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Trying to calculate how <r> in the Hydrogen atom changes with time

  1. Jun 26, 2014 #1
    I am working on the Hydrogen atom and I was trying to calculate [tex]\frac{d<r>}{dt}[/tex] using [tex]\frac{d<r>}{dt} = \frac{i}{\hbar} <[\hat{H} , \hat{r}]>[/tex]. Here [tex]r = \sqrt(x^2 + y^2 + z^2)[/tex] and [tex]H = \frac{p^2}{2m} + V[/tex] where [tex]p^2 = -\hbar^2 \nabla^2 [/tex]. Now according to Ehrenfest's theorem <r> should behave classically and give me some equivalent of velocity, and indeed I do get something but it does't resemble velocity: [tex]\frac{-\hbar^2}{2m} (2\nabla r \nabla f + f \nabla^2 r)[/tex]

    where f is a test function.

    Steps:
    [tex][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 [/tex] [tex]= \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) [/tex]

    Am I doing something wrong?
     
    Last edited: Jun 26, 2014
  2. jcsd
  3. Jun 26, 2014 #2
    There is no f and it's the operator ##[\hat{H},\hat{r}]## . You can multiply with i/##\hbar##, calculate ##\nabla r## and try to find ##\hat{p}##. There is a term I must think where it comes from(perhaps symetrisation of the product of classical observables that do not commute in the quantum world)
    And remember ##r=\|\vec{r}\|## so classically ##\frac{dr}{dt}=...##
     
    Last edited: Jun 26, 2014
  4. Jun 26, 2014 #3
    user3, might you show your work so we can check it?
     
  5. Jun 26, 2014 #4
    just added the steps
     
  6. Jun 28, 2014 #5

    BvU

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    Dear folks, at the risk of making a fool of myself: if you are interested in r, why work in Cartesian coordinates? (moot question, I would say, except for the appearance of ##r = \sqrt{x^2 + y^2 + z^2}\ ## which makes me worry).

    Then: if you wonder about ##\frac{d<r>}{dt}\,## : that 's got little to do with "some equivalent of velocity". Only with the radius. Interesting for radiative transitions and so on, but I don't think you'll be able to work towards a connection with the classical case there ...

    For a central force problem (usually treated before the H atom), most textbooks manage to reduce the problem of solving the Schrödinger eqn via ##\Psi(\vec r) = R(r)\,Y(\theta, \phi)\ ## which follows from [H,L]=0 .

    Then the H atom treatment yields radial wave functions that yield constant <r>. Time independent!

    My textbook is Eugen Merzbacher, Quantum Mechanics, 2nd ed 1970 ...
     
  7. Jun 28, 2014 #6

    BvU

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    Adding to this: I'd like a real QM teacher to step in and make clear what people like user3 can use as a mental image of this electron that classically circles the nucleus as a particle in the ubiquitous atom depiction, but quantum mechanically is such a mind-boggling ##\Psi## beast.

    I mean, after a whole life of physics (HEP at that!) even the simplest excited state of the simplest atom is already beyond my down-to earth abstraction level. Grmpf...:tongue:
     
  8. Jun 28, 2014 #7
    You are right this is more useful.
    I was trying to follow the path of the OP.

    My turn to risk of making a fool of myself: r=cst is for time independent/ energy eigenstates, isn't it?(and V must be spherically symmetric if I remember correctly)

    Here I would say the case is more general.
     
    Last edited: Jun 28, 2014
  9. Jun 28, 2014 #8
    I find something like ##\frac{dr}{dt}=\frac{\vec{r}}{r}\cdot\frac{\vec{p}}{m}## for the classical system and ##\frac{d<r>}{dt}=\frac{1}{m}<\frac{\vec{r}}{r}\cdot \hat{\vec{p}}>-another\,term## for the quantum one.
     
    Last edited: Jun 28, 2014
  10. Jun 29, 2014 #9

    BvU

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    Oh boy, so Ehrenfest is holding out quite well ! Provided we make the other term disappear. Can you post what you have thus far, bloby?

    r=cst is for time independent/ energy eigenstates, isn't it? Yes, if you mean <r>; it's a number.

    (and V must be spherically symmetric if I remember correctly) Yes: [H,L]=0 gets it all rolling.
     
  11. Jun 30, 2014 #10
    ##\frac{d<r>}{dt}=\frac{-i\hbar}{2m}<2\frac{\vec{r}}{r}\cdot\nabla +\nabla^2 r> = \frac{1}{m}<\frac{\vec{r}}{r}\cdot \hat{\vec{p}}>-\frac{i\hbar}{m}<\frac{1}{r}>##

    (If r is an observable, Ehrenfest's theorem is holding, even if the result does not look like the classical)
    But are all those averaged operators observables? The i in the second term looks suspicious...
     
    Last edited: Jun 30, 2014
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