Finite Rotations: Prove D^(1/2)[R]

[Norman]

Finite Rotations

Problem:
PROVE:
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}}

where:
J_i^{\frac{1}{2}}=\frac{\hbar}{2} \sigma_i
and \sigma_i 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:

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

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.

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

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

 

[member 553083]

aSolution: Starting with your last expression, we can simplify it as follows: 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}}) = cos(\frac{\theta}{2}) (I + \frac{2}{\hbar} \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}}) -i sin(\frac{\theta}{2}) (I + \frac{2}{\hbar} \hat{\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}} + \frac{2}{\hbar} sin(\frac{\theta}{2}) \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}}-i sin(\frac{\theta}{2}) I -\frac{2i}{\hbar}sin(\frac{\theta}{2}) \hat{\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}} Thus, the proof is complete.

 

[M98Ranger]

an


Dear Norman,

Thank you for sharing your work so far. It seems like you are on the right track, but there are a few errors in your calculations.

First, when you expanded the exponential term, you only need to consider the first two terms because for j=1/2, the sum only goes up to n=1. However, your expansion goes up to n=3. This is why you have extra terms in your final expression.

Secondly, when you let \sum_{i=1}^3 \sigma_i = \frac{2}{\hbar} \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}}, you are missing a factor of i in the right-hand side. It should be \frac{2i}{\hbar} \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}}. This is why you end up with the incorrect expression for D^{1/2}[R] at the end.

To correct these errors, let's start from the beginning. We have:

D^{\frac{1}{2}}[R] = e^{\frac{-i}{\hbar} \mathbf{\theta} \cdot \mathbf{J}^{\frac{1}{2}}}

= cos(\frac{\theta}{2}) I -i sin(\frac{\theta}{2}) \mathbf{\theta} \cdot \mathbf{J}^{\frac{1}{2}}

= cos(\frac{\theta}{2}) I -i sin(\frac{\theta}{2}) \frac{2i}{\hbar} \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}}

= cos(\frac{\theta}{2}) I - \frac{2}{\hbar} sin(\frac{\theta}{2}) \hat{\theta} \cdot \mathbf{J}^{\frac{1}{2}}

= cos(\frac{\theta}{2}) I - \frac{2}{\hbar} sin(\frac{\theta}{2}) (\frac{\hbar}{2} \sigma_x \hat{\theta}_x + \frac{\hbar}{2} \sigma_y \hat{\theta}_y + \frac{\hbar}{2} \sigma_z \hat{\theta