Left-hand eigenvectors

Each eigenvalue X, has associated with it a column, or right-hand eigenvector r, and a row, or left-hand eigenvector s . The X are given by the roots of the equation PX — XI = 0 (I is the unit matrix). 

The eigenvector corresponding to Aq is called the dominant eigenvector 1q. All its components are strictly positive /jo > 0 (i = 1,2,.. . ). The left-hand eigenvectors are given by the rows of the inverse matrix L. ... 

Eor second-order properties derived from the perturbation, we have a particularly attractive method that uses the EOM eigenvectors to represent the perturbed wavefunctions. Since the right- and left-hand eigenvectors form a complete set, we know we can write the perturbed wavefunctions in the form. 

Here, /j and rj are the l" left- and the J right-hand eigenvectors of the non-Hermitian Hamiltonian H. The operator is represented on the space spanned by the manifold created by the excitations out of a Hartree-Fock reference determinant, including the null excitation (the reference function). When we calculate the transition probability between a ground state g) and an excited state ]e), we need to evaluate and The reference function is a right-... 

The matrix Q can now be transformed into a stochastic matrix, which will be descriptive of the restricted random walks rather than of their generation employing probabilities based on unrestricted walk models. The transformation is performed as follows Let Xt be the largest eigenvalue of the matrix Q, and let Sj be the corresponding left-hand side eigenvector (defined by SjQ = X ). Let A be a diagonal matrix with elements a(i,j) = (/) 8st = [ 1(1),. v,(2),..., (v)] and 8(i,j) is the... 

Exercise. W has one eigenvalue zero. Hence (4.11) can only be solved when the inhomogeneous term on the left is orthogonal to the left null eigenvector of W. Show that this condition is fulfilled. On the other hand, the solution is not unique show that the requirement (4.9) takes care of that. 

In these equations, H is the matrix representation, while Rk and Lk are vectors, Lk being a row vector and Rk being a column vector. The left- and right-hand eigenvectors can be normalized so that they share a biorthonormal relationship [24] ... 

The leading term of 1, which does not appear in T (cf. Eq. [29]), is required in order that the left- and right-hand eigenvectors have unit overlap with one another. Unlike the cluster operators, f , the operators act to the left on Therefore, it is convenient to define them as de-excitation operators (or, equivalently, as bra-state excitation operators),... 

The task of determining the left-hand ground state eigenvector of H is thus reduced to determining the amplitudes 1% . The ground state coupled cluster energy may then be written as... 

Biorthonormality of the left-hand and right-hand eigenvectors may be enforced such that... 

It should be noted that Eq. (26) does not contain an -dependent matrix element on its left-hand side as do the MPn EOM equations (Eq. (14)) or the equivalent GE equations. This is because the EOM-CC equations are usually not subjected to the operator-space partitioning that the GF and MPn-based EOM theories commonly employ. It should also be noted that the operators appearing on the left-hand side of Eq. (26) produce non-Hermitian matrices. As a result, there arise non-orthogonal left- and right-eigenvectors in solving Eq. (26). As discussed in Refs. [34,35], it is important to compute both sets of eigenvectors if one wishes to compute, e.g. photo-electron intensities. 

Table 1 Left- and right-hand eigenvectors of zero-order Hamiltonians in p-MCPT and u-MCPT...

Excitation energies are thus computed as poles of the dynamic polarizability, that is, as the values of co leading to zero eigenvalues on the left-hand side of the matrix of Eq. 1.12. In the framework of the above equations, an efficient fast iterative solution for the lowest eigenvalue/excitation energies can be attained [62]. Oscillator strengths can also be obtained by the eigenvectors of Eq. 1.12, as explained by Casida [17]. 

The eigenvalue analysis of (3.12) is performed using the HQR algorithm. The left- and right-hand eigenvectors of the original problem are then restored in the form... 

An eigenvalue and its associated eigenvector of the Hessian of p (a principal curvature and its associated axis) at a critical point define a onedimensional system. If the eigenvalue or curvature is negative, then p is a maximum at the critical point on this axis and a gradient vector will approach and terminate at this point from both its left- and right-hand side as illustrated in Fig. 2.6 for the case (1, — 1), a system of rank 1 and signature... 

The left relation in (2.54) implies that a )o and its true counterpart have the same length irrespective of how the former is normed. Hence, the mappings (AT, L) actually conserve the norms of a set of right eigenvectors. On the other hand, the right equation in (2.54) shows that... 

Comparing zero-order operators of Eqs. (2) and (3) one may observe that an advantage of Schmidt-orthogonalization is getting the zero-order Hamiltonian symmetric at least in the one-dimensional reference space spanned by 0). Left- and right-hand zero-order eigenvectors expressed in terms of determinants HF), and MR function 0) are listed in Table 1 for completeness. Detailed derivation of the reciprocal vectors has been shown in an earlier report [23],... 

Equating the left- and right-hand sides separately in each eigenvector direction 5delds the following imcoupled set of equations for each cj t). 

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