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Notes 4: the Zeeman Effect

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Notes 4: the Zeeman Effect Notes 4: The Zeeman Effect Pieter Zeeman in 1896 discovered that a magnetic field broadened the yellow D lines produced when sodium is placed in a flame. Later this broadening was shown to be a splitting of the spectral lines into as many as 15 components. The line splitting results from the effect of the magnetic field on the atomic energy levels. In 1897, Thomas Preston reported a similar effect to Zeeman but with more complicated results. This ‘anomalous Zeeman effect’ is now known to be due to half-integer total electron spin. The Zeeman effect is very useful in measuring the strengths of stellar magnetic fields. Hamiltonian for a charge in electric and magnetic fields The Hamiltonian for the hydrogen atom was found by adding the electrostatic potential energy operator to the kinetic energy operator. Because the Lorentz force does no work, it does not have a scalar potential energy. In classical Lagrangian mechanics, the velocity-dependent Lorentz force is accounted for by using a velocity-dependent potential, Uq=−⋅Av, where A is the vector potential related to the magnetic field by BA=∇× . The quantization of the classical Hamiltonian for a charged particle in electric and magnetic fields leads to the Schrödinger equation ∂ψ 1 2 i =(pA −+ qq) ϕψ, (4.1) ∂tM2 where ϕ is the scalar potential and p =−∇i . (4.2) For a uniform magnetic field, B0, the vector potential can be taken to be 1 Ar( ) =−× r B. (4.3) 2 0 Note that ∇⋅A =0.The Hamiltonian in equation (4.1) is such that 1 2 Hψ=(pA −+ qq) ϕψ 2M 1 =( −∇−iqAA) ⋅−∇−( iψ q ψ) + q ϕψ (4.4) 2M 22 2 iq q = − ∇ψ +(A ⋅∇)ψ + AA ⋅ ψ +q ϕψ . 22MM M Using vector identities, it can be shown for the vector potential in equation (4.3) that 1 1 AA⋅=⋅⋅−⋅⋅( rr)( B00 B) ( rB 0)( rB 0) , 4 (4.5) iqqqq (A⋅∇=×⋅=−⋅×=−⋅)ψ( rBpBrpBL) ψ( ) ψψ, MMMM22200 0 where L is the angular momentum operator. The Hamiltonian for a charged particle in a uniform magnetic field is then pq22 q Hq=−⋅+⋅⋅−⋅⋅+B0 L( rr)( B00 B) ( rB 0)( rB 0) ϕ. (4.6) 22MM 8 M Application to the Zeeman Effect in hydrogen Since the electron has a magnetic moment, an additional term has to be added to the Hamiltonian in equation (4.6). The ‘perturbation’ due to the magnetic field is then 2 1 ee Hg=B0 ⋅++( Le S) ( rr ⋅⋅−⋅⋅)( B00 B) ( rB 0)( rB 0) , (4.7) 28MM where ge ≈ 2 is the electron gyromagnetic ratio factor. It is instructive to consider the relative magnitudes of the terms that are linear and quadratic in the magnetic field. Suppose the hydrogen atom is in its ground state. Then the ratio of the quadratic term to the linear term is roughly eB a 2 00 , (4.8) 2 2 −10 where a00= 4πε (Me ) =0.529 × 10 m is the Bohr radius. The two terms will be roughly equal at field strength = = × 5 Beq 2 2.35 10 T. (4.9) ea0 This is much larger than the strongest magnetic fields produced on Earth by explosive field compression (~3000 T) but much less than the fields associated with magnetars (highly magnetic neutron stars), which 11 can be as large as 10 T. When B0 = Beq, the energy corresponding the magnetic perturbation to the Hamiltonian is eB 2 eq = = ~ 2 2,I p (4.10) M Ma0 where Ip is the ionization potential of hydrogen. Hence, at even the strongest ‘terrestrial’ fields, the magnetic effects can be treated as a perturbation. The hydrogen atom’s internal magnetic field due to ‘orbital’ motion of the electron (see equation 3.21) will be comparable to the external magnetic field when 2 1 e ≈=≈ α 2 BB0 23 eq 12.5 T. (4.11) 4πε 00Mc a The weak-field Zeeman effect From equation (4.11), we see that line splitting due to fine structure will be larger than the Zeeman effect 2 if BB0 α eq . The Zeeman effect in this weak-field case is treated by including the fine structure terms in the unperturbed Hamiltonian and using as the perturbation to the Hamiltonian e Hg1 =BL ⋅+( S). (4.12) 2M 0 e 2 The operators J and Jz commute with the fine structure Hamiltonian. They will commute with the Zeeman perturbation provided we choose the z-axis parallel to the external magnetic field, so that e H1 = B( L + gS). (4.13) 2M 0 z ez The good states theorem tells us we can use non-degenerate perturbation theory with nl jmj as the good states. To find the first order corrections to the energies, we need to find LSJ+ggee =+−( 1.) S (4.14) Since J is constant, L and S are constrained to be coplanar with J, as shown in the figure below. S J L Since the azimuthal angle of the plane relative to the direction of J is uniformly distributed, the expectation value of S is equal to its projection onto J, i.e. SJ⋅ SJ= J 2 We can evaluate SJ⋅ by noting L = J – S so that LJ22= −2.SJ ⋅+ S 2 3 Hence J22−+ LS 2 2 SJ⋅= =j( j +−1) ll( ++ 1) ss( + 1.) 22 It follows that SJ⋅ LS+ggee =+−11( ) JJ = g J . J 2 where g −1 j( j+−11) ll( ++) ss( + 1) g =1 + e (4.15) J 21jj( + ) is called the Landé g-factor. The corrections to the energies are 1 e E= Bgm= µ Bgm, 2M 00Jj B Jj where e −−24 −115 − µ = = 9.274×× 10 J T = 5.7881 0 eV T , B 2M is called the Bohr magneton. The Zeeman effect splits the n = 1, l = 0, j = ½ ground state of hydrogen in two, since mj = ±1/2. Since gJ = 1, the Zeeman contribution to the energy is ±µB B0. Problem 7.24 Consider the eight n = 2 states. Find the energy of each state under weak Zeeman splitting and construct a diagram to show how the energies evolve as B0 increases. Label each line clearly and indicate it slope. Solution: The table shows the relevant quantum numbers, the Landé g-factor, the energy for zero external field and the derivative of the energy with respect to field strength for all eight states. -1 l j mj gJ Efs (eV) dE/dB0 (eV T ) 0 1/2 1/2 1 -3.397556534 2.894 10-5 0 1/2 -1/2 1 -3.397556534 -2.894 10-5 1 1/2 1/2 -1/3 -3.397556534 -9.647 10-6 1 1/2 -1/2 -1/3 -3.397556534 9.647 10-6 1 3/2 3/2 4/3 -3.397511307 1.158 10-4 1 3/2 1/2 4/3 -3.397511307 3.859 10-5 1 3/2 -1/2 4/3 -3.397511307 -3.859 10-5 1 3/2 -3/2 4/3 -3.397511307 -1.158 10-4 4 How the energies evolve with field strength are shown below. The blue lines are for the l = 0 states and the red lines for the j = 3/2 states. -3.397 mj = 3/2 mj = 1/2 mj = 1/2 (eV) j = 3/2 m = -1/2 E j j = 1/2 mj = 1/2 mj = -1/2 mj = -1/2 m = -3/2 -3.398 j 0 1 2 3 4 B (T) 0 The strong-field Zeeman effect For the strong field Zeeman effect, fine structure is treated as a perturbation to a Hamiltonian that includes the interaction with the external magnetic field. Since ml and ms are quantum numbers for the unperturbed wave function, the Zeeman effect contribution to the energy is e E= B( m += gm) µ B( m + gm). (4.16) Z 2M 00l es B l es Even though the degeneracies in ml and ms have been lifted by the external magnetic field, the eigenstates of the unperturbed Hamiltonian are degenerate because their energies do not depend on the angular momentum quantum number, l. However, both H0 and the fine structure perturbation H1 commute with the operator L2, which can serve as the Hermitian operator A in the good states theorem. The eigenstates nlmls m are good states and can be used in non-degenerate perturbation theory to find the fine structure corrections to the energies. (Note since ge is not exactly equal to 2 and is an irrational number, there is no other degeneracy, contrary to what is stated in the textbook.) The relativistic correction is the same as in the zero external field case. For spin-orbit coupling 5 e2 1 1 =−⋅ HgSOC( e 1,) 223SL (4.17) 8πε 0 Mcr we need SL⋅=nlmls m SL ⋅ nlmls m = S xx L + S yy L + S zz L (4.18) =++SLxx SL y y SL zz. 1 Noting that LLxy= = 0 for eigenstates of Lz, we get 2 SL⋅=Sz L z = mm sl. (4.19) Putting the pieces together, we get e2 1 13=−⋅− EgSOC( e 1) 22 rSL 8πε 0 Mc (4.20) e2 12 = − 2 ( ge 1) 22 3 3 mmsl 8πε 0 M c nl( l++12)( l 1) a Setting ge = 2, on adding the relativistic correction, we get the fine structure contribution to be α 2 3 l( l+−1) mm EEn10= − 2.ls (4.21) fs 2 n n 4 ll( ++12)( l 1) For l = 0, the term in the square brackets is indeterminate. It can be shown that for l = 0, 2 10α 3 Efs = Enn − .
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