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 $z\rightarrow0$?

  • 2
    $\begingroup$ Where did $z=0$ come from? Aren't you integrating over all space? Which surely includes regions where $z<0.$ $\endgroup$ – Timaeus Jun 28 '15 at 12:58
  • $\begingroup$ oh right!! I forgot about that completely, because I was using spherical coordinates all the time.. thx!! $\endgroup$ – 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.

P. S. (a joke, but also quite not). Never ever read the book by Landau and Lifshitz if you want to derive something. After many and many years of physics and mathematics in all possible flavours I am still unable to understand any single line they write (but it can just be me, of course).


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