4 Reciprocal Lattice

4 Reciprocal lattice

Reciprocal vectors and the basis of the reciprocal vectors were first used by J. W. Gibbs. Round 1880 he made used of them in his lectures about the vector analysis ([1], pp. 10–11, 83). In structure analysis the concept of the reciprocal lattice has been established by P. Ewald and M. v. Laue in 1913, at the very begining of the discipline [2]. The reason was to facilitate calculations in analytic geometry of linear forms in coordinate systems with non–orthonormal basis, which must inevitably be used when crystals of lower symmetry (non–cubic) are investigated. Since that time the reciprocal lattice belongs to the basic concepts of crystallography, solid state physics and other disciplines. The above mentioned classicists and also authors of textbooks and reference books define the basis + vectors ~a s of the reciprocal lattice in an algebraic way (see relations 4.2(1) below or Appendix C). Then, however, the textbooks and reference books declare the reciprocal lattice to be the Fourier transform of a lattice. The proof of that statement is however almost always missing. (Probably the only exception is the book by Guinier [3], pp. 85–87). The reason for that may lie with the fact that the proof is far from being simple. The fact that the reciprocal lattice is the Fourier transform of a lattice is, however, important not only for the study of cystalline solids but also for presentation of the Fourier series of periodic functions of two and more variables that are not periodic in orthogonal directions, for the formulation of the sampling theorem in dimensions N ≥ 2 etc. In this chapter we first give that algebraic definition of the reciprocal lattice (section 4.2) and then we submit the proof that reciprocal lattice is the Fourier transform of a lattice function (section 4.3). At that, we obtain the expression for the Fourier series of general lattice functions (section 4.4), i.e. even of such ones which characterize lattices that are periodic in non–ortogonal directions only.

4.1 The lattice function

The N–dimensional lattice with basis vectors ~ar, r = 1, 2,...,N, constituted by points is characterized by the so called lattice function X X f(~x) = δ ~x − n1~a1 − n2~a2 − · · · − nN~aN = δ ~x − ~x~n , (1) ~n ∈ inf ~n ∈ inf where

~x~n = n1~a1 + n2~a2 + ··· + nN~aN (2) denotes a lattice vector and the symbol ~n ∈ inf implies that all the components n1, n2, . . . , nN of the multiindex ~n acquire all integer values. Hence, the lattice function (1) is a N–multiple series of the Dirac distributions of N variables. Let us form the matrix

a11 a12 ··· a1N

a21 a22 ··· a2N A = ka k = , (3) rs ......

aN1 aN2 ··· aNN the rows of which consist of the coordinates ars of basis vectors ~ar of the lattice in a system of coordinates with orthogonal basis (~e1,~e2, . . . ,~eN ). Let us consider the variable ~x and the multiindex ~n to be column matrices formed by coordinates xr in that orthogonal basis and by integers nr, respectively. Then, we may express the lattice function in the form X f(~x) = δ~x − AT ~n. (4) ~n ∈ inf

The determinant of the matrix A is called the outer product of the vectors ~ar, r = 1, 2,...,N, (see e.g. [4], p. 95). ~a1,~a2, . . . ,~aN = det A. (5) Its absolute value does not depend on the choice of the orthonormal basis and deﬁnes (cf. e.g. [9], p. 216) the volume VU of the N–dimensional parallepiped (unit cell), the edges of which are the basis vectors ~ar of the lattice: 2 4 RECIPROCAL LATTICE

|det A| = VU . (6) The independence of the absolute value of the outer product on the choice of the orthonormal basis (~e1,~e2, . . . ,~eN ) is obvious from the square of that product:

2 2 T T ~a1,~a2, . . . ,~aN = det A = det A det A = det AA = det k~ar · ~ask = det G. (7)

The symbol AT denotes the transposed matrix of the matrix A and

~a1 · ~a1 ~a1 · ~a2 ··· ~a1 · ~aN

~a · ~a ~a · ~a ··· ~a · ~a 2 1 2 2 2 N det G = (8) ......

~aN · ~a1 ~aN · ~a2 ··· ~aN · ~aN is the Gram determinant of the basis vectors of the lattice whose independence on the coordinate system is evident. If the dimension N = 1, 2 and 3, the absolute value of the outer product has the meaning of length, area, and volume, respectively. For N ≥ 4 it is also reasonable to consider the absolute value of the outer product [~a1,~a2, . . . ,~aN ] as a volume because it has properties attached to the volume ([4], p. 95, [9], pp. 216–217):

(i) [~a1,~a2, . . . ,~aN ] = 0 if and only if the vectors forming the outer product are linearly dependent. (ii) If one of the vectors involved in the outer product is multiplied by a number α then the outer prouduct is multiplied by the number α.

(iii) [a1, a2, . . . , aN ] ≤ a1a2 ··· aN , where ar means the length of the vector ~ar.

4.2 Algebraic deﬁnition of reciprocal lattice In the middle of the last century the International Union of Crystallography recommended to deﬁne the + basis vectors ~a s of the lattice reciprocal to the lattice with the basis vectors ~ar by scalar products

+ ~ar · ~a s = K δrs, r, s = 1, 2,...,N, where K is the so called reciprocal constant (see [6], p. 12), the value of which can be chosen as the case may be. The recommendation of the Union has, however, not been accepted and in crystallography, material science and related technological branches K = 1 (see e.g. [3], p. 386, [5], p. 63) is consistently used, whereas in solid state physics, surface physics, etc. K = 2π is chosen (see e.g. [7], p. 62, [8], p. 87). (Emotional discussions have been carried on which of the two eventualities is the right one (cf. [10], p. 62)). We will follow here the customs of crystallography, though it slightly complicates the formula expressing the Fourier transform of the lattice function. + Thus, we deﬁne the basis vectors ~a s of the reciprocal lattice by scalar product

+ ~ar · ~a s = δrs, r, s = 1, 2,...,N. (1) It is evident from the definition that the reciprocal lattice to the reciprocal lattice is the original lattice + and that the basis vector ~a s of the reciprocal lattice is orthogonal to all the basis vectors ~ar of the original lattice with exception of the vector ~as. Definition (1) is an implicit definition of the basis vectors of the reciprocal lattice; it is not a formula + expressing the vectors ~a s in terms of the basis vectors ~ar of the original lattice. To get such an explicit expression we rewrite N 2 equations (1) in the matrix way

T A A+ = I, (2) where

+ + + a 11 a 12 ··· a 1N

a + a + ··· a + + + 21 22 2N A = a st = . . . . (3) . . .. . + + + a N1 a N2 ··· a NN 4.2 Algebraic deﬁnition of reciprocal lattice 3

+ is the matrix the rows of which are coordinates of basis vectors ~a s of the reciprocal lattice in the orthonormal system of coordinates and I is the unit matrix. It is evident from (2) that

+ −1T + (−1) A = A , i.e. a st = a ts . (4) Expressed by words: If the rows of the matrix A are coordinates of the basis vectors of the original −1 (−1) lattice, then the columns of the inverse matrix A = a st are the coordinates of the basis vectors of the original lattice:

N N X X s+t det Ast ~a + = a+ ~e = (−1) ~e , s = 1, 2,...N, (5) s st t det A t t=1 t=1

where det Ast is minor formed from det A by striking out the s–th row and t–th column. Expression (5) ~ can be written in a more simple way. Let us form the matrix As by replacing the s–th row of the matrix A by vectors ~e1,~e2, . . . ,~eN of the orthonormal basis. Then the sum

N X s+t ~ (−1) det Ast ~et = det As t=1 ~ is the Laplace development of the det As along the elements of the s–th row and the expression (5) for + the basis vectors ~a s of the reciprocal lattice takes a particularly simple form

det ~A ~a + = s , s = 1, 2,...,N. (6) s det A The disadvantage of these expressions is that they are related to an orthonormal system of coordinates. + In fact, the expression (6) provides the coordinates of the vector ~a s in terms of the coordinates art of vectors ~ar in an orthonormal system of coordinates. It is true that this orthonormal system may be arbitrary. In spite of that we will aim at an expression which does not require the use of any orthonormal system of coordinates. Fortunately in E3 the outer product is

det A = ~a1 · (~a2 × ~a3) and ~ ~ ~ det A1 = ~a2 × ~a3, det A2 = ~a3 × ~a1, det A3 = ~a1 × ~a2. Hence, the relations (6) are the well known expressions of the basis vectors of the reciprocal lattice straight by the basis vectors of the lattice:

+ ~a2 × ~a3 + ~a3 × ~a1 + ~a1 × ~a2 ~a 1 = , ~a 2 = , ~a 3 = . (7) ~a1 · (~a2 × ~a3) ~a1 · (~a2 × ~a3) ~a1 · (~a2 × ~a3) However, this happy disposition seems to be limited to N = 3 only. To be free from the necessity to use an orthonormal system of coordinates also in the case of a general dimension N, we multiply the numerator and denominator of (6) by the determinant det AT . In the denominator we get det A det AT = ~ T ~ ~ det G and in the numerator det As det A = det Gs, where det Gs is the determinant made up by replacing the elements of the s–th row in the Gram determinant by the basis vectors ~a1,~a2, . . . ,~aN of the lattice. + In this way we get the basis vectors ~a s of the reciprocal lattice expressed straight by the basis vectors ~ar of the lattice rather then by their coordinates:

det ~G ~a + = s , s = 1, 2,...,N. (8) s det G Note: The expressions (8) can be derived without any use of an orthonormal system of coordinates and + the matrix A: We resolve the vectors ~a s in the basis formed by the basis vectors of the lattice

+ ~a s = αs1~a1 + ··· + αsN~aN , s = 1, 2,...,N, (9)

and multiply each of these decompositions successively by vectors ~ar, r = 1, 2,...,N. Taking into account the deﬁnition (1) we obtain 4 4 RECIPROCAL LATTICE

αs1~a1 · ~ar + ··· + αsN~aN · ~ar = δrs, r, s = 1, 2,...,N. (10)

For a certain s, the system (10) represents N linear algebraic equations for N coeﬃcients αs1, αs2, ..., αsN . The coeﬃcient determinant of this system of linear algebraic equations is the Gram determinant 4.1(8). The right–hand side of the equations (10) is zero, if r 6= s, and one, if r = s. Cramer’s rule then provides a very simple solution of the sytem of equations (10): det G α = (−1)s+t st , t = 1, 2,...,N, (11) st det G

where det Gst is minor of the element ~as · ~at in the Gram determinant. On inserting solutions (11) in (9), we get

N ~ 1 X s+t det Gs ~a + = (−1) ~a det G = , s det G t st det G t=1 in acordance with (8). Let us use (8) to get expressions for basis vectors in E1 and E2. In E1 we obtain

~a ~a + ~a + = , ~a = . (12) a2 (a +)2

In E2 it is

~a ~a a2 ~a · ~a 1 2 1 1 2 ~a · ~a a2 ~a ~a + 2 1 2 + 1 2 ~a 1 = , ~a 2 = , a2 ~a · ~a a2 ~a · ~a 1 1 2 1 1 2 2 2 ~a2 · ~a1 a2 ~a2 · ~a1 a2 i.e.

a2~a − (~a · ~a )~a a2~a − (~a · ~a )~a ~a + = 2 1 1 2 2 , ~a + = 1 2 1 2 1 . (13) 1 2 2 2 2 2 2 2 a1a2 − (~a1 · ~a2) a1a2 − (~a1 · ~a2) On the contrary

+2 + + + + +2 + + + + a2 ~a 1 − ~a 1 · ~a 2 ~a 2 a2 ~a 2 − ~a 1 · ~a 2 ~a 1 ~a1 = , ~a2 = . (14) +2 +2 + +2 +2 +2 + +2 a1 a2 − ~a 1 · ~a 2 a1 a2 − ~a 1 · ~a 2

In E3 the traditional expressions (7) are much more simple than (8) and this simplicity invites us to make use of them for two–dimensional lattices, too. However, in E2 the vector product is not deﬁned. This diﬃculty is usually bypassed by adding the unit vector ~n to the two basis vectors ~a1, ~a2 of the two–dimensional lattice in such a way that vectors ~a1, ~a2, ~n form the three–dimensional right–handed basis. Then the basis vectors of the reciprocal lattice are calculated according to (7) (see e.g. [8], p. 87):

+ ~a2 × ~n + ~n × ~a1 ~a 1 = , ~a 2 = . (15) |~a1 × ~a2| |~a1 × ~a2| Equivalence of expression (15) and (13) is easy to see, if we write the unit vector ~n as the fraction

~a × ~a ~n = 1 2 , (16) |~a1 × ~a2| insert it into (15) and use the identities ~a×(~b×~c) = (~a·~c)~b−(~a·~b)~c,(~a×~b)·(~c×d~) = (~a·~c)(~b·d~)−(~a·d~)(~b·~c). We get

2 + ~a2 × (~a1 × ~a2) a2~a1 − (~a1 · ~a2)~a2 ~a 1 = 2 = 2 2 2 , (~a1 × ~a2) a1a2 − (~a1 · ~a2)

2 + (~a1 × ~a2) × ~a1 a1~a2 − (~a1 · ~a2)~a1 ~a 2 = 2 = 2 2 2 , (~a1 × ~a2) a1a2 − (~a1 · ~a2) 4.3 Reciprocal lattice and the Fourier transform of the lattice function 5

in accordance with (13). (As to the unimportant third basis vector of the reciprocal three–dimensional + basis is concerned, it is evident from (16) that ~a 3 = ~n.)

4.2.1 Example: Reciprocal lattice to the primitive rectangular lattice in E2

Let us calculate the basis vectors of the reciprocal lattice to primitive rectangular lattice in E2 with the ratio of the lengths of the basis vectors 1:2 and with the longer side of the unit cell in the vertical 2 2 direction. Let thelength of the basis vectors ~a1, ~a2 be a1 = a, a2 = 2a (see Figure 1(a)). Then a1 = a , 2 2 a2 = 4a , ~a1 · ~a2 = 0. and expressions (13) give the basis vectors of the reciprocal lattice in the form + ~a1 + ~a2 ~a 1 = a2 , ~a 2 = (2a)2 . The basis vectors of the reciprocal lattice thus have the same direction as the + 1 + 1 basis vectors of the original lattice and their magnitudes are a1 = a , a2 = 2a . The reciprocal lattice is again primitive rectangular lattice, but the ratio of the primitive cell sides is 2:1, i.e. the longer side has horizontal direction (see Figure 1(b)). If, for some reason, we do not choose orthogonal basis vectors of our rectangular lattice but soume other ones we get, of course, the same reciprocal lattice. However, 0 0 the basis vectors specifying it are diﬀerent. For example, let us choose the basis vectors ~a1, ~a2 shown in 02 2 02 2 0 0 2 Figure 1(c). Obviously, a1 = 8a , a2 = 5a , ~a1 · ~a2 = 6a , and expressions (13) give the following basis 0+ 5 0 3 0 2 0+ 3 0 0 2 vectors of the reciprocal lattice ~a1 = 4~a1 − 2~a2 /a , ~a2 = − 2~a1 + 2~a2 /a . Their magnitudes are 0+ √ 0+ √ a1 = 5/2a, a2 = 2/a (cf. Figure 1(d)).

4.2.2 Example: Reciprocal lattice to the centred rectangular lattice in E2

Now we ﬁnd the reciprocal lattice to the centred rectangular lattice in E2. Let the ratio of the sides of the rectangular unit cell be 1:2 and let the longer side have the vertical direction. We choose the non– 2 2 2 2 2 orthogonal basis vectors ~a1, ~a2 shown in Figure 2(a). Then a1 = a , a2 = 5a /4, ~a1.~a2 = a /2. According + 5 1 2 + 1 2 to (13) the basis vectors of the reciprocal lattice are ~a 1 = 4~a1 − 2~a2 /a ,~a 2 = − 2~a1 + ~a2 /a . Their + √ + magnitudes are a1 = 5/2a, a2 = 1/a (cf.Figure 2(b)). Hence, the reciprocal lattice is again a centered rectangular lattice, but with the side ratio 2:1, i.e. with the longer side in the horizontal direction.

4.3 Reciprocal lattice and the Fourier transform of the lattice function We will calculate the Fourier transform of the lattice function 4.1(1) three times: First in E1 (Section 4.3.1). In this case the calculation of the Fourier integrals gives the Fourier transform having the shape of the Fourier series. This Fourier series is simultaneously a geometric series whose ratio has the absolute value equal to one. The sum of the geometric series is the series of the Dirac distributions which is proportional to the lattice functions of the reciprocal lattice. The calculation in E2 (Section 4.3.2) takes advantage of the quoted properties of the one–dimensional lattice function. Nevertheless, it requires also several artifices typical for deriving of the Fourier transform of more–dimensional lattice function. The artifices are, however, geometrically illustrative. The calculation in EN (Section 4.3.3) is formally identical with the calculation in E2. In all the cases we will find that the Fourier transform of the lattice function is proportional to the lattice function of the reciprocal lattice with the reciprocal constant K = 2π/k.

4.3.1 The Fourier transform of the lattice function in E1 In one–dimmensional case the lattice function has the form

∞ X f(x) = δ(x − na) (1) n=−∞ and it represents a periodic distribution of the Dirac distributions of single variable with a period a (see Figure 3(a)). Thus, we may formally expand it into the Fourier series

∞ X x f(x) = c exp i 2π h . h a h=−∞

It is noteworthy that in the case of lattice function (1) all the coeﬃcients ch have the same value: 6 4 RECIPROCAL LATTICE

Figure 1: Two–dimensional primitive rectangular lattice (a), (c) whose basis vectors are chosen in diﬀerent ways and its reciprocal lattice (b), (d).

a/2 1 Z x c = f(x) exp −i 2π h dx = h |a| a −a/2

a/2 ∞ 1 Z X x = δ (x − na) exp −i 2π h dx = |a| a n=−∞ −a/2

a/2 1 Z x 1 = δ (x) exp −i 2π h dx = . |a| a |a| −a/2 Therefore, the Fourier series of the lattice function has the form

∞ ∞ ∞ X 1 X x 1 X δ (x − na) = exp i 2π h = exp i 2π h a+ x , (2) |a| a |a| n=−∞ h=−∞ h=−∞ 4.3 Reciprocal lattice and the Fourier transform of the lattice function 7

Figure 2: Two–dimensional centered rectangular lattice (a) with basis vectors ~a1, ~a2 specifying a primitive lattice and its reciprocal lattice (b).

x which is geometric series of the ratio equal to complex unity exp i 2π a . This is an important result which will be used when calculating the Fourier transform of more– dimensional lattice function. Therefore, we prepare a needful formula and, according to (2), we give the formula for the sum of geometric series of the ratio q = exp(ibx):

∞ ∞ X 2π X 2π exp (ibhx) = δ x − n . (3) |b| b h=−∞ n=−∞ Now, we can calculate the Fourier transform of the one–dimensional lattice function either straight from its deﬁnition (1) or from its Fourier series (2). We choose the ﬁrst possibility. By interchanging the order of integration and addition and using the sifting property of the Dirac distribution we get the Fourier tranform of the lattice function (1) in the form of the Fourier series:

" ∞ # Z ∞ X F (X) = A δx − na exp−ikXx dx = −∞ n=−∞ ∞ X Z ∞ = A δx − na exp−ikXx dx = n=−∞ −∞ ∞ ∞ X X = A exp−ikXna = A expikXna. (4) n=−∞ n=−∞

The Fourier series (4) of the Fourier transform of the lattice function (1) is geometric series of the ratio exp(ikXa). According to (3) it is the Fourier series of the function

∞ ∞ ∞ X X 2π h 2π X 2π h F (X) = A 2π δkXa − 2πh = A 2π δ X − ka = A δ X − . k a |ka| k a h=−∞ h=−∞ h=−∞

Thus, we conclude that the Fourier transform of the one–dimensional lattice function is proportional to 2π the one–dimensional lattice function with the period ka :

( ∞ ) ∞ ∞ X 1 X 2π h 1 X 2π FT δx − na = δ X − = δ X − ha+ . (5) B|a| k a B|a| k n=−∞ h=−∞ h=−∞ The lattice function (1) and its Fourier transform (5) are shown in Figure 3. 8 4 RECIPROCAL LATTICE

f(x) F(X) 1 1 B| a |

-a 0 a 2a 3a x - 2π 0 2π 22π 32π X ka ka ka ka (a) (b)

Figure 3: Linear lattice 4.3.1(1) with the period a, (a) and its Fourier transform 4.3.1(5), (b).

4.3.2 The Fourier transform of the lattice function in E2 In two–dimensional space the lattice function has the form

∞ ∞ X X f(~x) = f(x1, x2) = δ ~x − n1~a1 − n2~a2 =

n1=−∞ n2=−∞ ∞ ∞ X X = δ x1 − n1a11 − n2a21, x2 − n1a12 − n2a22 . (6)

n1=−∞ n2=−∞ By direct calculation of the Fourier tranform of this function

∞ ZZ " ∞ ∞ # 2 X X 2 F (X~ ) = A δ ~x − n1~a1 − n2~a2 exp −ikX~ · ~x d ~x = −∞ n1=−∞ n2=−∞ ∞ ∞ 2 X X = A exp −ikX~ · (n1~a1 + n2~a2) =

n1=−∞ n2=−∞ ∞ ∞ 2 X X = A exp ikX~ · (n1~a1 + n2~a2) (7)

n1=−∞ n2=−∞ we obtain the Fourier tranform expressed again by the Fourier series, this time, of course, by the double series. It is, however, evident that it may be rewritten as the product of two simple series:

∞ ∞ 2 X X F (X~ ) = A exp ik(X~ · ~a1)n1 exp ik(X~ · ~a2)n2 . (8)

n1=−∞ n2=−∞

Each of them is geometric series with the ratio exp(ikX~ · ~a1) and exp(ikX~ · ~a1), respectively. According to (3) they are the Fourier series of functions

∞ ∞ 2π X 2π 2π X 2π δ X~ · ~a − h and δ X~ · ~a − h . (9) |k| 1 1 k |k| 2 2 k h1=−∞ h2=−∞ The Fourier transform (8) thus takes the form

2 ∞ ∞ 2π X 2π X 2π F (X~ ) = A2 δ X~ · ~a − h δ X~ · ~a − h . (10) k 1 1 k 2 2 k h1=−∞ h2=−∞ Before going on with processing of (10), we will look up geometrical meaning of simple series in this expression (see Figure 4). Evidently, the equations

2π ~ai 2π X~ · ~ai = hi, i.e. X~ · = hi, hi = 0, ±1, ±2,... k ai kai 4.3 Reciprocal lattice and the Fourier transform of the lattice function 9

2π ka + 2 a2 a2 + a1 a1

2π ka1

Figure 4: To the derivation of the Fourier transform of the lattice function in E2. In the upper part of the ﬁgure there is a lattice with basis vectors ~a1, ~a2. The lower part of the ﬁgure shows the corresponding + + reciprocal lattice with basis vectors ~a 1 , ~a 2 .

2π represent in E2 a system of stright–lines perpendicular to the vector ~ai and having the spacing . The kai product of the series in (10) then represents the system of the Dirac distribution of two variables with non–zero values at points X~ speciﬁed by conditions

~a1 2π ~a2 2π X~ · = h1, X~ · = h2, h1, h2 = 0, ±1, ±2,..., a1 ka1 a2 ka2

i.e. at points

2π X~ = h ~a + + h ~a + k 1 1 2 2

2π constituting the two–dimensional reciprocal lattice with the reciprocal constant k . Of course, we can get this result just by algebraic treatment of (10), without using the geometrical interpretation. We will do it now and we will even complete the result. For this purpose we rewrite the product of two simple series in (10) as the double series of the Dirac distributions of two variables. The Fourier transform (10) takes then the form

∞ ∞ 1 X X 2π 2π F (X~ ) = δ X~ · ~a − h , X~ · ~a − h . (11) B2 1 k 1 2 k 2 h1=−∞ h2=−∞

If we consider the argument of the Dirac distributions in (11) to be the row matrix we may process it

further. The variable X~ is taken as the row matrix the elements of which are coordinates X1, X2 in an ~ orthonormal basis and the multiindex h is the row matrix with integers h1, h2 as elements. 10 4 RECIPROCAL LATTICE

~ 2π ~ 2π ~ ~ 2π X · ~a1 − h1, X · ~a2 − h2 = X · ~a1, X · ~a2 − h1, h2 = k k k

a11 a21 2π = X1,X2 − h1, h2 = a12 a22 k 2π = X~ AT − ~h = k 2π −1 = X~ − ~h AT AT = k 2π = X~ − ~h A+ AT . (12) k

The Fourier transform (11) with the argument of the Dirac distributions processed in this way can be further adjusted and brought back from the matrix representation to the vectorial one which does not depend on the choice of the orthonormal basis:

∞ ∞ 1 X X 2π F (X~ ) = δ X~ − ~h A+ AT = B2 k h1=−∞ h2=−∞ ∞ ∞ 1 1 X X 2π = δ X~ − ~h A+ = B2 | det A| k h1=−∞ h2=−∞ 1 1 X 2π + + 2π + + = 2 δ X1 − h1a 11 + h2a 21 ,X2 − h1a 12 + h2a 22 = B VU k k ~h ∈ inf 1 1 X ~ 2π + + = 2 δ X − h1~a 1 + h2~a 2 . (13) B VU k ~h ∈ inf

Here VU = | det A| is the area of the unit cell of the original lattice (cf. 4.1(6)). Hence, the Fourier 2 transform (13) of the lattice function (6) is proportional (with the proportionality factor 1/(B VU )) to the lattice function of the reciprocal lattice with the reciprocal constant K = 2π/k.

4.3.3 The Fourier transform of the lattice function in EN

The same result as in E2 is obtained also in a general case of the lattice function in EN . Algebraic and analytic processing is the same as in E2, only the geometric notion is missing. We calculate the Fourier transform of the lattice function 4.1(1):

( ) X F (X~ ) = FT δ (~x − n1~a1 − n2~a2 − · · · − nN~aN ) = ~n ∈ inf

∞ Z Z N X N = A ··· δ ~x − n1~a1 − n2~a2 − · · · − nN~aN exp −ikX~ · ~x d ~x = −∞ ~n ∈ inf

N X = A exp −ikX~ · (n1~a1 + n2~a2 + ··· + nN~aN ) = ~n ∈ inf N X = A exp ikX~ · (n1~a1 + n2~a2 + ··· + nN~aN ) . (14) ~n ∈ inf

Expression (14) represents the Fourier series of the N–dimensional lattice function in the form of N– multiple Fourier series. The series can be factorized and — by the interchange of the order of summation and multiplication — rewritten as the product of N simple geometric series: 4.4 The Fourier series of the lattice function in EN 11

N N X Y h i F (X~ ) = A exp ik X~ · ~aj nj = ~n ∈ inf j=1

N ∞ N Y X h i = A exp ik X~ · ~aj nj . (15)

j=1 nj =−∞

Each of the simple geometric series in (15) can be expressed (according to (3)) as simple series of Dirac distributions of single variable

N N ∞ 2π Y X 2π F (X~ ) = AN δ X~ · ~a − h . k j k j j=1 hj =−∞ On the analogy of (11) the product of simple series of the Dirac distributions of single variable can be expressed as the N–multiple series of the Dirac distributions of N variables:

1 X 2π 2π F (X~ ) = δ X~ · ~a + h ,..., X~ · ~a + h . BN 1 k 1 N k N ~h ∈ inf The argument of these Dirac distributions can be considered to be the row matrix and the treatment similar to (12) can be performed:

1 X 2π F (X~ ) = δ X~ AT − ~h = BN k ~h ∈ inf

1 X 2π −1 = δ X~ − ~h AT AT = BN k ~h ∈ inf

1 1 X 2π = δ X~ − ~h A+ . (16) BN | det A| k ~h ∈ inf

The argument of the Dirac distributions can be expressed — similarly to (13) — in terms of basis vectors of the reciprocal lattice and the use of 4.1(6) leads to the Fourier transform of the lattice function in the form

~ 1 1 X ~ 2π + + + F (X) = N δ X − h1~a 1 + h2~a 2 + ··· + hN~a N = B VU k ~h ∈ inf 1 1 X ~ 2π ~ = N δ X − X~h , (17) B VU k ~h ∈ inf where

~ + + + X~h = h1~a 1 + h2~a 2 + ··· + hN~a N (18) is the lattice vector of the reciprocal lattice. The Fourier transform (17) of the lattice function in its most general shape is then proportional (with N proportionality factor 1/(B VU )) to the lattice function of the reciprocal lattice with the reciprocal constant K = 2π/k.

4.4 The Fourier series of the lattice function in EN While deriving the Fourier transform of the lattice function we have got the transform in the form of the Fourier series. In one–dimensional case it is the expression 4.3(4), in two–dimensional case 4.3(7) and in N–dimensional case 4.3(14). In all these cases the exponents of the summation contain the scalar 12 REFERENCES

~ product of the variable X and of the lattice vector ~x~n of the lattice, not of the reciprocal lattice, as could be expected at the Fourier transform. Now the question arises how does it look like the Fourier transform of the lattice function. For the one–dimensional case we have found it and the expression 4.3(2) shows that in the exponent of the summands there is the product of the variable x and of the reciprocal lattice vector ha+. If we restrict ourselves in the N–dimensional case to the lattice functions with mutually orthogonal basis vectors it would be possible to factorize them, to calculate the Fourier series in each variable and then to put again the individual factors together. Unfortunately it is not possible to proceed this way in the general case when the vectors ~ar are not orthogonal. But, fortunately enough, the Fourier transform of the lattice function 4.1(1) can be obtained in any case by the inverse Fourier transform of the Fourier transform of the lattice function in the from 4.3(17):

X δ ~x − n1~a1 − n2~a2 − · · · − nN~aN = ~n ∈ inf     −1 1 1 X ~ 2π + + + = FT N δ X − h1~a 1 + h2~a 2 + ··· + hN~a N = B VU k  ~h ∈ inf 

∞ Z Z 1 X ~ 2π + + + ~ N ~ = ··· δ X − h1~a 1 + h2~a 2 + ··· + hN~a N exp ikX · ~x d X = VU k −∞ ~h ∈ inf

1 X + + + = exp i 2π h1~a 1 + h2~a 2 + ··· + hN~a N · ~x . (1) VU ~h ∈ inf

The Fourier series of the lattice function is frequently used for expressing the electron density in crystalline solids ([5], p. 169, [6], pp. 353 and following), for deriving the sampling theorem of functions of several variables when non–orthogonal sampling net is used [11], and elsewhere.

References

[1] Gibbs J. W.: Elements of Vector Analysis. Tuttle, Morehouse and Taylor, New Haven 1881–4. [2] Ewald P.: Historisches und Systematisches zum Gebrauch des ”Reziproken Gitters” in der Kristall- strukturlehre. Zeitschrift für Kristallographie 93 (1936), 396–398. [3] Guinier A.: X–Ray Diffraction In Crystals, Imperfect Crystals, and Amorphous Bodies. W. H. Freeman and Co., San Francisco 1963. [4] Cechˇ E.: Základyanalytickégeometrie I. Pˇr´ırodovˇedeckévydavatelstv´ı,Praha 1951. [5] Giacovazzo C. et al.: Fundamentals of Crystallography. International Union of Crystallography, Oxford University Press 1992. [6] Henry N. F. M., Lonsdale K. (eds.): International Tables for X-Ray Crystallography. Vol. 1. The Kynoch Press, Birmingham 1952. [7] Kittel Ch.: Uvod´ do fyziky pevnýchlátek. Academia, Praha 1985. [8] Lüth H.: Surfaces and Interfaces of Solid Materials. 3rd ed. Springer Verlag, Berlin 1995. [9] Gantmacher F. R.: Tˇeorija matric. 4. izd. Izdatˇel’stvo Nauka, Moskva 1988. [10] Kittel Ch.: Introduction to Solid State Physics. 4th ed., John Wiley, Inc., New York 1971. [11] Petersen D. P., Middleton D.: Sampling and Reconstruction of Wave – Number – Limited Functions in N–Dimensional Euclidean Spaces. Information and Control 5 (1962), 279–323.