States in the Hydrogen Atom that are not allowed

[guyvsdcsniper]

Homework Statement

Show that for any two l = 0 states, these transitions are not possible.

Relevant Equations

expectation value

I am a little lost on how to approach this problem.

What I know is the following:
The r vector is in terms of x y and z hat.
I know my two l=0 states can be the 1s and 2s normalized wave function for Hydrogen.

Should I be integrating over dxdydz?

 

[Orodruin]

What is significative of the $l=0$ states as functions of the spherical coordinates?

 

[guyvsdcsniper]

What is significative of the $l=0$ states as functions of the spherical coordinates?

l=0 states mean the orbitals are the shape of a sphere. Ok that's illuminating, so I should be integrating in spherical coordinates, r²sinθdrdθφ. So r should be from 0→∞.

I feel like I am still missing something. If I multiply this out its going to be a mess.

 

[Orodruin]

l=0 states mean the orbitals are the shape of a sphere. Ok that's illuminating, so I should be integrating in spherical coordinates, r2sinθdrdθφ. So r should be from 0→∞.

So the states have spherical symmetry. This means they are even under $\vec r\to-\vec r$. What about $\vec r$?

 

[guyvsdcsniper]

So the states have spherical symmetry. This means they are even under $\vec r\to-\vec r$. What about $\vec r$?

Ok I agree with your first two statements, that makes sense to me. If they are spherical they are symmetric.

Now what about $\vec r$.
Well I know cos is an even function and sin in an odd function, so would I neglect $\hat x$ and $\hat y$ and would only care about the $\hat z$ component of $\vec r$?

 

[Orodruin]

Now what about r→.
Well I know cos is an even function and sin in an odd function, so would I neglect x^ and y^ and would only care about the z^ component of r→?

Don’t care about the coordinates. What happens to $\vec r$ when $\vec r \to -\vec r$? (Not a trick question!)

 

[guyvsdcsniper]

Don’t care about the coordinates. What happens to $\vec r$ when $\vec r \to -\vec r$? (Not a trick question!)

Its points in the opposite direction right?

 

[Orodruin]

Its points in the opposite direction right?

So it picks up a minus sign and therefore is odd with respect to that symmetry. What does this mean fir the integrand? What about the integration volume?

 

[guyvsdcsniper]

So the integrand then is negative, and the integration volume would just be r²dr?

 

[Orodruin]

So the integrand then is negative, and the integration volume would just be r²dr?

No. Let’s try something different:

What is the value of the following integral and why?
$$
\int_{-1}^1 e^{-x^2} \sin(x) dx
$$

 

[guyvsdcsniper]

So it picks up a minus sign and therefore is odd with respect to that symmetry. What does this mean fir the integrand? What about the integration volume?

0 because we are integrating symmetric bounds over an odd function.

 

[Orodruin]

0 because we are integrating symmetric bounds over an odd function.

So, keeping this in mind, what about the integral you have?

 

[guyvsdcsniper]

ohhhh I think I see it. Since I have that sinθ (and that's its odd) in my dV, and we have spherical symmetry or even bounds this whole integral should just be zero.

 

[Orodruin]

ohhhh I think I see it. Since I have that sinθ (and that's its odd) in my dV, and we have spherical symmetry or even bounds this whole integral should just be zero.

No. The value of $\theta$ only runs over where the sine is positive. Again, try not thinking of coordinates here. Just consider the integrand and the volume. If you must, use regular Cartesian coordinates.

 

[guyvsdcsniper]

No. The value of $\theta$ only runs over where the sine is positive. Again, try not thinking of coordinates here. Just consider the integrand and the volume. If you must, use regular Cartesian coordinates.

Im kind of lost now.
So we have symmetric bounds over an odd function i get that. I am stuck here

 

[Orodruin]

Im kind of lost now.
So we have symmetric bounds over an odd function i get that. I am stuck here

That is all you need to know. The integrand is odd over a symmetric region.

 

[guyvsdcsniper]

That is all you need to know. The integrand is odd over a symmetric region.

So this proves the answer is zero because for an odd function, the integral over a symmetric interval equals zero?

 

[Orodruin]

Yes.

 

[guyvsdcsniper]

No. The value of $\theta$ only runs over where the sine is positive. Again, try not thinking of coordinates here. Just consider the integrand and the volume. If you must, use regular Cartesian coordinates.

I see what you were saying now in regards to this and can see how my thinking is wrong.

I understand the problem now and thank you for your help with this problem I took a lot away from it.