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I had been asked to prove the conservation of Quantum Laplace–Runge–Lenz Vector:

$$\hat{A}=\frac{i}{\hbar}\left[\hat{p},\,\frac{1}{2}\hat{L}^{2}-k\left|\hat{r}\right|\right]=\frac{1}{2}\left(\hat{p}\times\hat{L}-\hat{L}\times\hat{p}\right)+k\frac{\hat{r}}{\left|\hat{r}\right|}.$$

But unlike the case of classical mechanics, even after a lot of tries I found no beautiful/short way to show it's commutation with Hamiltonian, the calculation are quit lengthy opposite to my first expectations, but something tells me that there is a shorter way, so dose anybody managed to find a simple way or trick to show that

$$\left[\hat{H},\hat{A}\right]~=~0?$$

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You should define $H$. – Fabian Dec 21 '12 at 8:09
What you mean I should define it? it's known that this applies to central forces. – TMS Dec 21 '12 at 15:07
the Laplace-Runge-Lenz(-Pauli) vector does not commute with the Hamiltonian for a general central force field. It is only a specific Hamiltonian which it commutes with and this Hamiltonian should be a part of the question (in the end $k$ relates to some parameter in this Hamiltonian). – Fabian Dec 21 '12 at 21:15

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It's three lines of calculations. There's no shorter way as the statement is somewhat nontrivial. It is equivalent, via Noether's theorem, to the conservation of the direction of the elliptical orbits in the Coulomb/Kepler potential. With the $1/r$ potential, there's no "precession". The conservation of the vector has additional consequences, such as the $SO(4)$ enhanced symmetry of the hydrogen atom (click for an article with many related calculations, including a sketch of yours). But those few lines needed to prove $[H,A]=0$ have to be gone through.

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Thx, Actually I'm familiar with $SO(4)$ relation to this vector, but I have been asked to prove it for people that doesn't know Group theory yet, for that I was looking for some short proof (if any) by just playing with usual commutation relations. – TMS Dec 21 '12 at 7:34
Dear TMS, the straightforward proof from the commutators/Poisson brackets isn't really hard. Do you need to show it step-by-step? There are various tiny tricks that speed it up a little bit - symmetry, antisymmetry, the vanishing bracket [L,H] and of course you have to use the rules for [A,BC] etc. To get the full commutator right is a bit more demanding on the accuracy and time than to get the right Poisson bracket although the result is "the same". – Luboš Motl Dec 21 '12 at 7:57
No actually I already did all that staff, but the result isn't hard but really long even considering that $L$ is also integral of motion, and while it takes at most 7 lines in classical case, it took here about two pages, for that I though maybe there was some tricky short way, that it. – TMS Dec 21 '12 at 15:05
The LRL vector is much more nonlinear and special than the angular momentum. It is inevitable that the calculation is longer. – Luboš Motl Dec 21 '12 at 20:22

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