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Motion in Bound Electrons

  1. Sep 29, 2010 #1
    An electron in an orbital of an atom has some energy and some momentum. In some ways it can be considered "orbiting" but it is not really moving in a classical sense.

    I've heard more than once the explanation that electrons in high orbits move close to the speed of light and that this effects the properties of the material (e.g. the distinctive color of gold), but this doesn't ring true to me. The optical properties have to do with the difference in energy levels, and that's it. Right?

    But that got me thinking... what about time dilation? Does the electron experience less time because it has momentum, even though the wave is stationary? A more concrete example might be an exotic atom where a muon replaces an electron. The proper time of the muon is reflected in the decay time. Does a bound muon decay slower than a free stationary one, with the effect increasing with energy level?
     
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  3. Sep 29, 2010 #2

    alxm

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    Well, it's "moving" in the quantum-mechanical sense - which becomes the classical sense in the limiting case. At least so long as you interpret quantum-mechanical "momentum" as actual momentum. (which AFAIK only the deBB fans don't) I wouldn't consider them 'orbiting' in any sense though; since they're entirely capable of having zero angular momentum, and thus moving in-and-out rather than round-and-round, if you force a classical perspective on it.
    Optical properties have to do with a lot of factors. To begin with what kind of molecule or crystal the atom is in. E.g. nanoparticles of metallic gold will have http://www.webexhibits.org/causesofcolor/images/content/9_diameter_of_gold.jpg" [Broken] of different colors, depending on particle size. The yellow color is merely the limit at macroscopic (i.e. 'infinite') sizes. And for molecules there's polarizability and dipole moments and selection rules etc.

    In any case, yes, there's a difference in energy levels, for a metal that's the band gap. Special Relativity affects every electronic level, and not equally since they don't have the same momentum. So I don't know why you don't think the relative energy levels wouldn't change. The band gap for gold is basically the difference between the occupied 5d and unoccupied 6p orbitals. In silver, immediately above gold in the periodic table, this is in the UV range. Whereas in gold it's shifted down into the blue range, hence giving the complementary yellow color. We know this is due to SR, because a non-relativistic calculation will result in gold's band gap also being in the UV range, and hence having a silvery color as well (as do most metals).
    http://www.springerlink.com/content/wtr11w113r22g346/"
     
    Last edited by a moderator: May 5, 2017
  4. Sep 29, 2010 #3
    the velocity of the electron in the ground state in the bohr model of hydrogen is alpha * c
    c/137

    the velocity of electrons in the ground state of higher Z atoms would be even higher.
     
  5. Sep 29, 2010 #4
    according to http://en.wikipedia.org/wiki/Bohr_model

    7d64c076128da9d6b2a46c5197f9e422.png

    Re is the energy of the ground state of hydrogen​

    ebb33c81bd38aeac6c90bf1504f2c86a.png

    For the ground state n=1​

    combining these 2 forumulas results in v being proportional to Z (I think)

     
    Last edited: Sep 29, 2010
  6. Sep 29, 2010 #5

    alxm

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    On further pondering, I realized the source of your confusion is that, if they "don't move", how could they have relativistic momentum? Well, they do move, once you accept QM motion as motion. I.e. as far as I'm concerned, and every quantum chemist I know, we talk about 'electronic motion' all the time, even though we clearly don't mean any sort of classical motion - Especially considering that most of the time we're talking about time-independent systems. The main reason for this is correlation. Since this question does pop up on a regular basis, I thought I'd give a thorough explanation for future reference. All about correlation:

    Consider a two-electron system (e.g. helium). What if we assume that the electrons aren't 'really' moving? I.e. they "see" the electrostatic repulsion of the other electron's density cloud, and this determines their kinetic energy, but that there is no explicit interdependence between their kinetic energies. I.e. each electron only 'sees' the Coulomb potential of the other (Pauli repulsion need not be taken into account for the potential of ground-state Helium)
    [tex]V_{ee} = \int \int \frac{\phi_1^*(\mathbf{r})\phi_2^*(\mathbf{r'})\phi_1(\mathbf{r})\phi_2(\mathbf{r'})}{|r-r'|}d\mathbf{r'}d\mathbf{r}[/tex]

    Since the nuclear-electronic potential of each electron is obviously independent of the other electron's location, and we're making the assumption that the respective "kinetic energy" terms of the Hamiltonian does not depend on both r and r', this is equivalent to assuming the total wave function is a product of single-electron wave functions.

    Due to fermion antisymmetry, in singlet He this is the same spatial orbital with opposite spins:
    [tex]\Psi(\mathbf{r},\mathbf{r'}) = \phi(\mathbf{r})\phi(\mathbf{r'})(\alpha(\sigma_1)\beta(\sigma_2) - \alpha(\sigma_2)\beta(\sigma_1))[/tex]
    (where alpha and beta are spin functions) However, the orthogonality of spin states leads to the pair probability distribution (probability of electron 2 being at r' with electron 1 at r) being:
    [tex]|\Psi(\mathbf{r},\mathbf{r'})|^2 = \phi^2(\mathbf{r})\phi^2(\mathbf{r'})[/tex]

    So, if electron 1 is in a location, say, [tex]\mathbf{r} = (1,0,0)[/tex] from the nucleus (at the origin), the probability density of electron 2 in a given spot (x,y,z) is [tex]\psi^2(1,0,0)\psi^2(x,y,z)[/tex].
    But - if electron 1 is at [tex]\mathbf{r} = (0,1,0)[/tex] from the nucleus, the probability density of electron 2 in a given spot (x,y,z) is [tex]\psi^2(0,1,0)\psi^2(x,y,z)[/tex].

    This is the same function! Since [tex]\psi^2(1,0,0) = \psi^2(0,1,0) = constant[/tex], because the wave function is spherically symmetrical. (But regardless of the exact wave function is, there will be some contour where the wave function has the same constant value, and hence multiple positions of electron 1 which will give the same probability-distribution for electron 2.)

    This is an obviously wrong consequence no matter what interpretation you use. If electron 1 is one one side of the nucleus, electron 2 should have a higher probability of being on the opposite side. But if you assume, as most do, that the kinetic-energy term of the Hamiltonian really does represent kinetic energy, the physical interpretation is straightforward: By assuming that the electrons interact merely as static "density clouds", you neglect the interdependence of the electronic kinetic energies. As a result of this, the electrons cannot 'avoid' each other by adapting their instantaneous motion to that of the other electron. The result is an uncorrelated wave function, and an overestimation of the total electronic energy as a result. This is dubbed the 'correlation energy', and determining it is the central problem of the quantum mechanics of atomic and molecular systems. Even though the motion is by no means classical, the correlation is directly analogous (in this interpretation) to correlation within the classical many-body problem.

    So even though electronic motion is not classical, even though the resulting probability density is stationary, there is a definite and measurable effect on their energy (~5%) due to correlation, and correlation is difficult to rationalize in any other terms than as an effect due to the dynamics of electronic motion. Now for an alternate derivation, that assumes nothing about the wave function, the electronic Hamiltonian for Helium is:
    [tex]\hat{H}_e = -\frac{1}{2}(\nabla^2_1 + \nabla^2_2) - \frac{Z}{|\mathbf{r_1}|} - \frac{Z}{|\mathbf{r_2}|} + \frac{1}{|\mathbf{r_{12}}|}[/tex]

    This is the form one usually sees, and it's deceptively simple, because it mixes coordinate systems by having a Cartesian kinetic-energy operator but spherical potential-energy operators. Now substitute Cartesian coordinates for the vectors [tex]r_1, r_2, r_{12}[/tex] in the kinetic energy operator. After multiple applications of the chain rule, some tedious algebra, and some nice cancellations, you get:

    [tex]\nabla_1^2 + \nabla_2^2 = \frac{\partial^2}{\partial r_1^2} + \frac{2}{r_1}\frac{\partial}{\partial r_1} + 2\hat{r_1}\cdot\hat{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} + \frac{\partial^2}{\partial r_2^2} + \frac{2}{r_2}\frac{\partial}{\partial r_2} + 2\hat{r_2}\cdot\hat{r_{12}}\frac{\partial^2}{\partial r_2 \partial r_{12}}
    +\frac{4}{r_{12}}\frac{\partial}{\partial r_{12}} + \frac{\partial^2}{\partial r_{12}^2}[/tex]

    Believe it or not, but this is a simplification. We've gone from six coordinates to five - the scalar lengths of [tex]r_1, r_2, r_{12}[/tex] and the r1-r12 and r2-r12 angles (represented in the dot products). It's a less compact form, for sure, but it illustrates how totally interdependent the kinetic energies and coordinates of the two electrons are. So I've shown that the potential between the two electrons is dependent on [tex]r_{12}[/tex], and that the kinetic energy is indeed independent on [tex]r_{12}[/tex] as well, then. Hence, the kinetic energies of the two electrons is totally interdependent, which is hard to rationalize in other terms than that they're 'moving' and in that motion, 'avoiding' each other.

    In short: In quantum-mechanics, the time-independence of a state does not mean the particles are not exhibiting the usual many-body dynamics of motion.
     
    Last edited: Sep 29, 2010
  7. Sep 30, 2010 #6
    I was thinking that the energy level is determined by how many waves fit to make a standing wave. Whether the resulting energy is purely kinetic, or it banks some as added mass to approach the speed of light, doesn't change the energy level that I started from.
     
  8. Oct 1, 2010 #7

    alxm

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    Could you elaborate on that, because I really have no idea what you mean?
     
  9. Oct 4, 2010 #8
    Don't worry about it.
     
  10. Oct 4, 2010 #9

    alxm

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    I'm not worried, I'm just trying to find out the source of your confusion.

    Maybe you're thinking about the Bohr model? That's the only context I can think of where they rationalize electronic energy as standing-waves that 'fit'. But the Bohr model does have both kinetic and potential energy terms as well.

    The Bohr-model is non-relativistic, as is the Schrödinger equation. So of course the energy would not change because of SR, then. But both have relativistic versions, the relativistic Bohr-Sommerfeld model and the Dirac equation, respectively. The former gives the same result as the latter for Hydrogen (being that like the original Bohr model, it only works for Hydrogenic atoms).

    Special-relativistic effects on electronic energy levels have been measured in every element, and were well known even before modern quantum theory; It's not restricted to heavy elements like gold. E.g. the hydrogen the ground state gets shifted by 0.18 meV.
     
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