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1

The derivative of $x f(x)$ is $x f'(x) + f(x)$, and here you're integrating $k \int_{-\infty}^\infty dx ~ x ~ f'(x)$ for some constant $k$, and some complicated function $f$. When you integrate this by parts, you raise $f' dx$ and lower $x$ to find:$$k \int_{-\infty}^\infty dx ~ x ~ f'(x) = k \left[x ~f(x)\right]_{-\infty}^{~\infty} - k \int_{-\infty}^\infty ...


4

Perhaps its a little clearer if you shorten the contents of the brackets (and lets drop the constants too): $$\frac{d\langle x\rangle }{dt} \propto \int _{-\infty} ^{\infty} x \frac{\partial }{\partial x} \left[ \ldots\right] dx$$ $$ = \int _{-\infty} ^{\infty} \frac{\partial }{\partial x} \left(x \left[ \ldots\right] \right)dx - \int _{-\infty} ^{\infty} ...


2

The standard procedure is the following: starting from $ \langle nlm|\,\partial^2_z\,| n'l'm'\rangle $ insert the identity operator with respect to the position basis $$ 1 = \int d\textbf{r} |\textbf{r}\rangle\otimes\langle \textbf{r}| $$ to have $$ \int d\textbf{r}\, \langle nlm\, |\,\partial^2_z\,|\textbf{r}\rangle\cdot\langle \textbf{r}|n'l'm'\rangle. ...



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