Finding Pauli matrices WITHOUT ladder operators

[Penguin]

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

Thanks!

 

[vanesch]

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

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.

 

[Penguin]

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.

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?

 

[humanino]

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

 

[humanino]

I give here an alternative way to find the Pauli matrices, which seems natural to me. Any [tex]U(2)[/tex] matrix can be parameterized by :
[tex]
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)
[/tex]

and this reduces in the subgroup [tex]SU(2)[/tex] to [tex]w+v=0[/tex] or :

[tex]
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)
[/tex]

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

[tex]
\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)
[/tex]


[tex]
\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)
[/tex]


[tex]
\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)
[/tex]


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