# Homework Help: Figuring out Bravais lattice from primitive basis vectors

1. May 23, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

Given that the primitive basis vectors of a lattice are $\mathbf{a} = \frac{a}{2}(\mathbf{i}+\mathbf{j})$, $\mathbf{b} = \frac{a}{2}(\mathbf{j}+\mathbf{k})$, $\mathbf{c} = \frac{a}{2}(\mathbf{k}+\mathbf{i})$, where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the usual three unit vectors along cartesian coordinates, what is the Bravais lattice?

2. Relevant equations

3. The attempt at a solution

There are seven different crystal systems and fourteen different Bravais lattices.

The common length of the primitive unit cell is $a$, so the crystal system is either cubic or trigonal (rhombohedral).

Furthermore, the basis vectors are oriented at $90°$ to each other, so the crystal system must be cubic.

Finally, the Bravais lattice is face-centred cubic, because, if the basis vectors originate from one corner of the primitive unit cell, then they point to lattice sites at the centre of three adjacent (to the corner) faces of the primitive unit cell.

2. May 23, 2015

### unscientific

It's an FCC. Why? (Hint: Draw out the vectors)

1. How many lattice points does FCC have, and what are their locations?
2. Using the info above, write out the lattice vectors.

3. May 23, 2015

### spaghetti3451

In my mind, I translated from the origin (at the corner of one primitive unit cell) by the the lattice vector $\mathbf{R} = n_{1} \mathbf{a} + n_{2} \mathbf{b} + n_{3} \mathbf{c}$, where I used $(n_{1}, n_{2}, n_{3}) = (0,0,1),$ $(n_{1}, n_{2}, n_{3}) = (0,1,0),$ and $(n_{1}, n_{2}, n_{3}) = (1,0,0)$. Each time, I found that the system is invariant under translation.

4. May 23, 2015

### unscientific

You are over-complicating things.

1. All three lattice vectors have the same length - What does this tell you about its structure? (Cubic, orthorhombic, tetragonal ...)

3. Now draw 5 more neighbouring unit cells.

4. What is the bravais lattice type?

5. May 23, 2015

### spaghetti3451

1. FCC has four lattice points. They are located one at the corner of a chosen unit cell, and the other three at the centres of each of the three faces which intersect at the corner.

2. The lattice vectors are $\mathbf{R} = n_{1} \mathbf{a} + n_{2} \mathbf{b} + n_{3} \mathbf{c}$, where
$(n_{1}, n_{2}, n_{3}) = (0,0,0)$, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.

6. May 23, 2015

### unscientific

Yes. Good. Now using the corner as (0,0,0) how do you get to the other three lattice points? Hence write down the lattice vectors.
Wrong. You just told me the lattice points. What are the lattice vectors?

7. May 23, 2015

### spaghetti3451

The lattice vectors are $\mathbf{R} = \mathbf{0}$, $\mathbf{R} = \mathbf{a}$, $\mathbf{R} = \mathbf{b}$, and $\mathbf{R} = \mathbf{c}$.

8. May 23, 2015

### unscientific

No. Using the corner as (0,0,0) how do you get to the other three lattice points? Hence write down the lattice vectors.

9. May 23, 2015

### spaghetti3451

In that case, the three lattice vectors are $\mathbf{R} = \mathbf{a}$, $\mathbf{R} = \mathbf{b}$, and $\mathbf{R} = \mathbf{c}$.

10. May 23, 2015

### unscientific

Oops. That's right. To ensure you get all the marks, draw out the cubes to show that the repeating motif is indeed an FCC.