( ). Expanding the square and keeping in mind that

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Milton Skinner
5 years ago
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1 One-electron atom in a Magnetic Field When the atom is in a magnetic field the magnetic moment of the electron due to its orbital motion and its spin interacts with the field and the Schrodinger Hamiltonian is Ĥ = P ˆ + e 2 A ( ) Ze2 4πε 0 r ˆµS i +W (r) ˆL i Ŝ where A is the vector potential, e is the proton charge and ˆµS = g e µ Ŝ the magnetic moment of the electron. We have also included the spin-orbit term since it splits the levels with non-zero orbital angular momentum. Lets first consider the ˆ vector potential term P + e 2 A ( ). Expanding the square and keeping in mind that the momentum is a differential operator, ˆP =, we have i P ˆ 2 + e 2 A 2 + e ˆP ( i A + eai ˆP ). The question is does ˆP commute with A? Let g be an arbitrary function and note that and so ˆ P i Ag = i α A α g = i ˆ P i A Ai ˆP = i i A ( A α α g + g α A α ) = Ai ˆPg + i g i A And so if we use the Coulomb gauge where i A = 0, ˆP and A do commute and we can write the Hamiltonian as Ĥ = P ˆ 2 Ze2 4πε 0 r + e Ai ˆP e 2 + A 2 +W (r) ˆL i Ŝ m or Ĥ = Ĥ 0 + e m Ai ˆP e 2 + A 2 +W (r) ˆL i Ŝ

2 Let s first consider the term linear in A. Since a constant magnetic induction is derivable from the vector potential A = ( r ) we can write 2 Ai ˆP = r i ˆP = 2 2 ε r ˆP αβγ β γ α = 2 ε r ˆP γβα γ β α = i r ˆP 2 and since the orbital angular momentum ˆL = r ˆP we have e Ai ˆP e = i ˆL m where we measure the angular momentum in multiples of. As noted previously the quantity e is called the ohr magneton, J µ and equals T J T Joule Tesla. The orbital magnetic moment operator can be written as ˆµL = µ ˆL and so the contribution to the Hamiltonian due to terms linear in the vector potential takes the familiar form ˆµL i, often called the Zeeman term. Lets now consider the term e2 A 2. We can write ( ( ) ) 2. If Ai A = A α A α = 4 ε r ε r = αβγ β γ αλρ λ ρ ( 4 δ δ δ δ βγ γρ βρ γλ )r β γ r λ ρ = 4 r 2 2 r i the magnetic field defines the z axis this becomes to the Hamiltonian is e rsin θ and the contribution 8m r 2 sin 2 θ. Note that the dimensions of e m are C i kg kg i S C = S and it s a frequency. One defines the Larmor frequency so the diamagnetic contribution to the Hamiltonian is sometimes written 2 mω 2 r 2 sin 2 θ. The total Hamiltonian can be written L Ĥ = H 0 + µ ( ˆL + ge Ŝ ) i + e2 2 8m r 2 sin 2 θ +W (r) ˆL i Ŝ ω as e L and

3 The term linear in is approximately 0.47 cm T while the quadratic term is approximately e2 a m = 5.0 cm 0 7 2, the ratio being 0 6. So even in a fairly T 2 large field of 0T the term quadratic in is negligible compared with the term linear in so we will consider the Hamiltonian Ĥ = Ĥ 0 + W (r) ˆL i Ŝ + Ĥ Zeeman with Ĥ Zeeman = µ ( ˆL + 2 Ŝ ) i = µ ( ˆLZ + 2ŜZ ) and with in the z direction. Note that the spin-orbit term is intrinsic to the atom while the Zeeman term depends on and can have any value. Let s now consider the effect of a magnetic field when the spin-orbit splitting is much larger then the Zeeman interaction. In this case we can perturb the fine structure terms independently. Since each spin-orbit level is 2j + fold degenerate we need to see if degenerate perturbation theory is needed. Accordingly we form the matrix of the perturbation for the 2j + degenerate states for a given j and we need Φ jmj ' l/ 2 Φ jm' l/ 2 ( ˆL z + 2Ŝz ) Φ jmj. We can write ˆL l/ 2 z + 2Ŝz = Ĵ + Ŝz and so z ( ˆLz + 2Ŝz ) Φ jml/ 2 = mδ m' m + Φ jm' l/ 2 Ŝ z Φ jml/ 2. Using the explicit form for the angular functions we have ± m Φ Ŝ Φ = δ 2l + jm' l / 2 z jml / 2 m' m and so the perturbation matrix is diagonal and each degenerate state shifts ( ) according to E jml ( Zeeman) = µ ± 2l + where one uses the upper sign when j = +/ 2 and the lower when j = / 2. Note that we can define an effective g factor for each l value g jl = ± 2l + and write the energy shift as

4 ( ) E j l ( Zeeman) = g j µ = g j cm. Note further that for the ns /2, with = 0 only the plus sign is relevant and so g = 2. With the Zeeman term the total energy is E(nj ) = E n + α 2 R Z 4 ( j( j +) ( +) 3/ 4) + g 2 n 3 +/ 2 jl µ ( )( +) Note that the Zeeman term is independent of the principal quantum number n. Let s first look at the splitting in the ns /2 level where the spin-orbit interaction is zero. Since g ij = 2 each state is shifted by cm and they are split by cm. This energy separation can be measured via ESR spectroscopy. A transition between the two = ± 2 levels, ±.4674cm can be induced by applying an oscillating electromagnetic field at right angles to. If the frequency of the radiation is ν in Hertz we have the equality hν = 2(0.4674)cm. Using h = cm sec and expressing in Gauss we have = ν G. Many ESR spectrometers have a fixed frequency (X band) around Hz so the resonance magnetic field for H is approximately 3320 G. Now consider the 0 case and in particular the 2P /2 levels. These are split by the magnetic field into 2 and 4 states respectively with the energies E(2P /2 ) = 0.36 cm and E(2P 3/2 ) = cm. These energies are shown schematically the figure below.

5 =3/ j=3/2 2P 3/ =/2 =-/2 =-3/2 E(cm - ) j=/2 2P / =/2 =-/2 The question is what is the strength of a weak field for the perturbation of the individual fine structure levels to be valid? Taking the energy of the 2P /2 level as zero places the unperturbed 2P 3/2 level at cm - (the fine structure separation), the lowest state = 3 / 2 at cm and the highest 2P /2 state at cm, the resulting separation being cm. Clearly the closer these two states, the less reliable is perturbation theory. When this separation is zero the field is 0.335T or 3350G so something considerably smaller than this, perhaps 0.0T would seem reasonable. Note that no such limit is imposed on the ns /2 states because there is no spin-orbit coupling and a field of 0.355T is perfectly acceptable! If we relax the condition that the field is week enough that it doesn t mix the spin orbit states of the 2P /2 then we must treat the perturbation as W (r) ˆL i Ŝ + Ĥ Zeeman

6 and the unperturbed states as the three 2p levels each with an α or β spin. Alternatively we may take the 2P /2 wavefunctions as the unperturbed states since they are also eigenfunctions of Ĥ 0 and that s what we will do since we already have all 6 of the diagonal elements of the perturbation matrix in this basis and these are in general R n Φ W (r) ˆL jmj i Ŝ + Ĥ Zeeman R n Φ jmj = α 2 R Z 4 ( j( j +) ( +) 3/ 4) + g 2 n 3 +/ 2 jl µ ( )( +) We then need to see how the Zeeman operator couples the 2P /2 states. We can do this in general by evaluating j = / 2,,,/ 2 µ ˆL z + 2Ŝz ( ) j = +/ 2,,,/ 2 and using the explicit form of the eigenfunctions we find this is diagonal in and equals µ / 2 +/ 2 which for = equals µ 3 3 / / 2 so only the = ±/ 2 states are coupled, the matrix element being 2 3 µ. If we order the unperturbed functions 2P /2 as / 2,/ 2, 3 / 2,/ 2, / 2, / 2, 3 / 2, / 2, 3 / 2,3 / 2 & 3 / 2, 3 / 2 the perturbation matrix is block diagonal with two 2x2 blocks and two x blocks. If we let A = α 2 R 48 & C = µ 3 the 2x2 blocks are A C with eigenvalues and

7 A C with eigenvalues and the uncoupled m ±3/2 energies,. We plot these 6 eigenvalues as well as the two 2s /2 states (energies 5A ± 3C ) as a function of. 3 2 n=2 levels of H in a magnetic field 2p + α weak field 2p 0 α =+3/2 =+/2 E(cm-) sα 2p - α 2p + β 2sβ =+/- /2 =-/2 intermediate field 2p 0 β 2p - β =-3/ (T)

β matrices defined above in terms of these Pauli matrices as

β matrices defined above in terms of these Pauli matrices as The Pauli Hamiltonian First let s define a set of x matrices called the Pauli spin matrices; i σ ; ; x σ y σ i z And note for future reference that σ x σ y σ z σ σ x + σ y + σ z 3 3 We can rewrite α &