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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 at 8:20
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$\langle n | [H,r] | n \rangle = \langle n| Hr - rH |n \rangle = E_n ( \langle r \rangle - \langle r \rangle) =0 $ –  user26143 Jun 27 at 8:21
    
@JohnRennie but why doesn't that show in my calculations? or are they irrelevant to begin with? –  user120404 Jun 27 at 8:42
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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 at 9:09
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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 at 9:21

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up vote 1 down vote accepted

By using an arbitrary test function you haven't included any information about the state of the electron. This means you are considering all possible states of an electron in a coulomb potential, so that formula you derived is true for all bound states of the hydrogen atom and all the unbound states of an electron scattering off the proton. It is true in the classical limit, where it should give you back the Kepler orbits, but it also true for states which are behaving distinctly quantum mechanically, notably the eigenstates for the hydrogen atom, which are stationary.

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)

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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 at 13:26
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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 at 14:42
    
2. Yes the corresponding classical problem would be something like a planet orbiting the sun. You can solve that by writing down Newton's second law and solving. Any book on classical mechanics will give an account of how to do that. –  By Symmetry Jun 27 at 14:51
    
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 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 at 15:05

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