Schrodinger equation in position representation

MatinSAR
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Homework Statement
Derive schrodinger equation in position representation.
Relevant Equations
Schrodinger equation
Hello.

General form of the equaion is ## i \hbar \dfrac {\partial}{\partial t} | \psi (t) \rangle = \hat H | \psi (t) \rangle ##
According to my book ## \hat H ## in position space is ## - \dfrac {\hbar ^2}{2m} \nabla^2 + V( \hat {\vec R} , t)## so potensial V is a function of operator ##\hat R## and time ##t##. But in another page it used ##\hat V(\vec r , t ) ## instead, and I cannot understand why.
 
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What are R and r in the text ?
 
anuttarasammyak said:
What are R and r in the text ?
##\hat{R}## is a linear Hermitian operator corresponding to the observable ##\vec r##.
 
Thanks. In position presentation, applying operator R is multiplying its eigenvalue parameter r. e.g. a harmonic oscillator
\hat{V}=V(\hat{\mathbf{R}})=\frac{k}{2}\hat{\mathbf{R}}^2=\frac{k}{2}\mathbf{r}^2=V(\mathbf{r})=\hat{V}(\mathbf{r})
 
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anuttarasammyak said:
Thanks. In position presentation, applying operator R is multiplying its eigenvalue parameter r. e.g. a harmonic oscillator
\hat{V}=V(\hat{\mathbf{R}})=\frac{k}{2}\hat{\mathbf{R}}^2=\frac{k}{2}\mathbf{r}^2=V(\mathbf{r})
Are you saying that ##\hat V(\mathbf r)## is same as ##V( \hat R)##?
 
Yes, in position representation.
 
anuttarasammyak said:
Yes, in position representation.
I've just seen your edit in post #4. It's clear. Thank you for your time.
 
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