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Ross Stewart Outline of Lecture Magnetic Neutron Scattering Ross Stewart Outline of Lecture • How does the neutron see magnetism? • Magnetic form factors • Magnetic neutron diffraction • Magnetic inelastic neutron scattering How do the neutrons see magnetism? Magnetic “dipoles” Magnetic Dipole Moment, μ by definition from classical electromagnetism µ = IA → A e → so in terms of orbital angular momentum, L e µ = L − 2m e Magnetic “dipoles” Magnetic Dipole Moment, μ Dirac (1928) also postulated an intrinsic angular momentum with associated magnetic moment (seen experimentally) → S e e Except that this time: µ = S − m e For particle with spin and orbital contributions: e µ = g J “g-factor” − 2m e J = L + S Magnetic moment defs. Angular momenta, are measured in units of ℏ, so we can write e µ = gµBJ where µB = − 2me Bohr Magneton The gyromagnetic ratio, is defined as the ratio of the magnetic dipole moment, to the total angular momentum e γ = g − 2m Therefore gµB = γ and µ = γ J − Magnetic properties of the neutron For neutrons, we assume a spin-only angular momentum, with eigenvalues of ±ℏ/2 z 1 “Spin-Up” = 1/2 | ↑ 0 s = 1/2 ms = ± 1/2 0 “Spin-Down” = -1/2 | ↓ 1 Magnetic properties of the neutron According to quantum mechanics, the “classical” angular momentum is replaced by an angular momentum operator, σ µ = γσ The components of σ for a spin-1/2 particle (the neutron) are 01 0 i 10 σ = 1 σ = 1 σ = 1 x 2 10 y 2 i −0 z 2 0 1 − Pauli spin matrices Magnetic properties of the neutron Some gyromagnetic ratios - (all spin-1/2) Electron: 1.76 x 105 MHz / T Proton: 267 MHz / T Neutron: 183 MHz / T So the neutron moment is around 960 times smaller than the electron moment e Nuclear magnetons: µN = 2mp μp = 2.793μN μn = 1.913μN So we have for the neutron µ = γ µ σ n − n N where γn = 1.913 Neutron magnetic interaction Differential neutron cross-section The sum of all processes in which - the state of the scatterer changes from λ to λ` - the wavevector of the neutron changes from k to k` (where k` lies in solid angle dΩ) - the spin-state of the neutron changes from s to s` ∂2σ = Wk,λ,s k ,λ ,s ∂Ω∂E → k in dΩ 2 2 k mn = k, λ,s Vm k, λ,s δ(Eλ Eλ + ω) k 2π2 | | − “Fermi’s Golden Rule” Neutron magnetic interaction Vm is the potential felt by the neutron due to a magnetic field → BL - field due to orbit V = µ B m · → Bs - field due to spin Neutron magnetic interaction Field due to spin and orbital moments is ˆ ˆ µ0 µe R 2µB p R B = Bs + BL = ×2 ×2 4π ∇× R − R Magnetic vector → potential A, of a dipolar Biot-Savart Law for a field due to electron single electron of linear → spin moment momentum, p Evaluating the spatial part of the transition matrix element i j iQ rj ˆ ˆ ˆ k Vm k e · Q sj Q + pj Q | | ∝ × × Q × → where ℏQ is the momentum transfer Q = (k k) − Neutron magnetic interaction Summing over all unpaired electrons, we find j ˆ ˆ ˆ ˆ k Vm k Q M(Q) Q = M(Q) M(Q) Q Q = M (Q) | | ∝ × × − · ⊥ j where M⊥ is the perpendicular component of the Fourier iQ r transform of the magnetisation, in the sample M (Q )= M (r)e · dr Qˆ M M Qˆ Qˆ · Qˆ M Qˆ × × M Qˆ × Neutrons only see the components of the magnetisation perpendicular to the scattering vector Average over neutron spin Having calculated the spatial part of the matrix element we have to calculate the spin-part. If the neutrons are unpolarized - i.e. there are equal numbers of spin-up and spin-down, then we get 2 2 ∂ σ 2 k =(γr0) λ,s σ M λ,s δ(Eλ Eλ + ω) ∂Ω∂E k | · ⊥| − 2 k =(γr0) pλ λ M ∗ λ λ M λ δ(Eλ Eλ + ω) k | ⊥| · | ⊥| − λλ Collection of prefactors: 2 µ0 m µ0 e γµN 2µB 4π = γr0 where r0 = −4π 2π − 4π me r0 is called the “classical electron radius” and has a value of 2.818 x 10-15 m Magnetic form factors Scattering from a crystal th Assume nucleus at position Rjd representing the d atomic position th th in the j unit cell. re is the position of the e unpaired electron relative to the nucleus r = R jd + re by the definition of M(Q) M(Q)= exp(iQ r )s · i i i = exp(iQ R ) exp(iQ r )s · ld · e e e ld The matrix element λ M λ now becomes | ⊥| F (Q ) λ exp(iQ R )S λ | · jd jd| Magnetic structure factor Magnetic form factor Form factors, example F (S)=c (j (S) + c (j (S) + c (j (S) + .... 1 0 2 2 1 4 Tabulated in International Tables of Crystallography, Vol C spin only Dy3+ form factor orbital π Form factors • The magnetic form factor means that magnetic scattering is strongest at low Q (low angles for a cw neutron spectometer) • Nuclear scattering has no form factor - since nuclei are point-like on the scale of the neutron wavelength Form factor measurements Crystal structure P. Javorsky et al., Phys. Rev. B 67 (2003) 224429! UPtAl! Derived from Braggcrystal peakstructure positions/intensities! Magnetisation density Derived from inverse Fourier transform of F(Q) magnetic moment density! Form factors - moments 4 Lee et. a;. arXiv:1001.3658v13 Givord, et. al. J. Appl. Phys. 50(3), March 1979 µB to 0.879 µB (at the optimized theoretical minimum at larger and more difficult to measure scattering vectors. energy z As position). Table II shows the Fe d-orbital decomposed charges and moments. Although there are magnetic amplitudes deduced at 300 K andThe 4.2 similarityK are with bcc Fe is particularly evident in the variations among the occupied electronic orbitals, mag- reported in table II and shown in Fig. totalI(a) and 3d I(b)charge density within the muffin-tin sphere. This netic moments are fairly well distributed across all of the respectively. The difference between the two sets of orbitals. This would not be expected if one or more of data is striking. is 5.95 electrons for the experimental atomic positions in the orbitals were “localized”. Indeed, all of orbitals con- In the case of samarium, it is knownSrFe that2As the2, 6.02 electrons for the corresponding calculated tribute to a broad density of states. orbital and spin parts of the magnetization are opposite and nearly cancel out. As theirenergy spatial optimized AsB position and 5.83 electrons for bcc TABLE II: Fe d-orbital electron occupation and magnetic mo- distributions are very different,it resultsFe calculated in regions with the same muffin-tin radius. ment orbital decomposition. of space with a positive magnetization density and regions with a negative one. Consequently, the magnetic spin up spin dn Magnetic Moment :7 We have determined6 the magnetic form factor of Fe 10 C) (electrons) (electrons) (µB) structure factors FM' which are the Fourier coeffici- ents of the magnetization density, a maximum '::' ,,:\ exp zAs opt zAs exp zAs opt zAs exp zAs opt zAs in SrFe2As2 by neutron diffraction experiments and de- d 3.810 3.450 2.142 2.574 1.668 0.876 non located at sin 8/), = 0 (sin 8/), "'0.4 A-I), duced the Fe magnetic moment as 1.04(1)µB. We also dz2 0.719 0.618 0.495 0.513 0.224 0.105 leading to an unusual magnetic form factor. Moreover, , T 042 K dx2 y2 0.757 0.684 0.402 0.491 0.355 0.193 in smCos the very large exchange field andcalculated the crystal the magnetic form0 factor by first principles − 0 dxy 0.769 0.700 0.502 0.568 0.267 0.132 field mix the excited multiplets J=7/2 and J=9/2 into M electronic structure methods using both the experimen- 0 dxz 0.767 0.718 0.354 0.476 0.413 0.242 the ground multiplet J=5/2. This leads to large changes 00 6 0 - 0<'<-0 C'l.,..:. MN 0 ., 0 ;;; C'1NCO)'<tMq' ., N dyz 0.798 0.730 0.389 0.526 0.409 0.204 FIG. 2: (Color online) Spin density alongin the Fe-Asthe Sm direction magnetic moment and in its formtal factor and optimizedat As positions. WhileM., ""' the magnitude"' SmCo5 as calculated from LDA with the experimentalsmall atomicscattering posi- angles, thus allowing sensitiveof the calculated '.le- magnetic2 moments strongly depends on Figure 2 shows the spin density along Fe to As. It tion. Although As does not have a net magnetictermination moment of in exchange and crystal field parameters. indicates that although the spin density is strongly con- the antiferromagnetic structure, its spin densityThe distribution total Hamiltonian representing theexchange As position,inter- the normalized magnetic(a) form factorsf centrated inside Fe atoms, the spin density inside the As depends on the magnetic ordering of Fe.actions The inset and shows crystal field may be written as spin densities inside the Fe muffin-tin sphere along specific were remarkably similar with each other and also agreed muffin-tin sphere is affected by the surrounding Fe atoms directions demonstrating substantial anisotropy. through hybridization. With antiferromagnetic ordering well with the normalized magnetic form factors0000 from00 0 the0 o vN ... t") 100 0 <0 o (see Fig. 1), the As atoms do not have a net moment.
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