Matrix equation in SL(2,C)

  • Thread starter emma83
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  • #1
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Main Question or Discussion Point

Hello,

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

[tex]k0=
\left( \begin{array}{cc}
2 & 0 \\
0 & 0 \\
\end{array} \right)
[/tex]

[tex]k=
\left( \begin{array}{cc}
1+n_3 & n_- \\
n_+ & 1-n_3 \\
\end{array} \right)
[/tex]

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

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

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

Thanks a lot for your help!
 

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
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.
 
  • #3
33
0
Well, thanks Fredrik but now I really would like to know how to do "the rest" !
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6
I think you need to know the effect of [itex]A_k[/itex] on at least two linearly independent vectors to completely determine the matrix.
 
  • #5
33
0
Thanks, actually I found an answer for [tex]A_{k}[/tex] without the details of the calculation. I tried to compute [tex]A_{k}k_{0}A_{k}^{\dagger}[/tex] but I don't get [tex]k[/tex] as would be expected.

The proposed solution is:

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

The [tex]C[/tex] that appears is actually a constant in [tex]k[/tex] that I ignored for simplification in my first posting:

[tex]k=C
\left( \begin{array}{cc}
1+n_3 & n_- \\
n_+ & 1-n_3 \\
\end{array} \right)
[/tex]

Can somebody else try to see if this result is correct ?
 
  • #6
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
Without further restrictions, I don't think that there is a unique solution for [itex]A_k[/itex]. What happens when your [itex]A_k[/itex] is multiplied on the right by an element of the little group of [itex]k_0[/itex]?
 

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