Eigenvalues nondegenerate

E. Nondegenerate Eigenveetors of Hermitian Matriees are Orthogonal If two eigenvalues are different, then... 

There are now two possibilities the eigenvalue a, of A is either nondegenerate or degenerate. If a, is non-degenerate, then it corresponds to only one independent eigenfunction xpi, so that the function Bipi is proportional to... 

We next show that if the eigenvalue n of the number operator N is nondegenerate, then the eigenvalue n -h 1 is also non-degenerate. We begin with the assumption that there is only one eigenvector with the property that... 

Thus, all the eigenvectors (n + l)i) corresponding to the eigenvalue -h 1 are proportional to a ) and are, therefore, not independent since they are proportional to each other. We conclude then that if the eigenvalue n is nondegenerate, then the eigenvalue -h 1 is non-degenerate. 

Let the eigenvalue w be fixed and assume that fit is nondegenerate and unit-normalized. The restriction to nondegenerate eigenstates will be relaxed in Section V, but for now we consider only pure-state density matrices. The A -electron density matrix for the pure state fit is... 

Thus, let us assume that two different external potentials can each be consistent with the same nondegenerate ground-state density po- We will call these two potentials Va and wj, and the different Hamiltonian operators in which they appear and Ht,. With each Hamiltonian will be associated a ground-state wave function Pq and its associated eigenvalue Eq. The variational theorem of molecular orbital theory dictates that the expectation value of the Hamiltonian a over the wave function b must be higher than the ground-state energy of a, i.e.. 

Problem Write the eigenvalue problem (S9.1-15) in Dirac notation, and prove the fundamental properties (S9.1-16), (S9.1-17) of the eigenvectors and eigenvalues for Hermitian A, assuming the eigenvalues are nondegenerate (unequal). 

We can now show that the eigenfunctions for a molecule are bases for irreducible representations of the symmetry group to which the molecule belongs. Let us take first the simple case of nondegenerate eigenvalues. If we take the wave equation for the molecule and carry out a symmetry operation, / , upon each side, then, from 5.1-1 and 5.1-2 we have... 

We are interested in the solution of the time-independent Schrodinger equation (f is an eigenfunction and E a nondegenerate eigenvalue)... 

It is interesting to note that the density operator T has a structure that essentially captures the extreme configuration, i.e., the associated secular equation reveals a nondegenerate large eigenvalue XLp — (n — 1 )p2 and a small (n — 1) degenerate A.s = p2. Hence the density operator becomes... 

The first equation as in the nondegenerate case, can be satisfied by taking the correction to be orthogonal to the unperturbed ground state vector which is satisfied by any linear combination of vectors from the manifold related to the /-th degenerate eigenvalue. Inserting the required expansion we get ... 

In general the relaxation to equilibrium of E(t) is nonexponential, since the rate matrix in the master equation has an infinite number of (in principle) nondegenerate eigenvalues if there are an infinite number of states n). There are, however, two instances where the relaxation is approximately exponential. In the first instance one assumes that the initial nonequilibrium state has appreciable population only in the first two oscillator eigenstates, and further that k,. 0 k,, m and k0. t k0 m for m > 2. If one neglects terms involving these small rate constants, the master equation reduces to a pair of coupled rate equations for a two-level system ... 

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