 In this lecture, I will be calculating the neutrons scattering cross sections using Fermi Golden Rule. We discussed about, I introduced you to neutrons in the previous lecture. I mentioned all the properties, the desirable properties of neutrons that can be used for studying condensed matter structure and dynamics and I also showed you the various experimental facilities like reactors and spallation neutron sources that are being used but today I will give you the theoretical background of neutron scattering cross sections for neutrons using Fermi Golden Rule. So the basic concept of this that the sample is acting as a perturbation with its potential there are atoms molecules inside the sample and each of them do a perturbation for the incident neutron wave and sense it in some direction so either with the same energy or with change energy so it can be d sigma by d omega when it is elastic when it is elastic scattering or when it is inelastic that means there is an energy transfer then it will be d2 sigma d omega d inelastic scattering so this will encompass all sorts of scattering so fundamental premises are that we have a sample in which we have atoms molecules all microscopic units that comprise the sample and we have an incident neutron of energy EI and wave vector Ki the sample part of this wave and sense it in a direction which is defined by the final wave vector Kf and if it is inelastic then there is a different energy otherwise it will be the same energy so then that case in case of elastic scattering the magnitude of Ki and Kf will be same but only directions are different so this is the case of diffraction I let you mention it at the beginning because when we ring in the formalism of Fermi Golden Rule then I will use this elastic scattering but it is equally true for inelastic scattering which will be later so new direction new energy or new direction Ki is equal to Kf and the wave vector transfer is given by as I written here it is Ki minus Kf if I miss the vector sign then please excuse me sometimes I tend to miss while writing but they are all vectors basically these two they are vectors because they define the direction of the incoming neutron and the direction of the outgoing neutron and the Q is the momentum vector mode it's called a wave vector transfer often we also say momentum but it is not momentum h cross Q is the momentum transfer and Q for an elastic experiment if the scattering angle is 2 theta if the scattering angle is 2 theta then Q is given by 4 pi by lambda sin theta that the magnitude of Q for an elastic scattering in case of diffraction so diffraction is signified that the length of the momentum or the wave vector remains same and the Q value is given by 4 pi by lambda sin theta when 2 theta 2 theta is the angle of scattering so with this let me take you to the formalism so now in general expression what we are trying to define is let me just draw it for you so in the three-dimensional space let us call it the this is the direction of K prime K prime let us say K prime K prime the neutron comes in some direction K and then it is scattered in a direction so in this direction I can define a solid angle and we know that solid angle will be given by the component here so here if it is theta then this is sin theta this is this is cos theta this is sin theta and this angle is given by d phi d phi then the solid angle is given by sin theta d theta d phi and normally the nomenclature is d omega and what we are trying to evaluate is number of neutrals scattered per unit solid angle in the direction omega so this is d sigma by d omega that is what we will try to evaluate in our formalism the first this is the target this is the solid angle and we assume that we have put a detector here in our theoretical approach it is elastic so I am doing for elastic neutrals scattering so K and K prime only differ in direction their magnitude is the same and then if n is the incident number of neutrons per second then n d sigma d omega multiplied by d omega gives me the number of neutrons per second coming into the detector but the solid angle is given by sin theta d theta d phi now let me go to Fermi golden rule Fermi golden rule is about going from a straight psi K under the perturbation by the potential vr to a state psi K prime and then this is probability amplitude square of that will give me probability and then this is the density of states at K prime at energy e usually if it is in elastic then this e will differ but here e is same for the incident and the outgoing neutron and this gives me the probability of transition now let me just show you that for neutrons we do a box normalization you must be familiar with box normalization normalization because it is a plane wave we take and the plane wave it doesn't decay anywhere so we consider the plane wave is limited within a box of side l l so in that case the wave function psi K it's a plane wave function so with normalization component 1 by l to the power 3 by 2 normal textbook thing I write it psi K is e to the power ik dot r is a plane wave and the normalization component is l to the power 3 by 2 because we force the neutron to be inside a box of size l and then of course we can make l go to infinity so similarly psi K prime sorry here it should be psi K r and psi K prime r should be similarly 1 upon l to the power 3 by 2 e to the power ik prime r this is the plane wave that is coming in and the plane wave that is going out as I showed in this the plane wave which is coming in and the plane wave that is going out going out and then the these are the plane waves and if I often use size complex conjugate then it remains same only this I goes to minus side so this is the expression of the neutron wave coming in neutron wave going out and then in between I squeeze the potential V of r which is the interaction potential between the neutron and the target nuclei let us say number of arrangement of nuclei so this is the incoming neutron wave function and I showed you the psi K prime psi K prime r is also e to the 1 upon l to the power 3 by 2 e to the power ik prime dot r and this density of states I will evaluate for you so rho K prime at energy e is nothing but d 3 k I write it as d 3 k upon d e what is it d 3 k because I am trying to evaluate the density of states that means the number of states that are available at energy e here e is same at wave vector k prime so that is given by rho k prime e which is a density of state density of states it basically the probability depends on two things one that the probability amplitude quantum mechanically going from psi k to psi k prime under the action of potential and there should be a room for the neutron which is dictated by this density of states d 3 k is a small volume in k space just like real space it is k square d k and d omega is the direction which I described you earlier e is equal to h square k square by 2 m so d e will be equal to h square k by m and given 1 by d I have to put d here so it will be m by h square k so now I just substitute these two here so d 3 k by d e is given and also we have this d omega here so d 3 k is equal to sorry d 3 k absorbs d omega is equal to k square d k d omega and this is h square by k by m so it will be m by h square k d e this is expression for d 3 k by d omega so this goes to m k by h square d k by d e into d omega so this is m cross k by 12 k square and then sorry for the here and d omega now the question is that h square d omega what are the number of states here inside the box so now if the size is l then we know in one dimension there is one allowed k value in twice pi by l size now because it is a three dimensional box so I will have a small volume in k space which is twice pi by l cube which gives me one more so now in d 3 by d k d omega d 3 by d e d omega what I wrote here d 3 by d gives me the number of available numbers of k values that is given by the density and this volume so which is m k by h cross square which I found just now and how many number of allowed k values are there so this is the basic volume so that I have to multiply by density which is density will be 1 upon this so which is l cube upon twice pi whole cube so now I have this expression and this is the rho k prime e which I calculated just now this is the density and d 3 k by d gives me m k upon h square d omega d omega is the solid angle so now here if I put in 2 pi by h cross and now the wave function psi k psi k is given by 1 upon l to the power 3 by 2 to the power ik dot r and psi k star k prime star r is equal to 1 upon l to the power 3 by 2 to the power minus i k prime dot r so here then I have when I input this over here psi k prime vr psi k what I have got here is 1 by l to the power 3 by 2 into 1 by l to the power 3 by 2 then there is an integral of e to the power integral of e to the power minus i k prime dot r e to the power ik dot r and in between vr then d 3 r and then there will be this is the square of the whole thing square of the whole thing so I will get l cube here and this integral which I can also write e to the power minus i k prime dot r vr e to the power i k dot r which I can write in a bracket notation as this this this d 3 r is this is d 3 r dr so volume integral and I can write it in a bracket notation so now I have got l by 2 pi whole cube into 1 by l cube mk by h cross square then I have got k prime vk and square of that for my k to k prime the probability of transition from k to k prime so this will cancel now I have got and then when I talk about number of neutrons going in solid angle d omega d sigma this will be scaled with respect to the flux so then so here please note that this will be given by w k k prime divided by incident flux and p is equal to h cross k gives me incident flux is h cross k upon m square so now the number of neutrons going in solid angle d omega the number is d sigma is given by w k k prime divided by the incident flux and then excuse me so gives me so d sigma is given by m by inputting all those things twice by h cross square whole square into k prime vk in bracket notation square of that which I wrote there so this pre-factor and this so this is the expression and d sigma by d omega is m by twice pi h cross square whole square into k prime vk into square of that but this we can write in terms of a scattering amplitude this is scattering intensity per unit solid angle but we can write them in terms of a scattering amplitude defined as m by twice pi h cross square k prime vrk just not the square there and d sigma by d omega is given by square of fk k prime which is the this is the scattering amplitude so far I have not talked about the potential but I have reached a formal expression for d sigma by d omega where fk k prime k prime is given k prime vk which is nothing but the integral of e to the power minus i k prime dot r v of r then e to the power ik dot r these are the neutron wave functions this is the part having potential integrated over the entire space that is integrated over the entire sample gives me this so with this I have reached an expression for number of scattered neutrons per unit solid angle in a certain direction this is what I have reached so far now let me introduce the potential of interaction that is very important because without that we cannot go any further so now here that term comes from is pseudo potential why pseudo because the here I am considering the neutron and the nucleus interaction so these are strong interaction attractive interaction that is why potential will is deep it is extremely narrow why extremely narrow there is nothing like qualitative expression of extremely narrow what I mean is that the neutron wavelength neutron wavelength the thermal neutron wavelength wavelength is around angstrom it can be one angstrom it can be five angstrom it can be point one angstrom it is one angstrom but the extent of the nuclear potential is in femtometers which is around 10 to the power minus five angstrom now now I can say that this potential is infinitely narrow for this neutron of wavelength one angstrom and then I can represent the nuclear potential as a delta function so that is what form we did and that is why this is named after pharmaceutical as pharmaceutical potential so here so if you remember I had an expression before the scattering amplitude m by twice by h square so to offset that I will introduce this constant twice by h square by m so that they cancel with each other and it's a constant term I can do it but I also have a scattering amplitude b sigma and then the for one single scatter sitting at point r is a delta function so this is delta r minus r so this is the expression for the Fermi pseudo potential for a single scatter so now this scattering length is spin dependent and isopdependent because this is signifying this scattering amplitude signifies the interaction between a neutron and the nucleus so it is that's why it is it depends on the spin of the neutron and the spin of the nuclear protein I mean the nucleus and its orientation with respect to the neutron spin and that also depends on the isotope because the nuclear spin changes with isotope so it is isotope dependent spin dependent but one thing is there this is this so this now you can see if I write e to the power minus i q k prime dot r delta r because if there is only one nucleus I can prove it at origin so I can even write a delta delta r e to the power i k dot r d 3 r and then b sigma so this term gives me 1 because r equal to 0 and this is 1 so this gives me b sigma so d sigma by d and the constant factor m by twice pi h square square cancels with twice pi h square by m so d sigma by d omega is my b square and sigma is equal to 4 pi b square is my scattering scattering cross section cross section for a single scatter single scatter and then you can see that if I consider it classically it should have been pi b square but here because it is on solid angles it is 4 pi b square this is just a classical picture of what do we mean by scattering cross section in the beam path you have a disk of size b and then it is given pi b square the area of the disk here it is 4 pi b square for a single scatter but now let us take an assembly of scatters and what happens then so let us consider rigid lattice rigid lattice means I have taken out the dynamics from the lattice so it is a lattice at 0 degree kelvin 0 kelvin and there is no dynamics and the atoms are fixed at the sides so just now I said that for a single atom it is delta r minus r but now my atoms are fixed at this sides a general side I can call it a left side and then the potential is a sum of these delta functions and each side has got a scattering length of b l because I have not said what is the spin or what is the isotope at that side so in general it is a side dependent scattering length b l and delta r minus r l so this gives the sum over all the sides gives me the potential offered by the sample to the neutron in this neutron scattering experiment and now let me show you that now I wrote f k k prime equal to minus m by twice pi h cross square k prime v r k and d sigma by d omega is given by this and f k k prime is defined as this so now let me do this summation now v l is nothing but sum of delta functions so now earlier I put the one single scatter at the origin now I can do that so now my k prime v k is equal to e to the power i k prime dot r minus because this is complex conjugate bra and k this is complex conjugate and this is the wave function k prime minus i k prime dot r their vectors sorry again then there is a sum over l delta r minus r l because I have to add up over all the sides in the lattice that is sum them over l then e to the power i k dot r and then integral over d 3 l so this I can write as e to the power i k minus k prime dot r delta r minus r l d 3 r and I take the summation now summation over l and this is nothing but the wave vector transfer q so this is equal to sum over l e to the power i q dot now this r goes to rl rl summation over l and then sorry I forgot to put the bl over here the bl has to be there that is the most important term so bl summation over l so there is bl so this is bl so now my scattering amplitude if k k prime is sum over l bl e to the power i q dot rl so this is my scattering amplitude and this is the k prime v k