Stationary schrodinger equation (basic problems regarding the basis being used)

If you multiply on the left by <x| you get:

[tex] <x|H|Y> = E<x|Y> = E Y(x) [/tex]

Note H can't be pulled out of the bracket since it's an operator. Now, H is an operator on the Hilbert space, not directly on the space of wavefunctions, but of course these spaces are really the same, so there should be a way of acting H on the wavefunction, and there is. Inserting a 1 in the form [itex]\int dx' |x'><x'|[/itex] we find:

[tex] <x|H|Y> = \int dx' <x|H|x'><x'|Y> = \int dx' H(x,x') Y(x') [/tex]

where I've defined [itex]H(x,x') = <x|H|x'>[/itex]. If you work this out, you'll find something like:

[tex] H(x,x') = -\frac{\hbar^2}{2m} \delta''(x-x') + V(x) \delta(x-x') [/tex]

And then inserting this into the integral you find:

[tex] <x|H|Y> = -\frac{\hbar^2}{2m} Y''(x) + V(x) Y(x)[/tex]

which gives you the Schrodinger equation in the position basis.

 

Again, you need to be careful about the distinction between states and wavefunctions. An equation like:

[tex] <x|H|x'> = H <x|x'> [/tex]

makes no sense: the LHS is a number (depending on x and x') while the RHS is an operator, ie, a matrix on the infinite dimensional Hilbert space (also depending on x and x').

Now, what you said at the end of your last post is true. Specifically, we can represent the Hilbert space in a number of ways, the most common of which is the position basis, where there is a one-to-one correspondence between states and (square integrable) complex valued functions on the real numbers. In this basis, we can also represent Hilbert space operators (like H), typically by differential operators. For example, we know from experience that the Hamiltonian has the representation:

[tex] H |\psi> \leftrightarrow \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) [/tex]

(note the thing in paranetheses on the RHS is not H, but a differential operator that represents it in the position basis). What this equation says is that if the state [itex]|\psi>[/itex] is represented in the position basis by the function [itex]\psi(x)[/itex], then the state on the LHS of the above equation is represented by the function on the RHS. To put the last sentence in equations:

[tex] <x|\psi> = \psi(x) \Rightarrow <x|H |\psi> = \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) [/tex]

In the case we're concerned with, since [itex]<x|x'> = \delta(x-x')[/itex] (in other words, the wavefunction of the state [itex]|x'>[/itex] is [itex]\delta(x-x')[/itex]), we have:

[tex] <x|H|x'> = \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \delta(x-x') [/tex]

which is the formula I gave above.

You could also derive this from the more basic definition of the Hamiltonian (I'll put hats on things that are Hilbert space operators):

[tex] \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x}) [/tex]

With a little work, using the commutation relations [itex] [\hat{x}, \hat{p}] = i \hbar [/itex] and the fact that [itex]|x'>[/itex] is an eigenstate of the operator [itex]\hat{x}[/itex] with eigenvalue x', you can derive the same formula without using the position basis at all.