The Hamilton operator is often defined as
$$
\hat H = \frac{-\hbar^2 }{2m}\frac{d^2}{dx^2} + V(x)
$$
but shouldn't it rather be
$$\begin{aligned}
\hat H &= \int\int dxdx' |x\rangle \langle x|\hat H|x'\rangle \langle x'|\\
&= \int \int dx dx' |x\rangle \left[ -\frac{\hbar^2}{2m}\left(\frac{\partial}{\partial x'}\frac{\partial}{\partial x}\delta(x-x') \right) + V(x)\delta(x-x') \right ]\langle x'| \\
\hat H &= \int dx |x\rangle \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\langle x| \\
\hat H &= \int dx |x\rangle \langle x| \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\\
\hat H&=\hat 1\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]
\end{aligned}$$
Because it seems to me that only in this way we have
$$
\hat H |\psi\rangle = \int dx |x\rangle \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\psi(x)
$$
and
$$
\langle x|\hat H | \psi\rangle =\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\psi(x)
$$
Am I overly pedantic or even wrong? Is the distinction between
$$
\hat 1 \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ] = \hat H
$$
and
$$ \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]$$ unnecessary/unreasonable or should it be made?
Best Answer
You are not pedantic, you are just wrong! $$ \hat H= \hat p^2 /2m+ V(\hat x). $$ Recall $$ \hat p= -i\hbar \int\! dx ~|x\rangle \partial_x \langle x| ,\\ \hat x= \int\! dx ~|x\rangle x \langle x|~, \qquad \langle x| x'\rangle = \delta (x-x'),\\ \leadsto \qquad \hat p^2= -\hbar^2 \int\! dx ~|x\rangle \partial_x^2 \langle x| , $$ so that $$\begin{aligned} \hat H &= \int\int\! dxdx' ~|x\rangle \langle x|\hat H|x'\rangle \langle x'|\\ &= \int \int\! dx dx' ~|x\rangle \delta(x-x') \left( - \frac{\hbar^2}{2m} \partial_{ x'}^2 + V(x') \right )\langle x'| \\ &= \int\! dx ~~|x\rangle \left ( -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right )\langle x| ~~. \end{aligned}$$
Your teacher surely has taught you that the operator in the parenthesis is the hamiltonian in the x-representation! Thus, acting on a ket, $$ \hat H |\psi\rangle = \int\! dx ~~|x\rangle \left ( -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ) \psi(x) ~~. $$