# Why is z-axis so special in hydrodgen problem?

1. Dec 16, 2011

### AlonsoMcLaren

Why is z-axis so special in hydrogen problem? If we rotate the z-axis, will Lz still have zero uncertainty?

2. Dec 16, 2011

Staff Emeritus
It's not. We simply label the most important axis in a given problem with a z.

3. Dec 16, 2011

### AlonsoMcLaren

So what will happen after we rotate the coordinate system? Will Lz still have zero uncertainty?

4. Dec 16, 2011

### jewbinson

What do you mean? The eigenvalue of Lz has 0 uncertainty only if you measure/observe the eigenvalue of Lz for a given operator... before a measurement is made there is uncertainty. Measurements of Lz usually involve finding the eigenvalues of Lz. Possible outcomes are +-mh, where h=h_bar is Plank's constant, and m is an integer called the "quantum number" (or maybe something else, can't remember). m cannot take all integers: it is limited by a positive number l so that the possible values of m are: -l, -l+1, ..., l-1, l. l can be either an integer or a half-integer.

The S.E. doesn't care what a fundamental particle looks like: the important thing is that there is invariance under coordinate transformation (i.e. translation and rotation). This means is that the z axis is no more special than the x or y axis.

5. Dec 17, 2011

Staff Emeritus
If you rotate the coordinate system, the axes change. What else can happen?

The physics doesn't care about the coordinate system. The coordinate system is a human invention. By convention, we pick z to be the most important axis. We didn't have to, but it is easiest to communicate if we all adopt the same convention.

6. Dec 19, 2011

### AlonsoMcLaren

So, say there's a (2,1,0) hydrogen atom sitting there and we know that some direction is the direction of the axis of the "dumbbell". Then if we randomly pick an axis as z-axis, it's nearly impossible that the z-axis we picked is exactly the same as the axis of "dumbbell".

Therefore, the choice of z-axis is not arbitrary. But how can we know in advance that the z-axis we're going to choose is the correct one? (By saying that the z-axis is correct I mean that it is exactly the same as the axis of dumbbell)

Also, it seems to me that in the solution of the hydrogen problem, the choice of z-axis is completely arbitrary....

7. Dec 20, 2011

### Morgoth

using the commutors:
[Li,Lj]=ihεijk Lk
and heisenberg's uncertainty equation for two operators A,B:
ΔA ΔB= (1/2) | < [A,B]> |
you end up that the angular mommentum cannot be measured with 0 uncertainty in all axis.

So we choose one axis on which we measure the angular mommentum.
AFTERWARDS:
you need to write the rest as a sum of the eigenvectors of your chosen mommentum. So by the time you choose your reference axis to be z, then x,y axis get the uncertainty. If you choose to measure x, you will get the same values as you got for z (because of the symmetry of space), however the uncertainty goes to y,z and y,z values also must be written as a superposition of the eigenvectors of x.

It is just more easily to have a global reference, and this came to be z.

Last edited: Dec 20, 2011
8. Dec 20, 2011

### Morgoth

Now for rotations to see what happens you have to say that your new wavefunction -or more conveniently your new state is:

|Ψ>R = U(φ,n) |Ψ>0

where U is the rotation operator, for a rotation of angle φ around the n vector and |Ψ>0 is let us say your initial state (the one written in the eigenvectors of Lz).
After you see how the Psi is being transformed, you can know what changes.

9. Dec 20, 2011

### kith

I think you have a fundamental misconception, but I'm not shure what exactly your problem is.

In order to say that the state of an atom is (2,1,0), you need to have a well-defined z-axis, because the last number refers to Lz. In general, you don't know which ml-state your atom is in until you have measured Lz. In order to measure it, you have to define a z-axis by aligning the measurement apparatus in a distinct way. If you rotate your system or your measurement apparatus, you can calculate the new state of the system or the new observable you are measuring in terms of the old z-axis.

In all cases, the z-axis is well-defined.

10. Dec 20, 2011

### nonequilibrium

I think you adressed the correct misconception.

I think it's rather a common misconception, at least for people not specialized in physics (or not specialized far enough). I remember from my basic chem class that electrons in atoms were simply assigned a (n,l,m) value, not making clear that to define this, you first have to define a z-axis and that it will only be in such a (n,l,m) state if it the z-component of the angular momentum was measured; otherwise it would be in a super position of such states. This was never made clear to me or my fellow students.

11. Dec 20, 2011

### Antiphon

You choose the z axis along the dumbbell because the math is easier to work out.

Once you have the solution, you can tilt it into another coordinate system.

The coordinate system is chosen by you, not the atom.

BTW, you don't have to line up the z axis with the dumbbell. But you will really wish you had by the time you get to the answer.