Property of the adjoint operator

[ehrenfest]

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]

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]

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]

((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]

I see. So, $$<(cA)x|y> = c*<Ax|y> = c*<x|A^{\dag}y> = < x|c*A^{\dag}y>$$.

 

[ehrenfest]

What about the property $$(PQ)^{\dag} = Q^{\dag}P^{\dag}$$? This one seems a bit more difficult.

 

[radou]

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 ) = ...