I need an explanation behind this calculation

[Demon117]

When expressing $exp(\frac{i\sigma\cdot \widehat{n}\phi}{2})$ as a series expansion, why can we make the assumption that $(\sigma\cdot \widehat{n})^{2}=I$?

Someone I asked on campus showed me this and I don't quite understand its implications (partly because I don't quite understand permutation tensors):

$\sigma_{i}n_{i}\sigma_{j}n_{j}=n_{i}n_{j}(\delta_{ij}I+i\epsilon_{ijk}\sigma_{k})=(\widehat{n}\cdot \widehat{n})I+0=I$

The middle two steps are really foreign to me, but I get the component parts in front and the last part. Could someone explain in further detail why the middle two parts exist?

 

[mathfeel]

When expressing $exp(\frac{i\sigma\cdot \widehat{n}\phi}{2})$ as a series expansion, why can we make the assumption that $(\sigma\cdot \widehat{n})^{2}=I$?

Someone I asked on campus showed me this and I don't quite understand its implications (partly because I don't quite understand permutation tensors):

$\sigma_{i}n_{i}\sigma_{j}n_{j}=n_{i}n_{j}(\delta_{ij}I+i\epsilon_{ijk}\sigma_{k})=(\widehat{n}\cdot \widehat{n})I+0=I$

The middle two steps are really foreign to me, but I get the component parts in front and the last part. Could someone explain in further detail why the middle two parts exist?

Pauli matrices have the following IMPORTANT properties that you probably should verify and memorize:
$\sigma_j^2 = I\,,$ $[\sigma_i, \sigma_j] = 2 i\epsilon_{ijk} \sigma_k \,,$ $\{\sigma_i, \sigma_j\} = 2\delta_{ij} I$

Now, from the second step:
$n_i n_j \sigma_i \sigma_j = \frac12 n_i n_j (\sigma_i \sigma_j + \sigma_j \sigma_i) = \frac12 n_i n_j \{\sigma_i, \sigma_j\} = n_i n_j \delta_i I = I$

What the person you used is this:
$\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k$

 

[Demon117]

Pauli matrices have the following IMPORTANT properties that you probably should verify and memorize:
$\sigma_j^2 = I\,,$ $[\sigma_i, \sigma_j] = 2 i\epsilon_{ijk} \sigma_k \,,$ $\{\sigma_i, \sigma_j\} = 2\delta_{ij} I$

Now, from the second step:
$n_i n_j \sigma_i \sigma_j = \frac12 n_i n_j (\sigma_i \sigma_j + \sigma_j \sigma_i) = \frac12 n_i n_j \{\sigma_i, \sigma_j\} = n_i n_j \delta_i I = I$

What the person you used is this:
$\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k$

Wow, that is actually very helpful. Thank you. But I wonder, is the last portion $\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k$ necessary? It looks as though you can get by without that from the second step. Thanks again.