Exploring Spin Angular Momentum in Relativistic Quantum Mechanics

In summary, when an electron is boosted in a particular direction, its spin angular momentum changes, but its orbital angular momentum does not change.
  • #1
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 [itex]|m,+\rangle[/itex] being an eigenstate of [itex]J_z[/itex] , after boosting(x direction) it to the state [itex]|p,+\rangle[/itex], is no longer an eigenstate of [itex]J_z[/itex], because of the nonzero commutation relation [itex][J_z,K_x][/itex]. This fact is not surprising per se, because [itex]J_z[/itex] is total angular momentum, and we can't expect [itex]|p,+\rangle[/itex], 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 [itex]L_z=(r\times p)_z[/itex], spin part [itex]S_z=\frac{1}{2}\sigma_z[/itex](setting [itex]\hbar=1[/itex]). And an spin up eigenstate of [itex]S_z[/itex] 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)
[tex]J_z=\int{d^{3}x\psi^{\dagger}[(x\times(-i\nabla))_z+\frac{1}{2}\Sigma_z]\psi}[/tex]
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 [itex]|p,+\rangle[/itex] we don't see an eigenstate(as a reminder [itex]|p,+\rangle[/itex] is related to [itex]|m,+\rangle[/itex] 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 [itex]|p,+\rangle[/itex] is not an eignestate of [itex]S_z[/itex] in relativistic QM
(2)We should redefine [itex]S_z[/itex] so that [itex]|p,+\rangle[/itex] is an eigenstate.
(3)We should redefine [itex]S_z[/itex] by some other reason, and [itex]|p,+\rangle[/itex] is not necessarily an eigenstate.
So which one is correct? And what should we take as the starting point of defining [itex]S_z[/itex]?

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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  • #2


The Wikipedia article on the Pauli-Lubanski vector answers your questions.
 
  • #3


IIRC, unlike the nonrelativistic case, there's no longer an invariant distinction between spin and orbital angular momenta in the relativistic case.

(Sorry, don't have time right now to dig out a reference. Maybe later if no one else gives one.)
 
  • #4


In either case, the invariant distinction between spin and orbital angular momentum is that spin is defined to be the angular momentum in the rest frame.
 
  • #5


Bill_K said:
In either case, the invariant distinction between spin and orbital angular momentum is that spin is defined to be the angular momentum in the rest frame.
Doesn't it need to be bit more general than that? (The above doesn't work in the massless case.)
 
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  • #6


Thanks for the replies, but I'm a bit busy recently, I'll check out Pauli-lubanski later in detail.
 

1. What is spin angular momentum in relativistic quantum mechanics?

Spin angular momentum is a fundamental property of particles in quantum mechanics, which describes their intrinsic rotational motion. In relativistic quantum mechanics, spin angular momentum is considered as a component of the total angular momentum of a particle, along with its orbital angular momentum.

2. How is spin angular momentum measured in relativistic quantum mechanics?

In relativistic quantum mechanics, spin angular momentum is measured using the spin operator, which is a mathematical tool that represents the spin of a particle. The eigenvalues of the spin operator correspond to the possible values of spin angular momentum that can be observed.

3. Can particles have half-integer spin in relativistic quantum mechanics?

Yes, according to the principles of relativistic quantum mechanics, particles can have half-integer spin values, such as 1/2, 3/2, and so on. This is in contrast to classical mechanics, where only integer spin values are possible.

4. How does spin angular momentum affect the behavior of particles in relativistic quantum mechanics?

Spin angular momentum plays a crucial role in determining the properties and behavior of particles in relativistic quantum mechanics. It affects the interaction between particles, their energy levels, and their magnetic moments, among other things.

5. Can particles with zero mass have spin angular momentum in relativistic quantum mechanics?

Yes, according to the principles of relativistic quantum mechanics, particles with zero mass, such as photons, can have spin angular momentum. This is because spin is an intrinsic property of particles and is not dependent on their mass.

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