Why does dot product equal to one? (Pauli spin matrices)

Question

I was reading these lecture notes (NB: PDF):

For spin-1/2, the rotation operator $$ R_\alpha^{(s)}(\mathbf n)=\exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right) $$ can be written as an explicit $2\times2$ matrix. This is accomplished by expanding the exponential in a Taylor series: \begin{align} \exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right)&=1-\frac{i\alpha}{2}\left(\vec\sigma\cdot\hat{\mathbf n}\right)+\frac1{2!}\left(\frac{i\alpha}{2}\right)^2\left(\vec\sigma\cdot\hat{\mathbf n}\right)^2\\&\quad-\frac1{3!}\left(\frac{i\alpha}{2}\right)^3\left(\vec\sigma\cdot\hat{\mathbf n}\right)^3+\cdots \end{align} Note that $$ \left(\vec\sigma\cdot\hat{\mathbf n}\right)^2=\left(\vec\sigma\cdot\hat{\mathbf n}\right)\left(\vec\sigma\cdot\hat{\mathbf n}\right)=\hat{\mathbf n}\cdot\hat{\mathbf n}+i\sigma\left(\hat{\mathbf n}\times\hat{\mathbf n}\right)=1 $$ Thus, the Taylor series becomes \begin{align} \exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right)&=1-\frac{i\alpha}{2}\left(\vec\sigma\cdot\hat{\mathbf n}\right)+\frac1{2!}\left(\frac{i\alpha}{2}\right)^2\left(\vec\sigma\cdot\hat{\mathbf n}\right)^2\\&\quad-\frac1{3!}\left(\frac{i\alpha}{2}\right)^3\left(\vec\sigma\cdot\hat{\mathbf n}\right)^3+\cdots\\ &=\left[1-\frac1{2!}\left(\frac\alpha2\right)^2+\frac1{4!}\left(\frac\alpha2\right)^$+\cdots\right]\\&\quad-i\vec\sigma\cdot\hat{\mathbf n}\left[\left(\frac\alpha2\right)-\frac1{3!}\left(\frac\alpha2\right)^3+\cdots\right]\\ &=\cos\left(\frac\alpha2\right)-i\vec\sigma\cdot\hat{\mathbf n}\sin\left(\frac\alpha2\right) \end{align}

However, the part I don't understand is:

$$ \left(\vec\sigma\cdot\hat{\mathbf n}\right)^2=\left(\vec\sigma\cdot\hat{\mathbf n}\right)\left(\vec\sigma\cdot\hat{\mathbf n}\right)=\hat{\mathbf n}\cdot\hat{\mathbf n}+i\sigma\left(\hat{\mathbf n}\times\hat{\mathbf n}\right)=1 $$

Why is that equal to 1? Where do the dot-product and cross-product come from? Note that the $\sigma$ are Pauli spin matrices.

Answer 1

To show that $$ \left(\sigma\cdot\mathbf{n}\right)^2=\mathbf n\cdot\mathbf n+i\sigma\cdot\left(\mathbf n\times\mathbf n\right)\tag{1} $$ consider writing the above as \begin{align} \left(\sigma\cdot\mathbf a\right)\left(\sigma\cdot\mathbf b\right)&=\sum_j\sigma_ja_j\sum_k\sigma_kb_k\\ &=\sum_j\sum_k\left(\frac12\{\sigma_j,\,\sigma_k\}+\frac12[\sigma_j,\,\sigma_k]\right)a_jb_k\\ &=\sum_j\sum_k\left(\delta_{jk}+i\epsilon_{jkl}\sigma_l\right)a_jb_k\tag{2} \end{align} where the 2nd line arises from using the anti-commutating and commutating relation for the matrices. In the third line, we have the Kronecker delta and Levi-Civita symbol. The result (1) follows from completing the math from (2) (that is, writing it in vector notation and replacing $\mathbf a$ and $\mathbf b$ with $\mathbf n$).

The remainder is to show that this is equal to 1. For that, the following two hints should suffice:

1. Note that for two vectors $\mathbf a$ and $\mathbf b$, $\mathbf a\times\mathbf b=-\mathbf b\times\mathbf a$. What requirement is needed if $\mathbf b=\mathbf a$: $\mathbf a\times\mathbf a=?$
2. For the unit vector, e.g. $\mathbf n=(1,\,0)^T$, what is the dot product?