Finite Rotations-QM

  • Thread starter Norman
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Finite Rotations

Problem:
PROVE:
[tex] D^{\frac{1}{2}}[R]=exp( \frac{-i}{\hbar} \mathbf{\theta} \cdot \mathbf{J}^{\frac{1}{2}} ) = cos(\frac{\theta}{2}) I-\frac{2i}{\hbar}sin(\frac{\theta}{2}) \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}} [/tex]

where:
[tex] J_i^{\frac{1}{2}}=\frac{\hbar}{2} \sigma_i [/tex]
and [itex] \sigma_i [/itex] is the appropriate pauli matrix. And I is the identity matrix.

here is what I have so far... I get so close but the solution is incorrect:

[tex] = e^{\frac{-i \theta}{2}} e^{\sum_{i=1}^3 \sigma_i} [/tex]
[tex] = (cos(\frac{\theta}{2})-i sin(\frac{\theta}{2})) e^{\sum_{i=1}^3 \sigma_i} [/tex]
[tex] = (cos(\frac{\theta}{2})-i sin(\frac{\theta}{2})) \sum_{n=0}^\infty \frac{(\sum_{i=1}^3 \sigma_i)^n}{n!} [/tex]

for j=1/2 the sum over n only needs to go from 0 to 2j (1) so the last line only pics up the first 2 terms.

[tex] = (cos(\frac{\theta}{2})-i sin(\frac{\theta}{2}))(I + \sum_{i=1}^3 \sigma_i)[/tex]
Now let:
[tex] \sum_{i=1}^3 \sigma_i = \frac{2}{\hbar} \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}} [/tex]
therefore:
[tex] D^{\frac{1}{2}}[R]= (cos(\frac{\theta}{2})-i sin(\frac{\theta}{2}))(I + \frac{2}{\hbar} \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}}) [/tex]

Now I know this is isn't correct... but it is sooo close that I am having a hard time finding where I went wrong and how else to get the cosine and sine terms to show up. Please help, I am horribly frustrated.
Thanks,
Norm
 
Last edited:

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