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Consider how the topic of scattering by periodic crystal structures is covered in most books:

  1. Start by treating the crystal structure as a periodic potential and use Fermi's Golden Rule to show that scattering amplitude is non-zero for a special condition.

  2. This is the Laue condition which is nothing but a restatement of the conservation of crystal momentum.

  3. A special case of Laue's condition is the Bragg's condition: $$2d\sin\theta=n\lambda$$

And Bragg's condition basically asks us to treat the lattice planes as capable of doing specular reflection and that the reflected rays then interfere with the interference condition for the maxima being the equation I wrote above.

The arguments from 1) to 3) do not present a clear picture or intuition of how the phenomenon is actually happening.

For starters, I would think that when a wave falls on the crystal structure, the atoms interact and act as sources of spherical waves. This itself makes the picture a complicated mess. How then such a simple argument of planes as reflectors account for this complicated mess?

Besides this, why is there a relation between the $G$ of the reciprocal space of the lattice which just has the information of the periodicity of the lattice to the actual periodic wave that hits the crystal?

I am sure that the only physical condition that is responsible for these relations to pop is the periodicity of the crystal structure but it seems magical how from periodicity we reach the simple Bragg's condition.

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    $\begingroup$ I mean, as far as I can tell, the planes-as-reflectors argument that leads directly to the Bragg condition is at most a heuristic (and not a direct physical explanation) that allows us to derive the correct expression (I'm pretty sure that the Bragg condition is equivalent with the Laue condition, not a special case) for the angular positions of interference maxima. On the other hand, I feel like the direct derivation of the Laue condition captures the essence of what's actually going on: incoming light hits atoms with different phases, and outgoing spherical waves interfere. $\endgroup$
    – march
    Apr 4, 2022 at 16:08
  • $\begingroup$ But the fact that outgoing spherical waves interfere in precisely a "special" manner that re-produces the heuristic Bragg's law because of the periodicity of the crystal is something that seems out of the hat to me. $\endgroup$
    – Lost
    Apr 4, 2022 at 16:12
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    $\begingroup$ Not entirely: each reciprocal lattice vector defines a set of lattice planes that are perpendicular to that reciprocal lattice vector. Under the assumption of elastic scattering (ingoing and outgoing frequencies and therefore wavelengths are the same), you can show (visually, geometrically) that if $\vec{k}_{\textrm{out}} - \vec{k}_{\textrm{in}} = \vec{G}$, then the the two $\vec{k}$ vectors make equal angles with the set of lattice planes, which is exactly the Bragg condition. I can write an answer based on this geometric fact, if that's what you're looking for. $\endgroup$
    – march
    Apr 4, 2022 at 16:16
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    $\begingroup$ Yeah, I'm still not sure what you're looking for, either. I usually start from the interference caused by two scattering centers. It's pretty easy to derive the Bragg condition assuming outgoing spherical waves, breaking that down to the normal thing of looking at the "extra distance" traveled by one of the waves. From there, one can say that the condition must hold for every pair of lattice points (because any deviation from that condition creates large changes across the lattice, leading to destructive interference), which results in the Laue condition. $\endgroup$
    – march
    Apr 4, 2022 at 16:59
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    $\begingroup$ Then, finally, one can relate that to the Bragg condition in the way that I outlined above. $\endgroup$
    – march
    Apr 4, 2022 at 16:59

2 Answers 2

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Start by treating the crystal structure as a periodic potential

Indeed, a material without such a structure won't diffract light the same way. But as I'll explain below, there's an intuition for why a simple model works, which may not technically be the same as an intuition to motivate the model, but it still counts as thinking like a physicist.

it seems magical how from periodicity we reach the simple Bragg's condition

Let me offer an analogy. If I solve the driven damped harmonic oscillation $\ddot{x}+\gamma\dot{x}+\omega_0^2x=e^{i\omega t}$ for all $\omega\in\Bbb R$, the solution scales as $e^{i\omega t}$. There is resonance at the $\omega$ that leads to the largest amplitude for complex $x$. Fourier transforms then let me solve for more general driving terms, even real and/or non-periodic ones. What's more, all it took to characterize the response to an arbitrary driving force is parameters $\gamma,\,\omega_0^2$. While damped oscillations are common in e.g. fluid mechanics, these parameters sweep all the complicated physical details under the rug.

Similarly, even without a detailed physical model of a crystal, we know an incoming wave $e^{i(\vec{k}\cdot\vec{x}-\omega t)}$ is simply rescaled, with resonance at suitable four-frequencies, and the result for something more general follows by a Fourier transform. But since it's a periodic crystal, we also know the resonance condition is of the form $k_j\propto n_j$, which fixes some details of Bragg's law. In one dimension, $n\lambda$ is some constant; all that remains is to explain why it's $2d\sin\theta$, or in your case why a very simple model would get it right.

If we do calculations with a complicated model, all physical detail is swept under the rug in the form of a limited number of parameters analogous to $\gamma,\,\omega_0$. Well, it's obvious what the physically relevant parameters here are: $d,\,\theta$. So if a simple tractable model says $n\lambda=2d\sin\theta$, then $n\lambda=2d\sin\theta$.

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  • $\begingroup$ A really nice way to look at it! Here's a follow-up: We say that the incoming wave is a plane wave. It interacts with the atoms which produce many interacting wavelets which recombine to give a reflected wave. Is the reflected wave also a plane wave? (I think so otherwise what's the point of saying a ray comes out at an angle $\theta$). If yes, then one might ask how come the spherical waves interact in the special fashion to give back a reflected plane wave. Is the following a good way to look at it? $\endgroup$
    – Lost
    Apr 6, 2022 at 5:39
  • $\begingroup$ Since the crystal is periodic, the adjacent atoms (sources of spherical waves) are essentially identical and so can just be considered as points sources on a plane wave and we already know that point sources on a plane wave create a plane wave and so done. $\endgroup$
    – Lost
    Apr 6, 2022 at 5:41
  • $\begingroup$ @Lost That's essentially a linearity argument, so a Fourier transform intuition favours it. $\endgroup$
    – J.G.
    Apr 6, 2022 at 8:59
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Derivation of the Laue condition

Here is a direct derivation (adapted from Ashcroft and Mermin) of the Laue condition that avoids the use of Fermi's Golden Rule. Consider the following picture, in which there are only two scattering centers. A plane wave of wavelength $\lambda$ is incident on the two scatterers, and the wave vector $\vec{k}$ (whose magnitude is $2\pi/\lambda$) is as shown.

enter image description here

We consider an outgoing direction $\hat{k}'$ and assume elastic scattering, so that $|\vec{k}| = |\vec{k}'|$. Now, at all points along the horizontal dashed line, the phase of the wave is the same, since this corresponds to one of the "planes" in the plane-wave. On the other hand, if the phases are the same at either end of the other dashed line, than the outgoing rays labeled $\vec{k}'$ will also have the same phase, and so when they meet at infinity, they will constructively interfere and make an interference maximum.

For that to occur, the extra distance traveled by the rays on the right must be an integer multiple of the wavelength larger than the distance traveled by the rays on the right, i.e. $$ n\lambda = d_1+d_2\,. $$ From the two triangles shown, we can see that $$ d_1 = -\hat{k}\cdot\vec{R}\,, $$ and $$ d_2 = \hat{k}'\cdot\vec{R}\,. $$ Thus, the condition for there to be an interference maximum in the direction of $\hat{k}'$ given an incident $\hat{k}$ is $$ n\lambda = d_1+d_2 = \hat{k}'\cdot\vec{R} -\hat{k}\cdot\vec{R}\,. $$ multiplying both sides by the magnitude $2\pi/\lambda$ of the two wave vectors, this becomes $$ 2\pi n = \vec{k}'\cdot\vec{R} -\vec{k}\cdot\vec{R} = \left(\vec{k}' -\vec{k}\right)\cdot\vec{R}\,, $$ and we can re-write this as $$ 1 = e^{i2\pi n} = e^{i\left(\vec{k}' -\vec{k}\right)\cdot\vec{R}}\,. $$ Finally, in order for there to actually be an interference maximum in this direction, this statement must be true for every pair of scatterers in the solid, because any slight deviation from this condition for one pair of scatterers becomes a large deviation from the condition for other pairs of scatterers, and in summing the contributions form all pairs, you necessarily get destructive interference. The last piece of the puzzle is to note that the separation vector $\vec{R}$ between any two lattice points is in fact a vector in the direct lattice, and so the condition above must be true for every direct lattice vector $\vec{R}$. By definition of the reciprocal lattice, this means that $\vec{k}'-\vec{k}$ is a reciprocal lattie vector.

Equivalence with the Bragg condition

The reciprocal lattice vector $\vec{G}$, given by $$ \vec{G} = \vec{k}'-\vec{k}\,, $$ "generates" a set of lattice planes perpendicular to $\vec{G}$, where the spacing is $d=2\pi/|\vec{G}|$. Using the following picture,

enter image description here

we can see that taking the blue vector to be $\vec{G}$, we are forced to draw $\vec{k}$ and $\vec{k}'$ as making equal angles with the lattice planes due to both the Laue condition and the elastic scattering condition ($k=k'$). This means that $$ \hat{G}\cdot\vec{k}' = k\sin\theta = -\hat{G}\cdot\vec{k}\,. $$ Dotting the Laue condition by $\hat{G}$, then, and using $k=k'=2\pi/\lambda$, we get $$ \frac{2\pi}{d} = \hat{G}\cdot\vec{G} = \hat{G}\cdot\left(\vec{k}'-\vec{k}\right) =k\sin\theta - (-k\sin\theta) =2\frac{2\pi}{\lambda}\sin\theta\,. $$ Rearranging, we get exactly $\lambda = 2d\sin\theta$ ($n\lambda$ corresponds to multiples $n\vec{G}$), which the Bragg condition.

A simplified picture of the scattered spherical waves

The derivation of the Laue condition above is heuristically justified by the following picture:

enter image description here

The plane wave reaches the two scattering centers at different "times", and so the phase of the two scattered wave is slightly different leading to the sequence of wave fronts (maxima) emanating from the scatterers as shown. There is a contour of maximally constructive interference along the line joining the points at which the wave fronts meet. If we imagine the two blue rays as "parts" of the waves emanating from the scatterers that intersect the red ray very far away, then these two rays are approximately parallel, and we can treat them as such (which is what we've done above). If you draw a line perpendicular to the three rays, you will see that the where this line intersects the blue rays, the blue rays are (approximately) in phase with each other. This approximation gets better farther away from the scattering centers; in fact, by the edge of the drawing shown, the approximation is already very good. This is the justification for using the condition $n\lambda = d_1+d_2$ above.

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