# Homework Help: Trying to calculate how <r> in the Hydrogen atom changes with time

1. Jun 26, 2014

### user3

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 <r> 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?

Last edited: Jun 26, 2014
2. Jun 26, 2014

### bloby

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
3. Jun 26, 2014

### MisterX

user3, might you show your work so we can check it?

4. Jun 26, 2014

### user3

5. Jun 28, 2014

### BvU

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 ...

6. Jun 28, 2014

### BvU

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:

7. Jun 28, 2014

### bloby

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
8. Jun 28, 2014

### bloby

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
9. Jun 29, 2014

### BvU

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.

10. Jun 30, 2014

### bloby

$\frac{d<r>}{dt}=\frac{-i\hbar}{2m}<2\frac{\vec{r}}{r}\cdot\nabla +\nabla^2 r> = \frac{1}{m}<\frac{\vec{r}}{r}\cdot \hat{\vec{p}}>-\frac{i\hbar}{m}<\frac{1}{r}>$

(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