Neutron Scattering

k[Ä-1 ]z 0.695
E
meV
,,[Ä] 9.0451 VE meV (3 .11)
E[meV ~t
81.8/2[Ä]
To give an exaniple, neutrons of wavelength a,
= 2.4 A have an energy of 14 .2 meV with a
magnitude ofthe neutron wave vector ofk = 2.6 A_ 1 .
To obtain solutions of the wave equation (3 .9) in matter, we reformulate the differential
equation by explicitly separating the interaction terni :
(A+k
2m V . IP = : x
2
(3 .12)
Here k denotes the wave vector for propagation in empty space . The advantage of this
formulation is that the solution of the left hand side are already known. They are the plane
waves in empty space . Equation (3 .12) is a linear partial differential equation, i . e. the
superposition principle holds : the general solution can be obtained as a linear combination of
a complete set of solution functions . The coefficients in the series are determined by the
boundary conditions . To solve (3 .12) one can apply a method developed for inhomogeneous
linear differential equations . For the moment, we assume that the right hand side is fixed
(given as x) . We deine a "Greensfunetion" by:
(A+k''~(r,r')=8(r-r')
(3 .13)
We can easily verify that a solution of (3 .l3) is given by :
G( :,
e iklr-r'I
z^') = (3 .14)
4mlr- r'I
3-6