# I need an explanation behind this calculation

1. Aug 31, 2011

### 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?

2. Sep 1, 2011

### mathfeel

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:
$\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k$

Last edited: Sep 1, 2011
3. Sep 1, 2011

### Demon117

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.