Solving Matrix Equations in SL(2,C) for Arbitrary Vectors

[emma83]

Hello,

Is the law for matrix multiplication in SL(2,C) the same as usual ? I try to solve the equation $$A_{k}k_{0}A_{k}^{\dagger}=k$$ where $$k_0$$ corresponds to the unit vector $$\{0,0,1\}$$ and $$k$$ is an arbitrary vector, i.e.:

$$k0=<br /> \left( \begin{array}{cc}<br /> 2 & 0 \\<br /> 0 & 0 \\<br /> \end{array} \right)$$

$$k=<br /> \left( \begin{array}{cc}<br /> 1+n_3 & n_- \\<br /> n_+ & 1-n_3 \\<br /> \end{array} \right)$$

If I try to solve for
$$A_k=<br /> \left( \begin{array}{cc}<br /> a & b \\<br /> c & d \\<br /> \end{array} \right)$$

this gives (where $$a*$$ is the conjugate of $$a$$):
$$A_{k}k_{0}A_{k}^{\dagger}=<br /> \left( \begin{array}{cc}<br /> 2aa* & 2ac* \\<br /> 2ca* & 2cc* \\<br /> \end{array} \right)$$

So this gives conditions on $$\{a,c\}$$ but can $$\{b,c\}$$ be arbitrary ? How do I solve this equation and obtain the expression of $$A_k$$ involving only $$n_+, n_-$$ and $$n_3$$ ?

Thanks a lot for your help!

 

[Fredrik]

I'm too lazy to think about the rest, but I multiplied the matrices together and got the same result you did, so that part seems to be OK.

 

[emma83]

Well, thanks Fredrik but now I really would like to know how to do "the rest" !

 

[gabbagabbahey]

I think you need to know the effect of $A_k$ on at least two linearly independent vectors to completely determine the matrix.

 

[emma83]

Thanks, actually I found an answer for $$A_{k}$$ without the details of the calculation. I tried to compute $$A_{k}k_{0}A_{k}^{\dagger}$$ but I don't get $$k$$ as would be expected.

The proposed solution is:

$$A_k=\frac{1}{\sqrt{2C(1+n_3)}}<br /> \left( \begin{array}{cc}<br /> C(1+n_3) & -n_- \\<br /> Cn_+ & 1+n_3 \\<br /> \end{array} \right)$$
i.e.:
$$A_{k}=UB$$
where:
$$U=\frac{1}{\sqrt{2(1+n_3)}}<br /> \left( \begin{array}{cc}<br /> (1+n_3) & -n_- \\<br /> n_+ & 1+n_3 \\<br /> \end{array} \right)$$
and
$$B=<br /> \left( \begin{array}{cc}<br /> \sqrt{C} & 0 \\<br /> 0 & \frac{1}{\sqrt{C}} \\<br /> \end{array} \right)$$

The $$C$$ that appears is actually a constant in $$k$$ that I ignored for simplification in my first posting:

$$k=C<br /> \left( \begin{array}{cc}<br /> 1+n_3 & n_- \\<br /> n_+ & 1-n_3 \\<br /> \end{array} \right)$$

Can somebody else try to see if this result is correct ?

 

[George Jones]

Without further restrictions, I don't think that there is a unique solution for $A_k$. What happens when your $A_k$ is multiplied on the right by an element of the little group of $k_0$?