# Property of the wave functions of a free particle

How can I show that the following holds?

$$\langle nlm\mid \partial_z^2\mid nlm\rangle=-\int_0^{4\pi}d\Omega\int_0^{\infty}drr^2\left|\partial_z\psi_{nlm}\right|^2$$

The wave functions of a free particle are: $\mid nlm\rangle=\psi_{nlm}$.

This conversion is stated in the Quantum Mechanics book by Landau, Lifshitz (Vol.3) on the bottom of the page 137. (https://books.google.de/books?id=neBbAwAAQBAJ&pg=PA137&lpg=PA137&dq=boundary+deformation+as+perturbation+spheric+infinite+potential+well&source=bl&ots=FiqcKgb76e&sig=kmhf0opstnXK3R3fBnVMMv5ZO2s&hl=de&sa=X&ei=QM2NVaeSB8nuUJyZuPgO&ved=0CC0Q6AEwAQ#v=onepage&q=boundary%20deformation%20as%20perturbation%20spheric%20infinite%20potential%20well&f=false)

The only hint that is given there is that integration by parts is used. So I tried: $$\langle nlm\mid \partial_z^2\mid nlm\rangle=\int d^3r\,\psi_{nlm}^\star \left(\partial_z^2\psi_{nlm}\right)=\int dx\int dy\int dz\, \psi_{nlm}^\star\left(\partial_z^2\psi_{nlm}\right)=$$ $$=\int dx\int dy\left[\left.\psi_{nlm}^\star \left(\partial_z\psi_{nlm}\right)\right|_{z=0}^\infty-\int dz\left|\partial_z\psi_{nlm}\right|^2\right]=$$ $$=\int dx\int dy\,\left[\psi_{nlm}^\star \left(\partial_z\psi_{nlm}\right)\right]_{z=0}^\infty-\int d^3r\left|\partial_z\psi_{nlm}\right|^2$$ Now the first term has to vanish? For the upper boundary: $\psi_{nlm}=R_{nl}Y_{lm}\overset{z\rightarrow\infty}{\rightarrow}0$ is trivial because $R_{nl}\propto e^{-r}$. But what happens for $z\rightarrow0$?

• Where did $z=0$ come from? Aren't you integrating over all space? Which surely includes regions where $z<0.$ – Timaeus Jun 28 '15 at 12:58
• oh right!! I forgot about that completely, because I was using spherical coordinates all the time.. thx!! – Andy Jun 28 '15 at 13:06

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 $$\int d\textbf{r}\, \langle nlm\, |\,\partial^2_z\,|\textbf{r}\rangle\cdot\langle \textbf{r}|n'l'm'\rangle.$$ The first contribution in the integral is the representation of the derivative operator onto the position space (given in terms of derivations); the second contribution is the wave function of the particle with respect to the position space (what books usually call the spherical harmonics of the form $\psi_{nlm}=R_{nl}Y_{lm}$). At this point just integrate using whatever rule you want and you will end up with the correct equation.