# Finding the matrix A such that, exp(sA) is in SU(2)

1. Oct 21, 2009

### wilco

1. The problem statement, all variables and given/known data

Finding the matrix A such that, exp(sA) is in SU(2)

2. Relevant equations

My attempt is in trying to solve

$$\left(e^{sA}\right)^{t} B \left(e^{sA}\right) = B$$

for A, where A is some 2x2 (complex?) matrix.

and B is the matrix representing the group of SU(2) matrices. Trouble is I'm not sure what B is, but have been using the matrix of the general form of SU(2)

B = [ $$\alpha, -\beta*; \beta, \alpha*$$], where * denotes the conjugate

3. The attempt at a solution

$$\left(e^{sA}\right)^{t} B \left(e^{sA}\right) = B$$

simplifying to the solving of

$$A^{t}$$B + B A = 0

which I'm not really having much success at doing. Anyone who knows more about this than me will see that I not really sure what I'm up to. Some help would be appreciated.

Thanks, in anticipation..

2. Oct 21, 2009

### Dick

I don't see what B has to do with anything. You just want S=exp(sA) to be is SU(2), right? That means S is orthogonal, i.e. S^+=S^(-1) (where ^+ is hermitian conjugate) and det(S)=1. For the first one, S^(-1)=exp(-sA) and S^+=exp(sA^+) and compare the expansions of the exponentials (I'm assuming you mean s to be real). For the second condition use det(exp(A))=exp(trace(A)).

3. Oct 21, 2009

### wilco

Thanks, Dick - much appreciated.

I got went down the track of using the approach with B as it was used on another example that I had available to me. Your approach seems like common sense ... now.

Am I right that det(exp(sA))=exp(trace(A)) so the s does not figure in that calc?

Therefore, I require, exp(trace(A)) = 1, hence A is traceless?

Also, as an aside, am I correct in thinking that if s was not real, say s=i, then I can show that exp(iA) is unitary if A is Hermitian, so any traceless Hermitian matrix would be in SU(2)?

Thanks again - I'll carry on working through what you've given me..

4. Oct 21, 2009

### Dick

Well, no, det(exp(sA))=exp(trace(sA)), you can't just leave the s out on one part. But if A is traceless, then sA is traceless. You reach the same conclusion. And, yes, if s=i then A must be Hermitian. Can you use the essentially the same argument to show if s is real, then A must be antihermitian? I.e. A^(+)=(-A)?

5. Oct 21, 2009

### wilco

Actually, I've just done the expansions of S^(-1)=exp(-sA) and S^+=exp(sA^+), and comparing these brings me to exactly that conclusion, that A^(+)=(-A), ie. A must be antihermitian.

Adding the requirement, det(exp(sA))=1, would leave me concluding that A must be any traceless, antihermitian matrix. Does that sound on track?

6. Oct 21, 2009

### Dick

I believe you.

7. Oct 21, 2009

### wilco

Got it! Brilliant ... thanks again for your time.

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