Very stupid question about Wigner's Theorem
Wigner's representation theorem says that any invertible transformation between rays of a Hilbert space that preserves transition probabilities can be implemented by a transformation on the Hilbert space itself, which is either unitary or antiunitary, depending on the particular transformation considered, right?
For example, you can find in any QM book that almost all symmetries are represented by linear operators, the only significant exception being time inversion, right? I present here the most trivial example I can imagine: a onedimensional Hilbert space. The two complex function of complex variable [tex]f(z)=z\qquad\textrm{and}\qquad g(z)=z^*[/tex] are, respectively, unitary and antiunitary, and they both induce the same transformation between rays of the Hilbert space [tex]H=\mathbb{C}[/tex], the identity transformation. I know the solution of this apparent paradox should be easy, but I really can't see it! I know the proof of the theorem, and it doesn't help! Any hint woukd be appreciated. 
Re: Very stupid question about Wigner's Theorem
Quote:
equivalent to the identity transformation? If you're saying the z (a complex number) is a complex multiple of z*, then effectively there's only one state (ray), and all (invertible) transformations are the identity transformation. The transformations are different if the Hilbert space is considered as a vector space, but coincide if we revert to a projective space of rays. Wigner's theorem doesn't say (iirc) that the unitary and antiunitary transformations must be distinct for every possible type of Hilbert space. Maybe try a slightly less trivial example of a 2D Hilbert space? 
Re: Very stupid question about Wigner's Theorem
Quote:
Quote:

Re: Very stupid question about Wigner's Theorem
Quote:

All times are GMT 5. The time now is 04:36 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums