Inelastic scattering Quick reminder of the Dirac-notation (also - - PDF document

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Chapter 2

Inelastic scattering

Quick reminder of the Dirac-notation (also called bra-ket notation). The quantum mechanical state of a particle is described as a vector |Ψ, the following properties apply: Ψk(x) = x|Ψk Ψ1|Ψ2 = Ψ2|Ψ1∗ Ψ| ˆ A|Ψ =

• Ψ∗ ˆ

AΨd3r (2.1)

2.1 The inelastic form factor

In this lecture we are interested in finding the form factor for inelastic scattering related to excitations of core electrons. This will later allow us to find the related cross sections which can be compared directly to the intensity we observe in the EELS experiment. We assume that we start out with an incident plane wave of electrons as in equation 1.11: Ψ(k0, r) = Aeik0·r (2.2) We say that before the scattering event, this electron is in the state |k0, and after the scattering event it is in the state |k. Likewise, the atomic electron starts out in state |α and ends up in |β. The initial and final states of the system as a whole can be described as a product of one-electron states: |k0, α = |k0|α |k, β = |k|β (2.3) The initial state of the core electron is one of the atomic states of the atom, e.g. 1s, 2s, 2p etc. If the core electron leaves the atom all together, then the final state is a new plane wave. The atom is then ionized. However, we are usually most interested in the case where the final state is a bound state, that is, the core electron transitions to another (empty) state centred on the same atom.

1There is a factor of 2π missing in the exponential of the equation below as compared to equation 1.1. What is

the reason for the difference, and is it relevant?

5

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6 Chapter 2. Inelastic scattering

Figure 2.1

The momentum and energy of the system as a whole is of course preserved, but the distribution

• f momentum and energy between the incident electron and the core electron is changed during

the scattering. This means that k0 = k. The table below is taken from the textbook and gives the notation that we will use. In the beginning, the two electrons are far away from each other, and we assume that there is no interaction between them. We label the incident electron as ’1’ and the core electron as ’2’. The Hamiltonian for this system is then as follows: H0 = − 2 2me ∇2

1 − 2

2me ∇2

2 + V (r2).

(2.4) Here, the terms have their usual meaning, and V (r2) is the potential of the atom at position r2

• f electron 2. The Schrödinger equation is

H0|k0, α = (E0 + Eα)|k0, α. (2.5) As electron 1 approaches the atom, two new interactions need to be taken into account: Coloumb interaction between the electrons, and the potential of the atom. This can be described as a perturbation Hamiltonian H′: H′ = e2 |r1 − r2| + V (r1) (2.6) The total Hamiltonian is then: H = H0 + H′ = − 2 2me ∇2

1 − 2

2me ∇2

2 +

e2 |r1 − r2| + V (r1) + V (r2) (2.7) Question: what about the motion of the atom? What would its contribution to the Hamiltonian be, and why can it be ignored 2? From time dependent perturbation theory we know that the wavefunction of the scattered electron is given by:

2− 2 2M ∇2. The motion is very small and the associated energy is also small.

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2.1. The inelastic form factor 7 |k = f(k, k0)|φk (2.8) where |φk is a spherical wave with wavevector k. This is very similar to the case for elastic scattering which we have studied previously, but with a different form factor: f(k, k0) = −me 2π2 β|k|H′|k0|α (2.9) Inserting the explicit expression for the perturbation Hamiltonian from equation 2.6, we get the following: f(k, k0) = −me 2π2 β|k| e2 |r1 − r2| + V (r1)|k0|α = −me 2π2

• β|k|

e2 |r1 − r2||k0|α + β|k|V (r1)|k0|α

• = −me

2π2

• β|k|

e2 |r1 − r2||k0|α + β|αk|V (r1)|k0

• = −me

2π2

• β|k|

e2 |r1 − r2||k0|α

• (2.10)

Question: what happened in the final step? Is this simplification valid for all types of scattering? To recap: we have investigated the inelastic scattering of an incident electron (electron 1) on an atomic electron (electron 2). The incident electron is initially in the state |k0 and the atomic electron in state |α. After the scattering, they are in states |k and |β respectively. We are interested in the amplitude f(k, k0) of the scattered spherical wave |k = f(k, k0)|φk. The final result from equation 2.10 was: f(k, k0) = −me 2π2

• β|k|

e2 |r1 − r2||k0|α

• (2.11)

Remembering the definitions in equation 2.1 we now change to integral form: f(k, k0) = −me 2π2

• β|k|

e2 |r1 − r2||k0|α

• = −mee2

2π2

∞

• −∞

∞

• −∞

Ψ∗

β(r2)Ψ∗ k(r1)

1 |r1 − r2|Ψk0(r1)Ψα(r2)d3r1d3r2 = −mee2 2π2

∞

• −∞

∞

• −∞

Ψ∗

β(r2)e−ik·r1

1 |r1 − r2|eik0·r1Ψα(r2)d3r1d3r2 (2.12) where we in the final line have inserted the explicit form of the plane wave as in equation 2.2 with A = 1 for simplicity. Substituting r = r1 − r2 ∆k = k − k0 (2.13) we get

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8 Chapter 2. Inelastic scattering f(k, k0) = −mee2 2π2

∞

• −∞

∞

• −∞

Ψ∗

β(r2)e−i(k−k0)·r1

1 |r1 − r2|Ψα(r2)d3r1d3r2 = −mee2 2π2

∞

• −∞

∞

• −∞

Ψ∗

β(r2)e−i∆k·r1

1 |r1 − r2|Ψα(r2)d3r1d3r2 = −mee2 2π2

∞

• −∞

∞

• −∞

Ψ∗

β(r2)e−i∆k·(r+r2) 1

|r|Ψα(r2)d3rd3r2 = −mee2 2π2

∞

• −∞

e−i∆k·r 1 |r|d3r

∞

• −∞

Ψ∗

β(r2)e−i∆k·r2Ψα(r2)d3r2

(2.14) The first integral in the final line of equation 2.14 is just the integral of a spherical wave over all

• f space.

∞

• −∞

e−i∆k·r 1 |r|d3r = 4π ∆k2 (2.15) We arrive at the final expression for the form factor: f(k, k0) = −mee2 2π2 4π ∆k2

∞

• −∞

Ψ∗

β(r2)e−i∆k·r2Ψα(r2)d3r2

= − 2mee2 2∆k2

∞

• −∞

Ψ∗

β(r2)e−i∆k·r2Ψα(r2)d3r2

(2.16) This is now the amplitude of the scattered wave. Using the definition of the Bohr radius a0 = 2/(mee2) we get: f(k, k0) = − 2 a0∆k2

∞

• −∞

Ψ∗

β(r2)e−i∆k·r2Ψα(r2)d3r2

(2.17)

2.2 The double-differential cross section

The differential cross-section for a scattering event with scattering vector ∆k is dσ(∆k) dΩ = |f(k, k0)|2 = 4 a2

0∆k4

• ∞
• −∞

Ψ∗

β(r)e−i∆k·rΨα(r)d3r

• 2

(2.18) This is now the likelihood that the excitation event in question (|α → |β) with energy E and scattering vector ∆k. Question: what is the total cross-section for scattering with this energy transfer? 3

3σ = 4π sphere dσ(∆k) dΩ

dΩ.

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2.2. The double-differential cross section 9

Figure 2.2: From F&H

We will not go very much further in the lectures, except to note that equation 2.18 gives the probability of exactly one scattering event. However, if there are several possible final states available to excitations with energy E, then the total probability needs to be scaled with the number of such states, N = ρ(E)dE, arriving at: d2σ(∆k, E) dΩdE = 4 a2

0∆k4 ρ(E)

• ∞
• −∞

Ψ∗

β(r)e−i∆k·rΨα(r)d3r

• 2

= 4 a2

0∆k4 ρ(E)

• β|e−i∆k·r|α
• 2

(2.19) In the Fultz & Howe textbook (ch. 5.4.3) they go on to reformulate this equation in terms of scattering angle, in stead of scattering vector. This is convenient as it is usually angles that we can measure and control in our experiment. They arrive at the expression: d2σ(φ, E) dφdE = 2π4 a2

0m2 2EαβT

• exp. scatt. angles
• φ

φ2 + φ2

E

ρ(E)

• sol. state eff.

atomic osci. str.

• Gαβ(∆k, E)

(2.20) Where the atomic part is described by the ’generalized oscillator strength’ Gαβ(∆k, E) = Eαβ 2me 2∆k2

• ∞
• −∞

Ψ∗

β(r)e−i∆k·r2Ψα(r2)d3r

• 2

(2.21) The Bethe-surface can now be calculated and plotted as a function of energy E and scattering angle φ, see figure 2.2. Below the onset energy Eαβ the GOS is zero, with a very sharply peaked intensity onset at E = Eαβ and for very small scattering angles φ. When we perform an EELS experiment, we use an entrance aperture into the EELS spectrometer which accepts an angular range from zero to β. We call this the collection half-angle. Even with a very small collection half-angle we will capture most of the intensity from the edge onset. With larger β we will also gather intensity from

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10 Chapter 2. Inelastic scattering the so-called Bethe-ridge, which corresponds to a ’classical’ electron-electron scattering obeying energy and momentum conservation 4.

2.3 The dipole selection rules

Going back to equation 2.17 we see that the amplitude of the scattered wave is determined by the matrix element M(∆k, E) = β|e−i∆k·r|α =

∞

• −∞

Ψ∗

β(r)e−i∆k·rΨα(r)d3r

(2.22) Inspecting the integrand in the final expression, we see that the operator e−i∆k·r can be expanded as a Taylor series e−i∆k·r = 1 + (−i∆k · r) + (−i∆k · r)2 2! + (−i∆k · r)3 3! + · · · (2.23) Keeping the scattering angle (vector) small ∆k ≪

1 r, we see that the series can be truncated

after the second term. In practice this can be achieved by using a small spectrometer entrance aperture in the experiment, thereby limiting the angle of the electrons allowed to contribute to the spectrum. The matrix element then reduces to M(∆k, E) ≈ β|1 − i∆k · r|α = β|1|α − β|i∆k · r|α = −β|i∆k · r|α (2.24) where we have used the fact that β|1|α = 0 due to orthogonality of atomic states.

Figure 2.3: Illustrations of the par- ity of the function xn. When the function has odd parity, its integral

• ver all space will be zero due to all

elements f(x)dx being canceled out by the element f(−x)dx = −f(x)dx.

Under these conditions, the probability of scattering is deter- mined by the integral

• Ψ∗

β(r)i∆k · rΨα(r)d3r which closely

resembles the corresponding integral for optical excitations. In this case, the so-called dipole selection rules apply, requir- ing that the angular momentum of the initial and final states differ by one ∆l(α → β) = ±1. To show that this is the case, we can study the parity of the integrand above. The parity of a function is related to how it behaves when we change the spatial coordinates to their negative values f(x, y, z) → f(−x, −y, −z) (i.e. inversion). The simplest illustration of this is the functions f(x) = xn for n ∈ N0, where f(−x) = f(x) for n even −f(x) for n odd (2.25) as illustrated in figure 2.3. We say that these functions are either even or odd, and that they have parity (−1)n (that is (1) or (−1)) respectively. To find the parity of the product

• f several functions, we simply multiply their parities. For

example, the product x2 · x5 = x7 has parity (−1)2 · (−1)5 = (−1)7 and is an odd function. In fact, this multiplicative

4Question: how do we control the collection half angle in practice in the microscope?

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2.3. The dipole selection rules 11 property of the parity holds in general: an odd function multiplied with and odd function gives an even function, an odd function multiplied with an even function gives an odd result, and the multiplication of two even functions result in a new even function. This is relevant to our problem since the integral

• f(x)dx vanishes if the integrand is an odd
• function. So what are the parity properties of the three functions that contribute to the integrand
• f equation 2.24? On inversion r → −r the operator i∆k · r takes its own negative value, and

therefore has negative parity. But what about the wave functions Ψα and Ψβ? A result from the central-force problem for atomic states is that their wave functions can be described in terms of a radial term and a spherical harmonic term [1]: ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ) (2.26) It can furthermore be shown that while the radial part has even parity, the spherical harmonic part has a parity given by: Ylm(π − θ, π + φ) = (−1)lYlm(θ, φ) (2.27) that is, the parity of the wave function is determined by the angular momentum quantum number

• l. Turning back to the integrand in equation 2.24 we see that its parity is:

(−1)l · (−1) · (−1)l′ (2.28) We see that of the angular momentum quantum numbers of the initial and final states are the same, we end up with an odd integrand, thereby giving a transition probability equal to zero. However, if the quantum numebers differ by 1, the integrand is even, and the result is a non-zero transition probability. That is, the allowed transitions are those for which ∆l = ±1.