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Permutation Group and Determinants (Dated: September 16, 2021)

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Permutation Group and Determinants (Dated: September 16, 2021) Permutation group and determinants (Dated: September 16, 2021) 1 I. SYMMETRIES OF MANY-PARTICLE FUNCTIONS Since electrons are fermions, the electronic wave functions have to be antisymmetric. This chapter will show how to achieve this goal. The notion of antisymmetry is related to permutations of electrons’ coordinates. Therefore we will start with the discussion of the permutation group and then introduce the permutation-group-based definition of determinant, the zeroth-order approximation to the wave function in theory of many fermions. This definition, in contrast to that based on the Laplace expansion, relates clearly to properties of fermionic wave functions. The determinant gives an N-particle wave function built from a set of N one-particle waves functions and is called Slater’s determinant. II. PERMUTATION (SYMMETRIC) GROUP Definition of permutation group: The permutation group, known also under the name of symmetric group, is the group of all operations on a set of N distinct objects that order the objects in all possible ways. The group is denoted as SN (we will show that this is a group below). We will call these operations permutations and denote them by symbols σi. For a set consisting of numbers 1, 2, :::, N, the permutation σi orders these numbers in such a way that k is at jth position. Often a better way of looking at permutations is to say that permutations are all mappings of the set 1, 2, :::, N onto itself: σi(k) = j, where j has to go over all elements. Number of permutations: The number of permutations is N! Indeed, we can first place each object at positions 1, so there are N possible placements. For each case, we can place one of the remaining N − 1 objects at the second positions, so that the number of possible arrangements is now N(N − 1). Continuing in this way, we prove the theorem. Example: For three numbers: 1, 2, 3, there are the following 3! = 6 arrangements: 123, 132, 213, 231, 312, 321. Notation: One can use the following “matrix” to denote permutations: ! 1 2 ::: k ::: N σ = σ(1) σ(2) ::: σ(k) ::: σ(N) The order of columns in the matrix above is convenient, but note that if the columns were ordered differently, this would still be the same permutation. An example of a permutation in this notation is ! 1 2 3 4 σ = 3 4 1 2 2 Another way if writing a permutation is to include only the second row. An example of a permutation in this notation is σ = (3412): Multiplication: We define the operation of multiplication within the set of permutations as (σ ◦ σ 0)(k) = σ(σ 0(k)). For example, if ! ! 1 2 3 4 1 2 3 4 σ 0 = σ = 3 4 1 2 2 4 3 1 then ! 1 2 3 4 σ ◦ σ 0 = : 3 1 2 4 Symmetric group: We can now check if these operations satisfy the group postulates ◦ 2 • Closure: σ σ 0 SN . The proof is obvious since the product of permutations gives a number from the set, therefore is a permutation. • Existence of unity I: this is the permutation σI (k) = k. 1 1 • Existence of inverse, i.e., for each σ there exists σ − such that σ ◦ σ − = I. Clearly, 1 the inverse can be defined such that if σ(k) = j, then σ − (j) = k. • Multiplications are associative: ◦ ◦ ◦ ◦ σ3 (σ2 σ1) = (σ3 σ2) σ1: Proof is in a homework problem. ◦ σ SN = SN : One important theorem resulting from these definitions is that the set of products of a single permutation with all elements of SN is equal to SN ◦ σ SN = SN : Proof: Due to closure, the only possibility of not reproducing the whole group is that two different elements of SN are mapped by σ onto the same element: σ ◦ σ 0 = σ 000 = σ ◦ σ 00: 1 Multiplying this equation by σ − , we get σ 0 = σ 00 which contradicts our assumption. f 1g f 1g σ − = SN : Another theorem states that σ − = SN . This is equivalent to saying 1 that σ and σ − are in one-to-one correspondence. Indeed, assume that there are two 3 ◦ ◦ 1 permutations that are inverse to σ: σ1 σ = I = σ2 σ. Multiplying this by σ − from the right, we get that σ1 = σ2. Transpositions: A transposition is the simplest possible permutation other than σI , i.e., a permutation involving only two elements: 8 > > σ(i) = j ! <> 1 2 ::: i ::: j ::: N τ = τij = (ij) = > σ(j) = i = : > 1 2 ::: j ::: i ::: N :> σ(k) = k for k , i;j Permutation as product of transpositions: One important property of permutations is that each permutation can be written as a product transpositions. To prove that any permutation can be written as a product of transpositions, we just construct such a product. For a permutation σ written as ! 1 2 ::: k ::: N σ = (1) i1 i2 ::: ik ::: iN f g first find i1 in the set 1;2;:::;N and then transpose it with 1 (unless i1 = 1, in which , case do nothing). This maps i1 1. Then consider the set with i1 removed, find i2, and transpose it with the number in the second position (it will be 2 if the first transposition did not affect this place). Continuing in this way, we get the mapping of expression (1) which proves the theorem. As an example, consider ! 1 2 3 4 σ = : 2 4 3 1 We first look for i1 = 2 and then transpose it with 1: τ τ 1234 !12 2134 !14 2431 where in the second step we looked for i2 = 4 and transposed it with the element in the second positions, i.e., with 1. Then we looked for i3 = 3 and did nothing, similarly with i4 = 1. Therefore, σ can be written as ◦ σ = τ14 τ12: Let us check explicitly that the right-hand side indeed gives σ σ(1) = τ14(τ12(1)) = τ14(2) = 2; σ(2) = τ14(τ12(2)) = τ14(1) = 4; σ(3) = τ14(τ12(3)) = τ14(3) = 3; σ(4) = τ14(τ12(4)) = τ14(4) = 1: The decomposition of a permutation into transposition is not unique as we can always add τijτij = 1. Parity of permutation: Although the number of transpositions in a decomposition is not unique, this number is always either odd or even for a given permutation. The proof of 4 − πσ this important theorem is given as a homework. Thus, ( 1) , where πσ is the number of permutations in an arbitrary decomposition, is always 1 or −1 for a given permutation and we can classify each permutation as either odd or even. We say that each permutation has a definitive parity. Parity of inverse permutation: One theorem concerning the parity of permutations is that πσ π 1 that (−1) = (−1) σ− , i.e., that a permutation and its inverse have the same parity. To prove it, first note that ◦ 1 1 ◦ 1 (σ1 σ2)− = σ2− σ1− ◦ which is obvious if we multiply on the left with σ1 σ2. We can write σ as ◦ ◦ ··· ◦ σ = τ1 τ2 τr: Then 1 1 ◦ 1 ◦ ··· ◦ 1 σ − = τr− τr− 1 τ1− : − 1 Thus, π = π 1 . Note that τ = τ = τ , but this is not needed for the proof. σ σ − ij− ij ji III. DETERMINANT × A Definition of determinant: For a general N N matrix with elements aij, the determinant is defined as a a ::: a 11 12 1N a a ::: a X jAj ≡ A 21 22 2N − πσ det = = ( 1) aσ(1)1aσ(2)2 ::: aσ(N)N (2) ::::::::: σ aN1 aN2 ::: aNN where the sum is over all permutations of numbers 1 to N and πσ is the parity of the permutation. There are several important theorems involving determinants that we will now prove. Determinant is invariant to matrix transposition: Show that that jAj = jAT j; which also means that the definition (2) can be written as X jAj − πσ = ( 1) a1σ(1)a2σ(2) ::: aNσ(N): (3) σ AT ≡ A f g Proof. Using notation 0 = aij0 , we can write the definition (2) as X X jAT j − πσ − πσ = ( 1) aσ0 (1)1aσ0 (2)2 ::: aσ0 (N)N = ( 1) a1σ(1)a2σ(2) ::: aiσ(i) ::: aNσ(N) σ σ 5 (since aij0 = aji). Note that the term with a given σ in the equation written above is not the same as the term with the same σ in Eq. (2) since aij , aji. Let us reshuffle the terms in the last equations in such a way that the factor with σ(i) = 1 is in the first position (we can do it since product does not depend on the order of factors). There must be one such a factor aiσ(i) since σ(i) runs over the numbers 1;2;:::;N. Denote this value of i in a ≡ given term by i1, i1 = i so that aiσ(i) ai11 and move this factor to the first position in the product to get a1σ(1)a2σ(2) ::: ai 1 ::: aNσ(N) = ai 1a2σ(2) ::: ai 1σ(i 1) ai +1σ(i +1) ::: aNσ(N) 1 1 1− 1− 1 1 Next, look for σ(i) = 2, denote i = i2, so that σ(i2) = 2, and move ai22 it the second position ≡ in the product. Going forward, we move the term with σ(i) = k = σ(ik), i.e., aiσ(i) aikk. Continuing, one eventually gets a1σ(1)a2σ(2) ::: aσ(N)N = ai11ai22 ::: aikk ::: aiN N : (4) 1 Since k = σ(ik), ik = σ − (k). Therefore, we can write X T π jA j = (−1) σ a 1 a 1 ::: a 1 ::: a 1 σ − (1)1 σ − (2)2 σ − (i)i σ − (N)N σ Therefore, if we sum all possible terms on the right-hand side of Eq.
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