Ashcroft and Mermin remark, at the bottom of page 91 that

As a general rule, face-centered and body-centered cubic Bravais lattice[s] are described in terms of a conventional cubic cell, i.e. as simple cubic lattices with bases.

So far, no problem. They go on to say:

Since any lattice plane in a fcc or bcc lattice is also a lattice plane in the underlying simple cubic lattice, the same elementary cubic indexing can be used to specify lattice planes.

It is the bolded part which I am asking about -- I can convince myself intuitively that this should be true, but cannot prove it. Also, does the converse hold? Is every lattice plane in the underlying simple cubic lattice also a lattice plane in the given fcc or bcc structure? Essentially, I am asking for a proof that, given a fixed bcc or fcc lattice, there is a one-to-one correspondence between lattice planes as described in bcc or fcc and as described in the underlying simple cubic lattice.

Finally, they conclude with:

In practice, it is only in the description of noncubic crystals that one must remember that the Miller indices are the coordinates of the normal [to the lattice plane] in a system given by the reciprocal lattice, rather than the direct lattice [because the reciprocal of simple cubic is simple cubic].

The conclusion, I suppose, is thus that whenever we see Miller indices used to describe planes in fcc or bcc, by convention these are not in fact coordinates in the corresponding reciprocal lattices bcc and fcc, respectively (as they would be by the strict definition of Miller indices), but are instead coordinate in the reciprocal simple cubic lattice (to the underlying simple cubic direct lattice used to describe the crystal)? I can accept this only if the one-to-one correspondence asked about above exists.

Thank you for any help you can provide!

  • $\begingroup$ Contributors of Matter Modelling SE could provide a detailed answer if the question is displaced. The statement is intuitive and could be surely demontrated. You can start by checking if there are intergers $n_i$ so that two translational vectors $R=n_1 a_1 + n_2 a_2 + n_3 a_2$ hold for SC, BCC and FCC ($a_i$ are primitive vectors). $\endgroup$
    – M06-2x
    May 17 at 23:36

1 Answer 1


It is not a rigorous proof, so maybe you has already considered it to be convinced.

3 point defines a plane.

If Points A and B are cubic vertex and C is in the centre of a cube (BCC) or in the centre of a face (FCC), the line AC and BC can be extended by the same length to find A' and C' respectively, and both are also vertex points by the lattice symmetry. So the plane ABC is also AA'BB', a plane containing only vertex, and belonging to the underlying single cubic lattice.

On the other hand, it is possible to take the points of the centre of the cubes (BCC) or the centre of the faces (FCC) and use it as vertex of the cubes. So the cases of A and B off the vertex and C in the vertex are in reality equal to described above.

A, B and C off the vertex is equal to all them on the vertex by the argument of the last paragraph.

So, there is no way to find a plane in FCC or BCC that are not also in the underlying single cubic lattice.


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