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I Electron Spin vs Domain Creation in Permanent Magnets

  1. Dec 20, 2016 #1
    I am having difficulty thinking through how a ferromagnetic substance is permanently magnetized.

    I think I understand that when for example a sample of iron is heated above it curie temperature and then exposed to an external magnetic field , the sample will become permanently magnetized. The process involves the creation of domains that include some finite number of iron atoms, where the the magnetic moment of each atom is aligned to other atoms in the domain. The domains themselves become aligned to the external magnetic field, with parallel moment domains becoming larger and anti parallel smaller. When the sample cools , the domains become pinned into this parallel alignment , and the sample has a permanent magnetic field of its own

    However each atoms magnetic moment is a result of the spin of its unpaired outer shell electron, which can be up or down. When many iron atoms all have their "free unpaired" electrons "up" there is an aggregate external magnetic moment to that sample. This is also true if all the electrons are "down"

    Before the sample above was taken above Curie temperature we would expect the distribution of atoms with "up" and "down" spins to be random , where the ups and downs were adjacent , and the net effect is to produce a neutral external magnetic moment.

    So my question:

    When an external magnetic field is applied to a "hot " sample , and domains are created, are individual atoms "moving" within the sample to "line up" into adjacent groups ( domains) based on their electron spin? Sort of like all red cars going onto the top floor of the parking ramp). Or is the external magnetic field changing the spin of the electrons themselves so that along the lies of force of the external field, all the electrons are turned into "ups"? ( sort of like cars along a line get painted red)
     
  2. jcsd
  3. Dec 20, 2016 #2
    The second. Spins flip much more easily than atoms can move. What happens is that above the Curie temperature thermal effects are prevalent, while below there is some spin-aligning interaction that favours spins to be parallel - usually it's exchange interaction (whose effect is non trivial: it can both cause spins to try to be parallel OR antiparallel, it depends on the material). So when you cool under the Curie temperature spins start aligning themselves: but like all phase transitions this starts from local ordered 'nuclei' - same way as a snowflake crystallises around a grain of dust - and grows from there. And since the quantisation axis for the spins is arbitrary in the absence of a sufficiently strong external magnetic field all the spins inside a domain know is that they need to be parallel - nothing about in which direction. Hence why domains grow with different spin directions and then when they clash they just stop each other's growth. Actual defects in the crystal will probably influence this process of 'magnetic nucleation'.

    If you were to cool the material down under Curie's temperature while immersed in a strong enough magnetic field you'd observe much better alignment - still probably not perfect, there would still be domains, but good enough. And after all for some materials, like iron, even exposure to a strong magnetic field below Curie's temperature is enough to align the domains and cause an overall magnetisation.
     
  4. Dec 23, 2016 #3
    Thank you Gan_Hope for your very cogent explanation.
    I do have two clarifying questions if you have a few minutes

    First,I am interested in hearing more about the process in the exchange interaction. Are we saying here that the external magnetic field is contributing energy that allows an exchange interaction to occur among the electrons in this system ( sample) so that spins flip until we have domains that include only all up or all down ( +1/2 or -1/2) spin states?

    Second , lets take the heat out of the discussion and just limit the analysis to a "cold" sample of iron.
    Does a "raw" and "cold" sample of iron have pre-existing domains that get reordered when a magnetic field is applied, or do domains get created in the presence of the field.

    I suppose I should be more clear about why I am asking. In the end I am trying to understand what is necessary to create a population of all spin up or all spin down electrons, and then what the difference in the quality and nature of the aggregate magnetic field would be. More basically , is there even a magnetic aggregate field produced by a "cloud" of free electrons all with the same spin state?
    Would welcome any reading references, I get a lot to read about EPR and Electron Spectroscopy, but little on free electron analysis other thann from the LASER community

    Cheers and thanks
     
  5. Dec 23, 2016 #4
    Not exactly. The exchange interaction does not need the external magnetic field to exist. What happens goes more like this. Imagine you have a lattice of metallic atoms, each with one unpaired electron in a d orbital. If the atoms are close enough these orbitals are going to overlap. Now, we know from Pauli's principle that the overall wavefunction must change sign upon an exchange of two electrons. But the overall wavefunction can be expressed as the product of the spatial wavefunction (a complex-valued field) and the spin wavefunction (a spinor, which can be represented as a single complex 2-vector for each electron). In the absence of any magnetic interaction (a good approximation in most cases) the spatial wavefunction will have its own form independently of spin. So all that remains to be seen is: what is the behaviour of this spatial wavefunction upon exchange of two neighbouring electrons? If it does not change sign, then the spins need to be antiparallel (because that way the global wavefunction will change sign - thanks to the fact that you just swapped the spins). If it does change sign, then the spins need to be parallel. This second case is what you find in ferromagnetic materials. Simple models like the Ising model and the Heisenberg spin model show you better how spins would order themselves, thermodynamically, in such a system, but these models all take J, the neighbour spin exchange interaction, as a parameter (assuming only first neighbour interactions as a form of approximation). In order to compute J and especially determine its sign, which makes the difference between ferro and antiferromagnetic behaviour, you need to solve the Schroedinger equation and compute the properties of the spatial wavefunction.

    Domains already exist. They merely are disordered because all exchange interaction determines, as said above, is a tendency for spins to be parallel or antiparallel: it says nothing about the direction along which they ought to be aligned, if no external constraints are present. Basically the spin Hamiltonian of the system is

    $$ \mathcal{H} = J\sum_{i,j}\mathbf{S_i}\mathbf{S_j} - \hbar\gamma\sum_{i}\mathbf{S_i}\mathbf{B} $$

    where J is your exchange coupling and the second term is the energy induced by the external field, if present. If ##\mathbf{B}=0## there's no preferential direction for the spins to align into. If you had a sea of free electrons all with aligned spins that would make for quite the magnetic field: but in practice there's no reason why that ought to happen with free electrons, where the tendency will instead be for all electrons to occupy the same states in up-down pairs. Ferromagnetism happens because you have a situation where electrons are localised enough that sharing orbitals is not an option, and the global spatial wavefunction happens to be antisymmetric upon exchange of neighbouring electrons (which I really don't know how to express in a more physical way... I can only think of it as a do-the-numbers-and-see-what-happens kind of behaviour). Approximate treatment usually involves factorising said wavefunction in a product of atomic orbitals localised at various lattice points... it's not optimal but it's as good as it gets without actually using numerical integration and DFT or some other method to solve the system.
     
    Last edited: Dec 23, 2016
  6. Dec 23, 2016 #5
    Hi

    Thanks again

    I need to absorb your comments and do some reading on the topic before I reply to one particularly interesting comment ( free electrons would form pairs with opposing spins) (( but not at first ?))
     
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