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Orbitals and Energies

  1. Jan 30, 2012 #1
    Hey everyone,

    So I've been a little bit confused about something, and I'd like some input on this:

    When I look at systems like the particle in a box, or the Hydrogen Atom, it's clear that the higher energy states of comparable shape have more nodes. This seems to always be true. For any orbital, the number of radial nodes increases with increasing principal quantum number and with nodes of a given principal quantum number, the number of nodes, and the energy increases as the angular momentum quantum number increases.

    In the particle in a box system, as the characteristic quantum number increases, the number of nodes, and the energy increases.

    In molecular orbitals of a molecular system, orbitals with more nodes have higher energy than those with fewer nodes (whether in the simple homonuclear H2 or something more complicated systems like Benzene).

    So it seems like nodes implies energy. If that's the case why? With electrons, I have heard (and can believe) rationalizations based on electrostatic attractions and repulsions. For instance, in the antibonding MO of H2, the antibonding orbital places most of the electron density by the nuclei which raises the electrostatic potential energy of the system. Even for a particle in a box, where the particle is an electron, I can see that because by creating nodes, you're concentrating electron density in smaller volumes.

    My contention with that explanation though is that it hinges on electrostatic potential energy and that the nodes creating energy thing seems more general. Am I wrong? It seems to me like the creation of nodes to some degree decreases uncertainty because the quantum particle is being more localized. If that happens, then doesn't the zero point energy and the corresponding energies increase?
     
  2. jcsd
  3. Jan 31, 2012 #2
    Remember what your hamiltonian is made of: a kinetic energy as well as a potential energy part. The potential energy component (admittedly the most difficult part to create, especially for polyelectronic systems) is all about attraction and repulsion of electrons and nuclei (using the example of a molecule like benzene). But the kinetic energy component is what your particle in a box (or even particle in free space) derives its energy levels from! So for the particle in a box there are no electrostatic interactions (as, for example, an electron does not repel itself) so there simply exists higher energy levels due to the increased "gradient squared" (more precisely the second derivative) with is what you have from your definition of the momentum operator. Therefore the momentum of the particle determines its energy. If you have a very smooth, flat wavefunction you are not going to have very a very high kinetic energy because the second derivative of that function is not going to be big when you compare it to a wavefunction which describes a system with a lot of nodes (which implies the particle is going to have a lot of momentum) which would have a large second derivative and therefore a large kinetic energy.

    So there is, as you pointed out, an electrostatic contribution to the total energy of the system, but there is also a kinetic energy contribution. As a chemist, I'm sad to say that the kinetic energy part of the hamiltonian is typically brushed over, but it is critically important because without the kinetic energy part you would not even have simple formation of H2 from 2 H atoms since electrostatics cannot give a good enough reason as to why you would bring together two electrons between two protons! To prove it to yourself just look at the wavefunctions for the bondiong and antibonding orbitals of the H2 molecule and you can see that the gradient is much lower for the bonding MO than even the atomic orbital of a hydrogen atom. I'm getting a bit off topic now...

    tl;dr more nodes => bigger second derivative => higher kinetic energy => higher total energy
     
  4. Jan 31, 2012 #3
    That more or less makes sense. The only thing I'm now not clear about (and this isn't particularly related to the original question) is the statement you made about an electron not repelling itself. In QM isn't the electron more or less smeared out over a region to give you something like a stationary charge density distribution? And from electrostatics, isn't the energy of a system with high charge density higher than one where a similar quantity of charge is spread out over a larger region?
     
  5. Feb 1, 2012 #4
    When I mentioned about an electron repelling itself I was making a phacetious reference to density functional theory (theory which uses the density to variationally determine the ground state energy) wherein one approach (the most used appraoch when dealing with DFT) you have an approximate function which describes the exchange and correlation behaviour of the electrons. If you compare this to another way of solving the many-electron problem called the Hartree-Fock method, which contain exact expressions for the Coulomb and exchange interactions, the self-interaction is exactly cancelled by those two portions of the approach. However, in DFT it is not, so you can have the situation where an electron repels itself (which is not a sensical situation). In my first post I was originally going to go off on a tangent about DFT and this self-interaction error but I then decided not to!

    The other thing you're asking (the electron being "smeared" (presumably you mean the wave packet) over space) is something I never really considered and my knowledge of basic electrostatics is unfortunately quite lacking. Nevertheless from what I remember about measuring charge you must have some test charge present in the system (by the definition used for an electrical potential) to measure the charge, but that would invalidate the assumption of, say, particle in a box since there is only one particle in the box!

    That answer is not very satisfactory but another one I have is that the idea of charge is a particle-based property, not a wave idea. Due to particle-wave-duality you have can effectively have your spread out wave representing a particle at some (ill-defined) point in space, but the charge is not encoded in the wave, it is a property of the particle which is measured relative to some interaction it has with something else in the system. I guess what I'm trying to get at is that for a particle in the box you don't have the concept of charge since there is not potential for the charge to interact with. For more complicated systems (eg molecules) I have absolutely no idea whether an electron could - in principle - repel itself but I wuold hazard a guess that self-repulsion is something that doesn't occur. Maybe there is a better answer than this but I don't know it.
     
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