# Least Action

Nontrivializing triviality..and vice versa.

## Constructing a vector from a spinor

The purpose of this post is to examine why the quantity

$\chi^\dagger \sigma_i \chi$

transforms like a vector under rotations. Here, $\chi$ is a 2 component spinor, and $\sigma_i$ is one of the 3 Pauli matrices. The explanation follows essentially the arguments put forth by Sakurai in chapter 3 of his book on quantum mechanics.

An arbitrary state ket in spin space can be written as

$|\alpha\rangle = |+\rangle\langle +|\alpha\rangle + |-\rangle\langle -|\alpha\rangle = \left(\begin{array}{c}\langle +|\alpha\rangle\\ \langle -|\alpha\rangle\end{array}\right) = \left(\begin{array}{c}c_{+}\\c_{-}\end{array}\right)$

where $\chi_{+} = |+ \rangle = \left(\begin{array}{c}1 \\ 0 \end{array}\right)$ and $\chi_{-} = |-\rangle = \left(\begin{array}{c}0\\1\end{array}\right)$ are the base “kets” (actually they are spinors, for the spin-1/2 case). This identification suggests that we can specify a general two component spinor as

$\chi = c_{+}\chi_{+} + \chi_{-}\chi_{-}$

Further, from Sakurai’s equations 3.2.30,

$\langle \pm|S_{k}|+\rangle = \frac{\hbar}{2}(\sigma_{k})_{\pm, +}$

$\langle \pm|S_{k}|-\rangle = \frac{\hbar}{2}(\sigma_{k})_{\pm, -}$

Therefore, we get the most important result:

$\langle S_k\rangle = \frac{\hbar}{2}\chi^{\dagger}\sigma_k\chi$

Physically, this equation implies that the right hand side equals the expectation value of the spin operator $S_k$ in spinor-space (or in more precise terms, in the basis generated by the set of all possible $\chi$‘s). So, the quantity that we are examining is just the expectation value of the operator $S_k$ (modulo a constant factor of $\hbar/2$).

The argument..

Now, Sakurai has shown that under a rotation of the state kets, the expectation value of $\langle S_k\rangle$ transforms like the component of a classical vector. That is,

$\langle S_k\rangle \longrightarrow \sum_{l}R_{kl}\langle S_l\rangle$

(This is equation 3.2.11 of Sakurai)

where $R_{kl}$ are the elements of the 3 x 3 orthogonal matrix R that specifies this rotation.

The corresponding rotation of the state kets is carried out by the operator $\mathcal{D}(R) = \exp{(i\frac{\phi}{2\hbar}\hat{n}\cdot\boldsymbol{S})}$.

Therefore, $\langle S_k\rangle$ transforms like a vector, and hence so does $\chi^\dagger \sigma_i \chi$.

Thanks to David Angelaszek for a long, patient and extremely useful discussion, which led to the resolution of this argument. And of course, thanks to JJ Sakurai for forcing us to read between the lines 🙂