Recent content by peaco99

No he doesn't. But if I accept that to be true, Then I can see it. BTW, may I ask your physics education level? Thanks.

No, Sakurai, or in my notes, is \hat{A}\hat{A}^\dagger ever defined. I just assumed because it is anti"unitary", it is either 1 or -1, the -1 maybe coming from "funny" properties of antiunitary operators. So I'm guessing -1 isn't an option then <\Psi|\hat{A}|\Psi> =...

Well \hat{A}\hat{A}^\dagger either equals +1 or -1 in either case i'd get: <\Psi|\hat{A}^\dagger|\Psi> = <\Psi|-\hat{A}^2\hat{A}^\dagger|\Psi> = <\Psi|-\hat{A}\hat{A}\hat{A}^\dagger|\Psi> = 1) for the -1 case: <\Psi|-\hat{A}\hat{A}\hat{A}^\dagger|\Psi> = <\Psi|\hat{A}|\Psi> which...

I don't know how you got \hat{I}= -\hat{A}^\dagger\hat{A} , this would imply that \hat{A}^2= \hat{A}^\dagger, but anyways if you use this all I get is (<\Psi|\hat{A}|\Psi>)^\dagger = (<\Psi|-\hat{A}^\dagger\hat{A}\hat{A}|\Psi>)^\dagger = (<\Psi|-\hat{A}^\dagger\hat{A}^2|\Psi>)^\dagger =...

I'm sorry I was equating the inverse with the adjoint(transpose conjugate). So my problem still becomes that when I get the transpose conjugate in the problem, I don't know how to handle it. Like I mentioned before, Sakurai and my class claims we will never defIne the way an antiunitary operator...

Are you saying K^2=A^2= -I? Because the problem only states that A^2=-I, and the problem never mention what A is, other than it being antiunitary, I brought in the K operator as an attempt at solving the problem. In regards to your hint, everytime I try and work with the inner product, or in...

No restrictions on psi, in fact it says to show this for all psi in the hilbert space. All the problem states is that A is anti-unitary, A^2= -I, and show that A|psi> is orthogonal to |psi>.

Homework Statement \hat{A} is an anti-unitary operator, and it is known that \hat{A^2}= -\hat{I}, show that |\Psi> is orthogonal to \hat{A}|\Psi> Homework Equations I know that \hat{A} can be represented by a unitary operator, \hat{U}, and the complex conjugation operator, \hat{K}...