Taking the Hamiltonian eigenvalue problem into position space?

Question

I'm having a hard time notationally understanding the relationship of:

$$ \hat{H} \vert{\psi}\rangle = E\vert \psi\rangle $$ and $$\hat{H} \psi(x) = E\psi(x) $$

Here's my thought process: Starting from the equation: $$ \hat{H} \vert{\psi}\rangle = E\vert \psi\rangle $$ $$ \hat{H} \int dx \vert x \rangle \langle x\vert \psi \rangle = E\int dx \vert x \rangle \langle x\vert \psi \rangle $$ We know that $$ \langle x\vert \psi \rangle = \psi(x)$$ in our position representation. Does this mean that $ \hat{H} $ in Dirac notation translates to $ \hat{H} \int dx \vert x \rangle $ in position representation? Something about this seems quite strange to me.

Answer 1

It's better to do it this way. Start with the expression without reference to any basis

$$\hat{H} \vert{\psi}\rangle = E\vert \psi\rangle$$

Then bring in $\langle x|$

$$\langle x|\hat{H} \vert{\psi}\rangle =\langle x| E\vert \psi\rangle$$

You are correct in having $\langle x|\psi\rangle=\psi(x)$, so the right hand side with scalar $E$ easily becomes $E\psi(x)$

On the left hand side we exploit our completeness relation $$\langle x|\hat{H} \vert{\psi}\rangle = \int\langle x|\hat{H}|x'\rangle\langle x' \vert{\psi}\rangle\,\text dx'= \int\langle x|\hat{H}|x'\rangle\psi(x')\,\text dx'$$

Since $\hat H =\hat H(\hat X,\hat P)$, the matrix elements of $\hat H$ in the position basis are given as $$\langle x|\hat{H}|x'\rangle=\delta(x'-x)H\left(x',\frac{\text d}{\text dx'}\right)$$

Therefore, we end up with

$$\langle x|\hat{H} \vert{\psi}\rangle =H\psi(x)$$

And so we have $$H\psi(x)=E\psi(x)$$

Answer 2

Note that, formally, the Hilbert space in the position representation is $L^2(\mathbb R^n)$, so that $|\psi\rangle$ is indeed an $L^2$-function.