Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

S-orbitals, angular momentum and Heisenberg uncertainty

  1. Feb 26, 2010 #1
    After about 20 years I'm going back and brushing up on quantum. I am confused about an electron in an s-orbital having zero angular momentum as well as how angular momentum in general relates to Heisenberg uncertainty. I probably have a few misconceptions as well as missing some key pieces. Any help you can provide to clear up my confusion is greatly appreciated.

    If an electron occupying an s-orbital has zero angular momentum wouldn't that infer that it has zero angular velocity? If so wouldn't this mean that the electron is at rest and now localized in space? This qualitatively seems troubling to me. I also seem to remember someone mentioning that electron spin is a relativistic effect and due to the electron moving at a velocity near the speed of light. If so, then how can an electron in an s-orbital have a +1/2 or -1/2 spin if it is stationary?

    A related but more general question about angular momentum, I have heard arguments using Heisenberg uncertainty to explain 1) why an electron doesn't crash into and stay localized at the nucleus and 2) why the ground energy state can't be equivalent to the zero point energy. It is argued that if the electron is stationary at the nucleus or had zero point energy one would know both the exact velocity (ie zero) and the exact position, and in turn this violates Heisenberg uncertainty. But for angular momentum if one localizes the electron to a specific orbital (doesn't have to be an s-orbital),[tex]\Delta[/tex]L = 0. The angular position [tex]\phi[/tex] of the electron can be anywhere between 0 and 2[tex]\Pi[/tex] but I thought I remember that in this case [tex]\Delta\phi[/tex] = [tex]\pi[/tex]/[tex]\sqrt{}3[/tex]. In this case the products of the uncertainties equals zero. Doesn't this violate Heisenberg uncertainty?
  2. jcsd
  3. Feb 26, 2010 #2


    User Avatar
    Science Advisor

    Actually, that is precisely how the situation is viewed in one interpretation of QM, namely the deBroglie-Bohm interpretation. In the Copenhagen interpretation (which is also called standard QM or SQM), the picture is more vague. Essentially, one is forced to restrict declarative statements to measureable quantities. We know a) the electron has a measurable kinetic energy, which is non-zero and b) the electron has a measurable orbital angular momentum, which is zero. The question of whether or not one can interpret this in terms of a classical-like motion of the electron is simply not addressed. Stated another way, we can observe that the probability of finding the particle around the nucleus has a particular form (the s-orbital), but we don't know anything about how the electron "moves" between the various points within that orbital.

    The spin represents an "intrinsic angular momentum" and is distinct from the "orbital angular momentum", which is the quantity that is zero.

    You have to be careful about interpretation here. The "position-momentum" form of the HUP that you are referencing deals with linear momentum, not angular momentum. The fact that we know the orbital angular momentum is zero says nothing about the linear momentum. If you break down the vector cross product representing angular momentum (it is r x p, where r and p are the 3-D vectors representing momentum and position), you will see that it is a sum of terms, and so it is consistent to have a non-zero linear momentum and position that produce cancelation, resulting in a angular momentum of zero. (In fact, the cross product of any two parallel vectors is always zero.)

    There are other forms of the HUP that relate to angular momentum, specifically, using the cross product form of angular momentum, and the standard "position-momentum" form of the HUP, you can show that there are also uncertainty relationships between the 3 cartesian components of angular momentum. That means that you cannot simultaneously know, say, both the x- and y-components of angular momentum with arbitrary precision.
  4. Feb 26, 2010 #3
    I was actually just talking to my professor about this.

    When we talk about the motion of a particle in some state, such as in the L = 0 state (s state), we are addressing the propagation of a wave function through time, which in the sum over histories formulation of quantum mechanics, corresponds to a distribution of paths through space and time.

    For the L = 0 state, these paths have, on average, 0 angular momentum, but each path is likely undergoing some angular motion.

    For a classical particle with no angular momentum (and some energy), we have a particle that oscillates around the origin. The quantum mechanical solution to the problem should be similar, and a typical path of a particle would likely be a path that jitters around a classical path.
  5. Feb 26, 2010 #4
    What is the zero orbital angular momentum? What is the spin? Please just imagine.

    If there was already a clear answer to your questions, you would have already found the answers in the textbooks and would not have asked here.

    I think that the questions like these will continue forever (including the interpretations of QM), as long as the QM continue. :smile:
    Because the QM(the quantum mechanics) always says "shut up and calculate" though it contains "vague" parts.

    For example, about the spin, there is a long, long discussion even now here (in this thread).
    Last edited: Feb 26, 2010
  6. Feb 26, 2010 #5


    User Avatar
    Science Advisor

    Yeah, I have heard that interpretation, and I really don't buy it without more development. In SQM, every result of a single measurement returns an eigenstate of the operator corresponding to the physical observable being measured. Since the s-orbital of an H-atom corresponds to an eigenstate of the angular momentum operator, every measurement will return the corresponding eigenvalue zero. If it were somehow representable as a superposition of trajectories with non-zero angular momenta, it certainly seems like those values would be returned by single measurements, and then the expectation value (average) of all the measurements would sum to zero. This does not happen, so the latter case does not correspond with physical reality as far as I can tell.
  7. Feb 26, 2010 #6
    Thanks so much for the reply. I prefer SQM and understand the reasoning behind not forcing a classical interpretation on a quantum perspective. But obviously I don't always practice it. I was thinking of the electron as being in continuous motion about the nucleus (not necessarily an orbit) and from there my confusion arose. Motion of an electron doesn't have much meaning from an SQM perspective. If we can't observe it, we really can't say anything about it.

    I took the HUP for linear momentum and position and substituted the terms for angular momentum and position. Namely, [tex]\Delta[/tex]L [tex]\Delta[/tex][tex]\phi[/tex] [tex]\geq[/tex] [tex]\hbar[/tex]/2. This illegal procedure threw me into the pit. The perils of a biologist waxing physics.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook