Timeline for Expectation value of $x,y,z$ for general $nlm$ state of hydrogen atom
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2019 at 10:30 | comment | added | denklo | You are wellcome. There is a little typo though: $\cos(\theta) \in (-1,1)$. | |
| Apr 16, 2019 at 10:22 | comment | added | ohneVal | Thank you very much for completing the argument @denklo and for the insight, I believe this is the type of arguments the OP was looking for. | |
| Apr 16, 2019 at 10:15 | comment | added | denklo | For $z$, why not just use the symmetry of $P^m_l(\cos(\theta))$.We have the volume element $dV = d\cos(\theta)d\phi dr$. $z = r \cos(\theta)$, thus the integrand becomes $$\propto (P^m_l(\cos(\theta)))^2\cos(\theta)d\cos(\theta)$$ where $\cos(\theta) \in (0,1)$. Since $P^m_l(x) = (-1)^{m+l}P^m_l(-x)$ it quickly follows $\langle z \rangle = 0$. | |
| Apr 16, 2019 at 9:26 | comment | added | Emilio Pisanty | Then do that, if you think that it is clearer than using the correct symmetry properties of the state. That argument is certainly not guaranteed to work - it will produce all sorts of combinations of the form $\langle \psi_{nlm} | x |\psi_{nlm'}\rangle$ with $m\neq m'$ which do not vanish, but which cancel out when added together. I do not find such a layer of obfuscation to be helpful, but if you want to edit your answer such that it covers the $\langle z\rangle$ case in that way, then that's your choice. | |
| Apr 16, 2019 at 9:23 | comment | added | ohneVal | I can change coordinates within the integral if it serves you better, then use the property of rotations of the harmonics which will produce just an annoying combination of harmonics with the same $\ell$ and opposite sign $m$'s which will end up in the same sort of integrals as above. | |
| Apr 16, 2019 at 9:17 | comment | added | Emilio Pisanty | No, that argument doesn't work - your state is already specified, and rotating the system would change the state. The hamiltonian is symmetric, but the eigenstates do not share its full symmetry. | |
| Apr 16, 2019 at 9:16 | comment | added | ohneVal | For $z$ one can rotate the coordinate system so that $z$ lies in the plane $\theta = \pi/2$ and the same argument holds, since the location of your "northpole" is completely arbitrary. | |
| Apr 16, 2019 at 8:58 | comment | added | Emilio Pisanty | This works for x and y, but not for z. | |
| Apr 16, 2019 at 8:32 | history | answered | ohneVal | CC BY-SA 4.0 |