# Finding Pauli matrices WITHOUT ladder operators

1. Oct 4, 2004

### Penguin

Does anyone know of an alternative way of calculating the Pauli spin matrices$$\mbox{ \sigma_x}$$ and $$\mbox{ \sigma_y}$$ (already knowing $$\mbox { \sigma_z}$$ and the (anti)-commutation relations), without using ladder operators $$\mbox{ \sigma_+}$$ and $$\mbox{ \sigma_- }$$?

Thanks!

Last edited by a moderator: Oct 4, 2004
2. Oct 4, 2004

### vanesch

Staff Emeritus
How about brute force ? Knowing that you need traceless hermitean 2x2 matrices, you put in the unknowns, write out all the equations anti-comm relations ... and solve ?

cheers,
Patrick.

3. Oct 4, 2004

### Penguin

Brute force was my initial plan :shy: problem is: Only using comm and anti-comm I get a whole bunch of possible solutions (like e.g. \sigma_x'=-\sigma_x=(0 & -1 \\ -1 & 0) and \sigma_y'=-sigma_y=(0 & i \\ -i \\ 0) ) also obeying these commutation relations.

I would like to restrict these solutions to the 'traditional' Pauli matrices... Am I forgetting some basic equations somewhere that 'll do just that?

4. Oct 4, 2004

### humanino

Any representation is as good as another !
You find one, and make a rotation to go to the one you want.

5. Oct 5, 2004

### humanino

I give here an alternative way to find the Pauli matrices, which seems natural to me. Any $$U(2)$$ matrix can be parameterized by :
$$M_{U(2)} = \left( \begin{array}{cc} e^{\imath u}\cos(\theta) & e^{\imath v}\sin(\theta)\\ -e^{\imath w}\sin(\theta) & e^{\imath (w+v-u)}\cos(\theta) \end{array} \right)$$

and this reduces in the subgroup $$SU(2)$$ to $$w+v=0$$ or :

$$M_{SU(2)} = \left( \begin{array}{cc} e^{\imath u}\cos(\theta) & e^{\imath v}\sin(\theta)\\ -e^{-\imath v}\sin(\theta) & e^{-\imath u}\cos(\theta) \end{array} \right)$$

Now as usual to find the generators, one differentiate with respect to each parameters, and takes the values near the identity :

$$\frac{\partial M}{\partial\theta} = \left( \begin{array}{cc} -e^{\imath u}\sin(\theta) & e^{\imath v}\cos(\theta)\\ -e^{-\imath v}\cos(\theta) & -e^{-\imath u}\sin(\theta) \end{array} \right)_{\theta=0,u=0,v=0} = \left( \begin{array}{cc} 0 & 1\\ -1& 0 \end{array} \right)$$

$$\frac{\partial M}{\partial u} = \left( \begin{array}{cc} \imath e^{\imath u}\cos(\theta) & 0\\ 0 & -\imath e^{-\imath u}\cos(\theta) \end{array} \right)_{\theta=0,u=0,v=0} = \left( \begin{array}{cc} \imath & 0\\ 0& -\imath \end{array} \right)$$

$$\frac{\partial M}{\partial w} = \left( \begin{array}{cc} 0 & \imath e^{\imath v}\sin(\theta)\\ \imath e^{-\imath v}\sin(\theta) & 0 \end{array} \right)_{\theta=0,u=0,v=0} = \left( \begin{array}{cc} 0 & 1\\ 1& 0 \end{array} \right)$$

But these are not the Pauli matrices, they differ by a factor $$-\imath$$. This is exactly what is done : the Pauli matrices define an arbitrary $$SU(2)$$ matrix by :
$$M_{SU(2)} =e^{\imath \vec{L}\cdot\vec{\sigma}\alpha/2}=\sigma_0\cos(\frac{\alpha}{2}) -\imath \vec{L}\cdot\vec{\sigma}\sin(\frac{\alpha}{2})$$ with $$\sigma_0$$ the identity, $$\vec{L}$$ a unitary vector directing the rotation axis, and $$\alpha$$ the rotation angle. By differentiating this near the identity, one recovers the correct $$-\imath$$ factor w.r.t. the previously calculated matrices.