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Unformatted text preview: PH 216, Problem set 2, Due 4/13 1. A mass m is attached by a massless rod of length l to a pivot p and swings in a vertical plane under the influence of gravity, with deflection angle from the vertical. (a) In the small angle approximation find the energy levels of the system. (b) Find the lowest order correction to the ground state energy resulting from inaccuracy of the small angle approximation. Solution: (a) Take the equilibrium position of the point mass as the zero point of potential energy. For small angle approximation, the potential energy of the system is 1 V = mgl(1  cos ) mgl2 , 2 and the Hamiltonian is 1 1 H = ml2 2 + mgl2 . 2 2 By comparing it with the onedimensional harmonic oscillator, we obtain the energy levels of the system 1 En = n + , 2 where = g/l. 1 H = mgl(1  cos )  mgl2 2  1 1 mg 4 mgl4 =  x, 24 24 l3 1 2 2 exp  x 1/2 2 (b) The perturbation Hamiltonian is with x = l. The ground state wave function for a harmonic oscillator is 0 = with = m . The lowest order correction to the ground state energy resulting from inaccuracy of the small angle approximation is E = 0H 0 =  As 8 1 mg 0x4 0. 3 24 l 0x 0 = 4 x4 exp(2 x2 )dx = 2 . 32ml2  3 , 44 E =  2 2. A diatomic molecule behaves like a rigid rotator with a moment of inertia I = M r0 , where M is the reduced mass and r0 is the distance between the atoms. The Hamiltonian can be approximated by: L2 H= . (9) 2I Assume that the molecule consists of two atoms of charge e, separated by a distance r0 . Suppose that a uniform electric field is present which points in a fixed direction. Compute the energy eigenvalues of the system, treating the electric field as a perturbation. Use second order perturbation theory. HINT: There is one hard integral. This can be done trivially, if you know the following ClebschGordon coefficient: 2 1/2  m2 m , 1 0   1 m , 1 =  . (2 + 1) Solution: In this problem, the unperturbed Hamiltonian is H (0) = L2 , 2I with corresponding unperturbed energy eigenvalues, Em
(0) 2 ( + 1) = , 2I = 0, 1, 2, 3, . . . m = ,  + 1, . . . ,  1, . (10) This, each energy level has a (2 + 1)fold degeneracy. In the presence of a uniform electric field, the Hamiltonian contains an additional term due to the interaction of the charges e with the electric field. The diatomic molecule can be viewed as an electric dipole with dipole moment d = er0 . Hence, the perturbed term of the Hamiltonian is given by H (1) = d E = erE cos , where is the angle between the electric dipole moment and the electric field. We denote the magnitude of the electric field by E to avoid confusion with the symbol for energy. To employ perturbation theory, we first observe that the following matrix elements, m H (1) m = 0 unless m = m and    = 1 . 9 (11) To prove this result, we note that cos = z/r. Thus, we examine m zm. Since z is the 0th component of a spherical tensor of rank 1, it follows from the WignerEckart theorem that m zm = 0 unless m = m and   = 0 or 1. Parity considerations allow us to conclude that m zm = 0 unless    is an odd integer. The latter follows from the fact that z is a parityodd operator and m is an eigenstate of parity with eigenvalue (1) . Hence, eq. (11) follows. Since H (1) cannot connect states of different m, it follows that one can employ nondegenerate perturbation theory to solve for the perturbed energy eigenstates and eigenvalues, as long as one does not sum over degenerate intermediate states. Since m H (1) m = 0, there is no firstorder energy shift. Using secondorder perturbation theory, Em = (er0 E)2
(2) m cos  m2
= Em  E m (0) (0) . Using eqs. (10) and (11), only two terms survive in the sum above, 2I(er0 E)2 m cos   1m2 m cos  + 1m2 (2) Em = + . 2 ( + 1)  ( + 1)( + 2) ( + 1)  (  1) (12) Our final task is to evaluate the two matrix elements above. First, we examine m cos  + 1 m = d Ym (, ) cos Y+1,m (, ) . (13) To perform this integral, we first write: 4 cos Y+1,m (, ) = Y10 (, )Y+1,m (, ) 3 2 + 3 = + 1 m; 1 0L M ; + 1 1 + 1 0; 1 0L 0; + 1 1YLM (, ) , 2L + 1 LM after making use of eq. (1736) on p. 365 of Baym, where we have employed Shankar's notation for the ClebschGordon coefficients. Inserting this result into eq. (13), the remaining integrals are easily evaluated using the orthonormality relations, d Ym (, )Y m (, ) = mm . The end result is then given by: 2 + 3 m cos  + 1m = + 1 m; 1 0 m; + 1 1 + 1 0; 1 0 0; + 1 1 . 2 + 1 This can be evaluated by using the following general result for the ClebschGordon coefficient, 2 1/2 L  M2 . L M , 1 0  L  1 M , L 1 =  L(2L + 1) 10 Hence, ( + 1)2  m2 + 1 m; 1 0 m; + 1 1 + 1 0; 1 0 0; + 1 1 = ( + 1)(2 + 3) and we end up with: m cos  + 1 m = Second, we examine m cos   1 m = d Ym (, ) cos Y1,m (, ) . 1/2 +1 2 + 3 1/2 , ( + 1)2  m2 . (2 + 1)(2 + 3) (14) But, there is no need to perform an independent computation. Simply note that m cos   1 m =  1 m cos m =  1 m cos m , since the matrix elements above are clearly real. Thus, we can use the results of eq. (14) with replaced by  1. Hence, 2  m2 m cos   1 m = . (15) (2  1)(2 + 1) Inserting eqs. (14) and (15) into eq. (12), and simplifying the resulting expression yields I(er0 E)2 [( + 1)  3m2 ] (2) . Em = 2 ( + 1)(2  1)(2 + 3) Hence, 2 ( + 1) I(er0 E)2 [( + 1)  3m2 ] + 2 2I ( + 1)(2  1)(2 + 3) Em Another way to write the same result is: 2 2 er0 E ( + 1)  3m (0) Em Em 1 + . (0) 2(2  1)(2 + 3) Em We expect perturbation theory to be accurate if the uniform electric field is weak, i.e. (0) er0 E Em . 3. Two nonidentical particles, each of mass m, are confined in one dimension to an impenetrable box of length L. What are the wave functions and energies of the three lowestenergy states (i.e. in which at most one particle is excited out of its ground state)? If an interaction potential of the form V12 = (x1  x2 ) is added, calculate to first order in the energies of these three lowest states, along with their wave functions to zeroth order in . 11 Solution: Both particles can stay in the ground state because they are not identical. The energy and wave function are respectively E11 = 2 2 2 x1 x2 , 11 = sin sin . mL2 L L L If one particle is in the ground state, the other in the first excited state, the energies and corresponding wave functions are E12 = E21 52 2 2 x1 2x2 , 12 = sin sin , 2 2mL L L L 52 2 2 2x1 x2 = , 21 = sin sin . 2mL2 L L L When both particles are in the singleparticle ground state, ie., the system is in the ground state, we have the energy correction E
(1) 4 = (11 , V12 11 ) = 2 L L sin 4 0 3 x1 dx1 = , L 2L and the wave function to zeroth order in 11 = 11 . When one particle is in the ground state and the other in the first excited state, the energy level is twofold degenerate and we have to use the perturbation theory for degenerate states. We first calculate the elements of the perturbation Hamiltonian matrix: 12 V12 12 dx1 dx2 = x1 2x1 sin2 dx1 = , L L L 0 12 V12 21 dx1 dx2 = 21 V12 12 dx1 dx2 = . L sin2 We then solve secular equation det L 4 = 2 L 21 V12 21 dx1 dx2 L  E (1) L L and obtain the roots (1) = 0, E L E+ = (1) 2 , L E = 0, 12 which are the energy corrections. The corresponding zeroth order wave functions are 1 12 = (12 21 ). 2 13 ...
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 Spring '09
 Energy, Gravity, Mass

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