Thank you for your reply. I am puzzled with what you "start with". We are dealing with the free particle, so H=\frac{m}{(2mE)^{1/2}} . If a state is an eigenstate of H, it should be an eigenstate of P^2, and thus an eigenstate of P. (Why? We know that any eigenstate of P is also eigenstate of P...
I do not understand why the number of states in the two interval must be same, through it leads to the desired result. But the point is, what is wrong with the derivation above? Isn't that the normal form of propagator?
It can not be a just difference in normalization, since \frac{m}{(2mE)^{1/2}} is the factor of the integrand. I am facing the same problem, any help is appreciated.
Hello.I am reading Sharkar too, but for the first time. Very glad to talk about sth on what I am trying to understand.
I think the answer to your question is yes. The two commuting hermitian operators have the same eigenbasis. Check p29, the active and passive tranformations. What we do is just...