Brillouin Zones. Physics 3P41 Chris Wiebe


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1 Brillouin Zones Physics 3P41 Chris Wiebe
2 Direct spce to reciprocl spce * = 2 i j πδ ij Rel (direct) spce Reciprocl spce Note: The rel spce nd reciprocl spce vectors re not necessrily in the sme direction
3 Lue Equtions Another wy of expression the diffrction condition k = G is by the Lue equtions, which cn be derived by tking the sclr product of k nd G with 1, 2, nd 3 1 k = 2πν 1 2 k = 2πν 2 3 k = 2πν 3 (Lue Equtions) These equtions hve simple interprettion: for Brgg reflection, k must lie on certin cone bout the direction of 1 (for exmple). Likewise, it must be on cone for 2 nd 3. Thus, Brgg reflection which stisfies ll three conditions must lie t the intersection of these three cones, which is t point in reciprocl spce.
4 Single crystl diffrction k Ewld sphere k Ech time k = G, reciprocl lttice vector, you get Brgg reflection. This is point of intensity t some 2θ ngle, nd some Φ ngle in spce Brgg reflections
5 Actul single crystl diffrction ptterns
6 Experimentl setup
7 Powder vs. Single crystls So, Brgg peks re usully t single spot in rel spce for single crystls. For powder smples, however, which re composed of mny tiny single crystls rndomly oriented, the G vectors exist on cone of scttering (think of tking the Ewld circle, nd rotting bout k). k G k
8 Another picture of powder xry diffrction G k 2θ k Whenever Brgg s lw is stisfied (2dsinθ = nλ), we get diffrction pek. Since there is usully some crystllite which hs it s plnes orienting in the proper direction, we get cone of scttering bout the ngle 2θ
9 Brillouin Zones A Brillouin Zone is defined s WignerSeitz primitive cell in the reciprocl lttice. To find this, drw the reciprocl lttice. Then, use the sme lgorithm s for finding the WignerSeitz primitive cell in rel spce (drw vectors to ll the nerest reciprocl lttice points, then bisect them. The resulting figure is your cell). The nice result of this is tht it hs direct reltion to the diffrction condition: k (1/2 G) = (1/2 G) 2 ½ G D WignerSeitz cell Point D in reciprocl spce Therefore, the Brillouin Zone exhibits ll wvevectors, k, which cn be Brggreflected by crystl
10 The First Brillouin Zone The Zone we hve drwn bove using the WignerSeitz method is clled the first Brillouin zone. The zone boundries re k = +/ π/ (to mke the totl length to side 2π/ in reciprocl spce). The 1 st Brillouin zone is the smllest volume entirely enclosed by the plnes tht re perpendiculr bisectors of the reciprocl lttice vectors drwn from the origin. Usully, we don t consider higher zones when we look t diffrction. However, they re of use in energybnd theory An exmple: Rectngulr Lttice
11 Higher Order Brillouin Zones
12 Reciprocl lttice to SC lttice The primitive trnsltion vectors of ny simple cubic lttice re: Using the definition of reciprocl lttice vectors: We get the following primitive trnsltion vectors of the reciprocl lttice: b b b = = = π π π 1 = x 2 = y 3 = z b 1 = (2π/)x b 2 = (2π/)y b 3 = (2π/)z This is nother cubic lttice of length 2π/
13 Reciprocl lttice to SC lttice The boundries of the first Brillouin zone re the plnes norml to the six reciprocl lttice vectors +/ b 1, +/ b 2, +/ b 3 t their midpoints: +/ (π/) 2π/ The length of ech side is 2π/ nd the volume is (2π/) 3
14 Reciprocl lttice to BCC lttice The primitive trnsltion vectors for the BCC lttice re: The volume of the primitive cell is ½ 3 (2 pts./unit cell) So, the primitive trnsltion vectors in reciprocl spce re: Wht lttice is this? 1 = ½ (x + y  z) 2 = ½ (x+y + z) 3 = ½ (x  y + z) b 1 = 2π/ (x + y) b 2 = 2π/ (y + z) b 3 = 2π/ (z + x)
15 Reciprocl lttice to BCC lttice This is the FCC lttice! So, the fourier trnsform of the BCC lttice is the FCC lttice (wht do you expect for the FCC lttice, then?) The generl reciprocl lttice vector, for integrls ν 1,ν 2, nd ν 3 is then: G = ν 1 b 1 + ν 2 b 2 + ν 3 b 3 = (2π/)[(ν 2 + ν 3 )x +(ν 1 + ν 3 )y + (ν 1 + ν 2 )z)] The shortest G vectors re the following 12 vectors, where choices of sign re independent: (2π/)(+/y +/ z) (2π/)(+/x +/z) (2π/)(+/x +/y)
16 Reciprocl lttice to BCC lttice This is the first Brillouin zone of the BCC lttice (which hs the sme shpe s the Wigner Seitz cell of the FCC lttice). It hs 12 sides (rhombic dodechedron). The volume of this cell in reciprocl spce is 2(2π/) 3, but it only contins one reciprocl lttice point. The vectors from the origin to the center of ech fce re: (π/)(+/y +/ z) (π/)(+/x +/z) (π/)(+/x +/y)
17 Reciprocl lttice to the FCC lttice The primitive trnsltion vectors for the FCC lttice re: 1 = ½ (x + y) 2 = ½ (y + z) 3 = ½ (z + x) The volume of the primitive cell is 1/4 3 (4 pts./unit cell) So, the primitive trnsltion vectors in reciprocl spce re: This is, of course, the BCC lttice b 1 = (2π/) (x + y  z) b 2 = (2π/) (x+y + z) b 3 = (2π/)(x y + z) The volume of this cell is 4(2π/) 3 in reciprocl spce
18 Reciprocl spce to the FCC lttice This is the first Brillouin zone of the FCC cubic lttice. It hs 14 sides bound by: (2π/)(+/x +/y +/ z) (8 of these vectors) 4π/ nd (2π/)(+/2x) (2π/)(+/2y) (2π/)(+/2z) (6 of these vectors)
19 Summry So, the reciprocl spce for simple cubic lttice is simple cubic, but the other cubic lttice (BCC, FCC) re more confusing: Lttice Rel Spce Lttice kspce The BCC nd FCC lttices re Fourier trnsforms of one nother bcc WS cell fcc BZ fcc WS cell bcc BZ
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