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ehrenfest
Jul21-07, 12:10 AM
The adjoint of an operator A is defined as an operator A* s.t.

<\phi|A\psi> = <A^{*}\phi|\psi>.

How would you use the properties of inner products (skew-symmetry, positive semi-definiteness, and linearity in ket) to show that (cA)* = c*P*


Note that I am using the conjugate and the adjoint symbol interchangeably. If anyone knows how to get a real adjoint symbol in LaTeX let me know.

matt grime
Jul21-07, 03:41 AM
what does \dag do? I assume that A and P are supposed to be the same letter... and then it is trivial. What have you attempted?

ehrenfest
Jul21-07, 12:27 PM
Yes sorry. A and P are supposed to be the same letter.

You could use the skew-symmetry property to show that:

<\phi|cA\psi> = <cA\psi|\phi>^{*}

and that

<cA^{\dag}\phi|\psi> = <\psi|cA^{\dag}\phi>^{*}

but I do not see how that helps.

radou
Jul21-07, 12:37 PM
((cA)x | y) = ...

Unless I'm mistaken, you have to use the definition of scalar multiplication with operators, and two peoperties of the inner product. In three (four) steps, you can show what (cA)* equals.

ehrenfest
Jul21-07, 01:46 PM
I see. So, <(cA)x|y> = c*<Ax|y> = c*<x|A^{\dag}y> = < x|c*A^{\dag}y> .

ehrenfest
Jul21-07, 02:28 PM
What about the property (PQ)^{\dag} = Q^{\dag}P^{\dag}? This one seems a bit more difficult.

radou
Jul21-07, 03:58 PM
What about the property (PQ)^{\dag} = Q^{\dag}P^{\dag}? This one seems a bit more difficult.

I wouldn't really call it difficult. Again, ( (PQ)x | y ) = ...