Operator Properties and Mathematical Groups

MIT OpenCourseWarehttp://ocw.mit.edu 5.04 Principles of Inorganic Chemistry II �� Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 2: Operator Properties and Mathematical Groups The inverse of A (defined as (A)–1) is B if A ⋅ B = E For each of the five symmetry operations: (E)–1 = E (E)–1 ⋅ E = E ⋅ E = E (σ)–1 = σ (σ)–1 ⋅σ = σ⋅σ = E (i)–1 = i (i)–1 ⋅ i = i ⋅ i = E n–m(Cnm)–1 = Cn (Cnm)–1 ⋅ Cnm = Cnn–m ⋅ Cnm = Cnn = E e.g. (C52)–1 = C53 since C52 ⋅ C53 = E (Snm)–1 = Snn–m (n even) (Snm)–1 ⋅ Snm = Snn–m ⋅ Snm = Snn = Cnn ⋅ σhn = E (Snm)–1 = Sn2n–m (n odd) (Snm)–1 ⋅ Snm =Sn2n–m ⋅ Snm = Sn2n = Cn2n ⋅ σh2n = E Two operators commute when A ⋅ B = B ⋅ A Example: Do C4(z) and σ(xz) commute? C4(z) σ(xz)(x1, y1, z1) C4(z)(x1, –y1, z1) σd ’ ∴ C4(z)σ(xz) = σd ’ z1 does not change with σd ’ 5.04, Principles of Inorganic Chemistry II Lecture 2 Prof. Daniel G. Nocera Page 1 of 6 … or analyzing with matrix representations, ⎡010⎤00⎡⎤0−10⎤1 0 C4(z) ⋅σ xz = σ d´ Now applying the operations in the inverse order, ⎡⎢0– ⎢⎢⎣⎥⎥⎥⎦⎥⎥⎥⎦⎢⎢⎢⎣⎥⎥⎥⎦100−10−1 00⋅=001010 01⎢⎢⎢⎣σ(xz) C4(z)(x1, y1, z1) σ(xz)(y1, x1, z1) σd ∴ σ(xz)C4(z) = σd … or analyzing with matrix representations, 0001010⎡⎤⎤⎤⎥⎥⎥⎦⋅⎥⎥⎥⎦⎥⎥⎥⎦1010000–=0100101⎢⎢⎢⎣⎡⎢⎢⎢⎣⎡⎢⎢⎢⎣10 0– 1 00 σ xz ⋅ C4(z) = σ d ∴ C4(z)σ(xz) = σd’ ≠ σ(xz)C4(z) = σd so C4(z) does not commute with σ(xz) 5.04, Principles of Inorganic Chemistry II Lecture 2 Prof. Daniel G. Nocera Page 2 of 6 A collection of operations are a mathematical group when the following conditions are met: closure: all binary products must be members of the group identity: a group must contain the identity operator inverse: every operator must have an inverse associativity: associative law of multiplication must hold (A ⋅ B) ⋅ C = A ⋅ (B ⋅ C) (note: commutation not required… groups in which all operators do commute are called Abelian) Consider the operators C3 and σv. These do not constitute a group because identity criterion is not satisfied. Do E, C3, σv form a group? To address this question, a stereographic projection (featuring critical operators) will be used: So how about closure? C3 ⋅ C3 = C32 (so C32 needs to be included as part of the group)C3 ⋅ σv = ? ∴ C3 ⋅σv = σv´ Thus E, C3 and σv are not closed and consequently these operators do not form a group. Is the addition of C32 and σv´ sufficient to define a group? In other terms, are there any other operators that are generated by C3 and σv? … the proper rotation axis, C3: 5.04, Principles of Inorganic Chemistry II Lecture 2 Prof. Daniel G. Nocera Page 3 of 6 C3C3 ⋅ C3 = C32C3 ⋅ C3 ⋅ C3 = C32 ⋅ C3 = C3 ⋅ C32 = E C3 ⋅ C3 ⋅ C3 ⋅ C3 = E ⋅ C3 = C3etc.∴ C3 is the generator of E, C3 and C32 note: these three operators form a group … for the plane of reflection, σv σv σv ⋅σv = E σv ⋅σv ⋅σv = E ⋅σv = σvetc. So we obtain no new information here. But there is more information to be gained upon considering C3 and σv. Have already seen that C3 ⋅σv = σv’ … how about σv ⋅ C3? ∴ C3 ⋅σv = σv” Will discover that no new operators may be generated. Moreover one finds –1 −1−1 −1(C )⎛⎜⎝σv ′⎞⎟⎠⎛⎜⎝σv ′′⎞⎟⎠σE−1C3 −1 2 v3 inverses ↓↓↓↓↓ ↓ ′ ′′ E C32 C3 σ v σ v σ v The above group is closed, i.e. it contains the identity operator and meets inverseand associativity conditions. Thus the above set of operators constitutes a mathematical group (note that the group is not Abelian). Some definitions:5.04, Principles of Inorganic Chemistry II Lecture 2 Prof. Daniel G. Nocera Page 4 of 6 Operators C3 and σv are called generators for the group since every element of the group can be expressed as a product of these operators (and their inverses). The order of the group, designated h, is the number of elements. In the above example, h = 6. Groups defined by a single generator are called cyclic groups. Example: C3 → E, C3, C32 As mentioned above, E, C3, and C32 meet the conditions of a group; they form a cyclic group. Moreover these three operators are a subgroup of E, C3, C32, σv, σv’,σv”. The order of a subgroup must be a divisor of the order of its parent group. (Example hsubgroup = 3, hgroup = 6 … a divisor of 2.) A similarity transformation is defined as: ν-1 ⋅ A ⋅ ν = B where B is designated the similarity transform of A by x and A and B are conjugates of each other. A complete set of operators that are conjugates to one another is called a class of the group. Let’s determine the classes of the group defined by E, C3, C32, σv, σv’,σv”… the analysis is facilitated by the construction of a multiplication table may construct easily using stereographic projections E–1 ⋅ C3 ⋅ E = E ⋅ C3 ⋅ E = C3 C3–1 ⋅ C3 ⋅ C3 = C32 ⋅ C3 ⋅ C3 = E ⋅ C3 = C3 (C32)–1 ⋅ C3 ⋅ C32 = C3 ⋅ C3 ⋅ C32 = C3 ⋅ E = C3 σv–1 ⋅ C3 ⋅σv = σv ⋅ C3 ⋅σv= σv ⋅σv’ = C32 (σv’)–1 ⋅ C3 ⋅σv’ = σv’ ⋅ C3 ⋅σv’ = σv’ ⋅σv’’ = C32 (σv’’)–1 ⋅ C3 ⋅σv’’ = σv’’ ⋅ C3 ⋅σv’’ = σv’’ ⋅σv = C32 ∴ C3 and C32 from a class 5.04, Principles of Inorganic Chemistry II Lecture 2 Prof. Daniel G. Nocera Page 5 of 6 Performing a similar analysis on σv will reveal that σv, σv’ and σv’’ form a class and E is in a class by itself. Thus there are three classes: E, (C3, C32), (σv, σv’, σv’’) Additional properties of transforms and classes are: no operator occurs in more than one class order of all classes must be integral factors of the group’s order in an Abelian group, each operator is in a class by itself. 5.04, Principles of Inorganic Chemistry II Lecture 2 Prof. Daniel G. Nocera Page 6 of 6