# Question about H atom orbitals!

1. Feb 13, 2013

### RedCliff

Hello, all,

I have a question about the wave functions of H atom: for p orbital, wave functions from the Schrodinger equation are p+1, p0, and p-1. In chemistry, people use the linear combination of p+1 and p-1 to generate px and py orbitals. The question is: p+1 and p-1 orbitals are donut shaped and px and py are dumb bell shaped. However, the shape of the orbitals should be the physical characteristic and should not be related to the mathematical treatment. Some textbook explains as: the combination of the degenerated wave functions is also the eigen function of the Schrodinger equation. This raises another question: it seems like the Schrodinger equation only gives us the energy levels, but never gives us the actual wave functions if there is degeneracy there. That means, we never know the real wave function and therefore shape of the H atom, or the probability at certain point according to Born interpretation. Then how did Bohr and Heisenberg think the quantum theory is complete?

Does it mean wave functions from Schrodinger equation are not complete? Let's forget about spins at this moment.

John

2. Feb 13, 2013

### ZapperZ

Staff Emeritus
Donut shape? p-orbitals? Where?

And BTW, here are the hydrogen wave function, up to n=3.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html

Not sure where you got the idea that the Schrodinger equation doesn't give you the wave function. Every physics undergraduate student can tell you that is wrong.

Zz.

3. Feb 13, 2013

### RedCliff

"Donut shape? p-orbitals? Where?"

Yes, if you plot contour of ψ*ψ of p+1 and p-1, it will give you donut shape. Check Figure 6.13 on pg. 153 in Levine's "Quantum Chemistry" (5th ed).

"Not sure where you got the idea that the Schrodinger equation doesn't give you the wave function. Every physics undergraduate student can tell you that is wrong."

I didn't say that Schrodinger equation doesn't give us the wave function. What I meant is that if those wave functions really describe the real H atom or not since liner combination of those functions will give completely different shapes as described above, if those functions are degenerated.

4. Feb 13, 2013

### ZapperZ

Staff Emeritus
But ψ*ψ is the probability density! What does that have anything to do with anything? It is not even something one measures!

Zz.

5. Feb 13, 2013

### RedCliff

"But ψ*ψ is the probability density! What does that have anything to do with anything? It is not even something one measures!"

The contour of equal ψ*ψ is used to define the shape of the orbitals. Check the same book and other quantum books (but I am not sure if they make it so clear as that in Levin's book)

6. Feb 14, 2013

### tom.stoer

This is correct.

To be more precise, it depends on he question you are asking. If you only know about a hydrogen atom in some energy eigenstate the Schrödinger equation only tells you which energy eigenvalues and which corresponding eigenfunctions are allowed. It doesn't tell you in which eigenstate you will find the atom.

If you know that an atom has been excited from 1s to the first excited state, then all you know is that the atom is in a state described by a linear combination of the 2s, 2p states. If you have more knowledge about the excitation (e.g. if you know the details of the photon state like momentum and polarization) you can derive more details regarding the final state of the atom, so you may be able to exclude certain states which are forbidden by selection rules related to angular momentum conservation.

But in principle you are right. If all you know is the eigenvalue, the state is only specific as an arbitrary unit vector in the subspace spanned by the eigenstates corresponding to this eigenvalue. So for n-fold degeneracy the subspace is n-dimensional.

EDIT:

b/c the 'incompleteness' you are mentioning is the same as in Newtonian mechanics. Specifying energy E and angular momentum L for the Kepler problem does neither fix the plane of the orbit (b/c the direction of L is left unspecified), nor the shape and the orientation of the orbit within the plane (b/c the Laplace-Lenz-Runge vector is left unspecified). So in order to determine the state of a physical system completely you must specify the values for a maximal set of observables. This is never done using an equation of motion but always via specifying the initial conditions (for this e.o.m.), experimentally via preparation or obsevation.

Last edited: Feb 14, 2013
7. Feb 17, 2013

### RedCliff

Hello, Tom,

Thanks for your posting and explanation! Your answers clear this basic question held in my mind for a long time! As I read the Atkins' textbook, "physical chemistry", in one place, it says wave functions contain all information on a particle, and in another place, it says px and py can be generated by the linear combination of p-1 and p+1. Then I asked, if wave functions contain all information, why the shape of orbitals can be changed arbitrarily by math manipulation of wave functions?

If I understand you correctly, basically, you said that: similar to translational movement in classical physics which shows if we want to know the current position of a particle, we have to know its original position (initial condition) and it momentum (given by Newton's law), in quantum, if we want to know the current state, we have to know the initial state in addition to those eigen states allowed by Schrodinger equation. Therefore, a H atom in p orbital could be the dumb bell shape or donut shape, depending on the initial condition.

However, this raises another question: why in molecular orbital theory, scientists always use px, py and pz to generate molecular orbitals. Nobody mentioned using p+1 and p-1. Is that due to the symmtry reason or both sets give the same resutls?

Thank you again!

PS: forgive my English since it is my 2nd language.

John

8. Mar 21, 2013

### RedCliff

Hello, Tom and all,

Can you guys answer my question in the last post? That is, why only px, py, pz orbitals are used in LCAO-MO theory?

Thanks!

John

9. Mar 21, 2013

### fzero

Yes, both sets give the same results. (px, py ,pz ) is a choice of basis for the L=1 states, while ( p+1 ,p0, p-1) is just a different choice of basis. We can always write the elements of one basis in terms of linear combination of the other basis. The molecular orbitals are also linear combinations of the atomic orbitals, which would could express in either basis:

$$\psi_i = \sum_{j=x,y,z} c_{ij} p_j = \sum_{j=-1,0,1} c'_{ij} p_j.$$

Here the matrices $c$ and $c'$ are just related by the change of basis.

10. Mar 21, 2013

### RedCliff

Hello, fzero,

Thank you very much!

I got it now!

John

11. Mar 22, 2013

### DrDu

This is not generally true. The main reason for using the real valued functions is that the hamiltonian for molecules is time inversion symmetric whence eigenfunctions may be chosen to be real valued. This changes once molecules are considered in external magnetic fields.

12. Mar 22, 2013

### DrDu

I don't quite agree. People are measuring the probability density e.g. from high precision x-ray crystal structures and compare them to quantum chemical densities, e.g. from density functional theory. See, e.g.,