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 &amp; 0 \\<br /> 0 &amp; 0 \\<br /> \end{array} \right)<br />

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

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

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

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

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 &amp; n_- \\<br /> n_+ &amp; 1-n_3 \\<br /> \end{array} \right)<br />

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?