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Landau quantization and magnetic translation

  1. Feb 15, 2010 #1
    I am trying to learn the integer quantum hall effect and have a pretty straightforward question.

    I understand that the normal translation group does not commute with the Landau Hamiltonian. Does this mean that if you have a state in the lowest Landau level (LLL) and apply the translation operator to it you end up outside of the Hilbert space of the Landau Hamiltonian? If so, is it the magnetic translation group that allows us to translate a particle and stay inside of the Hilbert space since it commutes with the Hamiltonian? Is that why the idea of a projection operator onto the Hilbert space of the LLL is useful? Is the magnetic translation group just the normal translation group with the projection operator applied to it?

    These questions are probably really simple but I am still in undergraduate quantum mechanics so I haven't learned a lot of these ideas in a formal setting yet and I just want to make sure I have the details and motivations correct.
     
  2. jcsd
  3. Feb 16, 2010 #2

    DrDu

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    I think the Landau hamiltonian actually is translation invariant, but the representation of the translation group that it belongs to is a projective one,and not a regular one that is, in this representation the product of two translation operators is equal to the translation operator for the combined translation up to a non-trivial phase factor. This phase corresponds to a change of gauge whence it is not directly observable. As a side effect, the generators of the translations do not commute any more which is the reason why interesting new effects arise.
    Projective representations arise quite often in QM, the phase factors not being of prime importance as the wavefunctions themselves are only defined up to a phase.
    So e.g. the representations of the rotation group generated by a spin 1/2 particle are projective ones ( e.g. two rotations by 180 degrees yield the identity up to a phase factor -1) and generally the representation of the Galilei group in non-relativistic QM is a projective one, with the phase related to the mass of the particle in this case.
     
  4. Feb 22, 2010 #3
    OK, I had another idea now that I found out my last one was faulty.

    Given a wavepacket A in the lowest Landau level (LLL), we apply a magnetic translation T to it, resulting in a wavepacket A'. The changes the phase of the wavepacket and, of course, translates the wavepacket. However, we have run into the issue that it is no longer clear what eigenstates A' is composed of. So, we project it onto the LLL, which doesn't change A' at all, but merely retrieves the coefficients associated with each eigenstate in the LLL.

    Is that the right idea?
     
  5. Feb 24, 2010 #4
    Ok, for posterity's sake for anyone searching the forum in the future, I asked my advisor and he said that it is often useful to project the Hamiltonian onto a subspace to deal with an effective Hamiltonian that keeps the spectral information of the "true" Hamiltonian but acts only on the subspace and that a lot of arguments that physicists cannot prove rigorously is because they neglect to do this, and it is in this arena that a lot of things become much easier to prove.
     
  6. Feb 25, 2010 #5

    DrDu

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    I don't understand your comment, but maybe we speak of different things. Under the Landau hamiltonian I understand the Hamiltonian for a single particle, or a set of non-interacting Fermions, in a constant magnetic field. You are asking about the symmetry properties of this hamiltonian, which is a mathematically well defined question and which has a rather simple answer in this case. Whether it is an effective Hamiltonian or not seems to be irrelevant at this point.
     
  7. Feb 25, 2010 #6

    turin

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    No, for two reasons.

    The LLL is only a subspace of the Hilbert space anyway. So, even if the translation does not commute with the Hamiltonian (and thus would produce some projection above the LLL), the translated state would still be in the Hilbert space; it would just then be a superposition of LLL and non-LLL states.

    However, I don't think that a translation will change the Landau level, will it? (Shifting the location of the minimum of the harmonic oscillator potential does not change the oscillation frequency.) So, I think that a translation preserves the LLL.

    BTW, you seem to be confusing the Hamiltonian with it's lowest eigenvalue (or the ground state). Also, I'm not sure that "Hilbert space of the Landau Hamiltonian" is a meaningful concept. There is some Hilbert space on which the Hamiltonian acts, but (especially since there is a possibility of degeneracy in the Landau levels) the Hilbert space is not necessarily completely defined by the (spectrum of the) Hamiltonian. I suppose it would actually be more meaningful to say "Hilbert space of the LLL", where the basis is given in terms of some combination of position and momentum states, the convenience of which would depend on your gauge choice.
     
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