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}) =