Chapter 7. The Eigenvalue Problem

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7. The Eigenvalue Problem, December 17, 2009 14 Section 7.2. Degenerate eigenvalues § 10 Introduction. The eigenvalues of a matrix A are the roots of the characteristic polynomial det[A − λI]. There is no law that prevents some of these roots from being equal. When two or more eigenvalues are equal, we say that they are degenerate and the corresponding eigenvectors are also called degenerate. This name suggests a certain moral depravity for these eigenvalues and I am sure that they find it offensive. I could not find who introduced this terminology. Mathematiciansaremorepoliteanduseinstead the terms ‘multiple’ and ‘single’ eigenvalues, which is simpler and friendlier. One would think that finding operators that have identical eigenvalues should be as rare as finding by accident polynomials with multiple roots. But this is not so. Degenerate eigenvalues are fairly frequent and here I use two examples to suggest why. § 11 Particle in a cubic box with infinite potential-energy walls. One of the simplest models in quantum mechanics is a particle in a rectangular parallelepiped whose walls are infinitely repulsive. The Hamiltonian is Ĥ = ˆK x + ˆK y + ˆK z (70) where ˆK x = − ¯h2 ∂ 2 (71) 2m ∂x 2 is the operator representing the kinetic energy of the motion in direction x; ˆK y and ˆK z are defined similarly. In classical mechanics, the total kinetic energy is p 2 2m = p2 x 2m + p2 y 2m + p2 z (72) 2m where p is the momentum vector and p x is the momentum in the direction x (and analogously for p y , p z ). The operator ˆK x is the quantum analog of p 2 x/2m. The total energy of the particle is discrete and is given by (see H. Metiu, Physical Chemistry, Quantum Mechanics, page 104) E n,j,k = ¯h2 π 2 2m n 2 ¯h 2 π 2 2 j + + ¯h2 π 2 2 k (73) L x 2m L y 2m L z
7. The Eigenvalue Problem, December 17, 2009 15 where n, j, k cantakeanyofthevalues n, j, k =1, 2, 3,... (74) Here L x , L y ,andL z are the lengths of the box edges. The ket representing the pure states (the eigenstates of Ĥ) isdenotedby |n, j, k. For example, the symbol |2, 1, 1 means the state with n =2,j =1, k = 1 and the energy E 2,1,1 = ¯h2 π 2 2m ⎡ ⎣ 2 2 ⎤ 2 1 1 2 + + ⎦ (75) L x L y L z Exercise 4 Write down the energy of the states |1, 2, 1 and |1, 1, 2. If L x = L y = L z , the states are not degenerate (except if accidentally 2 2 2 n L x + k L y + m L z = n 2 L x + k 2 L y + m 2 L z in which case |n, j, m is degenerate with |n ,j ,m ). But consider the case of the cubic box, when L x = L y = L z = L (76) In this case the states |2, 1, 1, |1, 2, 1, and|1, 1, 2 havethesameenergy, equal to ¯h 2 π 2 2mL 2 2 2 +1+1 (77) They are degenerate states. Note the connection between degeneracy and symmetry. If L x = L y = L z , there is no degeneracy, except perhaps by accident. However, if L x = L y = L z , the box is symmetric and almost all states are degenerate. Exercise 5 Is the state |1, 1, 1 degenerate How about |3, 2, 1 Enumerate the states having the same energy as |3, 2, 1.
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Chapter 7. The Eigenvalue Problem

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