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I need an explanation behind this calculation

  1. Aug 31, 2011 #1
    When expressing [itex]exp(\frac{i\sigma\cdot \widehat{n}\phi}{2})[/itex] as a series expansion, why can we make the assumption that [itex](\sigma\cdot \widehat{n})^{2}=I[/itex]?

    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):

    [itex]\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[/itex]

    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. jcsd
  3. Sep 1, 2011 #2
    Pauli matrices have the following IMPORTANT properties that you probably should verify and memorize:
    [itex]\sigma_j^2 = I\,,[/itex] [itex][\sigma_i, \sigma_j] = 2 i\epsilon_{ijk} \sigma_k \,,[/itex] [itex]\{\sigma_i, \sigma_j\} = 2\delta_{ij} I[/itex]

    Now, from the second step:
    [itex]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[/itex]

    What the person you used is this:
    [itex]\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k[/itex]
     
    Last edited: Sep 1, 2011
  4. Sep 1, 2011 #3
    Wow, that is actually very helpful. Thank you. But I wonder, is the last portion [itex]\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k[/itex] necessary? It looks as though you can get by without that from the second step. Thanks again.
     
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