamma?.