Decompose SL(2C) Matrix: Real Parameters from Complex

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TL;DR
How can an SL(2,C) matrix be decomposed into a product of a boost along the z-axis and a pure rotation?
Hi,

suppose I am given an SL(2C) matrix of the form ##\exp(i\alpha/2 \vec{t}\cdot\vec{\sigma})## where ##\alpha## is the complex rotation angle, ##\vec{t}## the complex rotation axis and ##\vec{\sigma}## the vector of the three Pauli matrices.
I would like to decompose this vector into ##\exp(i\beta/2 \vec{q}\cdot\vec{\sigma})\exp(\gamma\sigma_z)##, where now the rotation angle ##\beta##, axis ##\vec{q}## and the boost parameter ##\gamma## are all real.
Is there a non-brain damaged way to do this? This isn't homework related.

Thank you!
 
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Just a thought, and maybe a silly one as I think in terms of algebra and less in terms of physics.
##\vec{t}\cdot\vec{\sigma}## is a complex number, hence ##\exp(i\alpha/2 \vec{t}\cdot\vec{\sigma})## is of the form ##e^{x+iy}=e^x\cdot e^{iy}##. In case ##\vec{t}\cdot\vec{\sigma}## is real, which I assume, then ##\beta =2\alpha /(\vec{t}\cdot\vec{\sigma}), \gamma =0## and ##\vec{q}## any vector such that ##\vec{q}\cdot\vec{\sigma}=1.##
 
Hi fresh, thank you. No, t sigma is not real. I now saw that my initial assumption is incorrect. You can't decompose a general Lorentz transformation into a boost along the z axis and a rotation. However, you can decompose it into a boost (along a direction which has to be determined) and a rotation. This is in principle a polar decomposition. I found some information in be book by Sexl and Urbantke.
 
DrDu said:
Hi fresh, thank you. No, t sigma is not real. I now saw that my initial assumption is incorrect. You can't decompose a general Lorentz transformation into a boost along the z axis and a rotation. However, you can decompose it into a boost (along a direction which has to be determined) and a rotation. This is in principle a polar decomposition. I found some information in be book by Sexl and Urbantke.
This all sounds like the decomposition of a matrix into its toral (diagonalizable) and nilpotent (upper triangular) part. Within Lie algebras it is called Malcev decomposition or Jordan-Chevalley. I asssume the latter holds for groups, too.
 
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Of course, but this wouldn't provide much insight.
fresh_42 said:
This all sounds like the decomposition of a matrix into its toral (diagonalizable) and nilpotent (upper triangular) part. Within Lie algebras it is called Malcev decomposition or Jordan-Chevalley. I asssume the latter holds for groups, too.
Yes, it is similar. In this case it is the polar decomposition, i.e. the decomposition of a normal matrix into the product of a hermitian and a unitary one.
 
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