Zareon Posted November 5, 2006 Share Posted November 5, 2006 I've read somewhere that a unitary matrix U can be defined by the property: (1) U*=U^{-1} (* = hermitian conjugate) or by the fact that it preserves lengths of vectors: (2) <Ux,Ux>=<x,x> I have trouble seeing why they are equivalent. It's obvious to see that (1)=> (2): <Ux,Ux>=(Ux)*(Ux)=x*(U*U)x=x*x=<x,x> But not the other way around. I CAN prove it for real vector spaces, where U is an orthogonal matrix from the fact that <v,w>=<w,v>. Then I would do: <v+w,v+w>=<U(v+w),U(v+w)>=<Uv,Uv>+<Uw,Uw>+2<Uv,Uw>=<v,v>+<w,w>+2<Uv,U,w> and working out the left side gives <Uv,Uw>=<v,w>. and from this that the columns of U are orthonormal, since [math]<Ue_i,Ue_j>=<e_i,e_j>=\delta_{ij}[/math] But for a complex vector space where <v,w>=<w,v>* all the above gives is: Re(<Uv,Uw>)=Re(<v,w>). EDIT: made some mistakes Link to comment Share on other sites More sharing options...
the tree Posted November 5, 2006 Share Posted November 5, 2006 (1)[tex]U^\dagger=U^{-1}[/tex]On this forum, use [math], rather than [tex], so [math]U^{\dagger}=U^{-1}[/math], or better, use the and tags when all you need is the the sub/superscript features so: U†=U-1. Link to comment Share on other sites More sharing options...
Zareon Posted November 5, 2006 Author Share Posted November 5, 2006 Thanks the tree. Anyway, I've found the answer. I just had to evaluate <U(v+iw),U(v+iw)> to find that also Im(<Uv,Uw>)=Im(<v,w>). Link to comment Share on other sites More sharing options...
the tree Posted November 5, 2006 Share Posted November 5, 2006 No problem. I wish I could have responded to your question but to be honest I don't know anything about matrices, so I figured I'd just give you some general info. Link to comment Share on other sites More sharing options...
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