Matrix commutator

The parity matrix commutes with the first entropy matrix, gS = Sg, because there is no coupling between variables of opposite parity at equilibrium, (xiXj)0 = 0 if e,e - = — 1. If variables of the same parity are grouped together, the first entropy matrix is block diagonal. 

The odd expansion coefficients are block-adiagonal and hence c j I c I [g.3+ k3] = 0. This means that the coefficient of x on the right hand side is identically zero. (Later it will be shown that 0 and that could be nonzero.) Since the parity matrix commutes with the block-diagonal even coefficients, the reduction condition gives... 

The commutator of two square matrices is defined as [A, B] = AB — BA. If [A, B] = 0 the matrices A and B are said to commute. All diagonal matrices commute, every matrix commutes with itself, every matrix commutes with its inverse, and every matrix commutes with the identity matrix. If A and B are Hermitian, then [A, B] = 0 if and only if both matrices may be diagonalized by the same unitary matrix. This does not mean that every matrix that diagonalizes A will diagonalize B but that at least one such matrix exists that will diagonalize both. This relation may be used to determine the classes of matrices that can be diagonalized by a unitary transformation. Let A be an arbitrary matrix and define A + = (A -I- A )/2 and A = (A - AV2i. Then... 

Thus, the commutator [1-, I,] is zero, since the multiple of the unit matrix commutes with every 3X3 matrix. 

Theorem 1 If a matrix commutes with all the matrices of an irreducible representation, the matrix must be a multiple of the unit matrix ... 

At self-consistency, the density and Fock matrix commute. That is, for an SCF ... 

A unit matrix 1 is a special diagonal matrix. Every diagonal element has a value of unity. A unit matrix times any matrix (of appropriate dimension) gives that same matrix as product. That is, 1A = A1 = A. It follows immediately that the unit matrix commutes with any square matrix of the same dimension. 

The first condition, (4.31a), means that the matrices must be unitary. For matrices of rank 2, any unitary matrix can be expressed as a linear combination of the 2 x 2 unit matrix and the Pauli matrices. These four will not do, because even though the Pauli matrices anticommute, the unit matrix commutes with the Pauli matrices, and therefore the anticommutation requirement cannot be met. Thus, the rank of the a and P matrices must be greater than 2. Through some elementary matrix algebra, it may... 

The identity element is always in a class by itself, since the identity matrix commutes with every other matrix. If a group includes the inversion operator, this operator is also in a class by itself. 

In the MO ba.sis, therefore, the Fock matrix commutes with the density matrix at conveigence. 

It is more convenient to re-express this equation in Liouville space [8, 9 and 10], in which the density matrix becomes a vector, and the commutator with the Hamiltonian becomes the Liouville superoperator. In tliis fomuilation, the lines in the spectrum are some of the elements of the density matrix vector, and what happens to them is described by the superoperator matrix, equation (B2.4.25) becomes (B2.4.26). 

For a coupled spin system, the matrix of the Liouvillian must be calculated in the basis set for the spin system. Usually this is a simple product basis, often called product operators, since the vectors in Liouville space are spm operators. The matrix elements can be calculated in various ways. The Liouvillian is the conmuitator with the Hamiltonian, so matrix elements can be calculated from the commutation rules of spin operators. Alternatively, the angular momentum properties of Liouville space can be used. In either case, the chemical shift temis are easily calculated, but the coupling temis (since they are products of operators) are more complex. In section B2.4.2.7. the Liouville matrix for the single-quantum transitions for an AB spin system is presented. 

The normal rules of association and commutation apply to addition and subhaction of matrices just as they apply to the algebra of numbers. The zero matrix has zero as all its elements hence addition to or subtraction from A leaves A unchanged... 

It is helpful to remember that the element py is formed from the ith row of the first matrix and the jth column or the second matrix. The matrix product is not commutative. That is, AB BA in general. 

For a more complicated [B] matrix that has, say, n columns whereas [A] has m rows (remember [A] must have p columns and [B] must have p rows), the [C] matrix will have m rows and n columns. That is, the multiplication in Equations (A.21) and (A.22) is repeated as many times as there are columns in [B]. Note that, although the product [A][B] can be found as in Equation (A.21), the product [B][A] is not simultaneously defined unless [B] and [A] have the same number of rows and columns. Thus, [A] cannot be premultiplied by [B] if [A][B] is defined unless [B] and [A] are square. Moreover, even if both [A][B] and [B][A] are defined, there is no guarantee that [A][B] = [B][A]. That is, matrix multiplication is not necessarily commutative. 

This shows that, when we have found the correct electron density matrix and correctly calculated the Hartree-Fock Hamiltonian matrix from it, the two matrices will satisfy the condition given. (When two matrices A and B are such that AB = BA, we say that they commute.) This doesn t help us to actually find the electron density, but it gives us a condition for the minimum. 

Both in Eq. (8-149) and Eq. (8-147), we have written the function in the center of the integrand simply for ease of visual memory in fact both /(q) and F(q,q ) commute with all the B-operators and their positions are immaterial. The B-operators operate on vectors > in occupation number space, so that we can evaluate the matrix elements of F in occupation number representation, viz., Eq. (8-145), either from Eq. (8-147) or from Eq. (8-149). 

All other products of y-matrices can, by using the commutation rules, be reduced to one of these sixteen elements. The proof of their linear independence is based upon the fact that the trace of any of these matrices except for the unit matrix, I, is zero. If Tr is any one of these matrices, then rrr, generates again one of the T s, the unit matrix... 

Theorem B.—Any four-by-four matrix that commutes with a set of y is a multiple of the identity. 

The proof of this theorem follows from theorem A A four-by-four matrix that commutes with the y commuted with their products and hence with an arbitrary matrix. However, the only matrices that commute with every matrix are constant multiples of the identity. Theorem B is valid only in four dimensions, i.e., when N = 4. In other words the irreducible representations of (9-254) are fourdimensional. 

Summarizing, we have noted that the Heisenberg operators Q+(t) obey field free equations i.e., that their time derivatives are given by the commutator of the operator with Ha+(t) = Ho+(0) and that this operator H0+(t) is equal to H(t) = H(0). The eigenstates of H0+ are, therefore, just the eigenstates of H. We can, therefore, identify the states Tn>+ with the previously defined >ln and the operator [Pg.602]

The representation of these commutation rules is again fixed by the requirement that there exist no-particle states 0>out and 0>ln. The -matrix is defined as the unitary operator which relates the in and out fields ... 

If we restrict ourselves to the case of a hermitian U(ia), the vanishing of this commutator implies that the /S-matrix element between any two states characterized by two different eigenvalues of the (hermitian) operator U(ia) must vanish. Thus, for example, positronium in a triplet 8 state cannot decay into two photons. (Note that since U(it) anticommutes with P, the total momentum of the states under consideration must vanish.) Equation (11-294) when written in the form... 

Here it is taken into account that density matrix p, being a scalar, commutates with any rotation operator, and diq defined in Eq. (7.51) is used. After an analogous transformation, in master equation (7.51) there remains the Hamiltonian, which does not depend on e ... 

First consider the dipole operator (O = r). The matrix elements on rhs of eq. 17 are thus just the dipole transition moments, and the commutator becomes C = -ip. As the exact solution (complete basis set limit) to the RPA is under consideration, we may use eq. 10 to obtain... 

It is also of interest to study the "inverse" problem. If something is known about the symmetry properties of the density or the (first order) density matrix, what can be said about the symmetry properties of the corresponding wave functions In a one electron problem the effective Hamiltonian is constructed either from the density [in density functional theories] or from the full first order density matrix [in Hartree-Fock type theories]. If the density or density matrix is invariant under all the operations of a space CToup, the effective one electron Hamiltonian commutes with all those elements. Consequently the eigenfunctions of the Hamiltonian transform under these operations according to the irreducible representations of the space group. We have a scheme which is selfconsistent with respect to symmetty. 

The operation of matrix multiplication can be shown to be associative, meaning that X(YZ) = (XY)Z. But, it is not commutative, as in general we will have that XY YX. Matrix multiplication is distributive with respect to matrix addition, which implies that (X + Y)Z = XZ + YZ. When this expression is read from right to left, the process is called factoring-out [4]. 

If the product AB equals the product BA, then A and B commute. Any square matrix A commutes with the unit matrix of the same order... 

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