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Momentum in two different k-points

  1. Sep 6, 2015 #1
    For two different points [itex] k[/itex] and [itex] k\prime [/itex] in the Brillouin zone, can we have the same momentum? (By momentum I mean the real momentum, not the crystal momentum [itex]\hbar k[/itex] or [itex]\hbar k\prime[/itex]).
     
  2. jcsd
  3. Sep 11, 2015 #2

    DrDu

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    Sure, e.g. phonons all have true momentum=0, independently of the k value. For other particles, like electrons, doesn't have a well defined value in crystals, i.e. the electron is not in a momentum eigenstate. Hence, the momentum distribution of electrons with different k values will in general overlap, so that it is possible to obtain the same momentum when measuring it.
     
  4. Sep 11, 2015 #3
    I think maybe not for Bloch wave states, although I am not sure.
    If there is crystal momentum at some real momentum p would be shown in ##<n, k| p>##
    $$\psi_{nk}(x) \propto e^{ik\cdot x} u_n(x)$$
    $$<n, k| p> \propto \int dx e^{i(p/\hbar - k)x}u_{n}(x)$$
    This can be interpreted as a Fourier transform. Because ##u_n## is periodic it will only have Fourier components at certain frequencies.
    $$p/\hbar - k = G$$
    for any
    $$G = g_1\mathbf{b}_1 + g_2\mathbf{b}_2 + g_3\mathbf{b}_3,\, g_i \in \mathbb{z}$$
    with ##b_i## being reciprocal lattice vectors.
    Note that this does not mean that there is a momentum component at that value of ##k##, it just means there could be. Whether or not it does depends on ##u_n##.
    So, although each ##k## couples to many different values of momentum, it seems those values of momentum are not shared with others in the same B.Z. But you will not get the values of ##p##to overlap for two ##k## in the same BZ. I am trying to illustrate this idea with the following crude picture.
    tV0qByr.png
    You would have to go out of the BZ to get another value of k that corresponded to the same p. Once outside the BZ you could add a reciprocal lattice vector.
    $$p_0/\hbar = k_1 + \underbrace{0}_{G_1} = \underbrace{k_1 + \mathbf{b}_1}_{k_2} - \mathbf{b}_1 = k_2 + G_2 $$
    So
    $$k_2 =k_1 + \mathbf{b}_1$$ would also correspond to the same momentum ##p_0##. This was by using a different value of ##G##, ## G_2 = - \mathbf{b}_1 ##. But by adding ##b_1##, we left the BZ of ##k_1##.
     
    Last edited: Sep 11, 2015
  5. Sep 14, 2015 #4
    Yes, I think so but if we take the expectation value of momentum, namely p=m vgroup , where vgroup is proportional to the slope of energy band, it seems reasonable to have the same momentum at different points of BZ.
     
  6. Sep 16, 2015 #5
    I am confused. When an electron absorb a phonon, it receives both the energy and momentum of that phonon; and a phonon mode is characterized by its energy and momentum (wavevector), and all modes in one branch have their energy-wavevector characteristics face, correct?

    Second, as an electron is not in a momentum eigenstate, it would not be associated with any specific k value. But I agree with that the electron will take a k value once it is measured.
     
  7. Sep 17, 2015 #6

    DrDu

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    You are confusing true momentum p and crystal momentum (or wavevector) k. When an electron absorbs a phonon, the wave vector is conserved up to a lattice vector. True momentum is also conserved, but it is not carried by the phonon, but by the whole crystal.
     
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