# I am trying to calculate how $<r>$ in the hydrogen atom evolves with time

I am working on the Hydrogen atom and I was trying to calculate $\frac{d<r>}{dt}$ using $$\frac{d<r>}{dt} = \frac{i}{\hbar} <[\hat{H} , \hat{r}]>.$$ Here $r = \sqrt(x^2 + y^2 + z^2)$ and $H = \frac{p^2}{2m} + V$ where $p^2 = -\hbar^2 \nabla^2$. Now according to Ehrenfest's theorem should behave classically and give me some equivalent of velocity, and indeed I do get something but it does't resemble velocity: $\frac{-\hbar^2}{2m} (2\nabla r \nabla f + f \nabla^2 r)$

where $f$ is a test function.

Steps: $[H ,r]f = [\frac{p^2}{2m} + V , r]f = \frac{p^2(rf)}{2m} + Vrf - \frac{rp^2(f)}{2m} - rVf = \frac{1}{2m}[p^2 ,r]f = \frac{1}{2m}[-\hbar^2\nabla^2 , r]f = \frac{-\hbar^2}{2m}[\nabla^2 , r]f$$= \frac{-\hbar^2}{2m} (\nabla^2(rf) - r\nabla^2(f)) = \frac{-\hbar^2}{2m} (\nabla r\nabla f + r\nabla^2f + \nabla f \nabla r + f\nabla^2 r - r\nabla^2f) = \frac{-\hbar^2}{2m} (2\nabla r \nabla f + f \nabla^2 r)$

Am I doing something wrong?

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Why would $<r>$ change with time? The hydrogen atomic orbitals are time independant aren't they? – John Rennie Jun 27 '14 at 8:20
$\langle n | [H,r] | n \rangle = \langle n| Hr - rH |n \rangle = E_n ( \langle r \rangle - \langle r \rangle) =0$ – user26143 Jun 27 '14 at 8:21
@JohnRennie but why doesn't that show in my calculations? or are they irrelevant to begin with? – user120404 Jun 27 '14 at 8:42
Suggestion: Solve first the corresponding classical problem as a warm-up to the quantum mechanical problem to get intuition of what to expect. – Qmechanic Jun 27 '14 at 9:09
I have used $\langle n | H =E_n \langle n|$ and $H|n \rangle = E_n | n \rangle$, though my calculation only applies to eigenstate.... – user26143 Jun 27 '14 at 9:21

Given that the expression is so general, its not that surprising that you get a result that doesn't immediately resemble the classical result. Normally to take the classical limit of a system you have to think a bit about what is physically happening as you take that limit. In this case you would probably need to consider a superposition of a large number of very high $n$ states, i.e. states with a large energy and a large uncertainty in the energy. (You would probably need similar conditions on $l$ as well)
Ok I am self-studying Quantum Mechanics and sometimes things drop, so I am learning alot from all these comments and answers. But just to confirm what I understand, let me ask this last round of questions: 1-how can $<n|H = E_n <n|$? don't operators operate to the right? 2-Is the classical corresponding problem that of a planet orbiting the sun, for example?If yes, is it attacked this way: $E=0.5mv^2 - GmM/r = -GmM/2r$ ,then we solve for r and get the derivative? 3-If the equation I derived holds for Eigenstates, how can I further work on in to get the 0 we expect from a stationary state? – user120404 Jun 27 '14 at 13:26
for 1. Dirac notation is not ideal for this point, so we will use the notation of putting operators inside the kets. $\langle \phi| H |\psi\rangle = \langle \phi|H\psi\rangle$ The adjoint of an operator $H^\dagger$ is defined by $\langle H^\dagger \phi|\psi\rangle$, so it is the operator that operates to the left and gives the same result. An operator is Hermitian if $H = H^\dagger$. In QM we require the operators corresponding to observables be Hermitian, so in this case the operator can operate in either direction and give the same result. – By Symmetry Jun 27 '14 at 14:42
3. $f$ is the wavefunction of the state you are considering. To find the eigenstates, you solve the time independent Schrodinger equation $Hf = E_nf$. I don't see an obvious way of finding the eigenstates from what you already have, as I suspect that if you simply solve the equation obtained from setting the expression you have equal to 0 you would also get solution corresponding to solutions with a fixed mean $r$, but an angular dependence which changes over time. – By Symmetry Jun 27 '14 at 14:56
* realised there was a mistake in my answer to 1. it should should read "the adjoint of an operator $H^\dagger$ is defined by $\langle \phi | H\psi\rangle = \langle H^\dagger\phi | \psi\rangle$" – By Symmetry Jun 27 '14 at 15:05