Calculate <x^2> for Hydrogen Atom Ground State | Symmetry Method

In summary, the problem asks to find x^2 of the ground state of a hydrogen atom, without using any new integration. Part (a) of the problem is finding r and r^2 for the same state, and part (b) is finding <x^2> and <y^2> of the ground state. Due to spherical symmetry, electron is equally likely to be measured along any of the 3 axis. Thus, according to the student, r^2 is equal to 1/3 of expectation of r.
  • #1
YAHA
121
0

Homework Statement


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.


Homework Equations





The Attempt at a Solution


I am failing to see the symmetry he is talking about.
 
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  • #2
There is no angular dependence in the ground state. Then what can you say about the relation between
[tex]<x^2> , <y^2> [/tex] and [tex] <z^2> [/tex] ?
 
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  • #3
They add up to <r^2>, correct?
 
  • #4
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
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
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
YAHA said:
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)?

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,

[tex] \int_0^5 (y+2)^2 dy [/tex]

and

[tex] \int_0^5 (x+2)^2 dx [/tex]

Do they have the same values when evaluated or different values?
 
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  • #8
YAHA said:
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)?

Yes, that's what <spherical symmetry> of the ground state means.
 
  • #9
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.
 
  • #11
AlexChandler said:
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,

[tex] \int_0^5 (y+2)^2 dy [/tex]

and

[tex] \int_0^5 (x+2)^2 dx [/tex]

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

The problem with this analogy is that usually these integrals are evaluated in spherical coordinates, and in those coordinates, [itex]x, y[/itex] and [itex]z[/itex] 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
Right, r^2. :) Thanks for everyone's help.
 
  • #13
Steely Dan said:
The problem with this analogy is that usually these integrals are evaluated in spherical coordinates, and in those coordinates, [itex]x, y[/itex] and [itex]z[/itex] 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.

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
AlexChandler said:
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

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.
 

1. What is the basic concept behind the symmetry method for calculating for a hydrogen atom in its ground state?

The symmetry method is based on the idea that the wave function of an electron in a hydrogen atom can be separated into two parts: a radial component and an angular component. By considering the symmetries of each component, we can simplify the calculation of by reducing it to a one-dimensional integral.

2. How is the symmetry method different from other methods of calculating for the hydrogen atom?

The symmetry method is unique in that it takes advantage of the symmetries of the hydrogen atom's wave function, rather than relying on complex mathematical techniques or numerical approximations. This makes it a more elegant and efficient approach to calculating , especially for the ground state of the hydrogen atom.

3. Can the symmetry method be applied to other atoms or molecules?

Yes, the symmetry method can be extended to other atoms and molecules, as long as they possess a similar symmetry as the hydrogen atom. However, the specific calculations and techniques may vary depending on the complexity of the system.

4. What are the limitations of using the symmetry method to calculate for the hydrogen atom?

The symmetry method is most accurate for the ground state of the hydrogen atom. As the energy level increases, the symmetries become more complex and the method may not be as effective. Additionally, the method assumes a spherically symmetric potential, so it may not be applicable to systems with more complex potentials.

5. How does the symmetry method contribute to our understanding of the hydrogen atom?

The symmetry method provides a deeper understanding of the symmetries present in the hydrogen atom's wave function, which in turn helps us better understand the behavior and properties of the atom. It also allows for more efficient and elegant calculations, making it a valuable tool for studying the hydrogen atom and other systems with similar symmetries.

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