Why is Δx Equal to r in a Spherically Symmetric Potential?

[S_Flaherty]

There was a question on an exam I took earlier this year which gave a radial potential of the form V(r) = -A/r. In the first part of the question it asked what the uncertainty in x was (i.e. was is Δx?)

In my notes I have: Δx = r since <x> = 0. I don't remember what verbal explanation went along with this so I'm confused how this conclusion was reached. Can someone help explain this to me. I feel like this is something really simple that I'm just overlooking.

I know that from what's above that <x^2> should be r^2 since Δx = √(<x^2> - <x>^2).

 

[Simon Bridge]

Yeah: in general, the variance of an operator is the difference between the expectation of the square and the square of the expectation, vis. for operator A:

$\Delta A^2 = \langle A^2 \rangle - \langle A \rangle^2$

Trying to make sense of the question:
$\Delta x$ would be a tad pointless in this case since the potential is radial ... which stops the question from making sense. Do you mean $\Delta r$ ? The expectations $\langle r^2 \rangle$ and $\langle r \rangle$ would depend on the state wavefunction - is $\Delta r$ independent of this?

Note: what would the ground-state wave-function look like?

 

[S_Flaherty]

Yeah: in general, the variance of an operator is the difference between the expectation of the square and the square of the expectation, vis. for operator A:

$\Delta A^2 = \langle A^2 \rangle - \langle A \rangle^2$

Trying to make sense of the question:
$\Delta x$ would be a tad pointless in this case since the potential is radial ... which stops the question from making sense. Do you mean $\Delta r$ ? The expectations $\langle r^2 \rangle$ and $\langle r \rangle$ would depend on the state wavefunction - is $\Delta r$ independent of this?

Note: what would the ground-state wave-function look like?

Well the question asked to find the uncertainty in the Cartesian coordinate x, so Δx is in terms of r. I'm not sure what the ground-state wave-function looks like but from my notes on this problem it looks like one can simply determine what Δx just by looking at the potential. But I don't know what the specific relation between the two is in order to do that.

 

[Simon Bridge]

Well the question asked to find the uncertainty in the Cartesian coordinate x, so Δx is in terms of r.

... also in terms of $\theta$ and $\phi$.

I'm not sure what the ground-state wave-function looks like but from my notes on this problem it looks like one can simply determine what Δx just by looking at the potential. But I don't know what the specific relation between the two is in order to do that.

I think it is more that this is a well-known potential (it's hydrogenic) - you are supposed to know what the radial wavefunctions look like just as you should know those for the harmonic oscillator and the square wells. These are things you can look up to jog your memory ;)

That's why I asked if it was the uncertainty in r for the ground-state since that has a simple(ish) relation - the sort of thing one may expect students to rattle out in an exam.