Eigenvectors degenerate

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. 

The determination of the eigenvectors for degenerate eigenvalues is somewhat more complicated and is not discussed here. 

Kuz min et al. (15) pointed out a standard result of classical mechanics If a configuration of particles has a plane of symmetry, then this plane is perpendicular to a principal axis (19). A principal axis is defined to be an eigenvector of the inertial tensor. Furthermore, if the configuration of particles possesses any axis of symmetry, then this axis is also a principal axis, and the plane perpendicular to this axis is a principal plane corresponding to a degenerate principal moment of inertia (19). 

Next, find the pair of eigenvectors associated with the degenerate eigenvalue of -2 root one eigenvector one ... 

Finding the first eigenvector was not too difficult. The degenerate eigenvectors are... 

What about the two degenerate eigenvectors v(l) and v(2) Are they also orthonormal So far, we know that these two eigenvectors have the structure... 

If we also wish to make the two degenerate eigenvectors orthogonal to one another... 

Thun, under the general conditions stated above, p.m. is valid to the second approximation ivith respect to the eigenvalues and to the first approximation with respect to the eigenvectors in the degenerate case too, provided =... 

Proof. The eigenvectors of a Hermitian matrix, if normalized and non degenerate, obey the relation — flw and consequently we can write ... 

Theorem. If A is Hermitian, its eigenvalues are real and its eigenvectors are orthogonal to each other provided they correspond to non-degenerate eigenvalues. 

Note that eigenvector orthonormality can also be safely assumed for the degenerate case, cii = ap without loss of generality.)... 

Each such vibration (6.32) is called a normal mode of vibration. For each normal mode, the vibrational amplitude Aim of each atomic coordinate is constant, but the amplitudes for different coordinates are, in general, different. The nature of the normal modes depends on the molecular geometry, the nuclear masses, and the values of the force constants ujk. The eigenvalues m of U determine the vibrational frequencies the eigenvectors of U determine the relative amplitudes of the vibrations of the q, s in each normal mode, since Ajm / A im = Ijm/L- For H2° here are 9-6-3 normal modes, and the solution of (6.17) and (6.18) yields the vibrational modes shown in Fig. 6.1. For some molecules, two or more normal modes have the same vibrational frequency (corresponding to two or more equal roots of the secular equation) such modes are called degenerate. For example, a linear triatomic molecule has four normal modes, two of which have the same frequency. See Fig. 6.2. The general classical-mechanical solution (6.30) is an arbitrary superposition of the normal modes. 

Exercise. Suppose T decomposes into two blocks as in fig. 8. Show that in this case the eigenvector ps is not unique, the eigenvalue 1 is degenerate. The system has two sets of states between which no transitions are possible. How should (5.3) be adapted to this case ... 

Exercise. Let W have an eigenvector with nonnegative components, some of which are zero. Then W is reducible. Also when W has a degenerate eigenvalue having two eigenvectors with nonnegative components. 

The time evolution is determined by the full effective Hamiltonian H and not by the rate matrix T alone. One cannot therefore discuss the time evolution without reference to the matrix H. Say, however, H and T commute, [H, T] = 0. A simple condition that ensures this result is that the bound states are strictly degenerate. If H and T commute, the eigenvectors of T evolve in time independently of one another. In the basis of states defined by the N eigenvectors of T there will be K states that will decay by direct coupling to the continuum and N - K states that are trapped forever. An arbitrary initial state is a linear combination of the N eigenvectors of T and hence can have a trapped component. 

The eigenvalues q)2 in eq. (2) are not necessarily all distinct, so the index j will now be replaced by the double index ak where a labels the distinct eigenvalues of D(q) and k= 1, 2,. .., 1(a) labels the linearly independent (LI) eigenvectors associated with the degenerate eigenvalue a. In this notation,... 

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