Using the Baker-Campbell-Hausdorff Identity

[Mindstein]

Homework Statement

Given that U = e$^{-i*θ*\hat{n}*\vec{J}/hbar}$

Show that U$^{†}$$\vec{J}$U = $\hat{n}(\hat{n}*\vec{J}) - \hat{n}\times(\hat{n}\times\vec{J})cos(θ) + \hat{n}\times\vec{J}sin(θ)$

Homework Equations

The Baker-Campbell-Hausdorff Identity

The Attempt at a Solution

U$^{†}$$\vec{J}$U = e$^{-i*θ*\hat{n}*\vec{J}/hbar}$$\vec{J}$e$^{i*θ*\hat{n}*\vec{J}/h_bar}$

U$^{†}$$\vec{J}$U = $\vec{J} + [i*θ*\hat{n}*\vec{J}/hbar,\vec{J}]+[i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,\vec{J}]]/2! + ...$

I guess I just need some help on whether or not I am headed in the right direction.

 

[mark726]

Thank you for your post. Your attempt at a solution looks like it is on the right track. However, I would suggest using the identity [A,[B,C]] = [B,[A,C]] + [C,[A,B]] instead of the Baker-Campbell-Hausdorff identity. This will simplify your calculations and make it easier to reach the desired result.

Using this identity, we can write U^{†}\vec{J}U as:

U^{†}\vec{J}U = \vec{J} + [i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,\vec{J}]]/2! + ...

= \vec{J} + [i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,\vec{J}]]/3! + ...

= \vec{J} + [i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*θ*\hat{n}*\vec{J}/hbar,[i*θ*\hat{n}*\vec{J}/hbar,\vec{J}] + [i*