Can the Orbital Spin be Described by a Probability Distribution?

[SimonBennett]

I am trying to to teach myself the basics of quantum mechanics and need a bit of help. I am a non-mathematician so go easy on the maths in any reply.

I think I have understood the difference between the wave function, probability density and probability distribution in terms of describing atomic orbital structure. As far as I can see the probability distribution "spreads" the electron's properties through the orbital in line with levels of probability predicted by Schrodingers equation. Clearly this will apply to the electron's charge and mass, but does it also apply to its spin?

Also, is there any other property of the electron that can be meaningfully distributed throughout the orbital with the probability distribution?

 

[alxm]

An orbital is the wave function of a single electron. Its absolute square ($$|\psi(x,y,z)|^2$$) at some coordinates is the probability of finding the electron at those coordinates. So the electron does not have a single definite location - when not being measured directly. Since the electron is charged, this means that the charge distribution is 'smeared' the same way.

But neither mass or charge are quantum mechanical states. These are constant properties of the electron. The spin state (but not the fact that it has spin) is a quantum mechanical property, just like its energy state (which corresponds to an orbital) and its angular momentum, etc. They're represented by the quantum numbers n,l,m - which determine the orbital. The orbital says everything about the electron - except spin. Because in non-relativistic QM (& neglecting magnetic fields), spin has no effect on the orbitals.

So spin (up,down) just gets regarded as a completely independent property with no relationship to the energy and/or spatial probability distribution (orbitals). Now given the Pauli Principle, that no two electrons can have the same quantum numbers (i.e. be in the same state), this means that you can have two electrons in each orbital. (one for each of the two permitted spin states)

 

[SimonBennett]

Thanks very much, that is very helpful.

 

[SimonBennett]

I have been reflecting on your answer, and another question has occurred to me.

You say that the spin state of an electron is a quantum mechanical property, but not the fact that the particle has spin. Given that the electron can either exist in a spin up or spin down state, which is the quantum mechanical component of spin, and the spin itself which produces a magnetic field, is a non-quantum mechanical state, is it right to say that the magnitude of the electron's intrinsic magnetic field is smeared across the orbital in line with the probability distribution, but that we do not know the orientation of this magnetic field because we do not know its spin state?

 

[alxm]

Iis it right to say that the magnitude of the electron's intrinsic magnetic field is smeared across the orbital in line with the probability distribution, but that we do not know the orientation of this magnetic field because we do not know its spin state?

Well, as I said it's easier to think about these things as independent properties. Note that since the electron is charged, and a charged particle with angular momentum will give rise to a magnetic field, orbitals with a non-zero l have a magnetic moment (for which reason the m quantum number is the 'magnetic' quantum number). So you have both an orbital magnetic moment and a spin magnetic moment. These two interact (spin-orbit copuling) and you may also have a nuclear magnetic moment which interacts as well (nuclear-spin coupling).

But if you neglect these interactions (and usually do in nonrel. QM) then you're left with how these moments interact with an external field. With no external field, the energy levels for the various states are degenerate (e.g. all p-orbitals have the same energy) so you'll have a superposition of orbital and spin states, with no net magnetic moment. (Same goes for the case where a sub-shell is filled)

 

[SimonBennett]

Thanks again.