Department of Physics Quantum Mechanics II, Physics 570 Temple University Instructor: Z.-E. Meziani Solution of Second Midterm Examination Thursday November 09, 017 Problem 1. (10pts Consider a system of three non interacting particles that are confined to move in a one dimensional infinite potential of length a: V (x = 0 for 0 < x < a and V (x = for other values of x. Determine the energy, degeneracy and wave function of the ground state and the first excited state when the three particles are Figure 1: Particle distribution among levels for the ground state (GS, the first excited state (FES for a system of three noninteracting identical Bosons (left and Fermions(right moving in an infinite well. Each state of the fermions system is -fold degenerate due to the two possible orientations of the isolated fermion a Identical particles of spin zero. Since the particles are non-interacting, the Hamiltonian of the three particle is the sum of the Hamiltonians of each particle moving in a one dimensional infinite potential. The total energy of the system of three particles is additive and is given by: and the wave function by E n1,n,n 3 = π ( n ma 1 + n + n 3 (1 ψ n1,n,n 3 (x 1, x, x 3 = 8 ( a 3 sin n1 π ( a x n π ( 1 sin a x n3 π sin a x 3 where n 1, n and n 3 are the possible quantum numbers that each particle can carry. ( 1
If all three particles are identical bosons, the ground state is non degenerate and will correspond to all particles in the lowest state n 1 = n = n 3 = 1 with the corresponding wave function ψ (0 = ψ 1 (x 1 ψ 1 (x ψ 1 (x 3 = since ψ n (x i = /a sin (nπx i /a. E (0 = E 1,1,1 = 3 π ma (3 8 ( π ( π ( π a 3 sin a x 1 sin a x sin a x 3 In the first excited state (which is also non-degenerate we have two particles in ψ 1 (each with energy ɛ 1 = π /(ma and one particle in ψ (with energy ɛ = π /(ma = ɛ 1 : E (1 = ɛ 1 + ɛ = ɛ 1 + ɛ 1 = 6ɛ 1 = 3 π ( ma (5 Since the particles are identical we can no longer say which particle is in which state only that two particles are in ψ 1 and one is in ψ. Starting from the general antisymmetrized expression of three identical particles but avoiding to double count since particle 1 and are described by ψ 1 and one is described by ψ. ψ (1 = b Identical particles of spin 1/.! 3! [ψ 1(x 1 ψ 1 (x ψ (x 3 + ψ 1 (x 1 ψ (x ψ 1 (x 3 + ψ (x 1 ψ 1 (x ψ 1 (x 3 ] (6 If the three particles are identical spin 1 fermions, the ground state corresponds to the case where two particles are in the lowest state ψ 1 (with one having a spin up + and the other spin down, while the third particle is in the next state ψ (its spin can be either up or down ±. The ground state energy is E (1 = ɛ 1 + ɛ = ɛ 1 + ɛ 1 = 6ɛ 1 = 3 π ma (7 The ground state wave function is antisymmetric and is given by the Slater determinant ψ (0 (x 1, x, x 3 = 1 ψ 1 (x 1 χ(s 1 ψ 1 (x χ(s ψ 1 (x 3 χ(s 3 3! ψ 1 (x 1 χ(s 1 ψ 1 (x χ(s ψ 1 (x 3 χ(s 3 ψ (x 1 χ(s 1 ψ (x χ(s ψ (x 3 χ(s 3 (8 This state is twofold degenerate, since there are four different ways of configuring the spins of the three fermions (the ground state (GS shown in Fig. 1 is just one of the two configurations. The first excited state corresponds to one particle in the lowest state ψ 1 (its spin can be either up or down and the other two particles in the state ψ ( the spin of one is up and down for the other. As in the ground state, there are also two different ways of configuring the spins of the three fermions in the first excited state (FES; The state shown in Fig. 1 is one of the four configurations. ψ (1 (x 1, x, x 3 = 1 3! ψ 1 (x 1 χ(s 1 ψ 1 (x χ(s ψ 1 (x 3 χ(s 3 ψ (x 1 χ(s 1 ψ (x χ(s ψ (x 3 χ(s 3 ψ (x 1 χ(s 1 ψ (x χ(s ψ (x 3 χ(s 3. (9
The energy of these four states is given by E (1 = ɛ 1 + ɛ = ɛ 1 + 8ɛ 1 = 9ɛ 1 = 9 π ma (10 leading to an excitation energy of E (1 E (0 = 9ɛ 1 6ɛ 1 = 3 π /(ma. Problem. (10pts When a hydrogen-like atom is placed in a weak magnetic field B, the energy of interaction is described by the Zeeman Hamiltonian ĤZ Where g s is approximately equal to. Ĥ Z = µ B B ˆ L + gs µ B B ˆ S (11 a Derive an equation for the energy of the atom using first order perturbation theory. Assume that in the absense of the B, the wave functions for the atom are eigenfunctions of ˆL, Ŝ, Ĵ and Ĵz where ˆ J = ˆ L + ˆ S. If we choose the magnetic field along the Oz direction, the perturbing Hamiltonian may be written as: H int = µ B B(ˆL z + Ŝz = µ B B(Ĵz + Ŝz (1 where we used g s =. The first order correction to the energy is then given by: E (1 = L, S, J, M J µ B B(Ĵz + Ŝz L, S, J, M J (13 = l, 1/, l ± 1/, m µ B B(Ĵz + Ŝz l, 1/, l ± 1/, m (1 = µ B Bm + µ B B l, 1/, l ± 1/, m j Ŝz l, 1/, l ± 1/, m (15 The matrix element of Ŝz can be obtained by expressing the eigenfunctions common to ˆL, Ŝ, Ĵ and Ĵz in terms of the eigenfunctions common to ˆL, ˆL z, Ŝ, and Ŝz as given in the relation l ± m + 1/ l m + 1/ l ± 1/, m >= ± m 1/, 1/ > + m + 1/, 1/ > (16 l + 1 l + 1 giving the needed matrix element as: l, 1/, l±1/, m j Ŝz l, 1/, l±1/, m = Leading thus to 1 m (l±m+1/ l±m 1/ = ± (l + 1 (l + 1 (17 [ E (1 = µ B Bm 1 ± 1 ] = gµ B Bm (18 l + 1 For j = l ± 1/ The Landé-factor is thus found to be g = 1 ± [1/(l + 1] 3
b The valence electron of an alkali metal atom is excited to a p state. Into how many components is each of the levels split when a weak field B is applied? In the absence of the field there are two energy levels specified by (l = 1, s = 1/, j = 3/ and (l = 1, s = 1/, j = 1/. These are denoted by the spectroscopic symbols P 3/ and P 1/, respectively. When the field is applied the energy changes are given by [ E = µ B Bm j 1 ± 1 ] = gµ B Bm j (19 l + 1 The j = 3/ level ( P 3/ is split into four components since m j takes on four values. The j = 1/ level ( P 1/ is split into two components corresponding to m j = ±1/ c How large are the splittings in units of µ B B? The splittings depend on the value of the Landé-factor g(l, s, j. For the P 3/ level: For the P 1/ level: These splittings are shown in Fig. g = l + 1 + 1 l + 1 g = l + 1 + 1 l + 1 = 3 = 3 (0 (1 Figure : Splitting diagram You may use the following relation: l ± m + 1/ l m + 1/ l ± 1/, m >= ± m 1/, 1/ > + m + 1/, 1/ > ( l + 1 l + 1 Problem 3. (10pts
Consider two spin 1/ s S 1 and S, coupled by an interaction of the form a 0 e t τ S 1 S. At t =, the system is in the state +, (an eigenstate of S 1z and S z with the eigenvalues + / and / a Calculate P (+ + by using first-order time dependent perturbation theory. First, we write the state of the system at t = in the basis of Ŝ and Ŝz, namely { S, M } ψ( = +, = 1 [ 1, 0 + 0, 0 ] (3 where 1, 0 and 0, 0 are kets of the S, M basis. In 1 st -order time-dependent perturbation theory, we have P(+ + = 1 + e iωfit W fi (tdt (5 where ω fi and W fi (t and W fi (t = + W (t + We also express + in terms of { S, M in order for us to calculate the matrix element W fi (t, + = 1 [ 1, 0 0, 0 ] (6 This leads us to W fi (t = a 0e t τ = a 0e t τ = a 0e t τ [ ] ( 1, 0 0, 0 (S S1 S ( 1, 0 0, 0 ( (7 [ 1, 0 (S S 1 S 1, 0 0, 0 (S S 1 S 0, 0 ] (8 [ 3 ] ( 3 = a 0e t τ (9 P(+ + = 1 e iω fit Given that ω fi = 0 we can write the transition probability as ( a0 P(+ + = + a 0e t τ dt e t τ dt = a 0 πτ (30 (31 b Now assume that the two spins are also interacting in a static magnetic field B 0 parallel to Oz. The corresponding Zeeman Hamiltonian can be written H 0 = B 0 (γ 1 S 1z + γ S z (3 where γ 1 and γ are the gyromagnetic ratios of the two spins, assumed to be different. Assume that a(t = a 0 e t τ. Calculate P(+ + by first-order time-dependent perturbation theory. With fixed a 0 and τ, discuss the variation of P(+ + with respect to B 0. 5
Again, but now with ω fi 0 P(+ + = 1 + e iωfit W fi (tdt ω fi = E0 + E 0 + H 0 + = B 0 ( γ 1 H 0 + = B 0 ( γ 1 (33 (3 + γ + (35 γ + (36 As in a ω fi = B 0(γ 1 γ B 0( γ 1 + γ = B 0 (γ 1 γ (37 W fi (t = a 0e t τ (38 P(+ + = = = + e ib 0(γ 1 γ t a 0 e t τ dt a 0 + e ib 0(γ 1 γ t t τ dt ( a 0 B exp 0 (γ 1 γ πτ dt /τ (39 (0 (1 P(+ + = τ πa 0 ( B exp 0 (γ 1 γ τ ( We find a gaussian function describing the probability as a function of the magnetic field strength. Note that this expression for P(+ + in the limit B 0 0, agrees with part a. 6
Figure 3: Probability as a function of the magnetic field strength 7