Download Magnetic Neutron Scattering and Spin

Magnetic Neutron Scattering and Spin-Polarized Neutrons
Physical origin: potential of magnetic dipole moment of the neutron in magnetic
field generated by electron spins and orbital moments in the solid.
K
µn
K
µe
K
He
K
Specialize to contribution of electron spin to H e :
e=
K
K
µe = −2µ B se with µ B =
2me
g
e=
K
K
K
µn = − g n µ N sn ≡ −γµ N σ with µ N =
and γ = n = 1.913
2
2mn
K
K
K
se and sn are spin-½ operators, σ = 2sn is the neutron Pauli matrix whose
eigenvalues are ±1 .
Differential neutron scattering cross section for one electron at rest at
origin of coordinate system:
2
K
2
dσ ⎛ mn ⎞ K
K K
with H int = − µn ⋅ H e
=⎜
k f m f H int ki mi
2 ⎟
d Ω ⎝ 2π = ⎠
This follows from Fermi’s Golden Rule, as in the case of nuclear neutron
scattering. The only qualitative difference is that we have to keep track of the
neutron spin state m = ±1 in addition to the spatial part of its wave function
K
labelled by the quantum number k .
Vector potential of dipole field (from classical electromagnetism):
K K
K
µ 0 µe × r µ 0 K K 1
Ae =
=
µe × ∇ K
r
4π rK 3
4π
K
K K
µ K ⎛K
1⎞
H e = ∇ × Ae = 0 ∇ × ⎜ µe × ∇ K ⎟
r ⎠
4π
⎝
2
dσ ⎛ mn ⎞
1⎞ K
2 K
K K ⎛K
=⎜
2γµ N µ B ) k f , m f σ n ⋅∇ × ⎜ se × ∇ K ⎟ ki , mi
2 ⎟ (
d Ω ⎝ 2π = ⎠
r ⎠
⎝
1
2
Use trick to simplify this expression:
K
∞
dp ipK ⋅rK
K 1 i pK rK cos Θ
=
2
π
e
d
p
d ( cos Θ )
∫ K2
∫
∫e
p
−1
0
K K
∞
K sin p r 2π 2
= 2π ∫ d p
K K = K
p r
r
0
K
Here, p is an auxiliary variable without physical meaning.
K
KK
1⎞
1
dp K K K
⎛K
∇ × ⎜ se × ∇ K ⎟
= 2 ∫ K 2 ∇ × ( se × ∇ ) eip⋅r
r ⎠
2π
⎝
p
=
K
1
∫ pˆ × ( se × pˆ ) e
2π 2
KK
ip⋅r
K
dp
K
Q
K
KK
1⎞ K
1
K KK K
K
⎛
k f ∇ × ⎜ se × ∇ K ⎟ ki = 2 ∫ dre −iQ⋅r ∫ dp pˆ × ( se × pˆ ) eip⋅r
r ⎠
2π
⎝
K
= 4π Qˆ × ( se × Qˆ )
K ≡ se ⊥
K
The last line follows by doing the r -integration first and using
K
1
K i( pK −Q )⋅rK
K K
dr
e
=
δ
p
− Q)
(
∫
( 2π )3
collect all prefactors:
2⎛ µ ⎞
2
2
⎛ mn ⎞
2γµ N µ B ) ⎜ 0 ⎟ ( 4π ) = ( γ r0 )
⎜
2 ⎟ (
⎝ 2π = ⎠
⎝ 4π ⎠
−15
where r0 = 2.8 × 10 m is the “classical electron radius” that also appeared in the
Thompson cross section for x-ray scattering.
2
dσ
2
K K
= ( γ r0 ) m f σ ⋅ se ⊥ mi
dΩ
2
2
For an unpolarized neutron beam, one has to average over the neutron spin
states m . For convenience, take the neutron spin quantization axis ẑ to be
K
parallel to se⊥ . Then
K K
m f σ ⋅ se ⊥ mi = se ⊥ m f σ z mi
⎧ se ⊥ if m f = mi
=⎨
⎩0 otherwise
The cross section for a single electron at rest for an unpolarized neutron beam
is therefore
dσ
2 K
2
= ( γ r0 ) se ⊥
dΩ
K
K
with se⊥ the projection of the electron spin perpendicular to Q .
2
K
se
K
se⊥
Generalization for an atom
K 2
2
K 2
dσ
2
2
2
= ( γ r0 ) ηˆ⊥ f ( Q ) = ( γ r0 ) ⎡⎢1 − ηˆ ⋅ Qˆ ⎤⎥ f ( Q )
⎣
⎦
dΩ
K
K
K
1
K
M ( r ) e −iQ⋅r is the magnetic form factor and
where f ( Q ) =
∫
2µ B
K K
K
M ( r ) = M ( r )ηˆ is the magnetic dipole moment density contributed by the spin
(
)
and orbital motion of all unpaired electrons in the atom.
Generalization for collinear magnets
For collinear magnets, that is, magnetically ordered solids where all electron
magnetic dipole moments point either parallel or antiparallel to a single direction
η̂ , this expression can be generalized in analogy to the differential cross section
for x-ray or nuclear neutron scattering:
2
K iQK ⋅RK 2
dσ
2⎡
⎤
ˆ
= ( γ r0 ) ⎢1 − ηˆ ⋅ Q ⎥ ∑
( ± ) f RK ( Q ) e
dΩ
⎣
⎦ RK
(
= ( γ r0 )
2
)
⎡1 − ηˆ ⋅ Qˆ 2 ⎤ N ( 2π )
⎢⎣
⎥⎦
V0
(
)
3
∑
K
K K
K 2
FM ( K M ) δ ( Q − K M )
K
M
K
where K M are the magnetic reciprocal lattice vectors and
K
K iQK ⋅dK
K
FM ( Q ) = ∑
±
f
Q
( ) d( )e
K
d
The +K or – sign is chosen depending on whether the magnetic moment at lattice
site d is parallel or antiparallel to η̂ .
Examples:
1 One-dimensional ferromagnet
a
nuclear and magnetic unit cells identical ⇒ K M = K N =
⎛ dσ ⎞
⎜
⎟
⎝ dΩ ⎠N
2π
n, n integer.
a
f (Q )
⎛ dσ ⎞
⎜
⎟
⎝ d Ω ⎠M
2π
a
4π
a
6π
a
2π
a
Q
3
4π
a
2
6π
a
Q
2 One-dimensional antiferromagnet
a
magnetic unit cell twice as large as nuclear unit cell ⇒ K M =
FM
2
= f (Q )
2
Qa ⎧⎪4 f ( Q )
=⎨
2 ⎪0
⎩
+1 − eiQa = 4 f ( Q ) sin 2
2
2
⎛ dσ ⎞
⎜
⎟
⎝ dΩ ⎠N
2
π
a
n ≠ KN =
for n odd
for n even
f (Q )
⎛ dσ ⎞
⎜
⎟
⎝ d Ω ⎠M
2π
a
4π
a
6π
a
π
Q
2π
n
a
a
3π
a
5π
a
2
Q
3 Flux-line lattice in type-II superconductor
Magnetization in type-II superconductor
M
H C1
HC 2
H
0 < H < H C1 :
magnetic field completely screened out of superconductor by
shielding currents flowing around the perimeter (“Meissner effect”)
H C1 < H < H C 2 : magnetic field penetrates partially into superconductor in the form
of “flux lines”, which form a regular array (“flux line lattice”)
flux line
screening current
4
H
Φ0
h
where Φ 0 =
is the
2e
H
magnetic flux quantum
a ~ 500 Å for H = 1T
a~
a
x
The periodic magnetic field distribution generated by the flux line lattice inside
the superconductor can be revealed by magnetic neutron scattering. Because
the lattice constant a is large, the scattering angle Θ ~
λ
is small. Even for cold
2a
neutrons ( λ ~ 5Å ) and relatively high magnetic fields ( H ~ 1T ) , Θ is less than
1°. In order to obtain the resolution required to separate the Bragg reflections
from the unscattered beam, one uses a dedicated Small Angle Neutron
Scattering (SANS) diffractometer with a long distance (>10 m) between sample
and detector:
Small angle neutron diffraction patterns from flux lines lattice in a type-II superconductor
Note that the structure of the flux line lattice changes as a function of magnetic field
(M.R. Eskildsen et al., PRL 86, 320 (2001))
As we have seen, nuclear and magnetic neutron scattering have comparable
strengths. They can be distinguished experimentally through three different
methods:
1 The magnetic form factor reduces the intensity of magnetic Bragg reflections
K
with large K , whereas there is no form factor for nuclear scattering. Strong
K
reflections with large K must therefore be nuclear in origin.
2 Magnetic Bragg peaks vanish at the magnetic ordering temperature (the
Curie temperature TC for ferromagnets, or the Néel temperature TN for
antiferromagnets). Nuclear Bragg peaks vanish at the melting temperature,
which is typically larger than TN or TC.
3 The neutron spin operator does not appear the the cross section for coherent
nuclear scattering. The neutron spin state is therefore unaffected by nuclear
scattering. By contrast, magnetic neutron scattering can be (but does not
have to be) associated with a spin-flip of the neutron.
5
Magnetic Scattering of Spin-Polarized Neutrons
The most unambiguous method to discriminate between nuclear and magnetic
neutron scattering is to keep track of the spin state of the neutron before and
after the scattering event. In this case, the magnetic neutron scattering cross
section for a collinear magnet can be written as
K 2
K K
2
2
K
⎛ dσ ⎞
ˆ
γ
r
m
σ
η
m
F
Q
δ
Q
=
⋅
− KM )
(
)
(
)
(
∑
0
f
⊥
i
M
⎜
⎟
K
⎝ d Ω ⎠M
KM
The neutron
spin quantization axis ẑ can be determined by applying a magnetic
K
field H at the sample
K position. Depending on the relative orientation of the
K
ˆ
vectors H , η , and Q , the scattering process does or does not cause a spin flip
of the neutron.
Examples:
1 For a ferromagnet, the applied field determines both the neutron spin
K
quantization axis and the electron spin direction η̂ , so that ηˆ & H & zˆ . First,
consider a situation in which the field is applied perpendicular to the
scattering plane:
K
H ,ηˆ
ηˆ⊥ = ηˆ = zˆ
K
K
K
K
kf
⇒ m f σ ⋅ηˆ⊥ mi = m f σ z mi
ki
⎧ +1 if m f = mi = +1
⎪
= ⎨ −1 if m f = mi = −1
⎪0 if m ≠ m
f
i
⎩
K
Q
because m is an eigenstate of σ z
The scattering is therefore entirely non-spin-flip. However if the field is
applied in a general direction in the scattering plane, both spin-flip and nonspin-flip contributions can be observed:
K
ki
K
H ,ηˆ
K
K
kf
σ ⋅ηˆ⊥ = σ xη⊥ x + σ yη⊥ y + σ zη⊥ z
Because σ x and σ y can be written as
linear superpositions of raising and
lowering operators:
1
σ x = (σ + + σ − )
2
1
σ y = (σ + − σ − )
2i
K
the x- and y-components of σ ⋅ηˆ⊥ induce
spin flips of the neutron
K
Q
6
2 For an antiferromagnet, the external field does not affect η̂ , and the neutron
K
spin quantization axis is not necessarily parallel to η̂ . In particular, if H is
K
K
applied parallel to Q , the product σ ⋅ηˆ⊥ has no z-component, so that the
scattering is entirely spin-flip.
Production of Spin-Polarized Neutrons
In order to apply this method, one needs to produce a beam of neutrons in a
single spin state. This can be accomplished by two fundamentally different
methods.
1 Nucelar-magnetic interference effect
K
K
If K M = K N , both magnetic and nuclear scattering contribute to a Bragg
reflection. Take for example a Kferromagnet with one atom per unit cell; with a
field applied perpendicular to Q as in the example above:
K
K K
dσ
K
2
ˆ
b
r
f
Q
m
m
Q
=∑
+
γ
σ
⋅
η
δ
−K)
(
)
(
f
⊥
i
0
K
dΩ
K
(
=+1 for spin up
=−1 for spin down
)
K
K K
2
2
γ
γ
δ
b
r
f
Q
b
r
f
Q
Q
=∑
+
±
−K)
(
)
(
)
(
0
0
K
K
K 2
K K
b ± γ r0 f ( Q ) δ ( Q − K )
=∑
K
K
The scattering cross section is thus different for the two different spin states of
the neutron. This difference can be exploited to generate a spin-polarized
neutron beam. For the (100) reflection of the ferromagnet Cu2MnAl (“Heusler
K
alloy”), b ≈ r0γ f ( Q ) so that only spin-up neutrons are Bragg reflected:
source
sample
The beam incident upon the sample is therefore almost perfectly spin-polarized.
A variant of this method relies on total external reflection from a ferromagnetic
mirror. By analogy to the above argument, the critical angle for a ferromagnetic
film can be written as
4π p ( b ± γ r0 f )
ΘC ↑↓ ~ 2δ ↑↓ =
k
7
For ΘC ↓ < Θ < ΘC ↑ , only spin-up neutrons are reflected:
Θ
ferromagnetic mirror
nonmagnetic films
magnetic films
Neutron reflectivity from nonmagnetic and magnetic films
http://www.orau.org/council/02presentations/klose.pdf
Ferromagnetic “supermirrors” with larger critical angles (see chapter on neutron
reflectivity) are commonly used to produce spin-polarized neutrons in modern
neutron scattering instruments:
Polarized-beam reflectometer
http://www.orau.org/council/02presentations/klose.pdf
8
2 A method that has been developed in the past few years is the neutron spin
filter. A spin-polarized 3He gas absorbs neutrons through the reaction
n + 3 He → 4 He ( S = 0) → p + 3 H .
K
H
3
He
Because the reaction proceeds via an intermediate state with S = 0 , the
absorption cross section for neutrons with spins antiparallel to those of the 3 He
nuclei is therefore about three orders of magnitude larger than if neutron and
3
He spins are parallel.
9