Chapter 7. The Eigenvalue Problem

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7. The Eigenvalue Problem, December 17, 2009 16 In the Schrödinger representation, the state |n, j, k is (see Metiu, Eq. 8.31) x, y, z | n, j, k ≡ ψ n,j,k (x, y, z) 2 = sin L x nπx L x 2 L y sin jπy 2 kπz sin L y L z L z (78) For the example of the cube, x, y, z | 2, 1, 1 = 8 2πx L sin sin 3 L πy sin L πz L (79) and x, y, z | 1, 2, 1 = 8 πx L sin sin 3 L 2πy sin L πz L (80) A similar equation holds for x, y, z | 1, 1, 2. These three states have the same energy but they are different states. What is the physical difference between them Let us look at the kinetic energy in the x-direction: ˆK x x, y, z | 2, 1, 1 = − ¯h2 2m ∂x ψ 2,1,1(x, y, z) 2 = ¯h2 2π 2 ψ 2,1,1 (x, y, z) (81) 2m L I obtained the second line from the first by taking the second derivative of ψ 2,1,1 (x, y, z) givenbyEq.79. Weseethat|2, 1, 1 is an eigenstate of ˆKx , with the eigenvalue ¯h2 2π 2.Inthesameway,youcanshowthat|2, 2m L 1, 1 is an eigenstate of ˆK y and ˆK z , with the eigenvalues ¯h2 1π 2. 2m L The total energy is the sum of those three kinetic energies. In the state |2, 1, 1, the particle has higher kinetic energy in the x-direction. We can analyze in the same way the states |1, 2, 1 and |1, 1, 2. Table 1 gives the result. The three degenerate states have the high (i.e. ¯h2 2π 2) 2m L kinetic energy in different directions. Can we distinguish these three states by some measurement We can. The particle in the state |2, 1, 1 will emit a photon in a different direction than will a particle in |1, 2, 1 or |1, 1, 2. The degenerate states |2, 1, 1, |1, 2, 1, and|1, 1, 2 are different physical states, even though they have the same energy. ∂ 2
7. The Eigenvalue Problem, December 17, 2009 17 state K x K y K z E ¯h |2, 1, 1 2 2 2π ¯h 2 2 1π ¯h 2 2 1π 6¯h 2 π 2 2m L 2m L 2m L 2mL 2 ¯h |1, 2, 1 2 2 1π ¯h 2 2 2π ¯h 2 2 1π 6¯h 2 π 2 2m L 2m L 2m L 2mL 2 ¯h |1, 1, 2 2 2 1π ¯h 2 2 1π ¯h 2 2 2π 6¯h 2 π 2 2m L 2m L 2m L 2mL 2 Table 1: Kinetic and total energy We have also found a curious thing. The degenerate states |2, 1, 1, |1, 2, 1, and|1, 1, 2 are eigenstates of Ĥ but also of ˆKx , ˆKy ,and ˆK z . Is this a coincidence No. Except for accidental degeneracies, we always find that the states of an observable (H) that are degenerate are so because they are also eigenstates of other observables (K x ,K y ,K z ). Finally, note that ˆK x , ˆKy ,and ˆK z all commute with Ĥ. Isthisacoincidence No. We will prove in a later chapter that operators that commute have common eigenfunctions. Exercise 6 Take L x = L y = L and L z = L. Does the system have degenerate states Are these states eigenstates of ˆK x , ˆKy ,and ˆK z Howaboutthe case L z = L y = L z § 12 The hydrogen atom. The hydrogen atom is another system whose energy eigenstates are degenerate. Its energy is (see Metiu, Quantum Mechanics, page300) E n = − 1 μe 4 n 2 2(4π 0 ) 2¯h 2 (82) and the states |n, ,m are labeled by three quantum numbers: n controls the energy, controls the square of the angular momentum (essentially, this is the rotational energy), and m controls the projection of the angular momentum ontheOZaxis. Thestates|n, ,m are eigenvectors of three operators. These
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Chapter 7. The Eigenvalue Problem

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