How to Solve the Matrix Equation in SL(2,C) with Given Vector Conditions?

[emma83]

Homework Statement

Solve the equation $$A_{k}k_{0}A_{k}^{\dagger}=k$$ in SL(2,C) where $$k_0$$ corresponds to the unit vector $$\{0,0,1\}$$ and $$k$$ is an arbitrary vector, i.e.:

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

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

Homework Equations

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

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

The Attempt at a Solution

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!

 

[Jhero]

Thank you for your question. Solving this equation can be done by using the properties of SL(2,C) matrices and the given information about k_0 and k. Firstly, we know that A_k is an element of SL(2,C), which means that its determinant is equal to 1. This gives us the following condition:

ad-bc=1

where a, b, c, and d are the elements of A_k. We also know that A_kk_0A_k^{\dagger}=k, which means that the product of these matrices should be equal to k. This gives us the following equations:

2aa*+2ac*=1+n_3
2ca*+2cc*=1-n_3
2ab*+2ad*=n_-
2cb*+2cd*=n_+

From these equations, we can solve for a, b, c, and d, which will give us the expression for A_k in terms of n_+, n_-, and n_3. You can use substitution and elimination to solve for these variables. Once you have the expression for A_k, you can verify if it satisfies the given equation by plugging it back in and checking if it gives the same result as k.

I hope this helps. Good luck with your solution!