# <x^2> Hydrogen Atom

1. May 3, 2012

### YAHA

1. The problem statement, all variables and given/known data
The problem (4.13 (b) in Grifitths) asks to find <x^2> of the ground state of hydrogen atom. It asks to do so without new integration. Supposedly, there is some kind of symmetry and one can use the results from part (a) of the problem. Part (a) is finding <r> and <r^2> for the same state.

2. Relevant equations

3. The attempt at a solution
I am failing to see the symmetry he is talking about.

2. May 3, 2012

### AlexChandler

There is no angular dependence in the ground state. Then what can you say about the relation between
$$<x^2> , <y^2>$$ and $$<z^2>$$ ?

Last edited: May 3, 2012
3. May 3, 2012

### YAHA

They add up to <r^2>, correct?

4. May 3, 2012

### lj612

That's part of the answer, but that's always true no matter the symmetry of the problem. The key is in the following question: how do measurements along different axes compare when you have spherical symmetry?

5. May 3, 2012

### dextercioby

If you can't <see> it from the physical perspective, try from the mathematical one. Write down the 3 integrals of exp values for x^2, y^2 and z^2. Then prove they are equal one to another.

6. May 3, 2012

### YAHA

So does this means that due to spherical symmetry, electron is equally likely to be measured along any of the 3 axis? In other words, x y z are equally likely to contribute to the final vector (r)?

7. May 3, 2012

### AlexChandler

Not sure exactly what you mean by that. However, what you should do is set up the integrals needed to calculate <x^2>,<y^2> and <z^2>. Look for similarities between the integrals. Do they look the same or different? For example: consider the following integrals,

$$\int_0^5 (y+2)^2 dy$$

and

$$\int_0^5 (x+2)^2 dx$$

Do they have the same values when evaluated or different values?

Last edited: May 3, 2012
8. May 4, 2012

### dextercioby

Yes, that's what <spherical symmetry> of the ground state means.

9. May 4, 2012

### YAHA

Alright I understand it now. I did the integrals also and it worked. Expectations of all components are equal. They are 1/3 of expectation of r.

10. May 4, 2012

### dextercioby

Of r^2 of course.

11. May 4, 2012

### Steely Dan

The problem with this analogy is that usually these integrals are evaluated in spherical coordinates, and in those coordinates, $x, y$ and $z$ look quite different from each other. I don't think it would be obvious to someone looking at those three integrals that they have to be the same.

12. May 4, 2012

### YAHA

Right, r^2. :) Thanks for everyone's help.

13. May 8, 2012

### AlexChandler

The point is that they are exactly the same integrals. It doesn't matter if you use x or y or Ω or √ or ∏ to label the variable, it is simply the same integral. It is like asking if the solutions to the following equations will be the same or different:

λ+2=1

θ+2=1

it doesn't matter which variable you use, the solution is always -1

14. May 8, 2012

### Steely Dan

Yes, but your comment was to "look for similarities" in the integrals; what I took from that was the implication that since the integrals are identical in form in rectangular coordinates, one could conclude they all had the same value. But the integrals do not look identical in spherical coordinates, even though they represent the same quantity. So the idea of "looking for similarities" is only helpful insofar as the student already knows to represent the integrals in rectangular coordinates (at which point it is obvious). I simply suggested that most students do not think to make this switch in the first place, since they are taught the hydrogen atom in spherical coordinates to begin with (due to the above symmetry). Also, the symmetry relationship should be obvious (no offense intended to the OP, of course) without even having to write down the integrals -- if the ground state is spherically symmetric, then by definition no direction can be preferred. That statement alone is enough to conclude that the expectation value should be the same for any of the three variables.