干渉計の位相差計算¶

干渉計における位相$\phi$は屈折率$n$, 中性子波長$\lambda$, 相互作用距離$L$を用いて $$ \phi = 2\pi\frac{1-n}{\lambda}L $$ と書ける．屈折率$n$は散乱長$b$, 数密度$N$を用いて $$ n = \sqrt{1-\frac{bN\lambda^2}{2\pi}} \approx 1-\frac{bN \lambda^2}{2\pi} $$ と書ける．

散乱長は物質の有効ポテンシャル$U$を用いて $$ U = \frac{2\pi \hbar^2}{m_n}Nb $$ より， $$ \phi = 2\pi\frac{m\lambda L}{h^2}U $$ だから，測定される位相差$\Delta\phi$は有効ポテンシャルの差$\Delta U$ を用いて $$ \Delta\phi = 2\pi\frac{m\lambda L}{h^2}\Delta U $$ となる．

測定感度は中性子波長$\lambda$，相互作用距離$L$に加えて，識別可能な位相決定精度$\Delta\phi$を調整して行う．

$\Delta\phi$は$\mathrm{Mod}(2\pi)$で決まる．

In [19]:

import math
h = 4.135667696e-15 # eV*s
c = 299792458 # m/s
m = 939.5654133e6 # eV/c/c
###
#L   = 1.00 # m
#lam = 1e-9 #m
##for Si
#L   = 0.025 # m
#lam = 0.18e-9 #m
L   = 0.01 # m
lam = 0.3e-9 #m
delta_phi = 0.03

#delta_phi = (0.03 / 58)
#delta_phi = 2.*math.pi/100
delta_U = delta_phi/2./math.pi*h*h/(m/c/c)/lam/L
print("Length;  "+str(L) + " m")
print("Lambda;  "+str(lam*1e9) + " nm")
print("delta_U; "+str(delta_U*1e15) + " feV")

Length;  0.01 m
Lambda;  0.3 nm
delta_U; 2603.908907697105 feV

In [2]:

from scipy import constants as phys
import handcalcs.render

In [3]:

e    = phys.e
h    = phys.h/e
hbar = phys.hbar/e
pi   = math.pi
c    = phys.c
m    = phys.m_n*c**2/e
NA   = phys.N_A

In [4]:

%%render
#Parameter
e
h
hbar
pi
m
c
NA

\[ \begin{aligned} e &= 1.602 \times 10 ^ {-19 } \; &h &= 4.136 \times 10 ^ {-15 } \; &\mathrm{hbar} &= 6.582 \times 10 ^ {-16 } \; \\[10pt] \pi &= 3.142 \; &m &= 939565420.52 \; &c &= 299792458.0 \; \\[10pt] \mathrm{NA} &= 6.022 \times 10 ^ {+23 } \;\end{aligned} \]

数密度，位相差の一覧¶

In [5]:

def LAMBDA2ENG(lam):
    E   = h**2/(2.*(m/c**2)*lam**2) #meV
    return E

def ENG2LAMBDA(E):
    lam = sqrt(h**2/2/(m/c**2)/E)
    return lam

def AtomicDensity(rho,M):
    N = rho/M*NA*1e6 #/m^3
    return N

def PesudoPotenstial(b,N):
    U = (2*pi*hbar**2)/(m/c**2)*b*N #eV
    return U

def RefractionIndex(E, U):
    n = math.sqrt(1 - U/E)
    return n

In [6]:

%%render

U_BSE = 108.e-9
E = LAMBDA2ENG(1e-9)
a = 1 - RefractionIndex(E, U_BSE)

rho_Al = 2.7 #g/cm^3
M_Al   = 26.98
N_Al = AtomicDensity(rho_Al,M_Al)
b_Al = 3.449e-15 #fm
U_Al = PesudoPotenstial(b_Al,N_Al)
b = 1 - RefractionIndex(E, U_Al)

\[ \begin{aligned} U_{BSE} &= 1.080 \times 10 ^ {-7 } \; \\[10pt] E &= \operatorname{LAMBDA2ENG} { \left( 1 \times 10 ^ {-9 } \right) } &= 8.180 \times 10 ^ {-4 } \\[10pt] a &= 1 - \operatorname{RefractionIndex} { \left( E ,\ U_{BSE} \right) } = 1 - \operatorname{RefractionIndex} { \left( 8.180 \times 10 ^ {-4 } ,\ 1.080 \times 10 ^ {-7 } \right) } &= 6.601 \times 10 ^ {-5 } \\[10pt] \rho_{Al} &= 2.7 \; \;\textrm{(g/cm^3)} \\[10pt] M_{Al} &= 26.98 \; \\[10pt] N_{Al} &= \operatorname{AtomicDensity} { \left( \rho_{Al} ,\ M_{Al} \right) } = \operatorname{AtomicDensity} { \left( 2.7 ,\ 26.98 \right) } &= 6.027 \times 10 ^ {+28 } \\[10pt] b_{Al} &= 3.449 \times 10 ^ {-15 } \; \;\textrm{(fm)} \\[10pt] U_{Al} &= \operatorname{PesudoPotenstial} { \left( b_{Al} ,\ N_{Al} \right) } = \operatorname{PesudoPotenstial} { \left( 3.449 \times 10 ^ {-15 } ,\ 6.027 \times 10 ^ {+28 } \right) } &= 5.412 \times 10 ^ {-8 } \\[10pt] b &= 1 - \operatorname{RefractionIndex} { \left( E ,\ U_{Al} \right) } = 1 - \operatorname{RefractionIndex} { \left( 8.180 \times 10 ^ {-4 } ,\ 5.412 \times 10 ^ {-8 } \right) } &= 3.308 \times 10 ^ {-5 } \end{aligned} \]

屈折率と位相差の式¶

In [ ]: