# Taking the Hamiltonian eigenvalue problem into position space?

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.

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 $$H\psi(x)=E\psi(x)$$

• Maybe parenthesis would help. Last expression is (H psi)(x)= E psi(x) – lalala Nov 16 at 7:15
• @lalala I think it's fine. At the end $H$ is a differential operator acting on the function $\psi(x)$ and is the form the OP is asking about and seems to be familiar with. – Aaron Stevens Nov 16 at 7:18

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.