Download Introduction to RXS-CDW

Resonant X-ray Scattering
and CDWs.
0
Generic Scattering Experiment
Hard X-ray Scattering
Bragg Diffraction
• ℏ𝜔 ~ 10 to 100 keV.
• Sensitive to all electrons.
Resonant X-ray Scattering
Core
2p3/2
Soft X-ray Absorption
e.g. Cu L3,2 edge
Cu-L3
• Element-specific, orbital specific.
• Sensitive to the intermediate electronic
sates of the particular element you choose.
L2
The challenge is to understand what
“sensitive to intermediate states”
actually means.
So what does RXS measure?
arXiv search: resonant x-ray scattering
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Since 2015: 213
Today
• Thomson Scattering – Classical Approach
• Kramers-Heisenberg Equations – Quantum Approach
• Resonant X-ray Scattering Cross-Section
• CDW structure factor in RXS
• X-ray Absorption Spectroscopy
• Example in a ladder cuprate system: Sr14Cu24O41.
• Resonant Inelastic X-ray Scattering (on a future lecture).
4
Thomson Scattering
Classical Approach
5
Electromagnetic radiation
Electro-magnetic fields (aka photons, radiation, X-rays):
Propagation in free space
(Maxwellian Physics):
𝐸 = 𝑒𝐸0 𝑒 𝑖(𝑘.𝑟−𝜔𝑡)
1
𝐵 = 𝑘×𝐸
𝜔
Energy quantization
𝐸 = ℏ𝜔 = ℎ𝑓 =
ℎ𝑐
𝜆
hc = 12.4 keV Å
Quantum Description
𝐴 𝑟, 𝑡 =
𝜖,𝜅
ℏ
𝜖𝑎𝜖,𝜅 𝑒 𝑖
2𝜀0 𝑉𝜔𝜅
𝜅.𝑟−𝜔𝑡
†
+ 𝜖 ∗ 𝑎𝜖,𝜅
𝑒 −𝑖
𝜅.𝑟−𝜔𝑡
𝑝 → 𝑝 − 𝑞𝐴
Generic Scattering Experiment
Differential Scattering
Cross-section
Figure from Elements of Modern X-ray Physics by Jens Als-Nielsen & Des McMorrow
7
Thomson Scattering Classically – 1 electron
First consider the EM field scattering of a point particle
The incident radiation has an electric field:
𝐸𝑖𝑛 = 𝐸𝑥0 𝑒 −𝑖𝜔𝑡
The point particle will vibrate with the electric field. In
turn the point particle will radiate:
𝐸𝑟𝑎𝑑
𝑒2
𝑒 𝑖𝑘𝑅
𝑅 =−
𝐸
sin Ψ
4𝜋𝜖0 𝑚𝑐 2 𝑖𝑛 𝑅
Re-writing in terms of the Thomson scattering length
𝐸𝑟𝑎𝑑
𝑒 𝑖𝑘𝑅
= −𝑟0
sin Ψ
𝐸𝑖𝑛
𝑅
This is the result for the in-plane scattering.
In general:
𝐸𝑟𝑎𝑑
𝑒 𝑖𝑘𝑅
= −𝑟0
𝜀 ∙ 𝜀′
𝐸𝑖𝑛
𝑅
Figure from Elements of Modern X-ray Physics by Jens Als-Nielsen & Des McMorrow
8
Thomson Scattering Classically – 1 electron
Remembering the expression for the scattering
cross-section we obtain the Thomson differential
cross-section of an EM wave by a free electron.
𝑑𝜎
𝐼𝑠𝑐
=
= 𝑟02 𝜀 ∙ 𝜀′
𝑑Ω Φ0 ΔΩ
Figure from Elements of Modern X-ray Physics by Jens Als-Nielsen & Des McMorrow
9
Thomson Scattering Classically – 1 atom
Consider a charge distribution around the atom
with a number density ρ(𝒓). The phase difference
between two volume elements separated by 𝒓 is:
Δ𝜙 𝒓 = 𝒌 − 𝒌′ ∙ 𝒓 = 𝑸 ∙ 𝒓
So we can sum up the individual contributions from
the scattering fields:
−𝑟0 𝑓 0 𝑸 = −𝑟0
ρ(𝒓)𝑒 𝑖𝑸∙𝒓 𝑑𝒓
And 𝑓 0 𝑸 is the atomic structure factor.
So the differential cross-section for EM radiation scattering off an atom is:
Figure from Elements of
Modern X-ray Physics by Jens
Als-Nielsen & Des McMorrow
𝑑𝜎
∝ 𝑟02 𝑓 0 𝑸
𝑑Ω
2
10
Thomson Scattering Classically – a crystal
Each unit cell will have multiple atoms. If 𝒓𝒋 are the
positions of the atoms in the u.c., and each atom
has a specific structure factor 𝑓𝑗 (𝑸). Then the unit
cell structure factor is:
𝑓𝑢𝑐 𝑸 =
𝑗
𝑓𝑗 (𝑸)𝑒 𝑖𝑸∙𝒓𝒋
But we also need to sum over the lattice sites:
𝑓𝑐𝑟𝑦𝑠𝑡𝑎𝑙 𝑸 = 𝑓𝑢𝑐 𝑸
𝑒 𝑖𝑸∙𝑹𝒏
𝑛
The lattice basis vectors are :
𝑹𝒏 = 𝑛1 𝒂𝟏 + 𝑛2 𝒂𝟐 + 𝑛3 𝒂𝟑
𝑖𝑸∙𝑹
𝒏 to be large?
What is the condition for 𝑛 𝑒
And the crystal structure factor is order 1, unless the sum over complex numbers is
coherent, yielding order N=total number of unit cells. This is possible when
𝑸 ∙ 𝑹𝒏 = 2𝜋 × 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
Figure from Elements of
Modern X-ray Physics by Jens
Als-Nielsen & Des McMorrow
or...
𝑸=𝑮
11
Some practicalities and conventions
• The detector angle determines 𝑸.
• Then one needs to rotate the sample to project 𝑸 onto the desired sample axis.
𝑸
𝒌′
𝟐𝜽 detector angle
(not 2 times theta)
𝒄
𝒌
𝒂
12
Some practicalities and conventions
• In this figure the scattered photon needs to pass through the sample.
• This is called transmission geometry or Laue geometry.
𝑸
𝒌′
𝟐𝜽 detector angle
(not 2 times theta)
𝜽 : sample angle
𝒌
• In this example 𝑸 ∥ 𝒂. i.e. [H 0 0] Bragg peaks can be probed.
13
Some practicalities and conventions
• In this figure the scattered photon is reflected off the surface of the sample.
• This is called Reflection geometry or Bragg geometry.
𝑸
𝒌′
𝟐𝜽 detector angle
(not 2 times theta)
𝜽 : sample angle
𝒌
• In this example 𝑸 ∥ 𝒄. i.e. [0 0 L] Bragg peaks can be probed.
14
Some practicalities and conventions
• If the incoming polarization is in the scattering plane, we call this p scattering.
𝑸
𝒌′
𝟐𝜽 detector angle
(not 2 times theta)
𝝐
𝜽 : sample angle
𝒌
15
Some practicalities and conventions
• If the incoming polarization is perpendicular to the scattering plane, we call
this s scattering.
𝑸
𝒌′
𝟐𝜽 detector angle
(not 2 times theta)
𝝐
𝜽 : sample angle
𝒌
16
Quantum Approach
Kramers-Heisenberg equation(s)
17
Kramers-Heisenberg – one electron
We will work the simplest Hamiltonian involving the light-matter interaction:
𝐻 = 𝐻𝑒𝑙 + 𝐻𝑟𝑎𝑑 + 𝐻𝑖𝑛𝑡 = 𝐻0 + 𝐻𝑖𝑛𝑡
Where 𝐻𝑒𝑙 and 𝐻𝑟𝑎𝑑 are the unperturbed Hamiltonians and 𝐻𝑖𝑛𝑡 represents the
interaction term which is obtained as a result of the minimal coupling: 𝒑 → 𝒑 −
𝐻𝑖𝑛𝑡 =
𝑖
𝑒𝑨
𝑐
.
−𝑒
𝑒2
𝒑∙𝑨+𝑨∙𝒑 +
𝑨∙𝑨
2𝑚𝑐
2𝑚𝑐
And the quantized radiation field is well known:
𝐴 𝑟, 𝑡 =
𝜖,𝜅
ℏ
𝜖𝑎𝜖,𝜅 𝑒 𝑖
2𝜀0 𝑉𝜔𝜅
𝑘.𝑟−𝜔𝑡
†
+ 𝜖 ∗ 𝑎𝜖,𝜅
𝑒 −𝑖
𝑘.𝑟−𝜔𝑡
And we want to know the probability amplitude that the electron-radiation system will
transition from a state where the electron is in state A, and the photon in the state 𝑘,
before the scattering, into a state with the electron in state B and the scattered photon
in state 𝑘′.
18
Kramers-Heisenberg – one electron
We can work this out in 2nd order time-dependent perturbation theory (Dirac formalism).
1
If 𝑐𝑚 are the linear coefficients of the vector basis for 𝐻0 , the unperturbed Hamiltonian:
𝐻0 𝑢𝑘 = 𝐸𝑘 𝑢𝑘
And we can expand the solutions to the full Hamiltonian in the unperturbed basis:
𝐸 𝑡
−𝑖 𝑚
𝑐𝑚 (𝑡)𝑢𝑚 𝑒 ℏ
𝜓=
𝑚
Then, if we assume that at 𝑡 = 0, before the photon arrives, the unperturbed state is in a
state 𝑐𝑘 , then the probability it will be in a state 𝑐𝑚 at time 𝑡 > 0 is (to 1st order):
1
𝑐𝑚
1
𝑡 =
𝑖ℏ
𝑡
𝑑𝑡 ′
𝑚 𝐻𝑖𝑛𝑡
(𝑡 ′ )
𝑘𝑒
𝑖
𝐸𝑚 −𝐸𝑘 𝑡 ′
ℏ
0
To 2nd order it will be:
2
𝑐𝑚
1
𝑡 =
𝑖ℏ
𝑡
𝑡′
𝑑𝑡 ′
2
𝑛
0
𝑑𝑡” 𝑚 𝐻𝑖𝑛𝑡 (𝑡”) 𝑛 𝑒
𝑖
(𝐸𝑚 −𝐸𝑛 )𝑡 ′′
ℏ
𝑛 𝐻𝑖𝑛𝑡 (𝑡′) 𝑘 𝑒
𝑖
(𝐸𝑛 −𝐸𝑘 )𝑡 ′
ℏ
0
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Kramers-Heisenberg – one electron
• The 1st order term will turn out to be the elastic Thomson scattering.
• The 2nd order term will be the resonant scattering cross-section.
The scattering cross-section to second order is:
𝑑𝜎
𝜔′
2
= 𝑟0
𝑑Ω
𝜔
𝛿𝐴𝐵 𝜀 ∙
𝜀′
1
−
𝑚
𝑛
𝐵 𝒑 ∙ 𝜖′ 𝑛 𝑛 𝒑 ∙ 𝜖′ 𝐴
𝐸𝑛 − 𝐸𝐴 − ℏ𝜔
2
For a detailed derivation see the notes online.
Figure from Elements of
Modern X-ray Physics by Jens
Als-Nielsen & Des McMorrow
20
RXS – multiple electrons
Summing over each unit cell and summing over all the atoms in the unit cell...
𝑑𝜎
𝑑Ω
𝑒 𝑖𝑸∙𝑹𝒌
∝
𝑘
𝑟𝑒𝑠
𝑗
𝑓𝑗 (𝑸)𝑒
𝑖𝑸∙𝒓𝒋
𝑛
𝐵 𝒑 ∙ 𝜖′ 𝑛 𝑛 𝒑 ∙ 𝜖′ 𝐴
𝐸𝑛 − 𝐸𝐴 − ℏ𝜔
• There is a large enhancement term when photon hits an atomic resonance.
• The denominator selects a specific electronic state.
• The cross-section can be approximated by the sum of a non-resonant and a
resonant structure factor
𝑑𝜎
∝ 𝑓
𝑑Ω
2
= 𝑓𝑇ℎ𝑜𝑚𝑠𝑜𝑛 + 𝑓𝑟𝑒𝑠
2
If 𝐸𝑛 − 𝐸𝐴 = ℏ𝜔 then we are doing a scattering experiment that is sensitive to a very
specific electronic associated with such resonance.
21
2
CDW in RXS
Summing over each unit cell and summing over all the atoms in the unit cell...
𝑑𝜎
𝑑Ω
𝑒 𝑖𝑸∙𝑹𝒌
∝
𝑘
𝑟𝑒𝑠
𝑗
𝑓𝑗 (𝑸)𝑒
Option 1: Structural distortion.
𝑉(𝑥)
1D-chain - dimerization
𝑖𝑸∙𝒓𝒋
𝑛
𝐵 𝒑 ∙ 𝜖′ 𝑛 𝑛 𝒑 ∙ 𝜖′ 𝐴
𝐸𝑛 − 𝐸𝐴 − ℏ𝜔
Consider the position of each atoms is
slightly distorted due to a CDW:
𝑅𝑛 = 𝑛𝑎 + 𝑢𝑛
𝑢𝑛 = 𝑢𝑛 cos 𝑄𝐶𝐷𝑊 (𝑛𝑎)
Then the structure factor will be:
a
L=Na
𝑒 𝑖𝑄𝑅𝑛 ≈ 𝑒 𝑖𝑄𝑛𝑎 (1 + 𝑖𝑄𝑢𝑛 )
Then we end up with a term that looks like:
𝑒 𝑖𝑄𝑅𝑛 cos 𝑄𝐶𝐷𝑊 (𝑛𝑎) ∝
𝑛
So
𝑒 𝑖𝑄𝑅𝑛 𝑒 𝑖𝑄𝐶𝐷𝑊𝑛𝑎 + 𝑒 𝑖𝑄𝐶𝐷𝑊𝑛𝑎
𝑛
𝑑𝜎
𝑑𝜎
or
will
𝑑Ω 𝑟𝑒𝑠
𝑑Ω 𝑇ℎ𝑜𝑚𝑠𝑜𝑛
2
have peaks when 𝑄 =
2𝜋
𝑎
± 𝑄𝐶𝐷𝑊 .
22
CDW in RXS
Summing over each unit cell and summing over all the atoms in the unit cell...
𝑑𝜎
𝑑Ω
𝑒 𝑖𝑸∙𝑹𝒌
∝
𝑘
𝑟𝑒𝑠
𝑗
𝑓𝑗 (𝑸)𝑒
𝑖𝑸∙𝒓𝒋
𝑛
𝐵 𝒑 ∙ 𝜖′ 𝑛 𝑛 𝒑 ∙ 𝜖′ 𝐴
𝐸𝑛 − 𝐸𝐴 − ℏ𝜔
If for example either 𝑥 𝑛 ∝ 𝑒 𝑖𝑄𝐶𝐷𝑊 (𝑛𝑎) or
if the energy states spatially modulate
𝐸𝑛 − 𝐸𝐴 = 𝑣 = 𝑣0 + 𝛿𝑣 cos 𝑄𝐶𝐷𝑊 (𝑛𝑎)
we will have a CDW peak.
Option 2: Electronic Modulation
𝑉(𝑥)
1D-chain - dimerization
In the valence modulation case:
a
𝑓𝑟𝑒𝑠 𝜔, 𝑣 ≈ 𝑓𝑟𝑒𝑠 𝜔, 𝑣0 +
𝑓′𝑟𝑒𝑠 𝜔, 𝑣0 𝛿𝑣 cos 𝑄𝐶𝐷𝑊 (𝑛𝑎)
L=Na
Putting it together with the crystal summation:
𝑒 𝑖𝑄𝑅𝑛 cos 𝑄𝐶𝐷𝑊 (𝑛𝑎) ∝
𝑛
So
𝑒 𝑖𝑄𝑅𝑛 𝑒 𝑖𝑄𝐶𝐷𝑊𝑛𝑎 + 𝑒 𝑖𝑄𝐶𝐷𝑊𝑛𝑎
𝑛
𝑑𝜎
will
𝑑Ω 𝑟𝑒𝑠
have peaks when 𝑄 =
2𝜋
𝑎
2
± 𝑄𝐶𝐷𝑊 . ONLY IN THE RESONANT TERM.
23
X-ray Absorption
Spectroscopy
How do you choose the energy?
24
X-ray Absorption Spectroscopy
Fermi
energy
Soft X-ray Absorption
e.g. Cu L3,2 edge
Cu-L3
L2
Core state
i.e. 2p electron
• If the core hole is an n=1 state, this is called a K-edge, if n=2 an L edge, if n=3
an M edge, and so on.
• For example in an L-edge absorption of a transition metal one excites an
electron from a full 2p shell to a 3d state.
X-ray Absorption Spectroscopy
• The nomenclature is archaic: 2S+1LJ
• S is the spin, L the orbital angular momentum, and J is the
total magnetic momentum of the open shell. They are
determined by Hund’s rules.
1. For a given electron configuration, the term with lowest energy is also the term with
maximum S.
2. For a given multiplicity, the term with the largest value of the total orbital angular
momentum quantum number L has the lowest energy.
3. For a given term, in an atom with outermost subshell half-filled or less, the level with
the lowest value of the total angular momentum quantum number, J. If the
outermost shell is more than half-filled, the level with the highest value of J is lowest
in energy.
X-ray Absorption Spectroscopy
•
•
•
•
•
•
•
2𝑝6 3𝑑 9
2𝑝5 3𝑑10
→
Ground state: S=1/2, mL=2, and J=1/2+2=5/2.
Ground state: 2D5/2
Excited state: S=1/2, mL=1, and J=1+1/2=3/2
L3 edge: 2P3/2
Excited state: S=1/2, mL=1, and J=1-1/2=1/2
L2 edge: 2P1/2
L3
L2
Hund’s rules:
1. For a given electron configuration, the term with lowest energy is also the term with
maximum S.
2. For a given multiplicity, the term with the largest value of the total orbital angular
momentum quantum number L has the lowest energy.
3. For a given term, in an atom with outermost subshell half-filled or less, the level with
the lowest value of the total angular momentum quantum number J. If the outermost
shell is more than half-filled, the level with the highest value of J is lowest in energy.
X-ray Absorption Spectroscopy
X-ray Absorption Spectroscopy – e.g. Ce3+
Ce metal
3𝑑10 4𝑓 1 → 3𝑑 9 4𝑓 2
Ground state: S=1/2, mL=3, and J=2-1/2=5/2.
Ground state: 2F5/2
After absorption 2 open shells: 2D ⨂ 3H
Excited states with S=1/2: 2F5/2,7/2 2G7/2,9/2 2H9/2,11/2
2I
2
11/2,13/2 J13/2,15/2
• Excited states with S=3/2: 4F3/2,5/2,7/2,9/2
4G
4
4
5/2,7/2,9/2,11/2 H7/2,9/2,11/2,13/2 I9/2,11/2,13/2,15,2
4J
11/2,13/2,15/2,17/2
• Figure out the allowed transitions...
•
•
•
•
•
Incoming photon energy (eV)
B. T. Thole, et al. – PRB 1985
Hund’s rules:
1. For a given electron configuration, the term with lowest energy is also the term with
maximum S.
2. For a given multiplicity, the term with the largest value of the total orbital angular
momentum quantum number L has the lowest energy.
3. For a given term, in an atom with outermost subshell half-filled or less, the level with
the lowest value of the total angular momentum quantum number J. If the outermost
shell is more than half-filled, the level with the highest value of J is lowest in energy.
The intermediate state in an RXS cross-section
is extremely complicated. It involves an
electron just above the Fermi energy, under
the influence of a local core-hole potential,
which has a spin, and orbital moment.
The core hole is a very strong perturbation...
30
CDW in Sr14Cu24O41
Crystallization of Oxygen Holes.
31
•
•
•
•
•
In this paper the authors show how RXS can be used to see CDWs with extreme sensitivity.
This is an intrinsically hole-doped system with 6 holes per formula unit. 5.2 in the chain, 0.8 in the
ladder.
Proposed to be a “hole crystal”.
Need RXS to directly access the holes.
Holes are associated with the hybridized Cu 3d and O 2p orbitals: Use Cu L and O K edges.
Figure from G. Blumberg – Science 2002
32
Oxygen K-edge in a cuprate material
3dx2-y2
2px
2py
Doped holes in a copper-oxide create a prepeak in Oxygen K-edge
C.T. Chen PRL – 1991
33
Oxygen K-edge – cuprate polarization dependence
3dx2-y2
2px
2py
• Couple to Pz and Px,y orbitals differently.
C.T. Chen PRL – 1992
34
On and Off Resonance
• Peak appears near LL=0.2 rlu only on resonance.
35
Full photon energy dependence
• We infer that this superstructure is electronic in origin.
• At the Oxygen K-edge the peak is only visible near the Mobile Carrier Peak.
• This MCP only appears in hole-doped copper-oxide materials.
36
Full photon energy dependence
• We infer that this superstructure is electronic in origin.
• At the Oxygen K-edge the peak is only visible near the Mobile Carrier Peak.
• This MCP only appears in hole-doped copper-oxide materials.
37
Temperature Dependence
38