Hydrogen orbitals/Pauli Principle

[Inquisiter]

Ok, so when the Sch. eq. is solved for hydrogen (H-like atom), we get various eingenstates of energy and angular momentum. For n=2, l=1, we get m=1, m=0, m=-1. The 1,0,-1 represent the z component of l. We also talk of hydrogen as having (for n=2, l=1) three p orbitals. Am I correct that these three orbitals are not the eingenstates of the z component of l? From what I understand (and I don't understand this well), one of them is the eigenstate of the z comp. of l with eigenvalue 0 (the one which is parallel to the z axis), each one of the other two orbitals, however, are actually superpositions of m=1 and m=-1 eigenstates. So, for example, the y component of the angular momentum is 0 for the orbital which is parallel to the y axis, while the y component for the eigenstate of the z comp. of ang. mom. with m=1 is not defined. So I take it that the p orbital parallel to the y-axis is a superposition of TWO eingenstates (with eigenvalues 1 and -1) of the z comp. of l. Can we have superpositions of the three eigenstates of the z comp. of l which produce p orbitals which are not perpendicular to each other. Why are the orbitals perpendicular in an H atom? Is it simply because it minimizes electron-electron repulsion? Also, I'm confused by the Pauli exclusion principle. It says that no two fermions can be in the same quantum state. But if we have three eigenstates (as we do for n=2, l=1), can't we produce an infinite number of states (each one with the same energy) by superimposing the three eigenstates(each time with different coefficients)? So when we talk about quantum states as they relate to the Pauli principle, must these states be linearly independent or what?? I know that we can place only 6 electrons in n=2, l=1 state, not an infinite number of electrons.

Thanks

 

[vanesch]

Am I correct that these three orbitals are not the eingenstates of the z component of l? From what I understand (and I don't understand this well), one of them is the eigenstate of the z comp. of l with eigenvalue 0 (the one which is parallel to the z axis), each one of the other two orbitals, however, are actually superpositions of m=1 and m=-1 eigenstates.

I guess you talk about what is usually pictured in chemistry books :-)
Yes, in that case, the lobes around the x-axis and the lobes around the y-axis are superpositions of the m=1 and m=-1 states (I wouldn't be surprised that it is the |1> + i |-1> and the |1> - i |-1> combinations), because the solutions for m = -1, 0 and 1 have cylindrical symmetry around the z-axis !

In fact, because in the hydrogen atom, these are DEGENERATE eigenstates of energy, you have a 3-dim space in which you are free to choose a basis. You can choose the m = -1, 0, 1 basis, but you can just as well take any other (orthogonal) basis, and to make nice pictures in chemistry books, people like to put these lobes on the x, y and z axis.
After all, there's nothing special about the z-axis.

cheers,
Patrick.

 

[Inquisiter]

You can choose the m = -1, 0, 1 basis, but you can just as well take any other (orthogonal) basis, and to make nice pictures in chemistry books, people like to put these lobes on the x, y and z axis.
After all, there's nothing special about the z-axis.

Well, of course! I suppose my real question was why is it that the electrons "choose" to occupy orbitals which are perpendicular to each other. After all, can't they occupy the m=1, m=0, m=-1 eigenstates (which, as you said, have cylindrical symmetry around the z axis), rather than the three p orbitals pictured in chemistry books (which do NOT have cylindrical symmetry around the z axis)?

 

[vanesch]

After all, can't they occupy the m=1, m=0, m=-1 eigenstates (which, as you said, have cylindrical symmetry around the z axis), rather than the three p orbitals pictured in chemistry books (which do NOT have cylindrical symmetry around the z axis)?

Of course they can ! They can occupy even several orbitals at once, even of different energy. That's the superposition principle. In that last case, your atom is simply not in a stationary state (but it doesn't have to ; only, due to interaction with the EM field, it will quickly decay into its ground state, but that's a very subtle point, usually not touched upon in chemistry books or introductory quantum texts).
All this simply depends on the initial conditions, and on the subsequent interactions.

The orbitals shown in chemistry books just show you a POSSIBLE basis in the state space. They are not the only states an electron can be in, just a basis of those states. And as you know, there is some liberty in choosing a basis.

cheers,
Patrick.

 

[Inquisiter]

Of course they can ! They can occupy even several orbitals at once, even of different energy. That's the superposition principle. In that last case, your atom is simply not in a stationary state (but it doesn't have to ; only, due to interaction with the EM field, it will quickly decay into its ground state, but that's a very subtle point, usually not touched upon in chemistry books or introductory quantum texts).
All this simply depends on the initial conditions, and on the subsequent interactions.

The orbitals shown in chemistry books just show you a POSSIBLE basis in the state space. They are not the only states an electron can be in, just a basis of those states. And as you know, there is some liberty in choosing a basis.

Yes, of course! But when an atom (Neon for example) is in the GROUND state, are the three p orbitals perpendicular to each other, and if so, why? I was thinking that perpendicular p orbitals would be favorable because maybe they minimize electron-electron repulsion.

 

[vanesch]

Yes, of course! But when an atom (Neon for example) is in the GROUND state, are the three p orbitals perpendicular to each other, and if so, why? I was thinking that maybe perpendicular p orbitals would be favorable because maybe they minimize electron-electron repulstion.

Ah, but once we leave the HYDROGEN atom, and we have multiple electrons, you cannot say anymore that ONE electron is in ONE orbital, strictly speaking. It can be an approximation (using the Slater determinant method), but you strictly now have to use n-electron wave functions, which are NOT a priori a product (or a slater determininant). And then your orbitals loose their meaning all together.

cheers,
Patrick.

 

[Inquisiter]

Ah, but once we leave the HYDROGEN atom, and we have multiple electrons, you cannot say anymore that ONE electron is in ONE orbital, strictly speaking. It can be an approximation (using the Slater determinant method), but you strictly now have to use n-electron wave functions, which are NOT a priori a product (or a slater determininant). And then your orbitals loose their meaning all together.

Haha, I was afraid you'd say that! But seriously, Neon is typically pictured as having three perpendicular p orbitals. I assume that this somehow corresponds to reality. So, why are the p orbitals perpendicular?

 

[vanesch]

Haha, I was afraid you'd say that! But seriously, Neon is typically pictured as having three perpendicular p orbitals. I assume that this somehow corresponds to reality. So, why are the p orbitals perpendicular?

This doesn't correspond to reality ! It corresponds to chemistry book wisdom.
After all, neon atoms at rest are perfectly spherically symmetric !

cheers,
patrick.

 

[Inquisiter]

This doesn't correspond to reality ! It corresponds to chemistry book wisdom.
After all, neon atoms at rest are perfectly spherically symmetric !

WOW! Neon atoms are PERFECTLY SPHERICALLY SYMMETRIC??! I have not seen this mentioned ANYWHERE. I mean, I've skimmed through advanced quantum mechanics books, I might have missed it, I don't know. Ok, so I guess the question is, why is the neon atom sph. symmetric in it's ground state? Is it because this minimizes electron-electron repulsion or is there some other reason?

 

[vanesch]

WOW! Neon atoms are PERFECTLY SPHERICALLY SYMMETRIC??!

If the ground state is not degenerate, which it is not as far as I know, then the ground state has to have the symmetry of the Hamiltonian, which is spherically symmetric. If the ground states are degenerate, you can have spontaneous symmetry breaking, but this is not the case for a neon atom.

cheers,
Patrick.

 

[dextercioby]

Vanesch,the Neon atom is not spherically symetric.The H and He atoms are,for example.Can u see why?

Daniel.

 

[Inquisiter]

So does anyone know whether the neon atom is spherically symmetric or not?? I found the following statements on some chemistry exam on Google:

The nitrogen atom, with its half occupied 2p orbitals, has a spherically symmetric electron distribution. Use the explicit expressions for the angular parts of the wavefunctions to prove this assertion by showing that the total electron density,ψ2(2px)+ψ2(2py)+ψ2(2pz), is independent of angle. (b)The same approach leads to the conclusion that the neon atom, with its completelyfilled 2p orbitals, is spherically symmetric. Similarly, filled and half-filled d orbitals are spherically symmetric.

Also, can someone please answer my question about the Pauli exclusion principle. (It's at the top)

 

[vanesch]

Vanesch,the Neon atom is not spherically symetric.The H and He atoms are,for example.Can u see why?

I think I see why you think it is not spherically symmetric, namely because H and He only use explicitly spherical S orbitals, and not Neon.
But my point was exactly to point out the mistake in this reasoning, which amounts to saying that 3-dim Euclidean space is not spherically symmetric, because there is an X, a Y and a Z axis !

What I meant was: if you have the COMPLETE solution (including all the electrons) of the neon atom ground state, and you apply ANY rotation of that solution, you find back the SAME solution (up to a phase factor). Of course, you can write it out in any BASIS you want, but you will find the same solution.
The reason for that is simple (unless I'm making a mistake here): the ground state of the neon atom is NON-DEGENERATE. (if that's not true, then I'm wrong).

Now, you know that each EIGENSPACE E of an operator is a representation of the symmetry group of that operator, in that if |psi> is an element of E, and G is the symmetry group of the operator, then g |psi> is also an element of E, for each g in G.
In general, that doesn't mean that g |psi> = |psi> so, indeed, psi does NOT, in general, have to have the same symmetry group as G.
But if E is ONE-DIMENSIONAL, then there's not much choice for g |psi> ; only a phase factor.
So in the case of non-degeneracy, g |psi> = a(g) |psi>. So psi is an eigenvector of each of the group elements of G. So the state psi has G as a symmetry group.

Now, the hamiltonian for a neon atom is spherically symmetric. That means that its ground state (if it is non-degenerate, which I think it is) is spherically symmetric, no ?

That doesn't mean, of course, that each term of it needs to be so: that's your choice of writing it down.

cheers,
Patrick.