Rotation of spin projection different to general rotation?

[bananabandana]

Homework Statement

Prove, for the matrix $S = exp \bigg(-\frac{i}{\hbar}\mathbf{\hat{n}}\cdot \mathbf{\hat{S}}\bigg)$ (spin-rotation matrix), and for an arbitary vector $\mathbf{a}$ that:

$$ S^{-1} \mathbf{a} \cdot \mathbf{\hat{x}} S = a(-\theta) \cdot \mathbf{x} = a \cdot \mathbf{x}(\theta) $$

Homework Equations

The Attempt at a Solution

Let $\langle \psi| \mathbf{V} \rangle = \mathbf{V}_{0}$. Where $\mathbf{V}$ is some vector operator. Then (after Binney, Physics of Qunatum Mechanics,pg74):

Where $\boldsymbol{\alpha}$ is an axis of rotation, and let $\mathbf{R}(\boldsymbol{\alpha})$ be a rotation operator about that axis.

$$ \mathbf{R}(\boldsymbol{\alpha})\mathbf{V} = U^{\dagger}(\boldsymbol{\alpha})\mathbf{V}U(\boldsymbol{\alpha})$$

i.e as standard. Apart from that, I have no idea where to begin! Apparently this is based on some 'standard relation in tensor calculus (read, matrix manipulations here, I think) - but I have no idea what is happening!

Thanks

 

[mark726]

for posting in the forum! First, let's define some variables for clarity. Let $S = exp\bigg(-\frac{i}{\hbar}\mathbf{\hat{n}}\cdot \mathbf{\hat{S}}\bigg)$ and $\mathbf{a}$ be an arbitrary vector. We also have the spin-rotation matrix $S^{-1}$.

To prove the given statement, we need to show that $S^{-1} \mathbf{a} \cdot \mathbf{\hat{x}} S = a(-\theta) \cdot \mathbf{x} = a \cdot \mathbf{x}(\theta)$.

To do this, we can use the definition of the spin-rotation matrix $S$ and the rotation operator $R(\boldsymbol{\alpha})$. We can write $S$ as $S = R(\boldsymbol{\alpha})$, where $\boldsymbol{\alpha} = -\frac{i}{\hbar}\mathbf{\hat{n}}\cdot \mathbf{\hat{S}}$.

Next, we can use the relation mentioned in the post, $\mathbf{R}(\boldsymbol{\alpha})\mathbf{V} = U^{\dagger}(\boldsymbol{\alpha})\mathbf{V}U(\boldsymbol{\alpha})$, to rewrite the left-hand side of the given statement. We get:

$$S^{-1} \mathbf{a} \cdot \mathbf{\hat{x}} S = U^{\dagger}(\boldsymbol{\alpha})\mathbf{a} \cdot \mathbf{\hat{x}} U(\boldsymbol{\alpha})$$

Now, we can use the properties of the rotation operator to simplify this expression. We know that the rotation operator $U(\boldsymbol{\alpha})$ satisfies the following relation:

$$U(\boldsymbol{\alpha})\mathbf{\hat{x}} U^{\dagger}(\boldsymbol{\alpha}) = \mathbf{\hat{x}}(\theta)$$

where $\theta = |\boldsymbol{\alpha}|$. Using this relation, we can rewrite the above expression as:

$$U^{\dagger}(\boldsymbol{\alpha})\mathbf{a} \cdot \mathbf{\hat{x}} U(\boldsymbol{\alpha}) =