Erik B. Karlsson, 2014-09-29
Entanglement creation in Compton scattering of neutrons on protons and its
possible consequences
Proofs of equations, etc.
Eq. (1):
robs = (q/m) × m/(|q|p) = (1/p ) Å
(1)
In low energy neutron scattering, momentum is conventionally expressed in Å-1 :
k = p = mv;
1 Å-1 = 1010 m-1 corresponds to p = 1010 = 1.054 10-24 kgm/s
Eq. (4):
 = 32/[M(E + W)] + 2/M(E + |Ws|) = t + 
(4)
The total cross section  is 20.36 barn.
The singlet part is (for E<< |Ws|)]; Ws = 66 keV
 = 2/M |Ws|) = 
= 19.7 10-28 m2 = 19.7 barn
≈ 95% of total 

Eq. (5):
k cot(kr0 + ) = -√(mWs)
(5)
First estimate: For k = 100 Å-1, r0 = 2 10-15 m = 2 10-5 Å: kr0 ≈ 2 10-3 <<1
Solve k cot  = -√(mWs) for small 
≈ tan = 1/ cot = – k /√(mWs)
In SI-units ( k i m-1 :
= 1.054 10-34 k /√(
= – 2.51 10-14 k [in m-1] = –2.5 10-4 k [in Å-1]
Eq. (10):
r ≤ r0
From eq. (3.a):
R0,k(r) = sin(kr0 +  sin(√(k2 + U0)r / kr cos sin(√(k2 + U0)r0
Here U0 = U(r) = V(r)×2
(3.a)
 = m/2
With V(r) = V0 = 2.2 MeV = 2.2 ×106 ×1.602 ×10-19 J and m = 1.67 10-27 kg
U0 = 2.2 ×106 ×1.602 10-19 ×1.67 ×10-27 Jkg = 5.89 ×10-40 Jkg, and
√(U0) = 2.43 ×10-20 kgm/s , whereas
k < 100 Å-1= 1.054 ×10-22 kgm/s, i. e. k << √(U0), or k2 << U0
Introduce: c1 = √(k2 + U0) ≈ √(U0) in
I1 = 0 r dr r2 R0,k(r)
= [sin  / k sin c1r0)]
0
∫ r dr r sin c1r =
= [sin  / k c12sin c1r0)] × [sin c1r – c1r × cos c1r]
where c12 = U0 .
Eq. (11):
r ≤ r0
I2 = 0 r dr r2 |R0,k(r)|2
= [sin  / k]2 0 ∫ r dr sin2(c1r ) =
= [sin2/ k2 2c1 sin2(c1r0)] × [c1r – sin c1r × cos c1r]
Eq. (13):
r ≤ r0
fk = I12/I2 = 2 (sin c1r – c1r×cosc1r)2/ c13 (c1r – sin c1r×cos c1r)
= 2 sin2 c1r × [ 1 – c1r × cot c1r ]2 / c13 [ c1r – sin c1r×cos c1r] =
(12)
= 2[ 1 – c1r × cot c1r ]2 / c13 [c1r/ sin2 c1r – cot c1r ] =
For r = r0,
fk = 2 [ 1 – c1r0 × cot c1r0 ]2 / c13 [ ( c1r0/ sin2 c1r0 – cot c1r0)] =
mutliply by r0 ,
= 2 r0 [ 1 – c1r0 × cot c1r0 ]2 / c13 [ ( c1r02/ sin2 c1r0 – r0 cot c1r0)] =
= 2 r0 [ 1 – c1r0 × cot c1r0 ]2 / c12 [ ( c12r02/ sin2 c1r0 – c1r0 cot c1r0)] =
Introduce F0 = c1r0 × cot c1r0 = √(k2 + U0) r0 × cot √(k2 + U0) r0
and put back c12 = (k2 + U0),
fk = 2r0 [ 1 – F0]2 / [(k2 + U0) (c12r02/ sin2 c1r0 – F0)]
(13)
This differs from ref. [1] in the term (c12r02/ sin2 c1r0) in the denominator,
which is missing in [1 (35)], and in that U0 is there neglected in comparison
with k2.
Eqs. (14):
r >> r0
Here, entanglement fidelity at r >> r0 is compared to th dr r2 |R0,k(r)|2
at valid at r0 . Since  << 1, cos≈
in eq. (3.b)
I1’(r)= r0 r dr r2 R0,k(r) =
= r0∫ r dr r sin(kr + ) / k ≈ (1/k) r0∫ r dr r sin(kr) =
= (1/k2)[< r cos (kr) – (1/k)sin(kr) – r0cos (kr0) – (1/k)sin(kr0)>] =
≈ (1/k2)[< r cos (kr) – r0 >]
(14.a)
valid for kr0 <<1 and k >> 1.
I’2 = r0∫ r dr dr r2 |R0,k(r)|2
≈ (1/k2) r0∫ r dr sin2(kr)
= (1/2k2) [< r – (1/k) sin(kr)cos(kr) – r0 + (1/k)sin (kr0)cos(kr0)>] =
≈ (1/2k2) [< r – r0 >] ≈ (1/2k2) (r – r0)
(14.b)
for k >>1 and kr0 <<1 . The development of entanglement for r > r0 is given
by the ratio
[I1’(r - r0)]2 / [(I2’(r - r0)] =1/k4)[< r cos (kr)> – r0 >]2 / [(1/2k2)(r – r0 )].
fk(r)/ fk(r0) = [I1’(r - r0)]2 / [(I2’(r - r0)] =
= (1/k4)[< r cos (kr)> – r0 >]2 / [(1/2k2)(r – r0 )] ≈ (2r0/k2)(r0/r)
(15)
In the last step, the average of the oscillating term < r cos (kr)> was put to
zero. For certain intervals of kr it would give rise to entanglement fidelities
largely exceeding that at r = r0 , which is unphysical. Its appearence is taken
as a mathemathical artifact connected to the definition of fk(r) in eq. (9).
A physical reason for neglecting cos (kr)-terms could be that neutron
Compton scattering experiments do not resolve oscillations when the spread
k×robs is large. Here k is determined by the angular opening

(≈ 0.1 rad) of the detector bank used in the
experiments.
k = (48.6/√2)
 = 34.4
 Å
(cf. Fig. 1)
k = 34.4 (sin / cos2×Å
= k tg ×
Below are numerical examples numerical examples for  = 60 and  ≈ 70
degrees:
For 60 degrees (k = 68.8 Å-1):
k×robs = 20.6
k ≈ 68.8 × 1.73 × 0.1 = 11.9 and k×robs = 3.57.
< cos(k×robs) > is the average over cosines in the interval 18.8 < k×robs < 22.4
radians, i.e. over an interval of 3.57 ×180/ =204 degrees. Split up in 0.2 rad
intervals:
< cos(k×robs) > = (1/18)( cos 18.8 + cos 19.0 + …….cos 22.2 + cos 22.4)=
Average of cos(kr) for k = 68.8 (60 degrees) and delta(kr) = 6, with
r = 0.3
>> x = linspace(18.8, 22.4);
>> y = (cos(18.8)+ cos(19.0) + cos(19.2) + cos(19.4) +cos(19.6) +
cos(19.8) +cos (20.0) + cos(20.2)+ cos(20.4)+ cos(20.6) + cos(20.8)+
cos(21.0) +cos(21.2)+ cos(21.4)+cos(21.6) + cos(21.8) + cos(22.0)
+cos(22.2) + cos(22.4))./20
-0.0847
Average of cos(kr) for k = 100 (70 degrees)and delta(kr) = 6, with r
= 0.3
x = linspace(27,33);
y = (cos(27)+ cos(27.5) + cos(28) + cos(28.5) +cos(29) + cos(29.5)
+cos (30) + cos(30.5)+ cos(31)+ cos(31.5) + cos(32)+ cos(32.5)
+cos(33))./13
-0.0052
Conclusion: The oscillation is only partially averaged out, but more so at 70 degrees
than at 60 degrees scattering angle.
Eqs. (18):
Start with E = × /t
where the passage time through the nuclear potential is (with v = p/m = 
k/m),
t = r / v = r × m/ k
(for k = 100
= 1.054 ×10-22 kgm/s and r = 1.2 fm = 1.2 ×10-15 m we
have t = 1.2 ×10-15 × 1.67 ×10-27 /1.054 ×10-22 = 2.1 ×10-20 s)
but  = 2.5 ×10-4 × k
so
E = 2.5 ×10-4 (2/r × m) k2
(eq. (5))
(18)
Units:
 = 2.5 ×10-4 × k is dimensionless when k is expressed
so [E ]= [1]×[]/ t] = [J]
With
t = 1.2 ×10-15/ 0.63×103 k = 1.90 ×10-18/k s
(with k in Å-1)
E = × /t = 2.5 ×10-4 × 1.054 ×10-34 k2 /1.90 ×10-18 J (with k in Å-1) =
= 1.39 × 10-20 k2 J
(with k in Å-1) =
= 1.39 × 10-20 /1.602× 10-19 k2 eV = 0.087 k2 eV (with k in Å-1)
Introduce
k = (48.6/√2)
 = 34.4
E = 0.087 × (3
 Å
eV = –1.03 ×102/cos2  eV
Numerical example:
k = 100 Å-1 : E = 0.087 × 104 = 870 eV
Eq. (19):
Start from
E = –1.03 ×102/cos2 

r0 = 2.5 ×10-5 Å
robs = 1/(p ×sin ) ≈ 0.33/(sin ) Å
(where p is measured in Å-1, i. e. robs in Å)
(E)obs = (E) (r0/robs) = (E) ×2.5 ×10-5 × (sin /0.33) =
– (1.03 ×102/cos2 ×2.5 ×10-5 sin /0.33
= – 0.0078 sin /cos2 





Repeat eq. (18) in SI-units:
E = 2.5 ×10-4 (2/r × m) k2
k i m-1 = 1010 Å-1
(18)
-1
 = 2.5 ×10-4 × 1.054 ×10-24 k = 2.6 ×10-28 k (with k i m )
r × m = 1.2 ×10-15 × 1.67 ×10-27 = 2.00 ×10-42 kg m
[2k2] = (1.054 ×10-34)2 ×10 40 k2
J2 s2m-2
E = 2.6 ×10-28 (2k2/(r × m)) in [J2 s2m-2 / m kg] = J
= 2.6 ×10-28 × (1.054 ×10 -34 )2× k2 1040 /2.00 ×10-42 J = 1.31x 10-16 J =
= 0.87 ×103 eV = 870 eV