Download Ross Stewart ISIS Neutron Facility Rutherford Appleton Lab, Didcot

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
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Medical ap
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Fundamental
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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