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Help with Reciprocal Space

  1. Oct 7, 2008 #1
    I am taking Solid State now and using Kittel as the textbook. Needless to say, I don't understand almost anything that's happening.

    I'm still stuck on Reciprocal Space here. If I have a lattice of atoms of spacing X, then in reciprocal space I get something like 2*pi/X spacing. My prof. explained that as being the momentum space since Reciprocal Space is basically a Fourier transform of the lattice.

    That only makes sense if X is a wavelength and therefore 2*pi/X * h-bar = momentum.

    But I don't understand why X would be a wavelength. I guess it would be the maximum wavelength between atoms? And then 2*pi/X * h-bar is the minimum momentum that can get transferred?

    I am really clueless.
  2. jcsd
  3. Oct 10, 2008 #2
    There are different ways to think about reciprocal space. For one, you can think of it as a purely geometrical thing. If you have a lattice with vectors [tex]\mathbf{a}_i[/tex], then the reciprocal lattice is defined by

    [tex]\mathbf{a}_i \cdot \mathbf{g}_j = 2\pi\delta_{ij}[/tex].

    It's a purely geometrical thing, and the above equation defines a lattice with vectors [tex]\mathbf{g}_i[/tex]. In mathematics this would probably be called a dual lattice, but in physics we call it a reciprocal lattice because of the units (g has units of 1 / length).

    Another way reciprocal space shows up is if you look at the solution to the Schroedinger equation for a periodic potential, Bloch's theorem says that the wavefunctions are a product of two periodic functions, and of the form

    [tex]\psi_{kn}(r) = u_{kn}(r) e^{i k \cdot r}[/tex]

    where k is the so-called pseudomomentum vector, which serves as a quantum number. k is restricted to the first Brillouin zone, where [tex]k = l_1 g_1 + l_2 g_2 + l_3 g_3[/tex] where the l's are restricted to the range [-0.5,0.5]. So k is restricted to wavelengths which are longer than a lattice vector. The other function [tex]u_{kn}(r)[/tex] is periodic within the unit cell, so if you expand it in planewaves, all the planewaves would be like [tex]n_1 g_1 + n_2 g_2 + n_3 g_3[/tex] where the n's are integers, thus these wavelengths are all the lattice constants divided by integers.
  4. Oct 10, 2008 #3
    X need not be a wavelength. X is the unit spacing in direct space and 2*pi/X is the unit spacing in inverse (momentum) space. The moentum space will be useful to you in determining certain optoelectronic properties.
  5. Oct 27, 2008 #4
    I am trying to explain your confusion, i believe i am right but if not i hope to be corrected by someone who follows this message. The reciprocal space vectors are very good way to define lattice planes in real space. For a group of lattice planes which are parallel to each other in real space, there are corresponding reciprocal space vectors which are perpendicular to the lattice planes, the shortest of one has length 2*pi/d, where d is the spacing between lattice planes. This comes from the fact that exp(iK*r) should be constant on the lattice planes and should be equal to 1 if r describes Bravis lattice vectors. Now just imagine a incident plane wave 'exp(ik*r)'on set of parallel planes. The only way exp(ikr) can have same value on all the planes is when they are separated by wavelength.
    I hope it makes sense :)

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