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Ryder, QFT 2nd Ed. Page 47, eqn (1.122)

  1. Aug 12, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex](1 - \frac{i}{2}\mathbf{\sigma\cdot\theta})\mathbf{\sigma}(1 + \frac{i}{2}\mathbf{\sigma\cdot\theta}) = \mathbf{\sigma - \theta\times\sigma}[/tex]

    2. Relevant equations



    3. The attempt at a solution
    At one point in this, I temporarily ignore the y and z components. I hope the notation is not too confusing.
    [tex](1 - \frac{i}{2}\mathbf{\sigma\cdot\theta})\mathbf{\sigma}(1 + \frac{i}{2}\mathbf{\sigma\cdot\theta}) = \mathbf{\sigma} + \frac{i}{2}(\mathbf{\sigma}(\mathbf{\sigma\cdot\theta}) - (\mathbf{\sigma\cdot\theta})\mathbf{\sigma})[/tex]
    [tex] = \mathbf{\sigma} + \frac{i}{2}(\sigma_x(\sigma_x\theta_x + \sigma_y\theta_y + \sigma_z\theta_z) - (\sigma_x\theta_x + \sigma_y\theta_y + \sigma_z\theta_z)\sigma_x) + ()[/tex]
    [tex] = \mathbf{\sigma} + \frac{i}{2}(\theta_x + 2i\sigma_z\theta_y - 2i\sigma_y\theta_z) - (\theta_x - 2i\sigma_z\theta_y + 2i\sigma_y\theta_z)) + ()[/tex]
    [tex] = \mathbf{\sigma} - 2(\sigma_z\theta_y - \sigma_y\theta_z) + ()[/tex]
    and finally, putting the y and z compenents back in:
    [tex](1 - \frac{i}{2}\mathbf{\sigma\cdot\theta})\sigma(1 + \frac{i}{2}\mathbf{\sigma\cdot\theta}) = \mathbf{\sigma - 2\theta\times\sigma}[/tex]
     
    Last edited: Aug 12, 2007
  2. jcsd
  3. Aug 12, 2007 #2

    dextercioby

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    [tex]\left[ \vec{\sigma}-\frac{i}{2}\left( \vec{\sigma}\cdot \vec{\theta}\right) \vec{\sigma}\right] \left[ 1+\frac{i}{2}\left( \vec{\sigma}\cdot \vec{\theta}\right) \right] =\vec{\sigma}+\frac{i}{2}\vec{\sigma}\left( \vec{\sigma}\cdot \vec{\theta}\right) -\frac{i}{2}\left( \vec{\sigma}\cdot \vec{\theta}\right) \vec{\sigma}+\mathcal{O}\left( \theta ^{2}\right) [/tex]

    [tex] =\left[ \sigma _{i}+\frac{i}{2}\left( \sigma _{i}\sigma _{j}-\sigma _{j}\sigma _{i}\right) \theta _{j}\right] \vec{e}_{i}+\mathcal{O}\left( \theta ^{2}\right)=\left( \sigma _{i}+\frac{i}{2}2i\varepsilon _{ijk}\sigma _{k}\theta _{j}\right) \vec{e}_{i}=\vec{\sigma}-\vec{\theta}\times \vec{\sigma}+\mathcal{O}\left( \theta ^{2}\right) [/tex]

    So Ryder was right this time.
     
  4. Aug 13, 2007 #3
    Thanks dextercioby. I do try hard to solve these matters before I post. It has taken me a month just to get to page 92. I cannot explain what mental blindness caused me to have trouble with this one though.
     
  5. Aug 13, 2007 #4

    dextercioby

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    I don't think it's mental blindness; it's just lack of exercise and at some points also the inspiration's missing. Just keep up the good work !!
     
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