# Shapes of electron orbitals

1. Feb 25, 2016

### Happiness

The probability distribution of the position of the electron of a hydrogen atom is related to the following polar plots

Suppose the electron is excited from the $1s$ orbital to the $2p_x$ orbital. Does it make sense to talk about the $2p_x$ orbital having a dumbbell shape pointing in the $x$ direction since the $z$ axis can be pointing in any direction? Shouldn't all orbitals be spherical when we consider there is an equal probability for the $z$ axis to point in any direction?

2. Feb 25, 2016

### blue_leaf77

What are shown in the picture above are actually the cross-section of $|Y_{lm}(\theta,\phi)|^2$ in the xz plane, so when viewed as a 3D distribution, the $2p_x$ orbital does not have a dumbbell shape.

3. Feb 25, 2016

### Staff: Mentor

Be careful that what you posted are pictures of the the spherical harmonics, which are complex functions, while $p_x$ is a real function, linear combination of spherical harmonics, and is not an eigenfunction of $\hat{L}_z$.

No, you can have a polarized sample. At equilibrium, you should have on average an isotropic distribution, but one can prepare an atomic sample in a given $l, m_l$ state. Note that the preparation aspect is going to define a laboratory z axis: the choice of axes is no longer arbitrary.

4. Feb 25, 2016

### Happiness

Must the spin angular momentum of an electron have a definite direction or could it be in a superposition of being in manydifferent directions? If so, how can we prepare such a superposition?

5. Feb 25, 2016

### Staff: Mentor

It is always in some superposition. If it is aligned along z in some coordinate system, it is in a superposition in any coordinate system where z' is not colinear with z.

6. Feb 25, 2016

### Happiness

So there must always exist exactly one coordinate system in which the spin is aligned along its z axis?

7. Feb 25, 2016

### Staff: Mentor

Yes. In the presence of an external field, you also have to account for spin precession, so it is possible that that coordinate system is a rotating one.

8. Feb 25, 2016

### DrDu

I just want to add that which orbital gets excited depends on the polarisation of the exciting electromagnetic field. The propagation direction defines one axis and the direction of the em field fixes another one. It is convenient to chose the quantization axes accordingly.

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