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1 1 Problem set 9 Handout: 1/24 Due date: 1/31 Problem 1 Prove that the energy to first order for the lowest-energy state of a perturbed system is an upper bound for the exact energy of the lowest-energy state of the perturbed system, that is, that E () + E (1) E. The ground state energy is given by E =< ψ H ψ > (1) and the first correction to the ground state energy is given by E 1 =< ψ H 1 ψ > (2) where the total Hamiltonian is given by H = H + H 1. We know from the variational theorem that using the zeroth-order wavefunction as a trial function for the total Hamiltonian will give an upper bound to the energy < ψ H ψ > E (3) since ψ is not the wavefunction associated with the Hamiltonian. Therefore, we can prove that < ψ H ψ > E (4) < ψ H + H 1 ψ > E (5) < ψ H ψ > + < ψ H 1 ψ > E (6) E + E1 E (7) Problem 2 An electron moves in a harmonic potential, V = 1 2 x2. What is the effect, to first order, on the energies of a perturbing electric field, Ĥ 1 = Fx? Explain your reasoning. [hint: you do not need to evaluate any integrals to answer the question.]. (8) 1

2 Since the particle moves in a Harmonic potential the wavefunction squared is symmetric. An electric field perturbation V 1 = Ex which is antisymmetric will have no effect since < ψ V 1 ψ >= dx ψ 2 V 1 = dxsymmetric antisymmetric = (9) Problem 3 Calculate the energy to first order of He + in its lowest-energy state. Use the hydrogen atom in its ground state as your zeroth-order approximation. Use atomic units. The zeroth order Hamiltonian is for the hydrogen atom H = r (1) whereas the Hamiltonian for the He + is given by H = r (11) Thus, the total Hamiltonian can be split as The energy to first order is then H = R = H 1 r = H + H 1 (12) E = E + E 1 = < 1s 1 1s > (13) r Evaluating the first-order correction to the energy we get E 1 = < 1s 1 r 1s >= 4π π exp( 2r)rdr = 4 1 = 1 (14) 22 The energy to first-order is then E = 1/2 1 = 3/2. Problem 4 A hydrogen atom in its ground state is perturbed by applying a uniform electric field, Fz, along the z-direction. 2

3 1. What is the first-order change in the energy? 2. For the first-order correction to the ground-state wavefunction, we can consider a 2s,1s and a 2pz,1s to be the coefficients for mixing in 2s and 2p z character. Write down the expression for these two coefficients. 3. Predict the signs of these coefficients. Explain your reasoning. The ground state wavefunction for the hydrogen atom is 1s = 1 π exp( r). a) The first-order change in the energy due to the electric field is then given by = F 1 π = F 1 π b) The two cofficients are r 3 exp( 2r)dr < 1s H 1 1s >= F < 1s r cosθ 1s > (15) r 3 exp( 2r)dr π π cosθ sin θdθ sym. antisym. 2π 2π dφ (16) dφ = (17) c (1) 2s,1s = < 2s H1 1s > E 2s E 1s = (18) c (1) 2p z,1s = < 2p z H 1 1s > E 2pz E 1s = (19) The first coefficient are zero due to symmetry, i.e. it cannot improve the energy due to perturbation along the Z since it is symmetric in all directions. The second coefficient is positive since the 2p z function is oriented along the direction of the perturbation and will therefore improve the wavefunction. Problem 5 The Hellman-Feynman theorem establishes a connection between the change in the energy and the change in the Hamiltonian of a system experiences an electric field perturbation. The theorem is given by de df = Ĥ (2) where E is the energy, F is the electric field strength and Ĥ is the total hamiltonian for the system. If we apply an electric field perturbation along the z-direction we can write the first-order hamiltonian as Ĥ (1) = ˆµ z F (21) 3

4 where ˆµ z is the electric dipole moment operator in the z-direction. 1) Use the Hellman-Feynman theorem to express the change in the energy due to an electric field in the z-direction in terms of the dipole moment operator. 2) Expand the energy of a system in a Taylor expansion relative to the energy, E, in the absence of the field. The expectation value of the dipole moment operator can also be written as < ˆµ z >= µ z + α zzf + (22) where µ z is the permanent dipole moment of the system and α zz is the polarizability. 3) Use this expansion to express the permanent dipole moment and polarizability in terms of derivatives of the energy with respect to the electric field. 4) Use non-degenerate perturbation theory to obtain an expression for the energy of the system in terms of the field strength to second order. 5) Finally, use this expression to identify the permanent dipole moment and polarizability of the system. [Hint: The final expression should be expression in terms of the electric dipole operator and the zeroth-order wavefunctions de df = Ĥ = Ĥ µ z F = µ z (23) E = E + E F E + (24) 2 2F2 de df = E + 2 E + = µ 2F2 z α zzf + (25) Therefore, we can identify the dipole moment and polarizability as µ z = E (26) α zz = 2 E 2 (27) 4

5 4. The energy expression from perturbations theory reads: E = E + E 1 + E 2 + (28) = E + < ψ () H 1 ψ () > + < ψ m () H 1 ψ () >< ψ () H1 ψ () E E m m = E < ψ () µ z ψ () > F + < ψ m () µ z ψ () >< ψ () µ z ψ () E E m m 5. We can therefore identify the terms as m > m > + (29) + (3) and µ z = E =< ψ() µ z ψ () > (31) α zz = 2 E = 2 < ψ m () µ z ψ () >< ψ () µ z ψ () 2 E E m m m > (32) Problem 6 A given unperturbed system has a doubly generate energy level for which the perturbation integrals have values of H 1 11 = 6a, H1 22 = 8a, and H1 12 = 2a, where a is a positive constant. We also know that the unperturbed wavefunctions are orthonormal. 1. Find the first-order correction to the energy in terms of the constant a. 2. Find the normalized correct zeroth-order wave functions. 1. The secular equation is det(h EI) = H1 aa E H 1 ba H1 ab H 1 bb E = 6a E 2a 2a 8a E = (33) Expansion of the secular equation gives the following quadratic equation E 2 14Ea + 44a 2 = (34) 5

6 for which the solution is E = 7a ± 2a = 11.47a, 2.52a. The coefficients for the lowest eigenvalue is then given by Using normalization we find (H 1 aa E1 )c 1 + H ab c 2 = (35) (6a 2.52a)c 1 + 2ac 2 = (36) c 1 =.57c 2 (37) 1 =.57 2 c c2 2 ==> c 2 =.867 (38) Therefore, the normalized wavefunction is ψ 1 =.49ψ a +.867ψ b (39) 6

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