# ch07 - CHAPTER 7 1(a The system under consideration has...

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CHAPTER 7 1. (a) The system under consideration has rotational degrees of freedom, allowing it to rotate about two orthogonal axes perpendicular to the rigid rod connecting the two masses. If we define the z axis as represented by the rod, then the Hamiltonian has the form H = L x 2 + L y 2 2 I = L 2 - L z 2 2 I where I is the moment of inertia of the dumbbell. (b) Since there are no rotations about the z axis, the eigenvalue of L z is zero, so that the eigenvalues of the Hamiltonian are E = h 2 l ( l + 1) 2 I with l = 0,1,2,3,… (c) To get the energy spectrum we need an expression for the moment of inertia. We use the fact that I = M red a 2 where the reduced mass is given by M red = M C M N M C + M N = 12 × 14 26 M nuc leon = 6 .46 M nuc leon If we express the separation a in Angstroms, we get I = 6 .46 × (1 .67 × 10 - 27 kg )(10 - 10 m / A ) 2 a A 2 = 1 .08 × 10 - 46 a A 2 The energy difference between the ground state and the first excited state is 2 h 2 /2 I which leads to the numerical result E = (1 .05 × 10 - 34 J . s ) 2 1 .08 × 10 - 46 a A 2 kg . m 2 × 1 (1 .6 × 10 - 19 J / eV ) = 6 .4 × 10 - 4 a A 2 eV 2. We use the connection x r = s in θ co s φ ; y r = s in θ s in φ ; z r = co s θ to write

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Y 1 ,1 =- 3 8 π e i φ sin θ =- 3 8 π ( x + iy r ) Y 1 ,0 = 3 4 π cos θ = 3 4 π ( z r ) Y 1 , - 1 = ( - 1) Y 1,1 * = 3 8 π e - i φ sin θ = 3 8 π ( x - iy r ) Next we have Y 2 ,2 = 15 32 π e 2 i φ sin 2 θ = 15 32 π (cos2 φ + i sin2 φ )sin 2 θ = 15 32 π (cos 2 φ - sin 2 φ + 2 i sin φ cos φ )sin 2 θ = 15 32 π x 2 - y 2 + 2 ixy r 2 Y 2 ,1 =- 15 8 π e i φ s in θ co s θ =- 15 8 π ( x + iy ) z r 2 and Y 2 ,0 = 5 16 π (3co s 2 θ - 1 ) = 5 16 π 2 z 2 - x 2 - y 2 r 2 We may use Eq. (7-46) to obtain the form for Y 2 , - 1 andY 2 , - 2 .
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• Spring '05
• mokhtari
• Mass, Trigraph, Excited state, ground state, 2s cs 4s, o2φ snφ+ isn

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