$Ad\circ\exp=\exp\circ ad$ and $e^{i(\theta/2)\hat{n}\cdot\sigma}\sigma e^{-i(\theta/2)\hat{n}\cdot\sigma}=e^{\theta\hat{n}\cdot J}\sigma$

Question

This question is inspired by my recent question How to prove $e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = e^{\theta \hat{n}\cdot J}\sigma$? with answer https://physics.stackexchange.com/a/819562/128186 and chatroom https://chat.stackexchange.com/rooms/153836/discussion-between-jagerber48-and-lpz.

We have one formula $$ \tag{1} U^{\dagger}(R)\sigma U(R) = e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot J)} = e^{\theta (\hat{n}\cdot J)} \sigma = R\sigma $$ Where $\sigma$ is a vector of the Pauli matrices (generators of $SU(2)$) and $J$ is a vector of the anti-symmetric generators of $SO(3)$.

There is another deeper formula involving Lie groups and Lie algebras: $$ \tag{2} \text{Ad}\circ \exp = \exp \circ \text{ad} $$ I am trying to understand the relationship between $(1)$ and $(2)$. Can $(2)$ be used to straightforwardly prove/derive $(1)$? I see the ingredients being something like

• $(2)$ holds generically for Lie groups/algebras and
• $\mathfrak{su(2)}$ is a representation of $\mathfrak{so}(3)$
• so somehow we can derive $(1)$ from $(2)$.

I'm very tempted to write $$ e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = \text{Ad}(\exp(-i(\theta/2)(\hat{n}\cdot \sigma)))(\sigma) $$ Then, using $(2)$, I'd want to equate this too \begin{align*} \exp(\text{ad}(+i(\theta/2)(\hat{n}\cdot \sigma)))(\sigma) \end{align*} but then I need to somehow relate this expression to $$ e^{\theta(\hat{n}\cdot J)}\sigma $$ and this step I do not know how to do. I suspect/wonder if it is related to the representation of $\mathfrak{so}(3)$ by $\mathfrak{su}(2)$.

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A few further steps: I wish I could show $$ \exp(\text{ad}(+i(\theta/2)(\hat{n}\cdot\sigma))) \stackrel{?}{=} \exp(\theta \hat{n}\cdot J) $$ Or, we would like to show $$ \text{ad}(+i(\theta/2)(\hat{n}\cdot\sigma)) \stackrel{?}{=} \theta \hat{n}\cdot J $$ Because $\text{ad}$ is linear we have $$ \text{ad}(+i(\theta/2)(\hat{n}\cdot\sigma)) = i(\theta/2)\text{ad}(\hat{n}\cdot\sigma) $$ So it seems if we could show $$ \text{ad}(\hat{n}\cdot\sigma) = -2 i \hat{n}\cdot J $$ that we would about have the answer.

Answer 1

The original question is really almost there. The answer is that we can almost prove $(1)$ directly from $(2)$. All we need in addition is a few facts about the pauli matrices $\sigma$ and the rotation generators $J$.

$$ \newcommand{\ad}{\text{ad}} \newcommand{\Ad}{\text{Ad}} $$

We seek to prove

$$ \tag{1} e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = e^{\theta (\hat{n}\cdot J)}\sigma $$ using $$ \tag{2} \Ad\circ \exp = \exp \circ \ad. $$ Note that $$ \Ad(X)(Y) = XYX^{\dagger}\\ \ad(x)(y) = [x, y]. $$ We can write $$ e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = \Ad(\exp(+i(\theta/2)(\hat{n}\cdot \sigma))(\sigma). $$ Using $(2)$, we then equate this to $$ \exp(\ad(+i(\theta/2)(\hat{n}\cdot \sigma)))(\sigma) $$ By linearity of $\ad$ we can write this as $$ \exp(+i(\theta/2)\ad(\hat{n}\cdot\sigma))(\sigma). $$ We would like this to be equal to the RHS of $(1)$: $$ \exp(\theta (\hat{n}\cdot J))\sigma. $$ This equality would hold if $$ \ad(\hat{n}\cdot \sigma)(\sigma) \stackrel{?}=-2i (\hat{n}\cdot J)\sigma $$ We will now prove this. \begin{align*} (\ad(\hat{n}\cdot \sigma)(\sigma))_j =& \ad(\hat{n}\cdot \sigma)(\sigma_j)\\ =& [\hat{n} \cdot \sigma, \sigma_j]\\ =& [n_i \sigma_i, \sigma_j]\\ =& n_i[\sigma_i, \sigma_j]\\ =& n_i 2i \epsilon_{ijk} \sigma_k\\ =& -2i n_i (J_i)_{jk} \sigma_k\\ =& -2i n_i (J_i \sigma)_j\\ =& -2i (n_iJ_i \sigma)_j\\ =& -2i ((\hat{n}\cdot J)\sigma)_j\\ =& (-2i (\hat{n}\cdot J) \sigma)_j \end{align*} where I have used $[\sigma_i, \sigma_j] = 2i \epsilon_{ijk} \sigma_k$ and $(J_i)_{jk} = -\epsilon_{ijk}$. So we see, indeed, that $$ \ad(\hat{n}\cdot \sigma)(\sigma) = -2i (\hat{n}\cdot J) \sigma $$ from which $(1)$ then follows.

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The one part of this proof I'm sketchy on is the implication that $$ \ad(+i(\theta/2)(\hat{n}\cdot \sigma))(\sigma) = \theta (\hat{n}\cdot J) \sigma $$ implies $$ \exp(\ad(+i(\theta/2)(\hat{n}\cdot \sigma)))(\sigma) = \exp(\theta (\hat{n}\cdot J)) \sigma $$ I find it a little funny because on the LHS we have an operator that acts on individual Pauli matrices and on the RHS we have a matrix that multiplies the Pauli matrix vector.

edit: This last implication that I was dubious about is no problem, see https://math.stackexchange.com/questions/4938616/operator-exponential-equality-question-does-x-sigma-y-sigma-imply-expx/4938648#4938648