Merzbacher QM exercise 9.8

1. Oct 23, 2014

tigigi

I think I almost get there, but something's not right, and need your help.

if F(ψa,ψb) is a complex valued (scalar) functional of the two variable vectors ψa and ψb with linearity properties. Show that F can be represented as an inner product F(ψa,ψb)=(ψa, Aψb), for every ψa,ψb and the previous equation defines a linear operator A uniquely

I write ψa and ψb in terms of basis {φk} and {φi}
ψa= ∑F(ψa,ψb)* φk
ψb=∑F(ψa,ψb)* φi Then
(ψa, Aψb)= (∑F(ψa,ψb)* φk, A∑F(ψa,ψb)* φi ) since ∑F(ψa,ψb)* is scalar, I can take them out
= ∑F(ψa,ψb)* ∑F(ψa,ψb) (φk, Aφi) let k=i and A=|φk> <φk|
= F(ψa,ψb)* F(ψa,ψb) <φk |φk> <φk|φk >

but then I don't get the functional F(ψa,ψb), instead I get its square.
if I write ψb=∑bi φi, everything else stay the same, I get either a functional times bk (or bi, the same) or I get
A= bk |φk> <φk|, but then A is not unique, bc bk can be anything.

What's going on? any suggestion? thanks

2. Oct 24, 2014

naima

Start with one basis φ and call F(φi,φj) $= a_{ij}$
Take two vectors X and Y then apply the linear properties of F.