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kof9595995
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By "spin" we sometimes refer to spin angular momentum, sometimes we refer to a specific representation of Lorentz group, in the following I'll refer to the former, otherwise I'll write "spinor representation"
Say an spin up(z direction) electron at rest, state vector |m,+\rangle being an eigenstate of J_z , after boosting(x direction) it to the state |p,+\rangle, is no longer an eigenstate of J_z, because of the nonzero commutation relation [J_z,K_x]. This fact is not surprising per se, because J_z is total angular momentum, and we can't expect |p,+\rangle, which is a plane wave, to be its eingenstate. However an interesting question arises: How should we divide the total angular momentum into spin part and orbital part?
In nonrelativistic QM it's pretty straightforward, orbital part L_z=(r\times p)_z, spin part S_z=\frac{1}{2}\sigma_z(setting \hbar=1). And an spin up eigenstate of S_z will stay an spin up eigenstate after any boost.
However, in relativistic case the division is not so straightforward, for a Dirac field, by Peskin page 60, eqn(3.111)
J_z=\int{d^{3}x\psi^{\dagger}[(x\times(-i\nabla))_z+\frac{1}{2}\Sigma_z]\psi}
It's very tempting to define the first term as the orbital part and second term as the spin part, but it might not be quite right, because if we apply the second term to |p,+\rangle we don't see an eigenstate(as a reminder |p,+\rangle is related to |m,+\rangle by a pure boost), i.e. the good old property we had in nonrel QM is lost, now there are 3 possibilities of this problem:
(1)Indeed |p,+\rangle is not an eignestate of S_z in relativistic QM
(2)We should redefine S_z so that |p,+\rangle is an eigenstate.
(3)We should redefine S_z by some other reason, and |p,+\rangle is not necessarily an eigenstate.
So which one is correct? And what should we take as the starting point of defining S_z?
PS: In case of confusion, I'm not asking about Wigner rotation, and I'm just asking if an electron is spin up in its rest frame, then after a pure boost is it still spin up? And in what sense?
Say an spin up(z direction) electron at rest, state vector |m,+\rangle being an eigenstate of J_z , after boosting(x direction) it to the state |p,+\rangle, is no longer an eigenstate of J_z, because of the nonzero commutation relation [J_z,K_x]. This fact is not surprising per se, because J_z is total angular momentum, and we can't expect |p,+\rangle, which is a plane wave, to be its eingenstate. However an interesting question arises: How should we divide the total angular momentum into spin part and orbital part?
In nonrelativistic QM it's pretty straightforward, orbital part L_z=(r\times p)_z, spin part S_z=\frac{1}{2}\sigma_z(setting \hbar=1). And an spin up eigenstate of S_z will stay an spin up eigenstate after any boost.
However, in relativistic case the division is not so straightforward, for a Dirac field, by Peskin page 60, eqn(3.111)
J_z=\int{d^{3}x\psi^{\dagger}[(x\times(-i\nabla))_z+\frac{1}{2}\Sigma_z]\psi}
It's very tempting to define the first term as the orbital part and second term as the spin part, but it might not be quite right, because if we apply the second term to |p,+\rangle we don't see an eigenstate(as a reminder |p,+\rangle is related to |m,+\rangle by a pure boost), i.e. the good old property we had in nonrel QM is lost, now there are 3 possibilities of this problem:
(1)Indeed |p,+\rangle is not an eignestate of S_z in relativistic QM
(2)We should redefine S_z so that |p,+\rangle is an eigenstate.
(3)We should redefine S_z by some other reason, and |p,+\rangle is not necessarily an eigenstate.
So which one is correct? And what should we take as the starting point of defining S_z?
PS: In case of confusion, I'm not asking about Wigner rotation, and I'm just asking if an electron is spin up in its rest frame, then after a pure boost is it still spin up? And in what sense?
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