[tex]Well, how do you go from an infinitesimal transformation to a finite transformation? You use the exponential map. An infinitesimal transformation on a field is generated by the generator of the corresponding symmetry group, which is an element of the Lie algebra of that group.
So to answer your question, you should write down in your convention an infinitesimal transformation, say that it's unitary, and then use the exponential map to make it a finite transformation. Then you should check if this expression is als unitary.
Write an infinitesimal transformation on a field phi as
[tex]
\delta\phi = \Theta \phi
[/tex]
where Theta is an element of the Lie algebra. Your assumption is that
[tex]
\Theta^{\dagger} = \Theta^{-1}
[/tex]
Now I use the exponential mapping to generate a finite transformation (formally you should take a sum over all generators, but let's keep this easy) with the convention of using an "i":
[tex]
\phi' = e^{i\Theta}\phi
[/tex]
Now calculate:
[tex]
[e^{i\Theta}]^{\dagger} = e^{-i\Theta^{\dagger}}
[/tex]
This finite transformation is unitary if
[tex]
e^{-i\Theta^{\dagger}} = e^{-i\Theta}
[/tex]
So in this case a unitary generator means a unitary group element.
Hope this helps :)
An infinitesimal rotation operator (for physicists) is Hermitian, not unitary.Hi
I have a question regarding unitary operators:
If an infinitesimal operation (such as a rotation) is unitary does this guarantee that a finite transformation will also be unitary?M
It does.I guess my question is:
I have a result for a finite transformation that is in fact a CP map
to understand the problem better I calculated the generators of the group
these appear to be hermitian which I thought would imply a unitary map
does this mean that my result is wrong? or is there a statement like
a finite map is unitary iff the corresponding infinitesimal map is unitary?
thanks
Yes, you're right. I'm mixing up things. But I hope my point was clear though ;)[tex]
e^{-i\Theta^{\dagger}} \left( e^{-i\Theta} \right)^{-1}= e^{-i\Theta^{\dagger} + i \Theta},
[/tex]
so the condition on [itex]\Theta[/itex] needed for unitarity of [itex]e^{i \Theta}[/itex] is [itex]\Theta = \Theta^\dagger[/itex] (hermiticity), not [itex]\Theta^{\dagger} = \Theta^{-1}[/itex] (unitarity).
The answer is no.Hi
I have a question regarding unitary operators:
If an infinitesimal operation (such as a rotation) is unitary does this guarantee that a finite transformation will also be unitary?
thanks
M