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Energy gap between energy bands in solid state physics
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[QUOTE="SpinFlop, post: 6048974, member: 650101"] For this simple 1D case, a wave traveling to the right will reflect to the left if it meets Bragg's condition, which corresponds to ##k=\frac{\pi}{a}##. Since Bragg scattering is an elastic process and we are constrained to 1D, the scattered wave will bounce directly backward with ##k' = -\frac{\pi}{a}##. This can also be worked out mathematically by applying your above stated Bragg equation ##\mathbf k' = \mathbf k + \mathbf G##. It is the superposition of the [I]same [/I]electron's initial and scattered wave function that sets up the standing waves. You can generate the standing waves from the initial and final plane wave function ##\exp(\pm i\pi x/a)##. This is how you get the standing waves of form: ##\psi(+) = \exp(i\pi x/a) + \exp(-i\pi x/a) = 2\cos(\pi x/a)## ##\psi(-) = \exp(i\pi x/a) - \exp(-i\pi x/a) = 2i\sin(\pi x/a)## If you want to visualize this more clearly in your mind in terms of classical standing waves, then the Brillouin zone boundary ##k = \frac{\pi}{a}## corresponds to a wave traveling down a string with wavelength ##\lambda = \frac{2\pi}{k} = 2a## where the lattice spacing ##a## is now just the distance between the endpoints of the string. ##\psi(+)## corresponds to both endpoints free (which sets up a node centered between the endpoints) while ##\psi(-)## corresponds to both endpoints fixed (which sets up nodes at the endpoints). Note that one end fixed and one end free is not allowed because this requires meeting the condition that ##a = \frac{n\lambda}{4}## where n is an odd integer. However, since Bragg's condition requires ##\lambda = \frac{2a}{m}## where m is any positive integer, then combining these conditions gives ##n = \frac{2}{m}##. Hence, no odd integer n exists. In higher dimensions nothing much has changed. As an elementary example consider an electron with ##\mathbf k = k_{x} \hat {\mathbf x} + k_{y} \hat {\mathbf y}## that hits a reciprocal lattice vector ##\mathbf G = -\frac{2\pi}{a} \hat {\mathbf x}##. Thus, we have that ##\mathbf k' = \mathbf k + \mathbf G = (k_{x}-\frac{2\pi}{a} ) \hat {\mathbf x} +k_{y}\hat {\mathbf y}##. From the Bragg condition ##k^{2} = |\mathbf k + \mathbf G|^{2}## we get that ##k_{x} = \frac{\pi}{a} ## an so taken together ##\mathbf k = \frac{\pi}{a} \hat {\mathbf x} + k_{y} \hat {\mathbf y}## ##\mathbf k' = -\frac{\pi}{a} \hat {\mathbf x} + k_{y} \hat {\mathbf y}## Note that the since ##\mathbf G## is along the x-direction, the Bragg condition required ##k_{x}## to be a specific value. However, ##k_{y}## is allowed to remain arbitrary. Thus, the diffracted beam in higher dimensions is not free to go anywhere, there is a direct bounce back (like in 1D) for the momentum projection along ##\mathbf G##. The upshot of this is that now when we put together the superposition of our initial and final plane waves we get that ##\psi(+) =\exp(ik_{y}y) \left( \exp(i\pi x/a) + \exp(-i\pi x/a)\right) = 2\exp(ik_{y}y)\cos(\pi x/a)## ##\psi(-) = \exp(ik_{y}y) \left( \exp(i\pi x/a) - \exp(-i\pi x/a)\right) = 2i\exp(ik_{y}y)\sin(\pi x/a)## So we see that an arbitrary phase is added into the standing wave solution. However, the electron density is given by the modulus and so this arbitrary phase disappears. [/QUOTE]
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Energy gap between energy bands in solid state physics
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