Let us begin by considering the system in its cyclic form (Fig. 6.1). Taking into account the symmetry properties of theCngroup it appears that the term hj j(k,k)is nil ifkis different fromkbecause thekandkvectors are associ- ated with different irreducible representations of theCngroup. In the following sections we will discuss how to use the possible symmetry operations of the system to predict when the interaction term(BO)j(k) | ˆh|(BOj(k) is nil.
Existence of an horizontalσhsymmetry plane
Here we will explore the consequences of the existence of a σh symmetry plane perpendicular to the Cn rotation axis. In this case, the Cnh group leaves the system invariant. Under such circumstances we must build a basis of Bloch orbitals, {(BO)j(k), j =1, . . . ,N0}, generated by orbitals, (ϕj)0,
If the cyclic system possesses a horizontal symmetry plane,σh, the inter- action term hj j(k) is nil provided that the(ϕj)0and(ϕj)0orbitals do not have the same properties with respect to theσhplane.
Existence of verticalσvsymmetry planes
Now let us assume that the system exhibits a vertical symmetry plane,(σv)0, containing the rotation axis, Cn, and going through theM0andMncells. The symmetry group for this system, generated by theCn (n even) rotation and by this symmetry plane,(σv)0, is the groupCnv(neven). This group contains n rotations Cnm (m=0,±1, . . . ,±(n−1),n) andn symmetry operations related to the vertical planes, which may belong to two different classes denoted{(σv)i ; i =0,1, . . . ,n−1}and{(σd)i ; i =0,1, . . . ,n−1}. The planes(σv)i and(σd)i go through the Mi and Mi−n cells, and the plane(σd)i
is the bisector of the(σv)iand(σv)i+1planes (Fig. 6.1).
We will use theCnvgroup to characterise the symmetry properties of the Bloch orbitals. Because of the existence of the (σv)0 and(σd)0 symmetry planes passing through the reference cell, we need to build a basis of Bloch orbitals,{(BO)j(k); j =1, . . . ,N0}generated by the(ϕj)0 orbitals, which are symmetric or antisymmetric with respect to either(σv)0or (σd)0. Obvi- ously, a(ϕj)0orbital can only be adapted to one of the two symmetry opera- tions.
We will start by projecting over the character table of groupCnvthe basis for the j representation, which is made up of the different ϕj orbitals {(ϕj)m,m=0,±1, . . . ,±(n−1),n}. The different Bloch orbitals(BO)j(k) will be labelled by means of the different irreducible representations of the Cnv group. As shown in Table 6.1, the character table for the Cnv group possesses four non-degenerate irreducible representations (A1,A2,B1,B2) and(n−1)doubly degenerate representations, taking into account the fact that the character table of the Cnv group may be deduced from that of the Cn group (see Table 3.1). The character of j is nil for all the rotations Cnm (m= ±1, . . . ,±(n−1),n)because none of theϕj functions is stable upon their action. Since the different symmetries may be grouped together into two different classes (nσvandnσd), we just need to estimate the char- acter ofj with respect to the(σv)0and(σd)0symmetries. These characters will be denotedχnσv(j)andχnσd(j). Only the(ϕj)0and(ϕj)n may be at the origin of the non-nullity of the character of j with respect to the (σv)0 and (σd)0 symmetries. In addition, the two orbitals(ϕj)0 and (ϕj)n
necessarily have the same character with respect to these symmetries. Conse- quently, the decomposition of thej representation depends on the symmetry
Table 6.1 The six initial lines reproduce the character table of groupCnv(neven). The four last lines are the characters of thejrepresentation for different possible cases.
Cnv E Cn C2n
. . . Cnn−1
C2 nσv nσd Cn−1 C−n2 Cn1−n
A1 Characters ofi=0of theCngroup(ka=0) 1 1
A2 Characters ofi=0of theCngroup(ka=0) –1 –1
B1 Characters ofi=nof theCngroup(ka=1/2) 1 –1 B2 Characters ofi=nof theCngroup(ka=1/2) –1 1 Ei(i= Characters ofi+−iof the
0 0
1, . . . ,n−1) Cngroup(ka= ±i/n) jif(ϕj)0
n 0 0 . . . 0 0 2 0
is symmetric w.r.t.(σv)0 jif(ϕj)0is
n 0 0 . . . 0 0 –2 0
antisymmetric w.r.t.(σv)0 jif(ϕj)0
n 0 0 . . . 0 0 0 2
is symmetric w.r.t.(σd)0 jif(ϕj)0is
n 0 0 . . . 0 0 0 –2
antisymmetric w.r.t(σd)0
properties of(ϕj)0with respect to(σv)0and(σd)0(i.e. the four last lines of Table 6.1).
Thus we can state that:
• if(ϕj)0is symmetric with respect to(σv)0
χnσv(j)=2 ⇒ j =A1+B1+
n−1 i=1
Ei (6.1)
• if(ϕj)0is antisymmetric with respect to(σv)0
χnσv(j)= −2 ⇒ j = A2+B2+
n−1 i=1
Ei (6.2)
• if(ϕj)0is symmetric with respect to(σd)0
χnσd(j)=2 ⇒ j =A1+B2+
n−1 i=1
Ei (6.3)
• if(ϕj)0is antisymmetric with respect to(σd)0
χnσd(j)= −2 ⇒ j =A2+B1+
n−1 i=1
Ei (6.4)
According to Table 6.1, a Bloch orbital associated with the point is a basis for the representations A1orA2, whereas a Bloch orbital associated with
BOj(k) ( j =1, . . . ,N0)associated with the samekavalue are a basis for the same irreducible representationEi(i =nka),whatever the nature of the(ϕj)0
orbital that generates it. Consequently, the fact that two Bloch orbitals may not interact at points other than and X is not related to these two planes.
According to eqns (6.1)–(6.4), atand X the interaction term between two Bloch orbitals hj j(k) may be nil because of these vertical symmetry planes.
For instance, at, the Bloch orbital BOj()generated by a(ϕj)0orbital that is symmetric with respect to(σv)0(i.e. ofA1symmetry) cannot interact with a BOj() Bloch orbital generated by a(ϕj)0 orbital that is antisymmetric with respect to(σv)0(i.e. of symmetry A2) or with a BOj() Bloch orbital generated by a(ϕj)0orbital that is antisymmetric with respect to(σd)0)(i.e.
of symmetryA2).
In conclusion, the existence of vertical symmetry planes,σv andσd, may be at the origin of the nullity of the interaction terms hj j(k) at theand X points. In contrast, for any other point, theσvandσd symmetry planes do not help in characterising the symmetry properties of the Bloch orbitals.
Using these results
We can now provide an illustration of these results by looking at the Ansystem in which A is an atom described bynsandnpz orbitals. Let us consider the interaction term hns,npz(k), between the Bloch orbitals BOns(k)and BOnpz(k) generated by the (ns)0 and(npz)0 orbitals, respectively, which are centred on the A atom of the reference cell. We denote as npz the np orbitals the symmetry axis of which is tangential to the circle (see Fig. 6.2a). Such a system possesses a horizontal symmetry plane, as well as vertical symmetry planes (σv)m (m=0, . . . ,n−1) going through the Am atoms and (σd)m
(m=0, . . . ,n−1)planes passing in between the Am and Am+1atoms (see Fig. 6.2b).
The existence of aσhsymmetry plane cannot lead to the nullity of hns,npz(k) because (ns)0 and(npz)0 are both symmetric with respect to the σh plane.
According to the conclusions of the previous section, consideration of the (σv)m and(σd)m symmetries can only lead to the prediction of the possible nullity of the hns,npz(k)interaction term at the and X points because the (ns)0and(npz)0orbitals are symmetric and antisymmetric, respectively, with respect to (σv)0 (see Fig. 6.2c). In contrast, at any other point these two Bloch orbitals may interact since these two symmetry operations do not apply (eqns (6.1) and (6.2)). This result may appear surprising because the(ns)0and (npz)0 orbitals behave differently with respect to the(σv)0 plane. To realise that the Bloch orbitals BOns(k) and BOnpz(k) may interact at any k point other thanand X we can calculate the interaction term hns,npz(k) using the procedure proposed in Section 3.2.2. Let us recall that:
Fig. 6.2
(a) Definition of thenpzorbitals; (b) Ansystem generated by an M0unit cell centred on the atom A0; (c) symmetry properties with respect to (σd)0of the atomic orbitals of the reference cell.
hns,npz(k)=n (ns)0
√(n)| ˆh|BOnpz(k)
(6.5) To facilitate the discussion, we will use the H¨uckel approach even if our purpose is completely general and strictly governed by symmetry. We will also use the linear representation of the system to simplify the estimation of the interaction between (ns√)0
n and the Bloch orbital BOnpz(k)(see Fig. 6.3).
Because of the occurrence of exp(i2πka) and exp(−i2πka) coefficients in the expression of the BOnpz(k) orbital, the interaction term hns,npz(k) (eqn (6.5)) can be estimated to be 2βi sin(2πka). Thus, hns,npz(k) is non- zero if ka is different from 0 or 1/2. This clearly shows the peculiarity of pointsand X. For these two points the coefficients associated with orbitals (npz)+1 and(npz)−1are equal, so that the interaction between the BOns(k) and BOnpz(k)orbitals is effectively nil. For all other points this is not the case:
the coefficients associated with thenpz orbitals of BOnpz(k) at the –1 and +1 positions, exp(−i2πka)and exp(i2πka), are not equal and thus hns,npz(k)is
Fig. 6.3
(a) Interaction termβns,npz between two adjacent orbitals(ns)0and(npz)1; (b) interaction between(ns√)0
n and the Bloch orbital BOnpz(k).
not nil. Loosely speaking we can say that the two coefficients exp(−i2πka) and exp(i2πka)break down the symmetry with respect to the 0 position.
Concluding remarks
We can now generalise the results discussed in the three previous paragraphs.
When taking into account the symmetry properties of a cyclic periodic system we must successively consider the points below.
• The rotations with respect to the main axis perpendicular to the circle.
Their consideration allows the construction of the Bloch orbitals. Their associatedkpoint characterises the symmetry properties of these func- tions with respect to the rotations of theCngroup.
• The possibleσh symmetry leaving invariant every cell. Only the Bloch orbitals possessing the same symmetry properties with respect to this operation may interact.
• The possible symmetry operations other than rotations of the Cn group that do not leave invariant every cell:
- at and X, only the Bloch orbitals possessing the same symmetry properties with respect to these operations may interact
- at any other point these symmetry operations may be ignored.