Least Action

Nontrivializing triviality..and vice versa.

Constructing a vector from a spinor

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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 🙂


Written by Vivek

November 13, 2010 at 16:13

Posted in Uncategorized

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