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Band diagram in real space vs reciprocal space

  1. Jan 16, 2008 #1

    can anybody rigurously explain the relationship between the band diagram in k-space (I think I understand this one) and the diagram in real space (as is often used to explain the p-n-junction).
  2. jcsd
  3. Jan 18, 2008 #2
    Diagram in real space shows the dependence of band edge E[tex]_{C}[/tex] (r) or E[tex]_{V}[/tex] (r) to the position r inside the system.
    This dependence is related to variation of carriers concentration with the position n(r), p(r); typical of inhomogeneous doped material (Ex: pn junction).
    In this system the total current density is the sum of two term: 1) drift contribute (proportional to external electric field) and 2) diffusional contribute (proportional to the gradient of carriers concentration: "Fick's Law").
    In equilibrium condition free carriers set up themselves in such a way to built up a spatial charge [tex]\rho[/tex][tex]\neq[/tex]0. This charge is due to fixed ionized doping impurities not compensated by free carriers distribution and generates an internal electric field contrasting diffusional current.
    This internal field is described by a potential function related to it by E(r)=-grad[tex]\varphi[/tex](r).
    If you solve Schrodinger's Equation for stationary state of one free particle with this potential term, you will obtain band structure of the inhomogeneous system.
    The solution is very simple if the variations of [tex]\varphi[/tex](r) potential are observable on length higher than primitive cell dimension.
    The eigenvalues are then translated by a quantity dependent of position E(r)=E(0)-e[tex]\varphi[/tex](r); where E(0) is the usual parabolic solution of Schrodinger's equation in a homogeneous system that you can display in k-space.
    In a homogeneous system [tex]\varphi[/tex]=cost and E(r)=E(0) everywhere, the bands are then flat; however the bands are bent. It' s important to note that differences on energy are not affected by the translation and so the energy gap.

    Please correct my grammatical mistakes !!!

  4. Jan 18, 2008 #3
    One could transform the k-space picture into x-space picture, by a Fourier transform, but to my understanding one uses a combined view for the actual picture.
    Basically one assumes the k-space picture at individual small localised cells, but these regions have to be large enough for k-space description to make sense (band diagrams are used for homogeneuos space only)
    A more simplyfied approach is to take the lowest free energy positions in the band picture and say that dispersion E(k) corresponds to spaces for free particles with momentum k.
    That's sort of the cell view described above.
  5. Jan 21, 2008 #4
    Yes, your interpretation - assuming k-space picture at small cells - is what I expected. But I think it is not easy to put this into a consistent mathematical form? I tried to interpred it along the same way as you do in accoustics when going from fourier transform to "windowed fourier transform". This also goes from a "fourier domain" to kind of a "fourier picture attached to each point of time". But I didnĀ“t succeed in adapting this procedure to the band diagrams.
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