Calculating <r> from <x> with no further calculation

[hmparticle9]

Homework Statement

Find $\langle x \rangle$ for an electron in the ground state of hydrogen from $\langle r \rangle$ with no further calculations.

Relevant Equations

$r^2 = x^2 + y^2 + z^2$

This is part of a larger, more difficult question. But this is the part I was a bit stuck with. $\langle r \rangle = \frac{3}{2}a$ and apparently $\langle x \rangle = 0$. I could arrive at this by plugging the numbers in but I would like to know how to come to this result without doing any further calculations?

 

[kuruman]

Consider using a symmetry argument.

 

[hmparticle9]

I am none the wiser. Without giving it away could you provide me with another nudge? I am trying to approach it by saying "suppose that <x> != 0" but no luck.

 

[PeroK]

Well, $x$ is frame dependent. So, I guess you have to know which side of the hydrogen atom is left and right.

 

[hmparticle9]

$\langle x \rangle$ and $\langle r \rangle$ are scalars. but $\langle \mathbf{r} \rangle = \langle (x,y,z) \rangle$ is a vector. I get that. I am sorry guys, i'm just not clicking with this.

 

[JimWhoKnew]

$\langle x \rangle$ and $\langle r \rangle$ are scalars. but $\langle \mathbf{r} \rangle = \langle (x,y,z) \rangle$ is a vector. I get that. I am sorry guys, i'm just not clicking with this.

I deleted a post above, not realizing you've already replied to it. I apologize.

$\langle \mathbf{r} \rangle = \langle x \rangle\hat{x}+\langle y \rangle\hat{y}+\langle z \rangle\hat{z}$ . Can you think of a direction it should point to?

 

[hmparticle9]

No worries, I am glad to have the help available.

I can't. I know that the ground state $\psi_{100} = \frac{1}{\sqrt{\pi a^3}} e^{r/a}$, hence we can infer symmetry in $x,y,z$. Also, $\langle r \rangle = \frac{3}{2}a$. I don't see how we can say that $\langle x \rangle =0$ from these facts.

I can't think of a direction.

 

[JimWhoKnew]

No worries, I am glad to have the help available.

I can't. I know that the ground state $\psi_{100} = \frac{1}{\sqrt{\pi a^3}} e^{r/a}$, hence we can infer symmetry in $x,y,z$. Also, $\langle r \rangle = \frac{3}{2}a$. I don't see how we can say that $\langle x \rangle =0$ from these facts.

I can't think of a direction.

What very important symmetry(ies) do both the Hamiltonian and the ground state of the Hydrogen atom possess?
(Beside LRL, something more obvious...)

 

[hmparticle9]

I am not sure. The Hamiltonian is an operator and the ground state of the Hydrogen atom is a function. They are both symmetric in the sense that $H = \frac{1}{2m}(p_x^2 + p_y^2 + p_z) +V$ ... ah I am not sure.

 

[JimWhoKnew]

In our context, a "symmetry" means that there is some kind of transformation that we can apply, such that nothing changes (invariance).
For a transformation $T$ , and the HA Hamiltonian, we can express the above conditions as$$THT^{-1}=H$$and$$T|\psi_{100}\rangle=|\psi_{100}\rangle \quad .$$Can you think of the (very important) transformations that satisfy these conditions?
(Note that the first condition implies $\left[T,H\right]=0$ , so $T$ must be a constant of motion)

 

[hmparticle9]

The identity matrix?

 

[PeroK]

No worries, I am glad to have the help available.

I can't. I know that the ground state $\psi_{100} = \frac{1}{\sqrt{\pi a^3}} e^{r/a}$, hence we can infer symmetry in $x,y,z$. Also, $\langle r \rangle = \frac{3}{2}a$. I don't see how we can say that $\langle x \rangle =0$ from these facts.

I can't think of a direction.

Do you think the expected value of $x$ is positive or negative? I.e. to the right of the atomic nucleas or to the left?

 

[hmparticle9]

The electron travels around the nucleus. I think the whole thing behind quantum mechanics is that we don't know exactly where the electron is, but we know where it is likely to be. For instance, the most probable value of $r$ for the electron in ground state is $a$ (the Bohr radius). Your argument is that if the electron travels around the nucleus then the expected value of $x$ is zero?

 

[PeroK]

Your argument is that if the electron travels around the nucleus then the expected value of $x$ is zero?

Not really. It's more fundamental than that.

You've never answered my question about left and right.

 

[hmparticle9]

Yes I am a little embarrassed because I think it is neither. The electron moves around the nucleus if you were to project its location on the $x$ axis then it should average to be zero.

 

[PeroK]

Yes I am a little embarrassed because I think it is neither. The electron moves around the nucleus if you were to project its location on the $x$ axis then it should average to be zero.

That's one way to look at it.

 

[PeroK]

Yes I am a little embarrassed because I think it is neither. The electron moves around the nucleus if you were to project its location on the $x$ axis then it should average to be zero.

You've probably struggled with this long enough. Let's assume that $\langle x \rangle$ is +ve; i.e., the expected value of $x$ is to the right of the nucleus. "To the right of the nucleus" depends on your choice of orientation of the x-axis. If you look at the atom from the opposite side, then left becomes right and right becomes left. It's the same equation however you set up your axes, so you must get the same numerical answer. In the new coordinate system $\langle x \rangle$ must still be +ve. But, that's now on physically the opposite side of the nucleus.

But, that implies a different physical solution depending on your choice of x-axis.

You can formalise this mathematically, but the basic physical symmetry argument is so important that you need to understand it.

If $\langle x \rangle \ne 0$, then you would need to have some physically asymmetry in your system. This would imply something in your Hamiltonian that depnds on $x$. So that, if you change the orientation of your x-axis, the equation you have to solve changes, and the solution changes. In the new coordinate system $\langle x \rangle$ would have to change sign.

 

[JimWhoKnew]

The identity matrix?

@PeroK's message in #17 worth repetition:

You can formalise this mathematically, but the basic physical symmetry argument is so important that you need to understand it.

The HA Hamiltonian and the ground state are spherically symmetric, which means they are invariant under all rotations (whose axis passes through the origin). They "look" the same in all directions, so how can $\langle \mathbf{R} \rangle$ "pick" a single preferred direction to point to? So $\langle \mathbf{R} \rangle=0$ , and hence $\langle X \rangle=0$ . If you'll think about it, it's pretty much the same as @PeroK's explanation.

Another way: Write down the integral for $\langle X \rangle$ in Cartesian coordinates. Try to convince yourself that the contribution from the volume element $dxdydz$ at any chosen point, say $\mathbf{x}_0$ , is canceled by the contribution at $(-\mathbf{x}_0)$ . It happens because $X$ is an odd operator and $\psi_{100} = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}$ is an even function (again, symmetry!).

Edit: corrected sign typo in $\psi_{100}$

 

[hmparticle9]

I am going to come back to this a few times today to make sure I fully understand everything.