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

Polar catastrophe?

  1. Feb 9, 2012 #1

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Hi all,

    As you may know, the interface of LaAlO3 and SrTi03 has received a lot of attention because of the presence of conducting electrons, superconductivity, and ferromagnetism.

    Because LaAl03 is a polar crystal, the polar catastrophe is often used as a first explanation for the presence of excess surface electrons. However, I think the carrier density at the interface, as measured by hall resistance, is much lower than the naive value from the polar catastrophe. Apparently, one possible explanation in this material is that Mott physics may localize the electrons.

    My interest is more general The polar catastrophe appears very generic: for example, it seems to apply to the free surface of LaAlO3 as well, yet apparently this free surface does not conduct.

    Does anyone know if there is a common mechanism or set of mechanisms that prevents one from seeing this surface conductivity in polar crystals in general e.g. disorder or mott physics? Or am I misinformed when I assume that such conducting behavior is uncommon (this is the impression I have from a few talks)?
     
  2. jcsd
  3. Feb 24, 2012 #2
    You should get a polar catastrophe if you cleave NaCl along (111). This is of course difficult to do because the resulting surface is charged, and thus you have to separate the opposite charges which takes a lot of energy.

    Large surface charges do not exist on open surfaces because they get neutralized by ions and charges from the atmosphere (or residual vacuum - you can always find a few electrons). Polar surfaces also tend to be instable and reconstruct.

    Interface conductivity has also been observed in SrTiO3 with an amorphous LaAlO3 layer on top of it. This indicates that the polar catastrophe is just one mechanism contributing to the effect, and probably not even the dominant one.
     
  4. Feb 24, 2012 #3

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    You're a brave soul, M Quack, after more than 700 views finally a reply. Thanks for taking the time.

    I'm happy with the idea that charged surfaces will be quickly neutralized by stray charged particles e.g. in air.

    I hadn't heard about amorphous LaAlO3, that sounds very interesting, is there somewhere I can read more about this?

    One thing I don't understand is what this term "reconstruction" means in this context. For example, suppose there really were very few free ions or electrons in your vacuum chamber, and suppose you did manage to cleave this thing. The system does have to neutralize itself somehow, but I don't see how a purely surface change can achieve this i.e. you need to move charge across the entire sample. Would a potential scenario be that the polar catastrophe does produce these surface charge layers with finite density mobile charges, but these layers then don't conduct for some other reason?
     
  5. Feb 24, 2012 #4
    Maybe reconstruction is not the right word in this context. You are right that you have to add or remove atoms/charges in order to neutralize the surface. There seem to be a few papers on the NaCl(111) surface, btw. It appears that this surface stabilizes by reducing the charge from Na+ to Na0.5+ - at least that's what I understood from glancing at the paper.

    As for the amorphous LaAlO3, I've just seen one talk. Not sure if the results are published. This is not my work, so to give credit where it belongs:

    Abstract title
    Metallic interfaces at amorphous oxide-SrTiO3 heterostructures
    Author
    MSc Kleibeuker, J. E., MESA Institute for Nanotechnology, Enschede, Netherlands (Presenting author)

    http://www.eventure-online.com/eventure/publicAbstractView.do?id=173497&congressId=5283
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook