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

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: Ψ k ( x ) = � x | Ψ k � � Ψ 1 | Ψ 2 � = � Ψ 2 | Ψ 1 � ∗ (2.1) � Ψ ∗ ˆ � Ψ | ˆ A Ψd 3 r A | Ψ � = 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 1.1 1 : Ψ( k 0 , r ) = Ae i k 0 · r (2.2) We say that before the scattering event, this electron is in the state | k 0 � , 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: | k 0 , α � = | k 0 �| α � (2.3) | k , β � = | k �| β � 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. 1 There is a factor of 2 π 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

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 of momentum and energy between the incident electron and the core electron is changed during the scattering. This means that k 0 � = 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: H 0 = − � 2 1 − � 2 ∇ 2 ∇ 2 2 + V ( r 2 ) . (2.4) 2 m e 2 m e Here, the terms have their usual meaning, and V ( r 2 ) is the potential of the atom at position r 2 of electron 2. The Schrödinger equation is H 0 | k 0 , α � = ( E 0 + E α ) | k 0 , 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 ′ : e 2 H ′ = | r 1 − r 2 | + V ( r 1 ) (2.6) The total Hamiltonian is then: H = H 0 + H ′ = − � 2 1 − � 2 e 2 ∇ 2 ∇ 2 2 + | r 1 − r 2 | + V ( r 1 ) + V ( r 2 ) (2.7) 2 m e 2 m e 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 2 M ∇ 2 . The motion is very small and the associated energy is also small.

2.1. The inelastic form factor 7 | k � = f ( k , k 0 ) | φ k � (2.8) where | φ k � 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 , k 0 ) = − m e 2 π � 2 � β |� k | H ′ | k 0 �| α � (2.9) Inserting the explicit expression for the perturbation Hamiltonian from equation 2.6, we get the following: e 2 f ( k , k 0 ) = − m e 2 π � 2 � β |� k | | r 1 − r 2 | + V ( r 1 ) | k 0 �| α � � � e 2 = − m e � β |� k | | r 1 − r 2 || k 0 �| α � + � β |� k | V ( r 1 ) | k 0 �| α � 2 π � 2 (2.10) � � e 2 = − m e � β |� k | | r 1 − r 2 || k 0 �| α � + � β | α �� k | V ( r 1 ) | k 0 � 2 π � 2 � � e 2 = − m e � β |� k | | r 1 − r 2 || k 0 �| α � 2 π � 2 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 | k 0 � 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 , k 0 ) of the scattered spherical wave | k � = f ( k , k 0 ) | φ k � . The final result from equation 2.10 was: � � e 2 f ( k , k 0 ) = − m e � β |� k | | r 1 − r 2 || k 0 �| α � (2.11) 2 π � 2 Remembering the definitions in equation 2.1 we now change to integral form: � � e 2 f ( k , k 0 ) = − m e � β |� k | | r 1 − r 2 || k 0 �| α � 2 π � 2 ∞ ∞ � � = − m e e 2 1 | r 1 − r 2 | Ψ k 0 ( r 1 )Ψ α ( r 2 )d 3 r 1 d 3 r 2 Ψ ∗ β ( r 2 )Ψ ∗ k ( r 1 ) 2 π � 2 (2.12) −∞ −∞ ∞ ∞ � � = − m e e 2 1 | r 1 − r 2 | e i k 0 · r 1 Ψ α ( r 2 )d 3 r 1 d 3 r 2 β ( r 2 ) e − i k · r 1 Ψ ∗ 2 π � 2 −∞ −∞ 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 = r 1 − r 2 (2.13) ∆k = k − k 0 we get

8 Chapter 2. Inelastic scattering ∞ ∞ � � f ( k , k 0 ) = − m e e 2 1 β ( r 2 ) e − i ( k − k 0 ) · r 1 | r 1 − r 2 | Ψ α ( r 2 )d 3 r 1 d 3 r 2 Ψ ∗ 2 π � 2 −∞ −∞ ∞ ∞ � � = − m e e 2 1 β ( r 2 ) e − i ∆k · r 1 | r 1 − r 2 | Ψ α ( r 2 )d 3 r 1 d 3 r 2 Ψ ∗ 2 π � 2 −∞ −∞ (2.14) ∞ ∞ � � = − m e e 2 β ( r 2 ) e − i ∆k · ( r + r 2 ) 1 | r | Ψ α ( r 2 )d 3 r d 3 r 2 Ψ ∗ 2 π � 2 −∞ −∞ ∞ ∞ � � = − m e e 2 e − i ∆k · r 1 | r | d 3 r β ( r 2 ) e − i ∆k · r 2 Ψ α ( r 2 )d 3 r 2 Ψ ∗ 2 π � 2 −∞ −∞ The first integral in the final line of equation 2.14 is just the integral of a spherical wave over all of space. ∞ � e − i ∆k · r 1 | r | d 3 r = 4 π (2.15) ∆ k 2 −∞ We arrive at the final expression for the form factor: ∞ � f ( k , k 0 ) = − m e e 2 4 π β ( r 2 ) e − i ∆k · r 2 Ψ α ( r 2 )d 3 r 2 Ψ ∗ 2 π � 2 ∆ k 2 −∞ (2.16) ∞ � = − 2 m e e 2 β ( r 2 ) e − i ∆k · r 2 Ψ α ( r 2 )d 3 r 2 Ψ ∗ � 2 ∆ k 2 −∞ This is now the amplitude of the scattered wave. Using the definition of the Bohr radius a 0 = � 2 / ( m e e 2 ) we get: ∞ � 2 β ( r 2 ) e − i ∆k · r 2 Ψ α ( r 2 )d 3 r 2 f ( k , k 0 ) = − Ψ ∗ (2.17) a 0 ∆ k 2 −∞ 2.2 The double-differential cross section The differential cross-section for a scattering event with scattering vector ∆k is � � 2 ∞ � � � d σ ( ∆k ) 4 � � = | f ( k , k 0 ) | 2 = β ( r ) e − i ∆k · r Ψ α ( r )d 3 r Ψ ∗ (2.18) � � a 2 0 ∆ k 4 dΩ � � � � −∞ 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 3 σ = � 4 π d σ ( ∆k ) dΩ. dΩ sphere

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 )d E , arriving at: � � 2 ∞ � � � d 2 σ ( ∆k , E ) 4 � � β ( r ) e − i ∆k · r Ψ α ( r )d 3 r = 0 ∆ k 4 ρ ( E ) Ψ ∗ � � a 2 dΩd E � � (2.19) � � −∞ 4 � � � 2 � � β | e − i ∆k · r | α � = 0 ∆ k 4 ρ ( E ) a 2 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: exp. scatt. angles atomic osci. str. � �� � d 2 σ ( φ, E ) 2 π � 4 φ � �� � = ρ ( E ) G αβ (∆ k, E ) (2.20) φ 2 + φ 2 a 2 0 m 2 d φ d E 2 E αβ T ���� E sol. state eff. Where the atomic part is described by the ’generalized oscillator strength’ � � 2 ∞ � � � 2 m e � � β ( r ) e − i ∆k · r 2 Ψ α ( r 2 )d 3 r Ψ ∗ G αβ (∆ k, E ) = E αβ (2.21) � � � 2 ∆ k 2 � � � � −∞ 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 E αβ the GOS is zero, with a very sharply peaked intensity onset at E = E αβ 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