Solution to Problem Set 3

First we will consider the scattering of a neutron by a single spin. If the spin is due to an electron, the magnetic dipole moment is m⃗ = 2μ_{B}S⃗, where S⃗ is the spin operator and can be represented by Pauli matrices as S⃗ = σ⃗/2. The vector potential due to a magnetic dipole moment m⃗ is,

A⃗(r⃗) = −\frac{μ_{0}}{4π} [m⃗ × ∇ (\frac{1}{r})] ,  (1)

from which the magnetic field can be calculated as,

B⃗(r⃗) = ∇ × A⃗(r⃗)
= \frac{μ_{0}}{4π} [\frac{3 (r̂ · m⃗) r̂ − m⃗}{r^{3}} + 4πm⃗δ^{3}(r⃗)] .  (2)

The magnetic moment of the neutron is m⃗_{N} = γ_{N}μ_{N} S⃗_{N}, and the magnetic interaction between the two is given by the Hamiltonian,

H = −m⃗_{N} · B⃗
= λS⃗_{N} · [∇ × \{S⃗_{e} × ∇ (\frac{1}{r})\}]
= λ (S_{N})_{i} ϵ_{imk}∂_{m}ϵ_{kjl} (S_{e})_{j} ∂_{l} (\frac{1}{r}) ,  (3)

where λ = 2μ_{B}γ_{N}μ_{N}μ_{0}/4π.

To calculate the scattering cross-section for the neutron with incident wave-vector k⃗ and final wave-vector k⃗′, we have to evaluate the matrix element ⟨ k|ϵ_{imk}ϵ_{kjl}∂_{m}∂_{l} (\frac{1}{r}) |k′ ⟩. This is the Fourier transform of the potential that the neutron experiences. The calculation is done using the trick (thanks to Chaikin & Lubensky),

\frac{1}{r} = ∫ \frac{d^{3}q}{(2π)^{3}} \frac{4π}{q^{2}} e^{−iq⃗·r⃗}.  (4)

Then,

⟨ k|ϵ_{imk}ϵ_{kjl}∂_{m}∂_{l} (\frac{1}{r}) |k′ ⟩ = ϵ_{imk}ϵ_{kjl} ∫ d^{3}re^{iq⃗·r⃗}∂_{m}∂_{l} (\frac{1}{r})
= ϵ_{imk}ϵ_{kjl} ∫ d^{3}re^{iq⃗·r⃗}∂_{m}∂_{l} ∫ \frac{d^{3}k}{(2π)^{3}} \frac{4π}{k^{2}} e^{−ik⃗·r⃗}
= −4πϵ_{imk}ϵ_{kjl} ∫ \frac{d^{3}k}{(2π)^{3}} \frac{k_{m}k_{l}}{k^{2}} ∫ d^{3}re^{i(q⃗−k⃗)·r⃗}
= 4π (q_{i}q_{j} − q^{2}δ_{ij}) \frac{1}{q^{2}} ,  (5)

where q⃗ = k⃗ − k⃗′.

The matrix element for the transition of the neutron from an initial state of wave-vector k⃗ and spin state |s ⟩ and the elctron in the initial spin state |σ ⟩ to corresponding final states k⃗′ and |s′ ⟩ for the neutron and |σ′ ⟩ for the electron is,

⟨ ks; σ|H|k′s′; σ′ ⟩ = 4πλ⟨ s|(Ŝ_{N})_{i}|s′ ⟩⟨ σ|(Ŝ_{e})_{j}|σ′ ⟩(q_{i}q_{j} − q^{2}δ_{ij})\frac{1}{q^{2}} .  (6)

In Born approximation the differential scattering cross-section for the neutron is given by,

(\frac{dσ}{dΩ})_{k⃗,s→k⃗′,s′} = (\frac{m_{N}}{2πh̄^{2}} )^{2} ∑_{σ,σ′} P (σ) |⟨ ks; σ|H|k′s′; σ′ ⟩|^{2} .  (7)

Here P (σ) is the probability of the electron to be in the spin state |σ ⟩. The summation over the electron spins can be done easily and we have,

∑_{σ,σ′} P (σ)⟨ σ|(Ŝ_{e})_{j}|σ′ ⟩⟨ σ′|(Ŝ_{e})_{k}|σ ⟩ = (\frac{1}{4})δ_{jk},  (8)

if the electron has an equal probability to be in up and down states. Then, combining equations (6), (7) and (8) we have,

(\frac{dσ}{dΩ})_{k⃗,s→k⃗′,s′} = (\frac{m_{N}}{2πh̄^{2}} )^{2} (4πλ)^{2}(\frac{1}{4}) (q^{2}δ_{ij} − q_{i}q_{j}) \frac{1}{q^{2}} ⟨ s|(Ŝ_{N})_{i}|s′ ⟩⟨ s′|(Ŝ_{N})_{j}|s ⟩.  (9)

If the neutron is initially in the state | ↑ ⟩ and finally in | ↓ ⟩, then the scattering cross-section is,

(\frac{dσ}{dΩ})_{k⃗,↑→k⃗′,↓} = (\frac{m_{N}}{2πh̄^{2}} )^{2} (4πλ)^{2} \frac{1}{q^{2}} \{⟨ ↑ |(Ŝ_{N})_{x}| ↓ ⟩⟨ ↓ |(Ŝ_{N})_{x}| ↑ ⟩(q^{2} − q_{x}^{2})
+ ⟨ ↑ |(Ŝ_{N})_{x}| ↓ ⟩⟨ ↓ |(Ŝ_{N})_{y}| ↑ ⟩(−q_{x}q_{y}) + ⟨ ↑ |(Ŝ_{N})_{y}| ↓ ⟩⟨ ↓ |(Ŝ_{N})_{x}| ↑ ⟩(−q_{y}q_{x})
+ ⟨ ↑ |(Ŝ_{N})_{y}| ↓ ⟩⟨ ↓ |(Ŝ_{N})_{y}| ↑ ⟩(q^{2} − q_{y}^{2})}
= (\frac{m_{N}}{2πh̄^{2}} )^{2} (4πλ)^{2}(\frac{1}{4})^{2} (1 + cos^{2} θ) .  (10)

Similarly, we can show that,

(\frac{dσ}{dΩ})_{k⃗,↑→k⃗′,↑} = (\frac{m_{N}}{2πh̄^{2}} )^{2} (4πλ)^{2}(\frac{1}{4})^{2} sin^{2} θ.  (11)

Now we consider, instead of a single spin, a regular array of spins in a lattice. Let the position of the spins be R⃗_{α}. The magnetic moment per unit volume of the sample is defined as,

m⃗(r′) = ∑_{α} m⃗_{α}δ(r′ − R⃗_{α}),  (12)

where m⃗_{α} is the magnetic moment of the spin at R⃗_{α}. The magnetic field created by the array of spins is,

B⃗(r⃗) = − (\frac{μ_{0}}{4π} ) ∇ × [m⃗_{α} × ∇ (\frac{1}{|r⃗ − R⃗_{α}|})]

= − (\frac{μ_{0}}{4π} ) ∫ d^{3}r′ ∇ × [m⃗(r′) × ∇ (\frac{1}{|r⃗ − r′|})] ,  (13)

where we have used equation (12) in the last step. The magnetic interaction between the neutron and the spins is given by,

H = λS⃗_{N} · [∫ d^{3}r′ ∇ × S⃗(r′) × ∇ (\frac{1}{|r⃗ − r′|})]
= λ ∫ d^{3}r′(Ŝ_{N})_{i}ϵ_{imk}∂_{m}ϵ_{kjl}Ŝ_{j}(r′) ∂_{l} (\frac{1}{|r⃗ − r′|}) ,  (14)

where S⃗(r⃗) is the total spin per unit volume of the array. The Fourier transform of the potential is similar to what we did before, and we have,

⟨ k|ϵ_{imk}ϵ_{kjl}∂_{m}∂_{l} (\frac{1}{|r⃗ − r′|}) |k′ ⟩ = 4π (q_{i}q_{j} − q^{2}δ_{ij}) \frac{1}{q^{2}} e^{iq⃗·r′}.  (15)

The relevant matrix element is,

⟨ ks; S|H|k′s′; S′ ⟩ = 4πλ⟨ s|(Ŝ_{N})_{i}|s′ ⟩⟨ S|Ŝ_{j}(q⃗)|S′ ⟩(q_{i}q_{j} − q^{2}δ_{ij})\frac{1}{q^{2}} .  (16)

Here |S ⟩ and |S′ ⟩ are the initial and final spin states of the array. S⃗(q⃗) is the Fourier transform of the spin of the array and is defined as,

S⃗(q⃗) = ∫ d^{3}rS⃗(r⃗)e^{ir⃗·q⃗}.

In the previous case, where there was a single spin, we considered elastic scattering of the neutron. This is usually the case, unless the spin has other degrees of freedom that the neutron can excite. However, now the neutron can exchange energy with the array. This is what happens in inelastic neutron scattering experiments. The scattering cross-section per unit energy, called the partial scattering cross-section is defined as,

(\frac{d^{2}σ}{dΩdE′} )_{k⃗,s,E→k⃗′,s′,E′} = \frac{k}{k′} (\frac{m_{N}}{2πh̄^{2}} )^{2} ∑_{S,S′} P (S) |⟨ ks; S|H|k′s′; S′ ⟩|^{2} δ(E_{S} − E_{S′} + h̄ω), (17)

where E_{S} is the energy of the array in the spin state |S ⟩ and h̄ω = E − E′, is the energy transferred from the neutron to the array. The delta function, which ensures total energy conservation during the scattering, can be represented as,

δ(E_{S} − E_{S′} + h̄ω) = ∫ \frac{dt}{2πh̄}e^{(i/h̄)(E_{S}−E_{S′}+h̄ω)t}.  (18)

The summation over the spin of the array gives,

∑_{S,S′} P (S)⟨ S|Ŝ_{j}(q⃗)|S′ ⟩⟨ S′|Ŝ_{l}(−q⃗)|S ⟩e^{(i/h̄)(E_{S}−E_{S′})t} = ∑_{S} P (S)⟨ S|Ŝ_{j}(q⃗, t)Ŝ_{l}(−q⃗, 0)|S ⟩
= χ_{jl}(q⃗, t),  (19)

where χ_{jl}(q⃗, t) is the dynamic spin-spin correlation function. The time integral from (18) Fourier transforms the correlation function,

∫ dtχ_{jl}(q⃗, t)e^{iωt} = χ_{jl}(q⃗, ω).  (20)

Using equations (16), (17), (19) and (20) we can write a general expression for the partial differential scattering cross-section as,

(\frac{d^{2}σ}{dΩdE′} )_{k⃗,s,E→k⃗′,s′,E′} = \frac{k}{k′} (\frac{m_{N}}{2πh̄^{2}} )^{2} \frac{(4πλ)^{2}}{2πh̄} \frac{1}{q^{4}} (q_{i}q_{j} − q^{2}δ_{i,j}) (q_{k}q_{l} − q^{2}δ_{k,l})
×⟨ s|(Ŝ_{N})_{i}|s′ ⟩⟨ s′|(Ŝ_{N})_{k}|s ⟩χ_{jl}(q⃗, ω).  (21)

Suppose the neutron is initially in the spin up state, and the detector can detect only down spins. Then, |s ⟩ = | ↑ ⟩, and |s′ ⟩ = | ↓ ⟩.In this case it is easy to see that in equation (21) indices i, k ≠ z. Doing a summation over the appropriate indices in equation (21), and after a bit of algebra we get the result,

(\frac{d^{2}σ}{dΩdE′} )_{k⃗,↑,E→k⃗′,↓,E′} = \frac{k}{k′} (\frac{m_{N}}{2πh̄^{2}} )^{2} \frac{(4πλ)^{2}}{2πh̄} (\frac{1}{4})\frac{1}{q^{4}} (q^{4}χ_{−+}(q⃗, ω) + q_{−}q_{+}q_{j}q_{l}χ_{jl}(q⃗, ω)
−q^{2}q_{+}q_{l}χ_{−l}(q⃗, ω) − q^{2}q_{−}q_{j}χ_{j+}(q⃗, ω)) ,  (22)

where, q_{+} = q_{x} + iq_{y} and q_{−} = q_{x} − iq_{y}.

If the neutron is in spin up state both initially and finally after the scattering, then |s ⟩ = |s′ ⟩ = | ↑ ⟩. In this case, we have i = k = ẑ in equation (21).

If both the initial and final neutron beam are unpolarized, then we have to do a sum over the neutron spins. Since,

∑_{s,s′} P (s)⟨ s|(Ŝ_{N})_{i}|s′ ⟩⟨ s′|(Ŝ_{N})_{k}|s ⟩ = (\frac{1}{4})δ_{ik},

now equation (21) can be simplified as,

(\frac{d^{2}σ}{dΩdE′} )_{k⃗,E→k⃗′,E′} = \frac{k}{k′} (\frac{m_{N}}{2πh̄^{2}} )^{2} \frac{(4πλ)^{2}}{2πh̄} (\frac{1}{4})\frac{1}{q^{2}} (q^{2}δ_{jl} − q_{j}q_{l}) χ_{jl}(q⃗, ω).  (23)

Suppose the spins are ordered in the ẑ direction, with an ordering wavevector Q⃗ (for ferromagnetic ordering Q⃗ = 0, while Q⃗ is finite for antiferromagnetic ordering). Then χ_{zz}(Q⃗) is peaked and the corresponding elastic scattering cross-section is large.