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Let $\sigma^0$ be the $2\times 2$ identity; $\sigma^1$, $\sigma^2$ and $\sigma^3$ the Pauli matrices:

$$ \sigma^1 = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix},\ \sigma^2 = \begin{pmatrix} 0 & -i \\ i & 0\end{pmatrix},\ \sigma^3 = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$

From a vector $\overrightarrow{v}$ having real components, we define the matrix

$$M = \overrightarrow{v}\cdot\overrightarrow{\sigma} = \sum\limits_{i=1}^3 v_i\sigma^i $$

that has the following eigenvalues

$$\mu_{\pm} = \pm|\overrightarrow{v}|=\pm\sqrt{v_1+v_2+v_3}$$

and the associated projectors

$$ P_{\pm} = \frac{1}{2}\left(\sigma^0\pm\frac{\overrightarrow{v}\cdot\overrightarrow{\sigma}}{|\overrightarrow{v}|}\right)$$

Which gives the following spectral decomposition:

$$ M = |\overrightarrow{v}|P_+ - |\overrightarrow{v}|P_-$$

If $\overrightarrow{w}$ is orthogonal to $\overrightarrow{v}$, i.e $\sum\limits_{i=1}^3v_iw_j=0$, I want to show the following relation:

$$ e^{\lambda\overrightarrow{v}\cdot\overrightarrow{\sigma}}\overrightarrow{w}\cdot\overrightarrow{\sigma} = \overrightarrow{w}\cdot\overrightarrow{\sigma} e^{-\lambda\overrightarrow{v}\cdot\overrightarrow{\sigma}}$$

I am tempted to use the spectral decomposition of $e^{\lambda\overrightarrow{v}\cdot\overrightarrow{\sigma}} = e^{\lambda |\overrightarrow{v}|}P_+ + e^{-\lambda |\overrightarrow{v}|}P_-$ but I'm stuck. Any idea how to prove the formula?

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This follows from the anticommutativity relation

$$(v\cdot\sigma)(w\cdot\sigma)=-(w\cdot\sigma)(v\cdot\sigma)$$

valid when $v\cdot w=0$. You can apply this term by term after expanding $\exp(\lambda v\cdot\sigma)$.

The general formula for multiplying sums of Pauli matrices is

$$ (v\cdot\sigma)(w\cdot\sigma)=(v\cdot w)\sigma^0+i(v\times w)\cdot\sigma $$

where $\times$ is the 3D cross product.

I only know this because of my familiarity with quaternions. In this setting, $\mathbb{H}$ is a 4D algebra whose elements are sums of scalars and 3D vectors. To multiply $(r+\mathbf{u})(s+\mathbf{v}$) one first uses the distributive property and then the relation $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$ which follows from inspecting the multiplication table for the basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$. If $S^3$ is the group of unit quaternions (which is a $3$-sphere, like $S^1$ in the complex numbers is a $1$-sphere i.e. a circle) then there is an isomorphism $S^3\to\mathrm{SU}(2)$ which extends to a (not onto) algebra homomorphism $\mathbb{H}\to M_2(\mathbb{C})$ given by $\mathbf{v}\mapsto v\cdot i\sigma$ (mixing up the notations a bit).

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  • $\begingroup$ Thanks for the answer. What do you mean by applying the anticommutativity relation term by term? I'm not sure on what I have to apply it $\endgroup$
    – Matt
    Jul 28, 2019 at 22:14
  • $\begingroup$ Expand $\exp(\lambda v\cdot\sigma)$ as an infinite series, then multiply it by $w\cdot\sigma$ with the distributive property. Can you simplify $(\lambda v\cdot\sigma)^n(w\cdot\sigma)$ with the anticommutativity relation? Or you can use Euler's formula, $$ \exp(\lambda v\cdot\sigma)=\cos(\lambda|v|)\sigma^0+\sin(\lambda |v|)\hat{v}\cdot\sigma $$ which follows from $(\hat{v}\cdot\sigma)^2=-\sigma^0$. $\endgroup$
    – runway44
    Jul 28, 2019 at 22:16
  • $\begingroup$ I suppose it's equal to $-(w\cdot\sigma)(\lambda v\cdot\sigma)^n$ but I don't manage to simplify the product with that. $\endgroup$
    – Matt
    Jul 28, 2019 at 22:25
  • $\begingroup$ No, think about it as $$ \underbrace{(\lambda v\cdot\sigma)\cdots(\lambda v\cdot\sigma)}_n (w\cdot\sigma). $$ Every time you slide a $\lambda v\cdot\sigma$ to the right across $w\cdot\sigma$ it introduces a $-$ sign, so you get $(-1)^n(w\cdot\sigma)(\lambda v\cdot\sigma)^n$, or in other words $(w\cdot\sigma)(-\lambda v\cdot\sigma)^n$. $\endgroup$
    – runway44
    Jul 28, 2019 at 22:28
  • $\begingroup$ Oh right, so by doing that it should give the infinite series of $exp(-\lambda v\cdot\sigma)$. Thank you for your help! Also do you have any reference for the derivation of the general formula that you mentionned earlier by any chances? $\endgroup$
    – Matt
    Jul 28, 2019 at 22:34

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