Atomic, molecular and laser physics PHY-331 Solution Set # 7 November 29, 2010 Zeeman effect and hyperfine interaction Solution 1. (a) Weak field The energies of hydrogen's n = 3 levels, when placed in a weak magnetic field, are shown in the Table. The correction to energy in a weak magnetic field is, ∆E = gJ µBmjBext where, j(j + 1) + 3=4 − l(l + 1) g = 1 + J 2j(j + 1) e~ µB = · 2me The energy corrections are relative to the respective fine-structure levels (defined by n; l; j) as shown in Figure 1. 1 Atomic, molecular and laser physics (5/2,5/2) = 3µΒΒ (5/2,3/2) ext = 9/5 µΒΒext (5/2,1/2) = 3/5 µΒΒext (5/2,-1/2) 3d 5/2 = − 3/5 µΒΒext (5/2,-3/2) = − 9/5 µΒΒext (5/2,-5/2) = 3µΒΒext (3/2,3/2) (3/2,3/2) = 2µΒΒext = 6/5 µΒΒext (3/2,1/2) (3/2,1/2) = 2/3 µΒΒext = 2/5 µΒΒext (3/2,-1/2) (3/2,-1/2) 2/5 Β ext 3p 3/2 = − 2/3 µΒΒext 3d 3/2 = − µ Β (3/2,-3/2) (3/2,-3/2) = − 2µΒΒext = − 6/5 µΒΒext (1/2,1/2) = µ ΒΒext 1/3 Β ext (1/2,1/2) = µ Β 3s 1/2 (1/2,-1/2) 3p 1/2 (1/2,-1/2) =−µ ΒΒext = − 1/3 µΒΒext (3/2,3/2) = 2µΒΒext (3/2,1/2) = 2/3 µΒΒext (3/2,-1/2) 2p 3/2 = − 2/3 µΒΒext (3/2,-3/2) = − 2µΒΒext (1/2,1/2) (1/2,1/2) = µ ΒΒext = 1/3 µΒΒext 2s 1/2 (1/2,-1/2) 2p 1/2 (1/2,-1/2) = −µ ΒΒext = − 1/3 µΒΒext (1/2,1/2) = µ ΒΒext 1s 1/2 (1/2,-1/2) Ε=−µ ΒΒext FIG. 1: Splitting of hydrogen's n = 3 energy levels when placed in a weak magnetic field. Above each state, the values of j; mj are specified explicitly. 2 Atomic, molecular and laser physics n l j mj gJ ∆Eweak 3 0 1=2 −1=2 2 −µBBext 3 0 1=2 +1=2 2 +µBBext − − µB Bext 3 1 1=2 1=2 2=3 3 µB Bext 3 1 1=2 +1=2 2=3 + 3 3 1 3=2 −3=2 4 −2µBBext − − 2µB Bext 3 1 3=2 1=2 4=3 3 2µB Bext 3 1 3=2 +1=2 4=3 + 3 3 1 3=2 +3=2 4 +2µBBext − − 6µB Bext 3 2 3=2 3=2 12=5 5 − − 2µB Bext 3 2 3=2 1=2 4=5 5 2µB Bext 3 2 3=2 +1=2 4=5 + 5 6µB Bext 3 2 3=2 +3=2 12=5 + 5 3 2 5=2 −5=2 6 −3µBBext − − 9µB Bext 3 2 5=2 3=2 18=5 5 − − 3µB Bext 3 2 5=2 1=2 6=5 5 3µB Bext 3 2 5=2 +1=2 6=5 + 5 9µB Bext 3 2 5=2 +3=2 18=5 + 5 3 2 5=2 +5=2 6 +3µBBext There are total 10 transitions which are possible from the n = 3 levels to the n = 1 levels. They are, 3p1=2 −! 1s1=2 4 transitions 3p3=2 −! 1s1=2 6 transitions: No transitions are possible between 3s and 1s, 3d and 1s. (b) Strong Field The Table of energies, for the hydrogen atom when placed in a strong magnetic field, is shown in Figure 2. The shift in energy for the strong field is, ∆Estrong = µB (ml + 2ms) Bext: This perturbation is applied before the spin-orbit interaction. The Fine-structure correction 3 Atomic, molecular and laser physics is, ( ) E2 m m − 4l(l + 1) ∆E = n l s + 3 ; and is applicable for l > 0 only: fs 2mc2 l(l + 1)(l + 1=2) n l ml ms ml + 2ms ∆EZeeman ∆Efs 3 0 0 −1=2 −1 −µBBext 0 3 0 0 +1=2 +1 +µBBext 0 2 − − − − En 3 1 1 1=2 2 2µBBext 4mc2 2 − − − En 3 1 0 1=2 1 µBBext 6mc2 2 − En 3 1 1 1=2 0 0 12mc2 2 − En 3 1 1 +1=2 0 0 12mc2 2 En 3 1 0 +1=2 +1 +µBBext 6mc2 2 En 3 1 1 +1=2 +2 +2µBBext 4mc2 2 − − − − 11En 3 2 2 1=2 3 3µBBext 15mc2 2 − − 43En 3 2 1 1=2 0 0 60mc2 2 − − − 7En 3 2 0 1=2 1 µBBext 10mc2 2 − − − 41En 3 2 1 1=2 2 2µBBext 60mc2 2 − 2En 3 2 2 1=2 +1 +µBBext 3mc2 2 − − − 11En 3 2 2 +1=2 1 µBBext 15mc2 2 − 41En 3 2 1 +1=2 +2 +2µBBext 60mc2 2 7En 3 2 0 +1=2 +1 +µBBext 10mc2 2 43En 3 2 1 +1=2 0 0 60mc2 2 11En 3 2 2 +1=2 +3 +3µBBext 15mc2 There is total 4 transitions which are possible from n = 3 level to n = 1 energy level. They are, 3p −! 1s 4 transitions: The selection rules used are ∆l = ±1; ∆ml = 0; ±1, (transition from 0 to 0 is not allowed) and ∆ms = 0. Solution 2. The distinct spectral lines resulting from 2p1=2 −! 1s1=2 are shown in the Figure 3. When placed inside a weak magnetic field, the correction to the energy relative to the fine-structure 4 Atomic, molecular and laser physics (2,1/2) = 3µΒΒext (-1,1/2) = 2 µΒΒext (0,1/2) (2,-1/2) = µ ΒΒext (1,1/2) (-1,-1/2) = 0 3d (0,-1/2) (-2,1/2) =−µ ΒΒext (1,-1/2) = − 2 µΒΒext (-2,-1/2) = − 3µΒΒext (1,1/2) (0,1/2) = 2 µΒΒext = µ ΒΒext (0,1/2) = µ ΒΒext (-1,1/2) (1,-1/2) = 0 3s (0,-1/2) 3p (0,-1/2) = −µ ΒΒext (-1,-1/2) =−µ ΒΒext = − 2 µΒΒext (1,1/2) = 2 µΒΒext (0,1/2) = µ ΒΒext (0,1/2) (-1,1/2) (1,-1/2) = µ ΒΒext = 0 (0,-1/2) 2s (0,-1/2) 2p =−µ ΒΒext = −µ ΒΒext (-1,-1/2) = − 2 µΒΒext (0,1/2) = µ ΒΒext 1s (0,-1/2) Ε=−µ ΒΒext FIG. 2: Splitting of hydrogen's n = 3 energy levels when placed in a strong magnetic field. Above each state, the values of ml; ms are explicitly specified. These lines further split up into different states (called fine-structure splitting), which are not shown in this diagram. levels is, ∆E = gJ µBmjBext 5 Atomic, molecular and laser physics where, j(j + 1) + 3=4 − l(l + 1) g = 1 + J 2j(j + 1) mj = j; j − 1;:::; −j + 1; −j; and e~ µB = · 2me (1/2,1/2) Ε=µ ΒΒext 3 2p 1/2 (1/2, - 1/2) Ε=−µ ΒΒext 3 υ (1/2,1/2) Ε=µ ΒΒext 1s 1/2 (1/2, - 1/2) Ε=−µ ΒΒext FIG. 3: The distinct spectral lines resulting from 2p1=2 −! 1s1=2 transitions. For each energy 0 level, values of j; mj are explicitly specified in brackets. The ∆E s mentioned are with respect to the degenerate levels. The wavenumbers corresponding to the 2p1=2 −! 1s1=2 transitions are, 2p(1=2;−1=2) −! 1s(1=2;−1=2) ν¯ 2µBBext 2p −! 1s − ν¯ + (1=2;+1=2) (1=2; 1=2) 3hc 2µBBext 2p − −! 1s ν¯ − (1=2; 1=2) (1=2;+1=2) hc 2µ B 2µ B 2p −! 1s ν¯ − B ext + B ext · (1=2;+1=2) (1=2;+1=2) hc 3hc The selection rules used are ∆j = 0; ±1; ∆mj = 0; ±1; ∆l = ±1. Similarly, for the transition 2p3=2 −! 1s1=2, the distinct spectral lines are shown in Figure 4. 6 Atomic, molecular and laser physics (3/2,3/2) Ε= 2µ ΒΒext (3/2,1/2) Ε= 2µ ΒΒext 3 2p 3/2 (3/2, - 1/2) Ε=−2µ ΒΒext 3 (3/2, - 3/2) Ε=−2µ ΒΒext (1/2,1/2) υ Ε= µ ΒΒext 1s 1/2 (1/2, - 1/2) Ε=−µ ΒΒext FIG. 4: The distinct spectral lines resulting from 2p3=2 −! 1s1=2 transitions. For each energy level, 0 values of j; mj are specified explicitly in brackets. Once again, the ∆E s are with respect to the degenerate levels. The wavenumbers corresponding to these transitions are, 2p(3=2;−3=2) −! 1s(1=2;−1=2) ν¯ 4µBBext 2p(3=2;−1=2) −! 1s(1=2;−1=2) ν¯ + (3hc ) 4µBBext 2p −! 1s − ν¯ + 2 (3=2;+1=2) (1=2; 1=2) 3hc 2µBBext 4µBBext 2p(3=2;−1=2) −! 1s(1=2;+1=2) ν¯ − + hc (3hc ) 2µBBext 4µBBext 2p(3=2;+1=2) −! 1s(1=2;+1=2) ν¯ − + 2 hc ( 3hc ) 2µ B 4µ B 2p −! 1s ν¯ − B ext + 3 B ext · (3=2;+3=2) (1=2;1=2) hc 3hc Solution 3. To draw the energy level diagram for the hydrogen atom (I = 1=2), first we need to know the values of F for different j. The total quantum number F is given by F = j + I; j + I − 1 ;::: jj − Ij. Therefore, For n=1 l = 0;I = 1=2; j = 1=2;F = 0; 1: 7 Atomic, molecular and laser physics For n=2 l = 0; 1;I = 1=2; j = 1=2;F = 0; 1 j = 3=2;F = 1; 2: For n=3 l = 0; 1; 2;I = 1=2; j = 1=2;F = 0; 1 j = 3=2;F = 1; 2 j = 5=2;F = 1; 2; 3: All these energy levels are shown in Figure 5.