# Miller Indices and the case of a cubic crystal

My textbook, Solid-State Physics, Fluidics, and Analytical Techniques in Micro- and Nanotechnology, by Madou, presents the following image and explanation in a section on x-ray diffraction and Laue equations:

... Laue’s equations can also be interpreted as reflection from the $$h,k,l$$ planes. From Figure 2.25 it can be seen that the spacing between the ($$hk$$) planes, and by extension between ($$hkl$$) planes, is given as$$^*$$:

$$d_{hkl} = \frac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_1}{h} = \frac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_2}{k} = \frac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_3}{l} \tag{2.27}$$

$$^*$$ Notice that in the case of a cubic crystal, Equation 2.27 can be simplified to give the distance between planes as in Equation 2.12.

Equation 2.12 referenced above is presented earlier in the textbook as follows:

Adjacent planes ($$hkl$$) in a simple cubic crystal are spaced a distance $$d_{hkl}$$ from each other, with $$d_{hkl}$$ given by:

$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \tag{2.12}$$

where $$a$$ is the lattice constant. Equation 2.12 provides the magnitude of $$d_{hkl}$$ and follows from simple analytic geometry.

The following comment is not clear to me:

$$^*$$ Notice that in the case of a cubic crystal, Equation 2.27 can be simplified to give the distance between planes as in Equation 2.12.

I was wondering if someone would please take the time to explain and show this.

EDIT:

In an attempt to provide further context, I am posting more of the textbook section surrounding this problem.

Constructive interference will occur in a direction such that contributions from each lattice point differ in phase by $$2\pi$$. This is illustrated for the scattering of an incident x-ray beam by a row of identical atoms with lattice spacing $$\mathbf{a}_1$$ in Figure 2.20. The direction of the incident beam is indicated by wave vector $$\mathbf{k}_0$$ or the angle $$\alpha_0$$, and the scattered beam is specified by the direction of $$\mathbf{k}$$ or the angle $$\alpha$$. Because we assume elastic scattering, the two wave vectors $$\mathbf{k}_0$$ and $$\mathbf{k}$$ have the same magnitude, i.e., $$2\pi/\lambda$$ but with differing direction. A plane wave $$e^{i k \cdot r}$$ is constant in a plane perpendicular to $$\mathbf{k}$$ and is periodic parallel to it, with a wavelength $$\lambda = 2\pi/\mathbf{k}$$ (see Appendix 2A). The path difference $$A_1B − A_2C$$ in Figure 2.20 must equal $$e\lambda$$ with $$e = 0, 1, 2, 3, \dots$$. For a fixed incident x-ray with wavelength $$\lambda$$ and direction $$\mathbf{k}$$, and an integer value of $$e$$, there is only one possible scattering angle $$\alpha$$ defining a cone of rays drawn about a line through the lattice points (see Figure 2.20).

Because crystals are periodic in three directions, the Laue equations in 3D are then:

$$\mathbf{a}_1 (\cos\alpha - \cos\alpha_0) = e \lambda$$ $$\mathbf{a}_2 (\cos\beta - \cos\beta_0) = f \lambda \tag{2.21}$$ $$\mathbf{a}_3 (\cos\gamma - \cos\gamma_0) = g \lambda$$

For constructive interference from a three-dimensional lattice to occur, the three equations above must all be satisfied simultaneously, i.e., six angles $$\alpha$$, $$\beta$$, $$\gamma$$, $$\alpha_0$$, $$\beta_0$$, and $$\gamma_0$$; three lattice lengths $$\mathbf{a}_1$$, $$\mathbf{a}_2$$, and $$\mathbf{a}_3$$; and three integers ($$e$$, $$f$$, and $$g$$) are fixed. Multiplying both sides of Equation 2.21 with $$2\pi/\lambda$$ and rewriting the expression in vector notation we obtain:

$$\mathbf{a}_1 \cdot (\mathbf{k} - \mathbf{k}_0) = 2 \pi e$$ $$\mathbf{a}_2 \cdot (\mathbf{k} - \mathbf{k}_0) = 2 \pi f \tag{2.22}$$ $$\mathbf{a}_3 \cdot (\mathbf{k} - \mathbf{k}_0) = 2 \pi g$$

with $$\mathbf{a}_1$$, $$\mathbf{a}_2$$, and $$\mathbf{a}_3$$ being the primitive vectors of the crystal lattice.

If we further define a vector $$\Delta \mathbf{k} = \mathbf{k} − \mathbf{k}_0$$, Equation 2.22 simplifies to

$$\mathbf{a}_1 \cdot \Delta \mathbf{k} = 2 \pi e$$ $$\mathbf{a}_2 \cdot \Delta \mathbf{k} = 2 \pi f \tag{2.23}$$ $$\mathbf{a}_3 \cdot \Delta \mathbf{k} = 2 \pi g$$

Dealing with 12 variables for each reflection simultaneously [six angles ($$\alpha$$, $$\beta$$, $$\gamma$$, $$\alpha_0$$, $$\beta_0$$, and $$\gamma_0$$), three lattice lengths ($$\mathbf{a}_1$$, $$\mathbf{a}_2$$ and $$\mathbf{a}_3$$), and three integers ($$e$$, $$f$$, and $$g$$)] is a handful; this is the main reason why the Laue equations are rarely referred to directly, and a simpler representation is used instead. The reflecting conditions can indeed be described more simply by the Bragg equation.

Further below we will learn that constructive interference of diffracted x-rays will occur provided that the change in wave vector, $$\Delta \mathbf{k} = \mathbf{k} − \mathbf{k}_0$$, is a vector of the reciprocal lattice.

Bragg’s law is equivalent to the Laue equations in one dimension as can be appreciated from an inspection of Figures 2.24 and 2.25, where we use a two-dimensional crystal for simplicity. Suppose that vector $$\Delta \mathbf{k}$$ in Figure 2.24 satisfies the Laue condition; because incident and scattered waves have the same magnitude (elastic scattering), it follows that incoming ($$\mathbf{k}_0$$) and reflected rays ($$\mathbf{k}$$) make the same angle $$\theta$$ with the plane perpendicular to $$\Delta \mathbf{k}$$.

The magnitude of vector $$\Delta \mathbf{k}$$, from Figure 2.24, is then given as:

$$|\Delta \mathbf{k}| = 2\mathbf{k}\sin(\theta)$$

We now derive the relation between the reflecting planes to which $$\Delta \mathbf{k}$$ is normal and the lattice planes with a spacing $$d_{hkl}$$ (see Figure 2.25 and Bragg’s law in Equation 2.20). The normal unit vector $$\hat{\mathbf{n}}_{hk}$$ and the interplanar spacing $$d_{hk}$$ in Figure 2.25 characterize the crystal planes ($$hk$$). From Equation 2.23 we deduce that the direction cosines of $$\Delta \mathbf{k}$$, with respect to the crystallographic axes, are proportional to $$e/a_1$$, $$f/a_2$$, and $$g/a_3$$ or:

$$e/a_1:f/a_2:g/a_3 \tag{2.25}$$

From the definition of the Miller indices, an ($$hkl$$) plane intersects the crystallographic axes at the points $$a_1/h$$, a_2/k, and a_3/l, and the unit vector $$\hat{\mathbf{n}}_{hkl}$$, normal to the ($$hkl$$) plane, has direction cosines proportional to:

$$h/a_1, k/a_2, \text{and} \ l/a_3 \tag{2.26}$$

Comparing Equations 2.25 and 2.26 we see that $$\Delta \mathbf{k}$$ and the unit normal vector $$\hat{\mathbf{n}}_{hkl}$$ have the same directions; all that is required is that $$e = nh$$, $$f = nk$$, and $$g = nl$$, where $$n$$ is a constant. The factor $$n$$ is the largest common factor of the integers $$e$$, $$f$$, and $$g$$ and is itself an integer. From the above, Laue’s equations can also be interpreted as reflection from the $$h,k,l$$ planes. From Figure 2.25 it can be seen that the spacing between the ($$hk$$) planes, and by extension between ($$hkl$$) planes, is given as:

$$d_{hkl} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_1}{h} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_2}{k} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_3}{l} \tag{2.27}$$

• Jan 30 '20 at 13:57

This answer uses vector algebra and the so-called reciprocal lattice to avoid trigonometry. There are probably other, more geometric ways to connect the Laue conditions to the lattice spacing.

$$d_{hkl} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_1}{h} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_2}{k} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_3}{l} \tag{2.27}$$

We use the Laue conditions to define the reciprocal lattice vector $$\mathbf{d}^*_{hkl}$$ and reciprocal lattice unit vectors $$\mathbf{a}^*_1, \mathbf{a}^*_2, \mathbf{a}^*_3$$. The reciprocal lattice point is defined as:

$$\mathbf{d}^*_{hkl} = \dfrac{\hat{\mathbf{n}}_{hkl}}{d_{hkl}},$$ i.e. it is perpendicular to the reflecting planes and has a magnitude of one divided by the lattice spacing. We can rewrite the three Laue conditions as:

$$h = \mathbf{d}^*_{hkl} \cdot \mathbf{a}_1$$ $$k = \mathbf{d}^*_{hkl} \cdot \mathbf{a}_2$$ $$l = \mathbf{d}^*_{hkl} \cdot \mathbf{a}_3$$

and define $$\mathbf{a}^*_1 = \mathbf{d}^*_{100}$$ $$\mathbf{a}^*_2 = \mathbf{d}^*_{010}$$ $$\mathbf{a}^*_3 = \mathbf{d}^*_{001}$$

The reciprocal space unit vectors have magnitudes that are the reciprocals of the magnitude of the corresponding real space unit vectors (dot product equals to one), and are perpendicular to the other two real space unit vectors (dot product is zero), as you can see if you apply the Laue conditions to, say, $$\mathbf{a}^*_1 = \mathbf{d}^*_{100}$$:

$$1 = \mathbf{d}^*_{100} \cdot \mathbf{a}_1$$ $$0 = \mathbf{d}^*_{100} \cdot \mathbf{a}_2$$ $$0 = \mathbf{d}^*_{100} \cdot \mathbf{a}_3$$

You can now express a reciprocal lattice point via the Miller indices and the reciprocal unit vectors:

$$\mathbf{d}^*_{hkl} = h \cdot \mathbf{a}^*_1 + k \cdot \mathbf{a}^*_2 + l \cdot \mathbf{a}^*_3$$

For the cubic case the three reciprocal unit cell vectors are mutually perpendicular just like the real space unit vectors, and they all have a magnitude of 1/a. We can calculate the magnitude of $$\mathbf{d}^*_{hkl}$$ as (using properties of the cartesion coordinate system, or Pythagoras' theorem):

$$|\mathbf{d}^*_{hkl}| = \frac{\sqrt{(h^2 + k^2 + l^2)}}{a}$$

To get $$d_{hkl}$$, you simply have to take the reciprocal.

Disclaimer

I called $$d_{hkl}$$ the lattice spacing. This is correct if $$h, k, l$$ have no common divisor. If they do, it is the lattice spacing divided by that common divisor. For example, the (1,2,3) reflection and the (10,20,30) reflection has the same reflecting plane, but a different diffraction angle.

• Thanks for the answer! So we have that $$|\mathbf{d}^*_{hkl}| = \sqrt{ \left( \dfrac{h^2}{a^2} + \dfrac{k^2}{a^2} + \dfrac{l^2}{a^2} \right) } = \dfrac{\sqrt{(h^2 + k^2 + l^2)}}{a}$$? Jan 22 '20 at 4:20
• @ThePointer Yes Jan 22 '20 at 5:03