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Two questions on solid state physics

  1. Sep 22, 2009 #1
    Hi all.

    I have two questions, which I hope you can help me with:

    1) In my book (Solid State Physics by Ashcroft and Mermin) it says on page 436 that:

    "When the wavevector k is equal to pi/a (where a is the lattice constant), then the motion in neighboring cells are 180 degrees out of phase. "

    This is when the basis consists of two ions. I cannot see what the argument is for this statement?

    2) At the boundary of the Brilliouin zone (i.e. k = pi/a), the group velocity is zero and the vibrational wave creates a standing wave. Here the wavelength is 2a, and with a Bragg angle of pi/2, the Bragg condition is satisfied:

    \theta = \frac{\pi}{2}, \,\,\,\, d = a, \,\,\,\, k = 2\pi/\lambda, \,\,\,\, n=1, \,\,\,\, \lambda = 2a \,\,\,\,\,\, \text{from}\,\,\,\,\,\, 2d\sin \theta = n\lambda

    But what does this mean? That there are standing waves being reflected between the ends of the Brillouin zone? If yes, then what is does this physically mean?
  2. jcsd
  3. Sep 22, 2009 #2
    Hi Niles.

    I don't have a copy of that text, but I assume we are talking lattice vibrations.


    [tex] k = \frac{\pi}{a},\qquad \lambda = \frac{2\pi}{\pi/a} = 2a[/tex]

    So the wavelength spans 2 unit cells and therefore the amplitude is always positive in cell 1 and negative in cell 2 (positive in cell 3, negative in cell 4 etc.), so the motion of atoms in neighbouring cells will be 180 degrees out of phase.

    Does that make sense?

    As for part 2, I'm not sure. But from above, if you consider this situation the modes with wavevector [tex]k = \pi/a[/tex] have nodes that fall on atomic sites that are equivilant by translational symmetry. So this wavevector corresponds to stationary points, for example, at (0,0,0) and (0,0,a).

    I believe that a standing wave requires the nodes be physical points, which for a crystal lattice are lattice sites, and so only this wavevector can be considered like that? (and maybe wavelengths of 4a 6a 8a etc. because odd numbers would have a "node" at a/2 which is not a lattice site, and any other value would fall at some other fraction).

    I am doing solid state physics as well, so my disclaimer is that this might not be correct at all. What do you think?
  4. Sep 23, 2009 #3
    #1) Yes, that does make sense. I don't know why I didn't see that.

    #2) I'm not quite sure myself. I just can't see what the physical significance is, when we say that the group velocity is zero. I.e., no energy is transferred between the boundaries, but again - so what?

    Which book are you using?
  5. Sep 23, 2009 #4
    Hi Niles.

    I think my comments for #2 were wrong or at least irrelevant ;)

    I am unsure of the site policy, but I googled it and got the result I linked to below. It seems to discuss it in the framework of balancing propogating waves leading to a standing wave (and not doing so away from the boundary) and address the zero group velocity.

    http://books.google.com.au/books?id...&q=brillouin zone group velocity zero&f=false

    I am doing my thesis in condensed matter without much background so I work from Kittel where I have the time.
  6. Sep 23, 2009 #5
    Yes, that link is good.

    I have read up to chapter 7 in Kittel, and I find the book to be good, but it does not explain everything in great detail. It seems that the reader has to tie many ends together.

    Do you know why waves are only reflected at the boundary of the Brillouin zone?
    Last edited: Sep 23, 2009
  7. Sep 23, 2009 #6
    The wavelength at this point satisfies Bragg's law at these values of k.
  8. Sep 23, 2009 #7
    Yes, exactly as I wrote in my very first post, and the angle here is pi/2, i.e. 90 degrees. But this implies that reflection always happens with 90 degrees and a wavevector which is pi/a, which surely is not the case. Where am I wrong?

    Thanks for helping.
  9. Sep 23, 2009 #8
    Why is it surely not the case?
  10. Sep 23, 2009 #9
    When light is reflected from a crystal plane, then it doesn't always reflect at 90 degrees.
  11. Sep 23, 2009 #10
    From what I understand, the two processes are separate completely. In one, the effect is bragg reflection, where the reflected wave from one plane interferes with the reflected wave from the next plane and causes an interference pattern. For this to be the case, the path difference must meet certain requirements (2d=nsin(theta)). In the case you mention, the wavelength is not of the correct size, so no interference occurs and there is no interference pattern formed.

    Within the lattice, only certain wavelengths meet the bragg requirements causing interference (of the electron with itself). This is why X-rays are used to measure inter-atomic spacing (they have the right wavelength for interference to occur).
  12. Sep 23, 2009 #11
    But it is of the right wavelength (see first post).

    Ok, this I need to understand. So when the electron in the lattice gets hit by the external light (i.e. external electric field), the elctron creates a standing wave at the zone boundary of the Brillouin zone. Then what happens? How does this standing wave get outside the lattice?
  13. Sep 23, 2009 #12
    I don't have the version of ashcroft and mermin that you do (mine's a translated version, so I can't check out the page number). I had kind of assumed that we were talking about electrons, hence my last post. It now looks like we're talking about phonons. Could you point out which part of the book you're at (chapter and section).

    If we're talking about phonons however, then they have a wavenumber too. When the wavenumber equals pi/a, then it will interfere with the reflected wave.

    I think I'm in danger of confusing you with the external light and electrons, so try to forget anything about that.

    The standing wave is set up by the superposition of the incident wave and the reflected wave. e.g. e^ikx + e^-ikx is proportional to cosine (a standing wave).

    "That there are standing waves being reflected between the ends of the Brillouin zone?"

    I don't understand what you mean by this. It sounds like you're confusing reciprocal space with real space. The Brillouin zone exists in reciprocal space, so the standing waves can't exist within it. The formation of standing waves means that [whatever it is that is reflected due to Bragg's law, eg electrons, phonons, etc] cannot propagate through the lattice and are effectively bound to the lattice. This explains why they gather a large effective mass.

    Does this make sense?
  14. Sep 23, 2009 #13
    With this question I am using Kittels book. But in Kittels book, it is shown that for crystal vibrations (i.e. phonons), then at the edge of the Brillouin zone, then there are standing waves. After this Kittel continues:

    "This situation is equivalent to Bragg reflection of X-rays: When the Bragg condition is satisfied a travelling wave cannot propagate in a lattice, but through successive reflections back and forth a standing wave is set up.
    The critical value K = pi/a satisfies the Bragg condition for theta = pi/2, K=2pi/lambda, n=2 and hence lambda = 2a."

    Can you please explain this to me (I think this is the easiest and most efficient way to solve our "problem")? Because this paragraph is my confusion.
  15. Sep 23, 2009 #14
    check out this applet : http://www.walter-fendt.de/ph14e/stwaverefl.htm

    1) Since the x-rays satisfy the bragg condition (at the right frequency), they are reflected.

    2) The superposition of the reflected part with the original (incoming wave) creates a standing wave and nothing propagates.

    If the frequency does not satisfy the Bragg condition then there is little or no reflection, so the wave passes unhindered through the lattice.
  16. Sep 23, 2009 #15
    First, a very nice applet. That made it visually clear.

    1) Ok, so when the X-rays satisfy the Bragg condition, a standing wave is created. If we refer to the applet: Is it the standing wave that is reflected in the lattice, i.e. when the X-rays scatter of the lattice planes, it is the standing wave we see?

    2) Look at this sentence again: "The critical value K = pi/a satisfies the Bragg condition for theta = pi/2, K=2pi/lambda, n=2 and hence lambda = 2a."

    Does this refer to phonons (i.e. crystal vibrations) or X-rays/electrons/neutrons etc..?
  17. Sep 24, 2009 #16
    The standing wave is made by the sum of the two waves. In the applet, the red wave is the incident one, the blue the reflected one and the black wave is the sum of the two and is the standing wave. As you can see, it doesn't travel like the other two waves. We obviously can only see the sum of the two waves, being unable to differentiate between which part is incident and which is reflected.

    In answer to the second question, I'm pretty sure it applies to both phonons and X-rays and electrons and neutrons, when they have the right wavelength. I'm not 100%, but 95%, so it'd be nice of someone else, who perhaps knows a little more than I, to just second this point.

    Just as a slight aside, it is the spacing between the lattice planes which causes the reflection. Increasing the distance increases the wavelength which is required. If you have studied thin films, then this can be used for anti-reflective coatings. It is the reason that puddles of water covered in a thin coat of oil reflect different colours. Essentially, the wavelength of certain light satisfies bragg's law for the depth of the oil on the water and so this wavelength is reflected (which wavelength is angle dependant [as it is in the solids we're talking about, so the situation is slightly complicated in 3D], so if you see an oil puddle and move your head around, the colours will change). So, if you could get layers spaced with the right width, you could theoretically reflect different colours of light at different viewing angles. This is essentially how some photonic crystals work.

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