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

I Cyclotron resonance and Landau Levels

  1. Aug 20, 2017 #1
    I'm trying to understand cyclotron resonance measurements of electron effective mass in intrinsic silicon. I need to understand the theory used to make the computations of effective mass in non-parabolic bands.

    A basic introduction to the cyclotron technique is here, but only for parabolic bands:
    http://large.stanford.edu/courses/2007/ap273/hellstrom1/

    In non-parabolic bands, carriers with increasing energy levels have increasing effective (classical) masses M; but I'm having a difficult time isolating exactly which carriers at what energies affect the cyclotron measurement of mass, and how.

    I am using this theory (K*p theory): https://arxiv.org/pdf/1701.07067.pdf

    See p.6 "At low temperatures the mass is usually measured at the Fermi energy within the band. As a consequence, one often plots the mass as a function of the electron density which determines the Fermi energy."

    Fermi Energy (or Fermi Level?) has two meanings that I know of; it's the energy of the topmost electrons in a densely filled band (cold electron gas). It's also the energy that marks the 50% probability of being occupied. In an intrinsic semiconductor (which is what I'm studying for silicon), Fermi level would be very near the middle of the bandgap; There are no carriers at that energy level in an E-k diagram ... therefore I'm not sure what the author means by saying that the mass is 'usually' measured at that level of energy.

    However, in many cyclotron experiments, the discussion follows a different tack:

    A.K.Walton, 1970, p.1415, journal of solid state physics, talking about Stradling and Ukhov's 1966 Cyclotron experiment:
    Which isn't very helpful, because I don't know what kind of "average" he is talking about and he doesn't cite a source that I have access to.

    A slightly less baffling comment is made in "Cyclotron resonance measurements of the non-parabolicity of the conduction bands in silicon and germanium", 1976, J.C.Ousset et al.
    Both comments seem to mean that cyclotrons measure the mass of the particular carrier having an E=kT from the conduction band's bottom, although I don't know what's special about that energy level. When I use generic density of states calculations, I come up with an estimate of kT/2 being the energy level that is occupied by the most electrons ; I naively expected that the mass of the most common carrier would dominate the results; but it clearly doesn't and I don't know why.

    Ousset used a 337μ exciting wavelength, so that's a frequency of 889.6 GHz, Therefore: ħω=hf ≅ 3.68 m_eV
    kT=3.68meV corresponds to a temperature T=42.7 Kelvin. The data measures mass from T=40K to 240K, but n~=1 at 40 Kelvin and only reaches n~=5 at 240 Kelvin. Ousset doesn't even seem to care that n isn't >> 1 in much of his data.

    How does anyone derive the result that peak absorption corresponds to the mass of a carrier with average thermal energy kT? What's the connection between Landau levels and a Fermi-Dirac (or Boltzmann approximation) carrier distribution?
     
    Last edited: Aug 20, 2017
  2. jcsd
  3. Aug 22, 2017 #2

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    Let me read the paper you quoted and get back to you, I did [itex]\vec{k}\cdot\vec{p}[/itex] years ago and worked thru the silicon theory while doing my dissertation.
     
  4. Aug 22, 2017 #3
    I appreciate the help. I suspect that the strong magnetic field of a cyclotron must be changing the fermi energy & level because it is uniformly warping the bands/molecular orbitals of the crystal in some way. eg: The idea's shown schematically on page 11 of the ARXIV article.

    But the energy levels in a semiconductor are very close together; so that if the schematic was accurate, it would hardly affect the conduction band probability distribution at all to shift electron energy states by half the difference of a nearest neighbor. The Landau level spacing of an actual cyclotron experiment, being in the milli-electron volt region, spans the energy space of a large number of undisturbed/nonmagnetic allowed energy states.
    Conceptually, I think the large number of allowed conduction band energy states in undisturbed silicon must be compressing into a lesser number of Landau levels. I can't quite picture how that will affect the probability distribution and fermi-energy. :)
     
    Last edited: Aug 22, 2017
  5. Aug 22, 2017 #4

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    if you apply a large enough magnetic field, you quantize the levels so they are not continuous.....
     
  6. Sep 9, 2017 #5
    Yes, and I still don't understand the connection between energy levels of Landau; eg: I don't know the how band theory connects to the lowest landau energy. Sspecifically, I still don't know why the cyclotron primarily measures electrons with an energy of kT rather than 3/2kT, or even 3kT; Those are all common thermal energies of great importance in free electron gasses.

    I am still trying to solve this problem myself. About the only thing I've figured out is that entire bands split into spin up and spin down non-degenerate pairs because of the application of a magnetic field. When I neglect atomic orbital distortions ... which I don't know how to compute anyway; I discover that depending on the spin, each band must spit into two bands having exactly the same density of states vs energy distribution; but halved in quantity everywhere. One half a band is uniformly raised in energy, while the other half is uniformly lowered in energy by the exact same amount.

    A band is raised in energy by a value (delta E) directly computed by the value of the cyclotron B field and the Bohr magneton; The raised energy half band will then loose electrons to the half band that was lowered in energy to fill up lower energy states; eg: as thermal agitation flips spins...

    So, I know that at least delta E's worth of electrons will eventually flip their spins. The reason is that the the density of states is always larger in the half band which containins more carriers. This is true both in undergraduate student's parabolic bands, and graduate student's KP hyperbolic bands.

    Here's a typical diagram from nature magazine of the net result: The topmost energy level for electrons, is equal on both half bands.

    srep19087-f3.jpg

    If the diagram is accurate, then delta E is around 10 milli electron volts for an 8 Telsa field in whatever secmiconductor they chose (but it really shouldn't matter much). It's not a large difference in energy even though the magnetic field is very strong.

    What's interesting, to me, is that they've drawn E_F as a level significantly inside the conduction band. In intrinsic silicon, the fermi level is generally computed to be inside the gap. That's a computation, however, which includes the statistics of electrons from both the conductance and the valence band. It's the average level of all electrons together. In the drawing, the fermi level appears to be drawn as if it was the 50% occupancy point of electrons within a given band. Either that, or the drawing is massively exaggerated for a doped semiconductor.

    Perhaps I didn't understand my undergraduate courses on semiconductor physics very well; or the idea of Fermi_level is somewhat abused in the literature. :D

    I would expect there to be two different E_F's one for electrons, and one for holes for the drawing shown by nature magazine.

    Is it possible that the ARXIV authors, and the early cyclotron experimenters, mean something like "quasi fermi level" with respect to what a cyclotron usually measures?
     
    Last edited: Sep 10, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Cyclotron resonance and Landau Levels
  1. Cyclotron resonance (Replies: 2)

Loading...