Exponential of Pauli spin matrices

  • Thread starter Happiness
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  • #1
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How do we get (6.265)?

Shouldn't we have

##exp(-i\frac{\alpha}{2}\hat{n}.\sigma)=\cos(\frac{\alpha}{2}\hat{n}.\sigma)-i\sin(\frac{\alpha}{2}\hat{n}.\sigma)##?

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Answers and Replies

  • #2
blue_leaf77
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By (6.241), you have ##(\mathbf{\sigma} \cdot \hat{n} )^2 = 1##. Use this identity in the Taylor expansions of ##\cos(\frac{\alpha}{2}\hat{n}.\sigma)## and ##\sin(\frac{\alpha}{2}\hat{n}.\sigma)## to reduce the higher powers.
 
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