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Neutron Scattering Ross Stewart ISIS Neutron Facility Rutherford Appleton Lab, Didcot Scattering experiments • In a scattering experiment, a beam of radiation is incident on a sample • The distribution of radiation scattered from the sample is measured • This is determined by the interaction potential of the radiation and the sample • The radiation must be coherent (either spatially or temporally or both) detector beam of particles sample Scattering experiments • “Diffraction” • assumes “elastic scattering” - no energy transferred to/from sample • measurement of crystal structures, atomic correlations in liquids/glasses • Inelastic scattering (spectroscopy) • energy transferred to/from sample • measurement of lattice vibrations (phonons), atomic diffusion, molecular modes • Small-angle scattering • diffraction at very small angles • measuring large objects - proteins, colloids, aggregates, nanoparticles, etc • Reflectometry • diffraction from a surface - specular or off-specular • measures depth profile of thin films, membranes, etc. Neutrons / X-rays • Neutrons and X-rays commonly used in scattering experiments Neutrons X-Rays Charge 0 0 Mass 1.675 ×10-27 kg 0 Spin 1/2 1 Magnetic Moment -1.913 μN 0 Momentum mv = ħk ħk Energy ½mv2 = ħ2k2/2m ħω = hc/λ Nuclear scattering O(r0 = 2.82 ×10-5 Å) (short range nuclear forces) r0 (E-field of photon and e) Magnetic scattering O(r0) (μN.Bdip) r0(ħω/mc2) (E,H-field of photon and e, μB) n 0.9997 → 1.0001 1→4 Φ/ΔΩ 1014 n/cm2/ster/s (60MW reactor) 1019 p/cm2/ster/s (60W lightbulb) Neutrons / X-rays • Neutrons and X-rays commonly used in scattering experiments So why neutrons? •No charge - penetrating •Weakly interacting (small perturbation) •Strong magnetic interaction (s = 1/2) •Strong scattering from light nuclei (e.g. H) Thermal neutrons (λ=1.8 Å, v = 2200 • λ ~ interatomic spacing • Ek ~ phonon excitations - developed alongside nuclear fission -1 ms ) Neutrons pictured as matter waves 1 (x, t) = p 2⇡ Z 1 i(kx w(k)t) A(k)e dk 1 Δx Neutron located “somewhere in Δx” must increase Δx to define λ better - or increase Δλ to define x better Heisenberg The longer the Δx, the larger the spatial and temporal coherence of the neutron - more monochromatic Neutrons are often considered as infinite plane waves Origins James Chadwick (1891–1974) Nobel prize, 1935 Origins Enrico Fermi (1901-1954) Nobel Prize, 1938 Discovered the transmutation of elements due to neutron irradiation Part of the team that first performed fission 2nd December 1942 first ‘atomic pile’ (CP-1) Origins Clifford Shull 1915 – 2001 Nobel Prize in 1994 Developed Neutron diffraction Discovered structure of ice and hydrides Worked out neutron cross-sections Discovered antiferromagnetism Bertram Brockhouse 1918 – 2003 Nobel Prize in 1994 Developed inelastic neutron scattering First measurement of phonon dispersion using a three-axis spectrometer Today Bio-sensors e Gre he c n tr y s i m Biolo gical s truct ures Medical ap plications Fundamental magnetism nt e m n o r i env e h t d n ya g r e n e , n Pollutio Today REPORTS σ′ A σ′ B C σ′ incident neutron polarization, the SF and NSF cross sections yield information on S yy(Q) and S zz(Q), respectively. We used a single crystal of Ho2Ti2O7 to map diffuse scattering in the h, h, l plane. Previous unpolarized experiments (20, 22) have measured the sum of the SF and NSF scattering, but in this orientation only the SF scattering would be expected to contain pinch points (26). Fi sca [s for tra Hi res 1( ca “magnetic monopoles” in Spin-Ice CuGeO3 - spin-1/2 chain D Fig. 2. Diffuse scattering maps from spin ice, Ho2Ti2O7. Experiment [(A) to (C)] versus theory [(D) to (F)]. (A) Experimental SF scattering at T = 1.7 K with pinch points at (0, 0, 2), (1, 1, 1), (2, 2, 2), and so on. (B) The NSF scattering. (C) The sum, as would be observed in an unpolarized experiment (20, 22). (D) The SF scattering obtained from Monte Carlo simulations of the near-neighbor model, scaled to match the experimental data. (E) The calculated NSF scattering. (F) The total scattering of the nearneighbor spin ice model. become equivalent to that of the near-neighbor model. T = 1.7 K should be sufficient to test this prediction because it is close to the temperature of the peak in the electronic heat capacity that arises from the spin ice correlations [1.9 K (20)]. In our simulations of the near-neighbor spin ice model (Fig. 2, D to F), the experimental SF scattering (Fig. 2A) appears to be very well described by the near-neighbor model, Neutron sources Neutron Sources Fission Reactor Spallation Source Neutron Sources Neutron Sources Neutron Sources Reactor sources (ILL) Nuclear Fission Institut Laue Langevin (57 MW) Rapport transparence et sécurité nucléaire – Institut Laue Langevin Grenoble, France, Built 1967 Safety Rod 235U fuel element Cold Source Neutron guides Biological shielding (“swimming pool”) Reflector tank (contains D2O) Control rod HFR “Swimming Pool” Cherenkov radiation Beam tubes/guides Spallation Sources (ISIS) pectrum (E ~1 MeV) + high energy tail Neutron spallation • 15-20 neutrons produced per proton • So, 4 x 1014 neutrons per pulse Target station 1 Figure 1: ISIS Target station 2 W1 W2 W3 W4 W5 W6 W7 W8 W9 E1 E2 E3 E4 E5 E6 E7 E8 E9 Extracted Proto n schematic lay 2 ayout of re 1. Beam 800 MeV Synchrotron 70 M 70 MeeV V H- - Linac H lin ac is shown in fig u ExtErxatrcatecte ddpPrroottoonnBbeeam am 1 RFQ RFQ Target Station Target Station ncluding TS-2 800 MeV proton synchrotron Extracted proton beam movie here m 50 pps to 40 pps. The schematic layout of ity including TS-2 is shown in figure 1. RFQ ISIS 70 MeV H- Linac Target Station 1 800 MeV Synchrotron Extracted Proton Beam m 50 pps to 40 pps. The schematic layout of ity including TS-2 is shown in figure 1. RFQ ISIS Ion Source • 70 MeV H- Linac Target Station 1 800 MeV Synchrotron Extracted Proton Beam • • H gas + Cs vapour in discharge H- ions produced Accelerated to 665 keV in RFQ accelerator m 50 pps to 40 pps. The schematic layout of 1. Tube Linac ity including TS-2 is shown in figure Drift RFQ ISIS 70 MeV H- Linac Target Station 1 • • • • 800 MeV Synchrotron Extracted Proton Beam Acceleration to 70 MeV Uses RF fields (202.5 MHz) Accelerated to 665 keV in RFQ accelerator Produces 200 μs pulses, 22 mA m 50 pps to 40 pps. The schematic layout of ity including TS-2 is shown in figure 1. RFQ ISIS Synchrotron • 70 MeV H- Linac • • • Target Station 1 800 MeV Synchrotron Extracted Proton Beam Electron stripped off on injection (AlO foil) 2 proton bunches formed in RF fields Accelerated to 800 MeV 10,000 turns completed target) pps uses a triplet structure to supply The a beam waist at of layout schematic pps. 40 to m 50 configuration of conducting coils and handling of target in both planes. In order to accommodate the issues were all carefully considered before producing beam size, Q(31 Q 36), VSM(4, 5) and 1. detailed designs. TS-2 is shown infinalfigure including itymaximum HSM(3, 4) are required to have a larger aperture than the other EPB-2 magnets. RFQ ISIS Extracted Proton Beam Diagnostics and Beam Control 70 • protection is provided by 15 gas ionisa Machine and(BLM) “septum” beam loss“kicker” monitors and magnets five intensity (b 2 pulses,(IM). 100 ns long analysis from the B • monitors Signal H- Linac MeVcurrent) Operates 50 Hz • beam trips the underathigh beam loss conditions, and IMs allow (4/5 beam pulses intensities to TS1, last atoneextraction, to TS2) du transmission and to target to be measured. B trajectories and widths are measured using 36 pr monitors and six position monitors distributed along length of EPB-2. Beams extracted to TS1 and TS2 using Other Considerations Target Station 1 Figure 4: Magnets EQ7, K2, septum, EHB4, Q1 and Q2. 04 Hadron Accelerators The realisation of EPB-2 has involved the conce efforts of many groups throughout ISIS and from con MeV 800well staff. As as construction of the new TS-2 build Synchrotron substantial civil engineering and building work has A15 High Intensity Acceler 2904 Extracted Proton Beam Target stations •EPB collides with target heavy metal •Neutrons produced by “spallation” •Beam power is 160 kW (TS1) Neutron moderators • Both fission and spallation sources produce high energy neutrons (E~1 MeV, λ=3 x 10-4 Å) • Neutron scattering requires “thermal” or “slow” neutrons (E ~ 25 meV, λ=1.8 Å at T=293 K) • Neutrons can be slowed down by elastic collisions with light atoms (e.g. H, D, C) Neutron moderators • Both fission and spallation sources produce high energy neutrons (E~1 MeV, λ=3 x 10-4 Å) • Neutron scattering requires “thermal” or “slow” neutrons (E ~ 25 meV, λ=1.8 Å at T=293 K) • Neutrons can be slowed down by elastic collisions with light atoms (e.g. H, D, C) • Moderators produce neutron beams with a Maxwellian distribution of energies, at the characteristic temperature SNS moderators Pulsed source moderators • • • • Proton pulse around 1 μs Several cms of material (e.g. water, methane, liquid H2) required to thermalise neutrons proton beam At 4Å, neutron speed is 1 cm / 10 μs : so moderator broadens the initial pulse Top decoupled poisoned H2! Moderators can be “coupled” to Reflector Pulsed-source time structure material (Be) around moderators Intensity! neutron beams λ=1Å" (“neutron friendly” units; meV, Å and K) λ=5Å" time (μs)" 100 20 cm ~2 k 2 81.81 E= = = kB T = 0.08617T 2 2mn λ=2Å" 0 coupled H2! 200 300! So a moderator peaked at 2 Å has a temperature of T = 237 K Pulsed-source time structures Time structure cold neutrons log(Intensity)! 10! ILL! 1! ISIS-TS1! 0.1! 0 20 40 ISIS-TS2! 80 60 80 time time(ms) (ms)" 100 100 120 120 ! Neutron Scattering Basics Scattering from a single nucleus kf ki Scattered spherical wave i = exp(iki z) f = Incoming plane wave Elastic scattering: 0 |ki | = |kf | = 6 |kf |, Inelastic scattering: |ki | = b exp(ikf r) r z 2⇡ ~ 2 E = ~! = (ki 2mn 2 2 kf ) Scattering from a single nucleus kf ki Scattered spherical wave i = exp(iki z) f b exp(ikf r) r = Incoming plane wave Probability densities: | i |2 = ⇤ = exp( iki z) exp(iki z) = 1 z 0 | 2 | f = ⇤ = b2 r2 Scattering length, b f • • • • • = b exp(ikf r) r The scattering length b, is taken to be constant for a particular nucleus (neglecting spin) In general, b is a complex number - imaginary part corresponds to neutron absorption (normally small) The minus sign indicates that a positive b corresponds to a repulsive scattering potential. Therefore for positive real scattering lengths, the scattered wavefunction is πshifted wrt. the incident wavefunction. A few nuclei have negative real scattering lengths corresponding to an attractive scattering potential, and zero phase shift Intensity of scattered beam is proportional to |b|2 Flux and cross-section • • The flux of a neutron beam is number of neutrons per unit area per second which is the probability density times velocity =| | v 2 • The total cross-section, σ is the number of neutrons scattered in all directions per second - normalised to the incident flux i f Giving the cross-section = | i| v = v 2 incident 2 b = | f |2 v = 2 v scattered r = f ⇥ 4⇡r2 i = b2 2 2 ⇥ 4⇡r = 4⇡b r2 • • • • Notes: The cross-section is the effective area of the nucleus viewed by the incident neutrons b is then the effective range of the scattering potential b is measured in fm (either fermi or femtometre = 10-15 m) σ is measured in b (barns = 10-28 m2) No nuclear theory of neutron-nuclear interaction - so b must be experimentally determined Differential cross-section • Can’t measure all neutrons scattered kf ki • • Detector ⌦ d The differential cross-section d⌦ is the number of neutrons scattered into solid angle dΩ per second - normalised to the incident flux times dΩ A ⌦= 2 r Solid angle is given by Therefore so for a single nucleus d d⌦ = 2 | f vA | i |2 v ⌦ = | | 2 2 | f r d = b2 = d⌦ 4⇡ (barns / steradian) (b sr-1) Scattering lengths and spin • • In general, b is dependent on spin-states of the nucleusneutron scattering system 2 possibilities, with scattering lengths b+ and b- b+ : Itot = Inuc + 1/2 with degeneracy 2Inuc + 2 b- : Itot = Inuc 1/2 with degeneracy 2Inuc • So for unpolarized neutrons and randomly oriented nuclear spins, the probability of scattering is proportional to I +1 p = 2I + 1 + • • I p = 2I + 1 The average scattering length, b̄ is given by (Inuc + 1)b+ + Inuc b b̄ = 2Inuc + 1 It’s useful also to calculate the standard deviation in the scattering lengths q b¯2 (b̄)2 = B p Inuc (Inuc + 1) where b+ b B= 2Inuc + 1 Example: H • Protons have spin-1/2 • Experimentally, it is found that the spin dependent scattering lengths for H are; b = 10.817 fm + b = 47.42 fm So the average scattering length is = b̄ = 3/2 ⇥ 10.817 3.74 fm 2 1/2 ⇥ 47.42 and the standard deviation in the scattering lengths is q b¯2 (b̄)2 10.817 + 47.42 p = ⇥ 3/2 2 = 25.22 fm Scattering from an assembly of nuclei 0 ki kf B rn A n scattering from nth nucleus: what is ΔRn ?: n Rn = = = ) k Rn = = At origin, detector is at distance R, and at nucleus n, path difference is wrt origin is ΔRn bn exp{ik(R + R + Rn ! ! An b0 ki .rn kf .rn k k (ki kf ).rn Q.rn where Q is the momentum transfer Q = (ki kf ) Rn )} Scattering from an assembly of nuclei 0 ki kf B rn A n scattering from nth nucleus: f = X n = = X bn exp(ikR) exp(iQ.rn ) R + Rn exp(ikR) X bn exp(iQ.rn ) -far field limit R recalling that the differential cross-section is we arrive at At origin, detector is at distance R, and at nucleus n, path difference is wrt origin is ΔRn d =| d⌦ X d = bn exp(iQ.rn ) d⌦ 2 2 2 | f R Fourier transforms • X d = bn exp(iQ.rn ) d⌦ 2 The differential cross-section is the square of the (discrete) Fourier transform of the scattering length (as a function of r) d S(Q) / b(r) d⌦ 2π/d d FT ..... ..... r ) ..... ..... • For a 1d crystal, the nuclear scattering is a series of Dirac delta-functions • For an infinite line of these, the Fourier transform is also a series of Dirac delta-functions in reciprocal space (momentum space) Q Coherent & Incoherent Scattering X d = bn exp(iQ.rn ) d⌦ 2 • Recall that for each nucleus, bn depends on • the element • the particular isotope • the relative spins of the nucleus and neutron • If the nuclei (and their spins) are uncorrelated then the differential cross section for one particular sample is the same as the average over many samples (with the same nuclear positions, rn) Ensemble averaging 2 types of nucleus, 50% pink - 50% white Ensemble average over random configurations will look like a single sample with one scattering length (the mean) Coherent & Incoherent Scattering X 2 d = bn exp(iQ.rn ) d⌦ X X = bn bm exp{iQ.(rn n rm )} m Taking the ensemble average we replace bn bm = bn bm If the nuclei are uncorrelated then bn b m = bn b m = b = b2 2 n 6= m n=m We can then write the differential cross-section as d 2X X = b exp{iQ.(rn rm )} + b2 d⌦ n m6=n 2XX 2 2 = b exp{iQ.(rn rm )} + b b | {z } n m | {z } Incoherent Coherent Scattering Scattering Coherent & Incoherent Scattering • Coherent Scattering • describes correlations between nuclei • peaks - due to interference - with phase factor Q.r • indicates the nuclear structure of the sample • (also the collective dynamics) • Incoherent Scattering • No information on structure (no Q.r dependence) • ‘flat background’ • (also the dynamics of single particles) • The total coherent and incoherent cross-sections are written coh = 4⇡b 2 inc = 4⇡(b2 2 b ) These values are tabulated for each isotope/element (http://www.ncnr.nist.gov/resources/n-lengths) Example: H • we found before that the average scattering length of H was -3.74 fm, and that the standard deviation was 25.22 fm • Therefore the coherent cross-section is H coh • = 175.8 fm2 = 1.758 barns (10-28 m2) And the incoherent cross-section is H inc • • = 4⇡( 3.74) 2 = 4⇡(25.22) 2 = 7993 fm2 = 79.93 barns (10-28 m2) So the incoherent scattering is dominant for H Since this incoherent scattering is entirely due to the nuclear spin, it’s termed spin-incoherent scattering • Example: Ti Ti metal has 5 nuclear isotopes Isotope % 46Ti 8.2 7.4 73.8 5.4 5.2 47Ti 48Ti 49Ti 50Ti bcoh = b binc = 4.93 fm 3.63 fm -6.08 fm 1.04 fm 6.18 fm q b2 b 2 0 fm 3.5 fm 0 fm 5.1 fm 0 fm First, we calculate the mean b, and standard deviation for the isotopes bTi coh = = bTi Iso inc 0.082 ⇥ 4.93 + 0.074 ⇥ 3.63 0.738 ⇥ 6.08 + 0.054 ⇥ 1.04 + 0.052 ⇥ 6.18 3.437 fm p = (0.082 ⇥ 4.932 + 0.074 ⇥ 3.632 + 0.738 ⇥ 6.082 + 0.054 ⇥ 1.042 + 0.052 ⇥ 6.182 ) = 4.526 fm 3.4372 So the coherent and incoherent cross-sections are Ti Coh = 4⇡0.34372 = 1.484 barns Ti inc = = 4⇡(0.45262 + 0.074 ⇥ 0.352 + 0.054 ⇥ 0.512 ) 2.86 barns (tabulated incoherent cross-sections are the sum of the isotope and spin parts) Neutron Diffraction from Crystals Cross-section for a crystal • Crystal lattice described by points displaced from the origin by a lattice vector l = n1 a + n2 b + n3 c where a, b, and c describe the unit cell of the crystal • The position of the nth nucleus, rn is • rn = l + d where d is the position of the nucleus in the unit cell Therefore, we can write the differential cross-section for the reciprocal lattice as X 2 d = bn exp(iQ.rn ) d⌦ 2 X X = exp(iQ.l) bd exp(iQ.d) l d Cross-section for a crystal d = d⌦ • X exp(iQ.l) l X 2 bd exp(iQ.d) d This cross-section is zero unless all the terms in the sum over lattice vectors are equal, i.e. the condition for constructive interference - also called the Laue Condition for scattering exp(iQ.l) = 1 for all l Reminder: definition of reciprocal lattice vectors is the set of vectors G which yield plane waves with the periodicity of a given Bravais lattice described by l exp{iG.(r + l)} = exp(iG.r) ) exp(iG.l) = 1 Laue Condition • Therefore the Laue condition for scattering is that the momentum transfer to the crystal (also called the scattering vector) coincides with a reciprocal lattice vector, G Q = G = ha⇤ + hb⇤ + lc⇤ • For a given crystal plane • For elastic scattering ki 2✓ “Scattering triangle” kf 2⇡ |Ghkl | = dhkl Q = ki Q kf M von Laue (1879-1960) Nobel Prize 1914 |Q| = 2k sin ✓ 4⇡ sin ✓ = Friedrich, Knipping, Laue, 1912 Bragg’s Law • Therefore we have for a given crystal plane 2⇡ dhkl ) • = 4⇡ sin ✓ = 2dhkl sin ✓ This is the condition for the appearance of a so-called Bragg peak in the differential cross-section Founders of crystallography Nobel Prize 1915 W H Bragg (1862-1942) W L Bragg (1890-1971) The Structure Factor • Writing out the differential cross-section for coherent Bragg scattering in full, we have X d = N 2| bd exp{i(ha⇤ + kb⇤ + lc⇤ ).(xd a + yd b + zd c)}|2 d⌦ d 2⇡ 2⇡ 2⇡ ⇤ ⇤ ⇤ (where a = c ⇥ a, and c = a⇥b) b ⇥ c, b = V V V therefore d = N 2 |Fhkl |2 d⌦ where the structure factor, Fhkl is given by Fhkl = X bd exp{2⇡i(hxd + kyd + lzd )} d with the sum over the d nuclear sites in the unit cell Example: Face-centred cubic lattice • In a fcc lattice there are 4 atoms at positions (0,0,0) (½,½,0) (0,½,½) and (½,0,½) fcc Fhkl = b (1 + exp{i⇡(h + k)} + exp{i⇡(k + l)} + exp{i⇡(h + l)}) • The exponential terms are either 1 or -1 depending on whether h, k and l are odd or even • fcc If h, k and l are all odd or all even, then Fhkl = 4b • fcc Otherwise, Fhkl =0 • Note that Fhkl is real for the fcc lattice. This is always the case for lattices with inversion symmetry - “centrosymmetric” • Specific systematic absences (like mixed odd/even hkl in fcc lattices) give further clues as to the point-group crystal symmetry Example: Si • Si has fcc symmetry but with a basis of two Si atoms at (0,0,0) and (¼,¼,¼) - diamond structure • This gives 8 atoms per unit cell, with a structure factor: Si Fhkl = bSi (1 + exp{i⇡(h + k)} + exp{i⇡(h + l)} + exp{i⇡(k + l)} ⇡ ⇡ + exp{i (h + k + l)} + exp{i (3h + 3k + l)} 2 2 ⇡ ⇡ + exp{i (3h + k + 3l)} + exp{i (h + 3k + 3l)} 2 2 which can be written as • Si Fhkl = bSi (1 + exp{i⇡(h + k)} + exp{i⇡(k + l)} + exp{i⇡(h + l)}) ⇣ ⌘ ⇡ ⇥ 1 + exp{i (h + k + l)} 2 This gives the extra condition that ½(h+k+l) must not be an odd number (2 Example: Si (powder) [111] [220] [311] [400] CH3-rubredoxin crystal Magnetic neutron scattering Neutron magnetic interaction Magnetic scattering length due to potential felt by the neutron in a magnetic field Vm = µ · B → Bs - field due to spin → BL - field due to orbit Neutron magnetic interaction • The expression for the magnetic scattering cross-section is more complex than for nuclear scattering ⇣ r ⌘2 X d 0 2 2 = g F (Q) ( d⌦ 2 ↵ prefactor = 0.073 barns Landé g-factor for magnetic ion ↵ Q̂↵ Q̂ ) X n Orientation factor: only spin components perpendicular to Q are involved exp(iQ · rn ) < S0↵ >< Sj > Spins summed with phasefactor (i.e. magnetic structure factor) Magnetic form-factor for scattering ion • Magnetic neutron scattering only occurs from components of the magnetisation perpendicular to the scattering vector Neutron magnetic interaction • The last part of the magnetic cross-section is the square of the perpendicular component of the Fourier transform of the magnetisation in the sample X ( ↵ ↵ M(Q) = Q̂↵ Q̂ ) Z X n exp(iQ.rn ) < S0↵ >< Sj >= M2? (Q) FT of real-space magnetisation M(r) exp(iQ.r)dr of sample (in continuous limit) Q̂ M (M.Q̂)Q̂ Q̂ ⇥ (M ⇥ Q̂) M ⇥ Q̂ M? = M (M.Q̂)Q̂ = Q̂ ⇥ (M ⇥ Q̂) Magnetic form factors • Since the distribution of magnetisation in real-space from unpaired electrons in a crystal is extended (order 1 Å) this means the Fourier transform (i.e. the scattering cross-section) will be modulated - this is called the magnetic form factor and is specific to each magnetic ion M(Q) M(r) 2π/d d ..... FT ..... r ) • The form-factor plays the same role as an aperture function in optics. The wider the “aperture” the more limited the Q-range of the diffraction pattern. Q Magnetic form factors • The shape of the form factor can generally by modelled by a series of spherical Bessel functions F (S) = c1 h(j0 (S)i + c2 h(j2 (S)i + c1 h(j4 (S)i + .... spin only orbital Dy3+ form factor π •The Fourier transform of the form factor is the real space magnetisation distribution around the magnetic ion Form factor measurements Crystal structure P. Javorsky et al., Phys. Rev. B 67 (2003) 224429! UPtAl! crystal peak structure! Derived from Bragg positions/intensities Magnetisation density Derived from inverse Fourier transform of F(Q) magnetic moment density! Form factor measurements Lee et. a;. arXiv:1001.3658v1 SrFe2As2 4 at larger and more di⇤cult to measure scattering vectors. The similarity with bcc Fe is particularly evident in the total 3d charge density within the mu⇤n-tin sphere. This is 5.95 electrons for the experimental atomic positions in SrFe2 As2 , 6.02 electrons for the corresponding calculated energy optimized As position and 5.83 electrons for bcc Fe calculated with the same mu⇤n-tin radius. We have determined the magnetic form factor of Fe in SrFe2 As2 by neutron diffraction experiments and deduced the Fe magnetic moment as 1.04(1)µB . We also calculated the magnetic form factor by first principles electronic structure methods using both the experimental and optimized As positions. While the magnitude of the calculated magnetic moments strongly depends on the As position, the normalized magnetic form factors were remarkably similar with each other and also agreed well with the normalized magnetic form factors from the Magnetic order in MnO MnO, 80 K, antiferromagnetic MnO, 300 K, paramagnetic = 2d sin ✓ New magnetic peaks C. G. Shull & J. S. Smart, Phys. Rev. 76 (1949) 1256 Magnetic order in MnO 2a a Nuclear unit cell Magnetic unit cell re Magnetic order in MnO How do we know that the moments lie in the (111) plane? reciprocal space (111) ( 1_ 1_ 1_ 222 ( ( - 3_ 1_ 1_ 222 ( (311) -5 3 1 ( _2 _2 _2 ( (220) moment direction (fcc lattice, hkl all even/odd) H. Shaked et al., Phys. Rev. B 38 (1988) 11901 Example - MgB2 = 2d sin ✓ MgB2 is a superconductor below 39K, and expels all magnetic field lines (Meisner effect). Above a critical field, flux lines penetrate the sample. Cubitt et. al. Phys. Rev. Lett. 90 157002 (2003) λ = ~2°! 10 Å d = 425 Å Inelastic neutron scattering re Inelastic scattering triangle • In general the incident and final wavevectors will have different absolute values |ki | = 6 |kf |, ~2 E = ~! = (ki2 2mn ~Q = ~(ki ki 2✓ kf energy transfer kf2 ) kf ) momentum transfer 2 2 2 Q = k + k Cosine rule: i f Q 2ki kf cos 2✓ ~2 ki2 giving, in terms of the incident energy, Ei = 2mn and the energy transfer ~! ~2 Q2 = 2Ei 2mn ~! 2 cos 2✓ • Each scattering event characterised by (Q,ω) p Ei (Ei ~!) ~! Inelastic scattering triangle (assume Ei is fixed) Ei +ve energy transfer: Neutron energy LOSS ~ki LOSS ki created excitation kf 2~ki ~Q -ve energy transfer: Neutron energy GAIN GAIN ki kf absorbed excitation re Inelastic scattering cross-section • If we can distinguish the energy of the scattered neutrons, we can measure the partial differential cross-section defined by the derivative of the differential cross-section with respect to energy @ d /d⌦ for Ef between E and E + = lim E!0 @⌦@E E 2 E •So the differential cross-section (measured in diffraction) is actually the sum over all possible final energies d = d⌦ Z 1 0 @2 dE @⌦@E which is not the same as strictly elastic scattering, given by d d⌦ elastic @2 = @⌦@E E=Ei •The full integral differential cross-section is called the total scattering re Inelastic scattering cross-section kf ki Detector ⌦ cross-section we have •For inelastic scattering the 2 2 @ @⌦@E = = | f | vf A | i |2 vi ⌦ kf | f |2 R2 ki •where the scattered wavefunction now must account for the time dependence of the nuclear positions f = therefore where S(Q, !) = exp(ikR) X X bn exp(iQ.rn R t n @2 kf = S(Q, !) @⌦@E ki XX t n bn exp(iQ.rn !t) 2 !t) “scattering function” re Principle of Detailed Balance We have to weight the neutron cross-section with the probability of scattering from an initial state and sum over these. This leads to the law of detailed balance S (Q, )=e /kB T S (Q, ) i.e. more likely to create a phonon than absorb one when at low temperatures by a Boltzmann factor S(Q, !) S(Q, !) kB T ⇠ ~!0 !0 !0 ! T =0 !0 !0 ! re Phonon Scattering • Scattering will occur at the characteristic momentum and energy for phonons to be created (or absorbed) by the neutron - i.e when ~Q = ~(kph + G) ~!n = ~!ph •Consider a static sinusoidal lattice distortion on a 1d chain Δx α x position of the nth nucleus is xn = nd + ↵ sin(kph nd) giving for the cross-section d d⌦ = = X X n bn exp(iQ.rn ) 2 2 bn exp {iQ(nd + ↵ sin(kph nd))} re Phonon Scattering • Expanding the exponential assuming small lattice displacements such that Q↵ ⌧ 1 d = d⌦ • The first term X X 2 bn [1 + iQ↵ sin(kph nd)] exp(iQnd) n 2 will peak at n which are the Bragg peak positions bn exp(iQnd) Q=m ✓ 2⇡ d ◆ (m integer) •The second term simplifies as X n 2 bn Q↵ [exp {i(Q + kph )nd} giving peaks at Q = m 2 ✓ 2⇡ d with intensity / b Q2 ↵2 ◆ ± kph exp {i(Q kph )nd}] d d⌦ Phonon Scattering 2⇡ d 4⇡ d 6⇡ d Q •Phonon is dynamic sinusoidal lattice distortion - cross-section is same as for static case - but conserving energy 2 @ when ~Q = ~(kph + G) and ~!n = ~!ph •Peaks in @⌦@E 2 2 b Q 2 2 2 • With intensity / b Q ↵ / !ph e.g. transverse phonons in lead Phonon Examples •Volume of S(Q, !) space for a full crystal plane •Taken using time-of-flight neutron inelastic scattering Bragg-peaks Dove et. al. ISIS, 2008 Calcite crystal Phonons: agreement with experiment & theory Phonon Examples Pho non disp ersion curv es as measurealong d for GaA dispersion measurements high s. Strauch & Dorner •Detailed (199 0). Thedirections lines give the result of ab initio calculations and show that symmetry the forces between the atoms are well understood. The letters below give the notausing tion for the symneutron metry dire three-axis inelastic scattering ctions or points. •Taken Note the calculated density-of-states on the extreme right. •Modelled by ab-initio calculations extracting force constants between atoms in the crystal Dorner et. al. ILL, 1990 1 meV = 8.065 cm-1 Spin-Waves Coherent - and propagating magnetic excitations in ordered magnets ~! = 4JS (1 cos(2qa)) = 2Dq for qa << 1 2 ω (4JS) 2 1 0 0 0.5 Q (2π/a) 1 The spin-wave stiffness D gives the strength of the exchange interaction, J between the magnetic ions Example: Rb2MnF4