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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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1 Answers1

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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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