Today’s Outline - October 26, 2016csrri.iit.edu/~segre/phys570/16F/lecture_18.pdf · Today’s...

Today’s Outline - October 26, 2016

• Bragg & Laue geometries

• Kinematical approach for many layers

• Dynamical diffraction theory

• Characteristics of the Darwin curve

• Asymmetric reflections

• Dumond diagrams

• Monochromators

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Wednesday, November 02, 2016

Friday, October 28 APS visit is cancelled!

C. Segre (IIT) PHYS 570 - Fall 2016 October 26, 2016 1 / 20

• Dumond diagrams

• Monochromators

• Dumond diagrams

• Monochromators

• Dumond diagrams

• Monochromators

• Dumond diagrams

• Monochromators

• Dumond diagrams

• Monochromators

• Dumond diagrams

• Monochromators

• Dumond diagrams

• Monochromators

• Dumond diagrams

• Monochromators

Kinematical vs. dynamical diffraction

The kinematical approximation we have discussed so far applies to mosaiccrystals. The size of the crystal is small enough that the wave field of thex-rays does not vary appreciably over the crystal.

For a perfect crystal, such as those used in monochromators, things arevery different and we have to treat them specially using dynamicaldiffraction theory.

This theory takes into account multiple reflections, and attenuation of thex-ray beam as it propagates through the perfect crystal.

Bragg & Laue geometries

symmetric

asymmetric

Bragg & Laue geometriesBragg

symmetric

asymmetric

symmetric

asymmetric

symmetric

asymmetric

symmetric

asymmetric

symmetric

asymmetric

Scattering geometry

Consider symmetric Bragg geometry

We expect the crystal to diffract in anenergy bandwidth defined by ∆k

This defines a spread of scattering vec-tors such that

ζ =∆Q

called the relative energy or wavelengthbandwidth

Q=mG(1+ζ)

Scattering geometry

ζ =∆Q

Q=mG(1+ζ)

Scattering geometry

ζ =∆Q

Q=mG(1+ζ)

Scattering geometry

ζ =∆Q

Q=mG(1+ζ)

Dynamical diffraction - Darwin approach

The Darwin approach treats a perfectcrystal as an infinite stack of atomicplanes.

Considering a single thin slab with elec-tron density ρ and thickness d � λ, thereflected and transmitted waves are

g =λr0ρd

sin θ

if the layer is made up of unit cells with

volume vc and structure factor FQ=0−−−→ Z ,

the electron density is ρ = |F |/vc andusing the Bragg condition, we can rewriteg as

S = −igT(1− ig0)T ≈ e−ig0T

g =[2d sin θ/m]r0(|F |/vc )d

sin θ=

2d2r0vc|F |

g =λr0ρd

sin θ

sin θ=

2d2r0vc|F |

g =λr0ρd

sin θ

sin θ=

2d2r0vc|F |

g =λr0ρd

sin θ

S = −igT

(1− ig0)T ≈ e−ig0T

sin θ=

2d2r0vc|F |

g =λr0ρd

sin θ

sin θ=

2d2r0vc|F |

g =λr0ρd

sin θ

sin θ=

2d2r0vc|F |

g =λr0ρd

sin θ

sin θ=

2d2r0vc|F |

g =λr0ρd

sin θ

2d2r0vc|F |

g =λr0ρd

sin θ

sin θ=

2d2r0vc|F |

since vc ∼ d3 then g ∼ r0/d ≈ 10−5

the transmitted beam depends on

g0 =λρat f

0(0)r0∆

sin θ

which can be rewritten

g0 =|F0||F |

where F0 is the forward scattering factorat Q = θ = 0

2d2r0vc|F |

g0 =λρat f

0(0)r0∆

sin θ

g0 =|F0||F |

2d2r0vc|F |

g0 =λρat f

0(0)r0∆

sin θ

g0 =|F0||F |

2d2r0vc|F |

g0 =λρat f

0(0)r0∆

sin θ

g0 =|F0||F |

2d2r0vc|F |

g0 =λρat f

0(0)r0∆

sin θ

g0 =|F0||F |

Kinematical Reflection

If we now extend this model to N layers we can use this kinematicalapproximation if Ng � 1.

Proceed by adding reflectivity from each layer with the usual phase factor

rN(Q) = −igN−1∑j=0

e iQdje−ig0je−ig0j = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ) where the x-rays pass through each layertwice

these N unit cell layers will give a recip-rocal lattice with points at multiples ofG = 2π/d we are interested in small de-viations from the Bragg condition:

ζ =∆Q

=∆λ

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

ζ =∆Q

=∆λ

rN(Q) = −igN−1∑j=0

e iQdje−ig0je−ig0j

= −igN−1∑j=0

e i(Qd−2g0)j

ζ =∆Q

=∆λ

rN(Q) = −igN−1∑j=0

e iQdje−ig0je−ig0j

= −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

where the x-rays pass through each layertwice

ζ =∆Q

=∆λ

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

ζ =∆Q

=∆λ

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

these N unit cell layers will give a recip-rocal lattice with points at multiples ofG = 2π/d

we are interested in small de-viations from the Bragg condition:

ζ =∆Q

=∆λ

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

ζ =∆Q

=∆λ

Multiple Layer Reflection

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

The term in the phase factor now be-comes

Qd − 2g0 = mG (1 + ζ)2π

G− 2g0

= 2π(m + mζ − g0

rN(Q) = −igN−1∑j=0

e i2π(m+mζ−g0/π

= −igN−1∑j=0

e i2πmje i2π(mζ−g0/π)

= −igN−1∑j=0

1 · e i2π(mζ−g0/π)

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

Qd − 2g0

= mG (1 + ζ)2π

G− 2g0

= 2π(m + mζ − g0

rN(Q) = −igN−1∑j=0

= −igN−1∑j=0

1 · e i2π(mζ−g0/π)

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

Qd − 2g0 = mG (1 + ζ)2π

G− 2g0

= 2π(m + mζ − g0

rN(Q) = −igN−1∑j=0

= −igN−1∑j=0

1 · e i2π(mζ−g0/π)

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

Qd − 2g0 = mG (1 + ζ)2π

G− 2g0

= 2π(m + mζ − g0

rN(Q) = −igN−1∑j=0

= −igN−1∑j=0

1 · e i2π(mζ−g0/π)

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

Qd − 2g0 = mG (1 + ζ)2π

G− 2g0

= 2π(m + mζ − g0

rN(Q) = −igN−1∑j=0

= −igN−1∑j=0

1 · e i2π(mζ−g0/π)

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

Qd − 2g0 = mG (1 + ζ)2π

G− 2g0

= 2π(m + mζ − g0

rN(Q) = −igN−1∑j=0

= −igN−1∑j=0

1 · e i2π(mζ−g0/π)

rN(Q) = −igN−1∑j=0

e i(Qd−2g0)j

Q=mG(1+ζ)

Qd − 2g0 = mG (1 + ζ)2π

G− 2g0

= 2π(m + mζ − g0

rN(Q) = −igN−1∑j=0

= −igN−1∑j=0

1 · e i2π(mζ−g0/π)

This geometric series can besummed as usual

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

[sin(πN[mζ − ζ0])

sin(π[mζ − ζ0])

]|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

This describes a shift of the Bragg peak away from the reciprocal latticepoint, the maximum being at ζ = ζ0/m

The kinematical approach now breaks down and we need to develop a newtheory for dynamical diffraction.

First, let’s explore how the intensity would vary using the kinematicalexpression

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

]|rN(ζ)|2 → g2

2 sin2(π[mζ − ζ0])

ζ0 =g0

=2d2|F0|πmvc

rN(Q) = −igN−1∑j=0

e i2π(mζ−g0/π)

|rN(ζ)| =

]|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

Diffraction in the kinematical limit

Recall that in thekinematical limit, thediffraction from manyatomic layers is givenby

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

|rN(ζ)|2 → g2

2(π[mζ − ζ0])2

Difference equation review

T ej+1-iϕ S ej+1

Tj+1 = e−ηe imπTj

Sj+1 = e−ηe imπSj

iη = ±√

(∆− g0)2 − g2

g =λr0ρd

sin θ, g0 =

|F0||F |

∆ = mπζ, ζ =∆λ

Where g0 is the absorption due to a single atomic layer, g is the reflectioncoefficient from a single atomic layer, and ∆ is the small deviation fromthe Bragg condition of the phase angle φ = mπ + ∆.

T ej+1-iϕ S ej+1

iη = ±√

(∆− g0)2 − g2

g =λr0ρd

sin θ, g0 =

|F0||F |

∆ = mπζ, ζ =∆λ

T ej+1-iϕ S ej+1

iη = ±√

(∆− g0)2 − g2

g =λr0ρd

sin θ, g0 =

|F0||F |

∆ = mπζ, ζ =∆λ

T ej+1-iϕ S ej+1

iη = ±√

(∆− g0)2 − g2

g =λr0ρd

sin θ, g0 =

|F0||F |

∆ = mπζ, ζ =∆λ

T ej+1-iϕ S ej+1

iη = ±√

(∆− g0)2 − g2

g =λr0ρd

sin θ, g0 =

|F0||F |

∆ = mπζ, ζ =∆λ

T ej+1-iϕ S ej+1

iη = ±√

(∆− g0)2 − g2

g =λr0ρd

sin θ, g0 =

|F0||F |

∆ = mπζ, ζ =∆λ

Reflectivity of a perfect crystal

In order to calculate the absolute reflectivity curve, solve for S0 and T0

using the solution and the recursive relations.

Tj+1 Sj+1

S1 = e−ηe imπS0

S0 = −igT0 + (1− g0)S1eiφ

S0 = −igT0

+ (1− g0)S0e−ηe imπe imπe i∆

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

T0≈ −ig

1− (1− ig0)(1− η)(1 + i∆)≈ −ig

ig0 + η − i∆=

iη + (∆− g0)

Tj+1 Sj+1

Sj = −igTj + (1− g0)Sj+1eiφ

S0 = −igT0

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

T0≈ −ig

1− (1− ig0)(1− η)(1 + i∆)≈ −ig

ig0 + η − i∆=

iη + (∆− g0)

Tj+1 Sj+1

S1 = e−ηe imπS0

Sj = −igTj + (1− g0)Sj+1eiφ

S0 = −igT0

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

T0≈ −ig

1− (1− ig0)(1− η)(1 + i∆)≈ −ig

ig0 + η − i∆=

iη + (∆− g0)

Tj+1 Sj+1

S1 = e−ηe imπS0

S0 = −igT0 + (1− g0)S1eiφ

S0 = −igT0

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

T0≈ −ig

1− (1− ig0)(1− η)(1 + i∆)≈ −ig

ig0 + η − i∆=

iη + (∆− g0)

Tj+1 Sj+1

S1 = e−ηe imπS0

S0 = −igT0 + (1− g0)S1eiφ

S0 = −igT0

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

T0≈ −ig

1− (1− ig0)(1− η)(1 + i∆)≈ −ig

ig0 + η − i∆=

iη + (∆− g0)

Tj+1 Sj+1

S1 = e−ηe imπS0

S0 = −igT0 + (1− g0)S1eiφ

S0 = −igT0

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

−ig1− (1− g0)e−ηe i2mπe i∆

≈ −igig0 + η − i∆

iη + (∆− g0)

Tj+1 Sj+1

S1 = e−ηe imπS0

S0 = −igT0 + (1− g0)S1eiφ

S0 = −igT0

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

T0≈ −ig

1− (1− ig0)(1− η)(1 + i∆)≈ −ig

ig0 + η − i∆

iη + (∆− g0)

Tj+1 Sj+1

S1 = e−ηe imπS0

S0 = −igT0 + (1− g0)S1eiφ

S0 = −igT0

[1− (1− g0)e−ηe i2mπe i∆

]= −igT0

T0≈ −ig

1− (1− ig0)(1− η)(1 + i∆)≈ −ig

ig0 + η − i∆=

iη + (∆− g0)

Darwin reflectivity curve

iη + ε=

ε±√ε2 − g2

, ε = ∆− g0 = mπζ − πζ0

Darwin reflectivity curve

iη + ε=

ε±√ε2 − g2

, ε = ∆− g0 = mπζ − πζ0

Standing waves

←− x = −1out of phase

x = +1 −→in phase

Standing waves

←− x = −1out of phase

x = +1 −→in phase

Absorption effects

Energy dependence

Polarization dependence

Harmonic suppression

The displacement of the Darwin curve varies inversely as the order, m, ofthe reflection.

The width varies as the inverse squared.

ζ0 =g0

2d2|F0|r0πmvc

ζD =2g

4d2|F |r0πm2vc

By tuning to the centerof a lower order reflec-tion, the high orders canbe effectively suppressed.

By tuning a bit off on the “high” side we get even more suppression. Thisis called “detuning”.

The displacement of the Darwin curve varies inversely as the order, m, ofthe reflection.

The width varies as the inverse squared.

ζ0 =g0

2d2|F0|r0πmvc

ζD =2g

4d2|F |r0πm2vc

The displacement of the Darwin curve varies inversely as the order, m, ofthe reflection. The width varies as the inverse squared.

ζ0 =g0

2d2|F0|r0πmvc

ζD =2g

4d2|F |r0πm2vc

ζ0 =g0

2d2|F0|r0πmvc

ζD =2g

4d2|F |r0πm2vc

ζ0 =g0

2d2|F0|r0πmvc

ζD =2g

4d2|F |r0πm2vc

Angular offset

We can calculate the angular offset by noting that the offset and widthhave many common factors.

Converting this to an angular offset.

ζ0 =2d2|F0|r0πmvc

ζD =4d2|F |r0πm2vc

ζoff =ζ0

|F ||F0|

∆θoff =ζD

|F ||F0|

tan θ

For the Si(111) at λ = 1.54056A

ωtotalD = 0.0020◦ ∆θoff = 0.0018◦

Angular offset

We can calculate the angular offset by noting that the offset and widthhave many common factors.

Converting this to an angular offset.

ζ0 =2d2|F0|r0πmvc

ζoff =ζ0

|F ||F0|

∆θoff =ζD

|F ||F0|

tan θ

For the Si(111) at λ = 1.54056A

ωtotalD = 0.0020◦ ∆θoff = 0.0018◦

Angular offset

We can calculate the angular offset by noting that the offset and widthhave many common factors. Converting this to an angular offset.

ζ0 =2d2|F0|r0πmvc

ζoff =ζ0

|F ||F0|

∆θoff =ζD

|F ||F0|

tan θ

For the Si(111) at λ = 1.54056A

ωtotalD = 0.0020◦ ∆θoff = 0.0018◦

Angular offset

ζ0 =2d2|F0|r0πmvc

ζoff =ζ0

|F ||F0|

∆θoff =ζD

|F ||F0|

tan θ

For the Si(111) at λ = 1.54056A

ωtotalD = 0.0020◦ ∆θoff = 0.0018◦

Angular offset

ζ0 =2d2|F0|r0πmvc

ζoff =ζ0

|F ||F0|

∆θoff =ζD

|F ||F0|

tan θ

For the Si(111) at λ = 1.54056A

ωtotalD = 0.0020◦

∆θoff = 0.0018◦

Angular offset

ζ0 =2d2|F0|r0πmvc

ζoff =ζ0

|F ||F0|

∆θoff =ζD

|F ||F0|

tan θ

For the Si(111) at λ = 1.54056A

ωtotalD = 0.0020◦ ∆θoff = 0.0018◦

Darwin widths

the quantities below the widths are f 0(Q), f ′, and f ′′ (forλ = 1.5405 A). For an angular width, multiply times tan θand for π polarization, multiply by cos(2θ).

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