Spin

We have just seen how one can create a state of arbitrary momentum by Lorentz transforming the solution for a particle at rest. In a similar manner, we can create a state of arbitrary spin polarized along the $\vec{s}$ direction by applying the rotation operator,

$$S = \exp \left( \frac{i}{2}\phi\vec{\Sigma}\cdot\vec{s} \right) ,$$ (5.198)

to the solution for a particle at rest and polarized in the $z$ direction. The defining relationship for such a state is

$$\vec{\Sigma}\cdot\vec{s}\omega(\vec{p}) = \omega(\vec{p}) ,$$ (5.199)

where the spinor $\omega(\vec{p})$ corresponds to a particle polarized along the direction of the unit vector $\vec{s}$. Remember that the spin operator is $\hat{S}=\frac{\hbar}{2} \vec{\Sigma}$ and thus the eigenvalue of the spin operator in this case is really $\frac{\hbar}{2}$, as we might expect.

Let $u(p,s)$ denote a spinor of positive energy, momentum $p^\mu$, and spin $s^\mu$. The spin vector is defined as

$$s^\mu = a^\mu_{\ \nu}\breve{s}^\nu ,$$ (5.200)

where $\breve{s}^\nu = (0,\hat{s})$ is the polarization unit vector in the rest frame and $a^\mu_{\ \nu}$ is a Lorentz transformation from the rest frame. Thus we also have $p^\mu = a^\mu_{\ \nu} \breve{p}^\nu$, where $\breve{p}^\nu = (m,0)$. This tells us that

$$s\cdot s = -1$$ (5.201)

and

$$p\cdot s = 0 ,$$ (5.202)

which is true in any frame since they are Lorentz scalars. Thus in the rest frame

$$\vec{\Sigma}\cdot\breve{s}u(\breve{p},\breve{s}) = u(\breve{p},\breve{s}) .$$ (5.203)

Let $v(p,s)$ denote a spinor of negative energy, with polarization $-\breve{s}$ in the reset frame. In this case,

$$\vec{\Sigma}\cdot\breve{s}v(\breve{p},\breve{s}) = -v(\breve{p},\breve{s}) .$$ (5.204)

Therefore in an arbitrary Lorentz frame

$$\omega^1(\vec{p}) = u(p,u_z) ,$$

$$\omega^2(\vec{p}) = u(p,-u_z) ,$$

$$\omega^3(\vec{p}) = v(p,-u_z) ,$$

$$\omega^4(\vec{p}) = v(p,u_z),$$ (5.205)

where $u^\mu_z$ is a 4-vector, which in the rest frame is $\breve{u}^\mu_z = (0,\breve{u_z}) = (0,0,0,1)$. An arbitrary spinor is thus specified by the sign of the energy, its momentum $p_\mu$, and polarization in the rest frame $\breve{s}_\mu$.