Geometric Structure Factor for Monatomic FCC lattice

I am trying to find the geometric structure factor and my work here is clearly wrong. I will put my wrong answer and then I will throw up the link to wikipedia for the correct answer, because I cannot tell the difference.

My attempt: BCC has a four atom basis. If x,y,z vectors are taken to be along the edges of the conventional cube like in the picture below (all credit due to Aschroft and Mermin) : But the basis are located at: Therefore the geometric structure factor is:

$$F_{hkl} = \sum_{j=i}^{N} f_j e^{i \Delta \vec{k} \bullet \vec{r_j} }$$

But because the structure is monatomic, $f_j = f$ for all j. Also, $r_j$ denotes the location of the jth atom in the cell (I.E. $r_1 = \vec{0}, r_2 = \vec{ a_1} ,r_3 = \vec{ a_2}, r_4 = \vec{ a_3}$ in the picture )

$\Delta \vec{k}$ is just some vector that is an element of Reciprocal vector space. That is, $$\Delta \vec{k} = h \vec{b_1} + k \vec{b_2} + l \vec{b_3}$$

but because $\vec{a_i} \bullet \vec{b_j} = 2 \pi \delta_{ij}$ , $F_{hkl}$ reduces to:

$$F_{hkl} = f [ e^0 + e^{i(h \vec{b_1} + k \vec{b_2} + l \vec{b_3}) \bullet \vec{a_1}} + e^{i(h \vec{b_1} + k \vec{b_2} + l \vec{b_3}) \bullet \vec{a_2}} + e^{i(h \vec{b_1} + k \vec{b_2} + l \vec{b_3}) \bullet \vec{a_3}} ] = f [ 1 + e^{i2 \pi h} + e^{i2 \pi k} + e^{i2 \pi l} ] = 4f$$

This answer is incorrect. The correct answer can be seen at this link to Wikipedia, scrolling down to fcc. From what I can tell, they must have defined their reciprocal lattice vector differently from me but I cannot see why.

• The sum (the last one) should be over the basis vectors, not over the lattice vectors! So, for example, $\vec r_2 = a/2 (\vec a_1 + \vec a_2)$ etc. – Lagerbaer Sep 17 '13 at 0:15
• Treat me dumb, I do not follow. – user28823 Sep 17 '13 at 0:17

As you write, the geometric structure factor is $$F_{hkl} = \sum_{j=1}^N f_j e^{i \delta k \cdot \vec r_j}$$

You also correctly state that $\vec{r}_j$ denotes the location of the $j$-th atom in the cell.

Now, you incorrectly say that $\vec{r}_2 = \vec{a}_1$, $\vec{r}_3 = \vec{a}_2$.

But $\vec{a}_1$ doesn't tell you where in the cell the 1st atom is, $\vec{a}_1$ tells you in what direction the next cell starts. The $\vec{a}_i$ are called the lattice vectors and are just there to define the cubic structure. They can be chosen the same for the simple cubic, the bcc and the fcc lattice.

What you need are the basis vectors. In a simple cubic lattice, you only have one basis vectors, $\vec{0}$. In the fcc lattice, we have four atoms per unit cell, and therefore we have four basis vectors, and those are the vectors that you have written down above: $\vec{0}, a/2 (\vec{x}+ \vec{y})$ and so on, where $\vec{x}$ is the unit vector in $x$-direction. In a cubic lattice, we have $\vec{a}_1 = a\vec{x}, \vec{a}_2 = a\vec{y}, \vec{a}_3 = a \vec{z}$, so your basis vectors read $\vec{r}_1 = \vec{0}, \vec{r}_2 = 1/2(\vec{a}_1 + \vec{a_2})$ etc.

Now if you insert those into the exponential and use the fact that $\vec{a}_i \cdot \vec{b}_j = 2\pi\delta_{ij}$ you should get the correct result.

• I understand what, I do not understand why; At least not completely. What is the difference? A vector is a vector is a vector. is it not? It seems all you did was make the transformation $\vec{x} \implies \vec{a_1}$. I know this is not true but I just don't see it. Thank you for your answer though. – user28823 Sep 17 '13 at 0:35
• The $\vec{a}_1$ are the lattice vectors. In a cubic lattice, they are $(1,0,0), (0,1,0)$ and $(0,0,1)$. These vectors define the translational symmetry of the crystal. The vectors $\vec{r}_n$ are the (crystallographic) basis vectors and, for an fcc lattice, are $(0,0,0), (0.5, 0.5, 0), (0.5, 0, 0.5)$ and $(0, 0.5, 0.5)$. The latter ones are the vectors you must insert into the sum! – Lagerbaer Sep 17 '13 at 1:30
• Also, you're mixing up the notation! At some point you call $\vec{a}_n$ the vectors of your four point basis. But for the basis vectors, $\vec{a}_i \cdot \vec{b}_j$ is not $2\pi \delta_{ij}$. – Lagerbaer Sep 17 '13 at 1:35
• If you define the Lattice vectors the way you have (cubic), then you can never get to the points inside the conventional cell. Isn't that the point of using the primitive vectors (and hence the primitive cell) to describe things? you can reach every point IN the lattice, ergo, every point that is a part of the basis? As a consequence, of your notation, why are you allowed to use fractional vectors? What happened to the primitive vectors? The "atomic" vectors of direct space? don't we measure everything with respect to them? This is what I do not understand. – user28823 Sep 17 '13 at 1:46
• But if you have primitive vectors, you have only one basis vector, and that would be $(0,0,0)$. If we define the lattice vectors like I did, then we can indeed not get to the points inside the conventional cell - if we only use the lattice vectors. That's where the basis vectors come in. With my cubic lattice vectors, each (conventional) cell contains 4 atoms, so I need 4 basis vectors. If you use primitive vectors, then the primitive cell contains only one atom, so you only get one basis vector. – Lagerbaer Sep 17 '13 at 5:11

If you look at the example in Aschroft and Mermin for the body centered cubic case (which is on page 105 in the version of the book I have) you'll see that you should use the simple cubic reciprocal vector, not the face-centered cubic reciprocal lattice vector.

Your answer is correct! The answer given in Wikipedia is also correct! The discrepancy in the requirements of (h,k,l) comes in your choice of the reciprocal lattice(RL) basis vectors. In Wikipedia, the RL basis vectors are derived from the non-primitive direct lattice basis vectors a1=ax, a2=ay,a3=az. Thus the (h,k,l) are the coefficients of the corresponding non-primitive RL basis vectors (also in x,y,z). Whereas in your case, the RL basis vectors are derived from the primitive direct lattice basis vectors, and the (h',k',l') are the coefficients of the corresponding primitive RL basis vectors which is different from the previous (h,k,l). When (h',k',l') are with reference to the primitive RL basis vectors, any combination of h'k'l' is allowed to give diffraction. However, when this (h',k',l') is converted into that (h,k,l), we can easily see the (h,k,l) can only be all even or all odd.