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$$ \left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle=E\langle x \mid E\rangle $$ is often referred to as the time-independent Schrödinger equation in position space. This equation also results from projecting the energy eigenvalue equation $$ \hat{H}|E\rangle=E|E\rangle $$ into position space: $$ \langle x|\hat{H}| E\rangle=E\langle x \mid E\rangle $$

How is $$ \left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle= \langle x|\hat{H}| E\rangle~?$$

Urb
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Kashmiri
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  • Well, what is $\langle x| P^2$ and $\langle x| V(X)$, with $P$ and $X$ operators? Then by linearity you'll find the answer for $H = P^2 + V$ (which I guess is assumed here anyway). – Tobias Fünke Feb 08 '22 at 08:15
  • I've just started studying QM, I don't know what is $\langle x| P^2$ I can substitute the differential operator in place of $p^$, but don't know how to go ahead. – Kashmiri Feb 08 '22 at 08:27
  • @By Symmetry , you wrote in that post : $\langle x|\hat{p}|\psi\rangle = -\imath\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle;.$ , why is that true? – Kashmiri Feb 08 '22 at 12:49

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Insert identity into $\langle x \vert H \vert E \rangle$ \begin{equation} \langle x \vert H \vert E \rangle = \int \textrm{d}x' \langle x \vert H \vert x'\rangle \langle x' \vert E \rangle. \end{equation} The position basis matrix element of the Hamiltonian is \begin{equation} \langle x \vert H \vert x'\rangle = \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} + V(x') \right] \delta(x-x'). \end{equation} The derivative operator acts on the delta function. After plugging this into the first line, integrate by parts twice. Then the integral eats the delta function and we obtain the usual Schrödinger equation.

loewe
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