Introduction#
Solid-state physics is the branch of condensed matter physics that investigates how macroscopic properties of solid materials like conductivity, magnetism, optical, thermal, ect. emerge from the microscopic quantum-mechanical interactions of insanely large numbers of atoms. Because analyzing each individual atom is impossible to do classically, it is more convenient to represent these groups of atoms as systems where these atoms pack densely together, typically organizing into regular, repeating periodic structures that we can analyze more easily. Semiconductors are a subset of the broader Solid-state physics pie (groups II through VI on the periodic table) that all can be combined in novel ways to create specific crystal lattices that have favorable properties, allowing us to construct useful electronic devices (e.g. diodes and transistors).
Crystal Lattices and the Unit Cell#
The smallest geometric volume that can effectively “tile” the entire volume of the lattice withoug any voids or overlaps is called a Unit Cell. You can think of it as being conceptually analogous to how photons are the smallest unit (quanta) of light in a very loose sense; the unit cell is the smallest fundamental building block of our lattice. Depending on how the unit cell is bounded relative to the lattice points, it is categorized as either a Primitive Unit Cell, which is the minimum-volume cell containing exactly one lattice point per cell, or the Conventional (Non-Primitive) Unit Cell, which is a larger volume chosen to simplify calculations accounting for more lattice points to leverage highly-symmetric systems. The two unit cells are illustrated below:
Insert graphic of both unit cells
Semiconductor physics uses both cells to address different classes of problems; for wafer level ASIC and layout macroscopic problems, the Conventional Unit Cell is typically used, whereas for microscopic quantum simulations of semiconductor lattices (using Quantum ESPRESSO for example), the Primitive Unit Cell yields much higher computational efficiency.
To see how these spatial frameworks translate into Fourier space, we can anchor our analysis in the conventional cubic cell. By utilizing its orthogonal Cartesian axes, we can systematically derive how the Face-Centered Cubic (FCC) real-space lattice transforms into its dual—the Body-Centered Cubic (BCC) reciprocal lattice.
Relationship of Face-Centered Cubic (FCC) and Body-Centered Cubic (BCC)#
We construct our unit cell (conventional) as shown below:
Insert graphic for unit cell cube showing vectors
Real-space lattice vectors $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$ give reciprocal lattice vectors defined by:
$$ \hat{\mathbf{b}_1} = 2\pi\,\frac{\hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3}}{\hat{\mathbf{a}_1}\bullet(\hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3})} \tag{1.1} $$$$ \hat{\mathbf{b}_2} = 2\pi\,\frac{\hat{\mathbf{a}_3}\times\hat{\mathbf{a}_1}}{\hat{\mathbf{a}_1}\bullet(\hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3})} \tag{1.2} $$$$ \hat{\mathbf{b}_3} = 2\pi\,\frac{\hat{\mathbf{a}_1}\times\hat{\mathbf{a}_2}}{\hat{\mathbf{a}_1}\bullet(\hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3})} \tag{1.3} $$These satisfy
$$\mathbf{a}_i\bullet\mathbf{b}_j = 2\pi \delta_{ij}$$by construction.
Primitive Vectors of FCC#
For an FCC lattice with conventional cubic lattice constant $a$, the primitive (rhombohedral) unit vectors — each pointing from a cube corner to a face center — are:
$$ \mathbf{a}_1 = \frac{a}{2}(\hat{y}+\hat{z}) \tag{1.4} $$$$ \mathbf{a}_2 = \frac{a}{2}(\hat{z}+\hat{x}) \tag{1.5} $$$$ \mathbf{a}_3 = \frac{a}{2}(\hat{x}+\hat{y}) \tag{1.6} $$Plug these values into the expressions for each reciprocal lattice vector ($\hat{\mathbf{b}_1}$, $\hat{\mathbf{b}_2}$, and $\hat{\mathbf{b}_3}$) and solve
$$ \hat{\mathbf{b}_1} = 2\pi \frac{\frac{a_0}{2}(\hat{y} + \hat{z})\times\frac{a_0}{2}(\hat{z} + \hat{x})}{\frac{a_0}{2}(\hat{x} + \hat{y})\bullet\left[\frac{a_0}{2}(\hat{y} + \hat{z})\times\frac{a_0}{2}(\hat{z} + \hat{x})\right]} $$Insert XYZ cartesian coordinate graph to illustrate each vector and visual RHR
We could use the standard equation for the cross product here ($AB\sin{\theta}$), but the visual intuition is important, so we will do it the long way:
Back out $\hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3}$:
$$ \hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3} = \left(\frac{a_0}{2}\right)(\hat{y}+\hat{z})\times\left(\frac{a_0}{2}\right)(\hat{z}+\hat{x}) $$$$ = \left(\frac{a_0}{2}\right)^2\big[\hat{y}\times\hat{z} + \hat{y}\times\hat{x} + \cancel{\hat{z}\times\hat{z}}^{\to 0} + \hat{z}\times\hat{x}\big] $$$$ \Rightarrow \quad \boxed{\hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3} = \left(\frac{a_0^2}{4}\right)\big[\hat{x} + \hat{y} - \hat{z} \big]} \tag{1.7} $$Now find $\hat{\mathbf{a}_1}\bullet \hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3}$:
$$ \hat{\mathbf{a}_1}\bullet \hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3} = \left(\frac{a_0}{2}\right)\big[\hat{x} + \hat{y}\big] \bullet \left(\frac{a_0^2}{4}\right)\big[\hat{x} + \hat{y} - \hat{z} \big] $$$$ = \left(\frac{a_0^3}{8}\right)\big[\cancel{\hat{x}\bullet\hat{x}}^{\to 1} + \cancel{\hat{x}\bullet\hat{y}}^{\to 0} + \cancel{\hat{x}\bullet-\hat{z}}^{\to 0} + \cancel{\hat{y}\bullet\hat{x}}^{\to 0} + \cancel{\hat{y}\bullet\hat{y}}^{\to 1} + \cancel{\hat{y}\bullet-\hat{z}}^{\to 0}\big] $$For intuition, the dot product $AB\cos{90^\circ} = 0$
$$ = \left(\frac{a_0^3}{8}\right)\big[1 + 0 + 0 + 0 + 1 + 0 \big] = \left(\frac{2a_0^3}{8}\right) $$$$ \Rightarrow \quad \boxed{\hat{\mathbf{a}_1}\bullet \hat{\mathbf{a}_2}\times\hat{\mathbf{a}_3} = \left(\frac{a_0^3}{4}\right)} \tag{1.8} $$We can now plug in equations $1.7$ and $1.8$ into $1.1$ and solve:
$$ \hat{\mathbf{b}_1} = 2\pi\frac{\cancel{\frac{a_0^2}{4}}(\hat{x}+\hat{y}-\hat{z})}{\cancel{\frac{a_0^3}{4}}^{\to a_0}} $$$$ \Rightarrow \quad \boxed{\hat{\mathbf{b}_1} = \frac{2\pi}{a_0}(\hat{x}+\hat{y}-\hat{z})} \tag{1.9} $$Repeat the process to back out $\hat{\mathbf{b}_2}$ and $\hat{\mathbf{b}_3}$. Start with $\hat{\mathbf{b}_2}$; first find $\hat{\mathbf{a}_3}\times\hat{\mathbf{a}_1}$
$$ \hat{\mathbf{a}_3}\times\hat{\mathbf{a}_1} = \left(\frac{a_0}{2}\right)(\hat{z}+\hat{x})\times\left(\frac{a_0}{2}\right)(\hat{x}+\hat{y}) $$$$ = \left(\frac{a_0^2}{4}\right)\big[\hat{z}\times\hat{x} + \hat{z}\times\hat{y} + \cancel{\hat{x}\times\hat{x}}^{\to 0} + \hat{x}\times\hat{y}\big] $$$$ \Rightarrow \quad \boxed{\hat{\mathbf{a}_3}\times\hat{\mathbf{a}_1} = \left(\frac{a_0^2}{4}\right)\big[-\hat{x} + \hat{y} + \hat{z} \big]} \tag {1.10} $$Similarly for $\hat{\mathbf{b}_3}$, find $\hat{\mathbf{a}_1}\times\hat{\mathbf{a}_2}$:
$$ \hat{\mathbf{a}_1}\times\hat{\mathbf{a}_2} = \left(\frac{a_0}{2}\right)(\hat{y}+\hat{z})\times\left(\frac{a_0}{2}\right)(\hat{z}+\hat{x}) $$$$ = \left(\frac{a_0^2}{4}\right)\big[\hat{x}\times\hat{y} + \hat{x}\times\hat{z} + \cancel{\hat{y}\times\hat{y}}^{\to 0} + \hat{y}\times\hat{z}\big] $$$$ \Rightarrow \quad \boxed{\hat{\mathbf{a}_1}\times\hat{\mathbf{a}_2} = \left(\frac{a_0^2}{4}\right)\big[\hat{x} - \hat{y} + \hat{z}\big]} \tag{1.11} $$We can now solve for $\hat{\mathbf{b}_2}$ by plugging in equations $1.8$ and $1.10$ into $1.2$:
$$ \hat{\mathbf{b}_2} = 2\pi \frac{\cancel{\frac{a_0^2}{4}}\big[-\hat{x} + \hat{y} + \hat{z} \big]}{\cancel{\frac{a_0^3}{4}}^{\to a_o}} $$$$ \Rightarrow \quad \boxed{\hat{\mathbf{b}_2} = \frac{2\pi}{a_0}(-\hat{x}+\hat{y}+\hat{z})} \tag{1.12} $$And for $\hat{\mathbf{b}_3}$:
$$ \hat{\mathbf{b}_3} = 2\pi \frac{\cancel{\frac{a_0^2}{4}}\big[\hat{x} - \hat{y} + \hat{z} \big]}{\cancel{\frac{a_0^3}{4}}^{\to a_o}} $$$$ \Rightarrow \quad \boxed{\hat{\mathbf{b}_3} = \frac{2\pi}{a_0}(\hat{x}-\hat{y}+\hat{z})} \tag{1.12} $$We now have our 3 reciprocal lattice vectors $\hat{\mathbf{b}_1}$, $\hat{\mathbf{b}_2}$, and $\hat{\mathbf{b}_3}$ that we can represent on the unit cell
Insert Illustration of lattice points
$$ \hat{\mathbf{b}_1} = \frac{2\pi}{a_0}(\hat{x}+\hat{y}-\hat{z}) \quad \Rightarrow \quad \boxed{\frac{2\pi}{a_0}(1, 1, -1)} $$$$ \hat{\mathbf{b}_2} = \frac{2\pi}{a_0}(-\hat{x}+\hat{y}+\hat{z}) \quad \Rightarrow \quad \boxed{\frac{2\pi}{a_0}(-1, 1, 1)} $$$$ \hat{\mathbf{b}_3} = \frac{2\pi}{a_0}(\hat{x}-\hat{y}+\hat{z}) \quad \Rightarrow \quad \boxed{\frac{2\pi}{a_0}(1, -1, 1)} $$
These three vectors are, by inspection, exactly the primitive vectors of a BCC lattice with conventional cubic lattice constant $4\pi/a_0$ — each vector points from a cube corner toward the body center of an adjacent octant, scaled by $2\pi/a_0$.
$\hat{\mathbf{b}_1}, \hat{\mathbf{b}_2}, \hat{\mathbf{b}_3}$ map to points that define the geometry of the reciprocal lattice — they are the basis vectors of the BCC lattice and linearly combine (through integer coefficients) to generate every lattice point of the BCC structure by symmetry.
The triangular plane formed by the vector magnitudes (points) of $\hat{\mathbf{b}_1}$, $\hat{\mathbf{b}_2}$, and $\hat{\mathbf{b}_3}$ reflects the 3-fold rotational symmetry about the directional point (1,1,1), which is the point normal to the triangular plane in the positive direction at the corner of the cube. Because BCC is the reciprocal of FCC, it encodes the full translational periodicity of the FCC structure in Fourier space — each reciprocal lattice vector corresponds to a family of FCC lattice planes, not an individual atom.
The FCC and BCC lattices are Fourier Transforms of one another (more succinctly Fourier Transforms of the lattice’s periodic density function), which is why they are treated as dual-pair structures in solid state physics.
Looking Ahead#
Now that we have our FCC/BCC relationship, we are ready to dive into Brillouin zones and band theory, and will explore Bloch’s theorem to become more comfortable with band gaps.