A QM problem on finding eigenvalues

[reilly]

HackaB Think about a slighly different version, say a spin dependent problem, with

H0(m,n) = S*S + V (|m><n| where S is spin (S*S has eigenvalues s(s=1), and the |m> states are Sz eigenstates. The problem specifies the potential to be proportional to a "projector", a single state projection operator. The only choice of m & n that conforms to the properties of a single-state projection operator is m=n. Generalize this reasoning, and get the delta function.
Regards,
Reilly Atkinson

 

[reilly]

By the way, Seratend's solution is the correct one for the non-diagonal version -- the projection operator is replaced by a transition operator. -- which is the equation and solution that Cooper used in his original Cooper Pair paper.
Regards,
Reilly Atkinson

 

[HackaB]

So you do not agree that <x|v>V<v|y> = kV exp(-k|x|)exp(-k|y|) ?

The only choice of m & n that conforms to the properties of a single-state projection operator is m=n. Generalize this reasoning, and get the delta function.

I'm sorry, but I don't follow this reasoning. |v> is not a "single state" in the x-representation. So I do not see how a delta function would arise. Could you show the mathematical steps to get it?

By the way, Seratend's solution is the correct one for the non-diagonal version -- the projection operator is replaced by a transition operator.

What is a transition operator? Is it similar to a projection operator?

 

[reilly]

HackaB -- I blew it with the Bessels -- too good to be true, I suppose. So with egg on my face, I retract the delta function, and remain,
Reilly Atkinson

 

[CarlB]

If the assistance this thread has given the OP, has not convinced him that quantum mechanics is impossibly difficult, I wonder what it would take.

Carl