2
$\begingroup$

I was reading these lecture notes (NB: PDF):

For spin-1/2, the rotation operator $$ R_\alpha^{(s)}(\mathbf n)=\exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right) $$ can be written as an explicit $2\times2$ matrix. This is accomplished by expanding the exponential in a Taylor series: \begin{align} \exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right)&=1-\frac{i\alpha}{2}\left(\vec\sigma\cdot\hat{\mathbf n}\right)+\frac1{2!}\left(\frac{i\alpha}{2}\right)^2\left(\vec\sigma\cdot\hat{\mathbf n}\right)^2\\&\quad-\frac1{3!}\left(\frac{i\alpha}{2}\right)^3\left(\vec\sigma\cdot\hat{\mathbf n}\right)^3+\cdots \end{align} Note that $$ \left(\vec\sigma\cdot\hat{\mathbf n}\right)^2=\left(\vec\sigma\cdot\hat{\mathbf n}\right)\left(\vec\sigma\cdot\hat{\mathbf n}\right)=\hat{\mathbf n}\cdot\hat{\mathbf n}+i\sigma\left(\hat{\mathbf n}\times\hat{\mathbf n}\right)=1 $$ Thus, the Taylor series becomes \begin{align} \exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right)&=1-\frac{i\alpha}{2}\left(\vec\sigma\cdot\hat{\mathbf n}\right)+\frac1{2!}\left(\frac{i\alpha}{2}\right)^2\left(\vec\sigma\cdot\hat{\mathbf n}\right)^2\\&\quad-\frac1{3!}\left(\frac{i\alpha}{2}\right)^3\left(\vec\sigma\cdot\hat{\mathbf n}\right)^3+\cdots\\ &=\left[1-\frac1{2!}\left(\frac\alpha2\right)^2+\frac1{4!}\left(\frac\alpha2\right)^$+\cdots\right]\\&\quad-i\vec\sigma\cdot\hat{\mathbf n}\left[\left(\frac\alpha2\right)-\frac1{3!}\left(\frac\alpha2\right)^3+\cdots\right]\\ &=\cos\left(\frac\alpha2\right)-i\vec\sigma\cdot\hat{\mathbf n}\sin\left(\frac\alpha2\right) \end{align}

However, the part I don't understand is:

$$ \left(\vec\sigma\cdot\hat{\mathbf n}\right)^2=\left(\vec\sigma\cdot\hat{\mathbf n}\right)\left(\vec\sigma\cdot\hat{\mathbf n}\right)=\hat{\mathbf n}\cdot\hat{\mathbf n}+i\sigma\left(\hat{\mathbf n}\times\hat{\mathbf n}\right)=1 $$

Why is that equal to 1? Where do the dot-product and cross-product come from? Note that the $\sigma$ are Pauli spin matrices.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

To show that $$ \left(\sigma\cdot\mathbf{n}\right)^2=\mathbf n\cdot\mathbf n+i\sigma\cdot\left(\mathbf n\times\mathbf n\right)\tag{1} $$ consider writing the above as \begin{align} \left(\sigma\cdot\mathbf a\right)\left(\sigma\cdot\mathbf b\right)&=\sum_j\sigma_ja_j\sum_k\sigma_kb_k\\ &=\sum_j\sum_k\left(\frac12\{\sigma_j,\,\sigma_k\}+\frac12[\sigma_j,\,\sigma_k]\right)a_jb_k\\ &=\sum_j\sum_k\left(\delta_{jk}+i\epsilon_{jkl}\sigma_l\right)a_jb_k\tag{2} \end{align} where the 2nd line arises from using the anti-commutating and commutating relation for the matrices. In the third line, we have the Kronecker delta and Levi-Civita symbol. The result (1) follows from completing the math from (2) (that is, writing it in vector notation and replacing $\mathbf a$ and $\mathbf b$ with $\mathbf n$).

The remainder is to show that this is equal to 1. For that, the following two hints should suffice:

  1. Note that for two vectors $\mathbf a$ and $\mathbf b$, $\mathbf a\times\mathbf b=-\mathbf b\times\mathbf a$. What requirement is needed if $\mathbf b=\mathbf a$: $\mathbf a\times\mathbf a=?$
  2. For the unit vector, e.g. $\mathbf n=(1,\,0)^T$, what is the dot product?
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.