Recent content by Excimer

Ah, you got in before me. Ok thanks both of you for helping me out. Cheers

If ξ is real then it equals it's own adjoint, and on page 28 Dirac remarks that if ξ is real than so is ξξ, so if you can take that at face value then ξ must commute with it's adjoint.

Compuchip, So you mean to say that the theorem should specify m > 1 since otherwise you allow the trivial solution? It might be a silly detail but I find things more confusing when they are not stated rigorously. In any case I think I get it now thanks to you.

Sorry the link didn't work, here it is again http://books.google.com.au/books?id=XehUpGiM6FIC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

By the way I do understand that because ξ is a real linear operator that the conjugate to ξ|P> is <P|ξ so that you can write <P|ξξ|P>, but I don't see that it follows this is equal to zero.

I'm continuing through Dirac's book, The Principles of Quantum Mechanics. You can view this as a google book in the link below. http://books.google.com.au/books?id=...page&q&f=false On page 28-29 he proves this Theorem: If ξ is a real linear operator and ξm|P> = 0 (1) for a...

Thanks BIll, You are referring to the previous page (27) where he says that when a linear operator is self adjoint it is called a 'real linear operator' (only because when the operator is a number and self adjoint then the conjuagte complex is real). Is that correct? So the term 'real'...

I am reading The Principles of Quantum Mechanics 4th Ed by Paul Dirac, specifically where he introduces his own Bra-Ket notation. You can view this book as a google book. http://books.google.com.au/books?id=XehUpGiM6FIC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false...