Transcript B.R. Martin. Nuclear and Particle Physics. Appendix A. Some results

B.R. Martin. Nuclear and Particle Physics.
Appendix A. Some results in Quantum mechanics.
A.1. Barrier penetration.
• Nucleons in nucleus behave not like particles, moving with momenta p but like waves with
de Broglie wavelength λB~h/p:
• Because p ~250 MeV/c < E =939 MeV it’s behaviour is described by non-relativistic
Quantum Mechanics (NR QM).
• 1900 M. Planck – blackbody radiation: emission of el-magn waves by the heated bodies:
light is not only a wave but also a particle (γ-particle) carried a portion of energy E=hν
• 1905 A. Einstein – photoelectric effect
• 1924 De Broglie - particles have also a wave origine - λB~h/p
(based on Einstein photoeffect and Compton scattering)
• 1927 Thompson , Davisson and Germer experiment showed the diffraction behaiviour
of the electrons with wavelength λB~h/p
M Born - the state of Quantum System is described by the wave function Ψ(x,t)
(or probability amplitude) which connected with probability P
to find the system in volume dV as
dP = | Ψ|2 dV where | Ψ|2 is probability density
P= ∫V dp = ∫V | Ψ|2 dV – total propability with normalization to unity:
∫V →∞| Ψ|2 dV=1
The wave function Ψ(x,t) is the state vector similar to the r(t) in classical mechanics
Superposition principle: the difference between bullets passed through two slits and
electrons scattered on two small slits is that one bullet always fly through only one slit wih probability P1 or P2
and in the case of two opened slits the detector will detect
P12=P1+P2
while electron behaves as a wave so the probability that two electrons will pass though two slits is the same
as in the case of waves with interferomety term P12=| ψ12 |2=|ψ1+ψ2|2 ≠ P1+P2=|ψ1|2+ |ψ2|2
i.e. In the case of microworld the superposition principle is valid not for probabilities but for probability amplitudes:
ψ12 =ψ1+ψ2
here P1 =|ψ1|2 (P2=|ψ2|2 ) is the probability to find electron in the detector if only 1st (2nd) slit is open.
It means that each electron feels the existance of two slits,
and the wave of one electron from 1st slit is interfering with the wave of the same electron of the other slit.
Superposition principle is valid not only for 2 but for many states
ψ =ψ1+ψ2 +ψ3 +…+ψN
and the particle can be described by the electro-magnetic wave packat with group velocity vgr=p/m=vpart
=> 1925 E Schroedinger QM : particle is not classical :
its momenta and position cannot be simultaneously
measured exactly but with some accuracy :
1927 Heisenberg uncertainty principle :
for position x and momentum p : ΔxΔpx< ћ/2 , or
for energy E and time t ΔEΔt < ћ/2 (E=hν) where ћ=h/2π – Planck’s constant or
for angular momentum L and its projection on z lz and on xy Δφ : ΔφΔlz < ћ/2
QM: Schrodinger equation
•
•
•
The Schrodinger equation plays the role of Newton's laws and conservation of energy
in classical mechanics - i.e., it predicts the future behavior of a dynamic system. It is a
wave equation in terms of the wavefunction ψ(x,p,t) which predicts analytically and
precisely the probability P= |ψ(x,p,t) |2 of the event.
The kinetic (T) and potential (V) energies are transformed into the Hamiltonian
which acts upon the wave function to generate the evolution of the wavefunction in
time and space.
T + V = E
The Schrodinger equation gives the quantized energies of the system and gives the
form of the wavefunction so that other properties may be calculated.
•
•
Operators in Quantum Mechanics
Associated with each measurable parameter in a physical system is a quantum
mechanical operator. Such operators arise because in quantum mechanics you are
describing nature with waves (the wavefunction) rather than with discrete
particles whose motion and dymamics can be described with the deterministic
equations of Newtonian physics. Part of the development of quantum mechanics
is the establishment of the operators associated with the parameters needed to
describe the system. Some of those operators are listed below.
•
It is part of the basic structure of quantum mechanics that functions of position
are unchanged in the Schrodinger equation, while momenta take the form of
spatial derivatives. The Hamiltonian operator contains both time and space
One dimensional Schrodinger equation
Nonrelativistic QM equation for spinless particle-wave packet is simply the energy conservation law:
p2
 V ( x)  Etot
- energy conservation
2m
  
p̂=-i (   )
x y z
2

i
  (
 2  V ) - time dependent Schrodinger eq.
t
2m
2
H  (
2m
 2  V )  E - if V=V(x) - time independent Schrodinger eq
 ( x, t )   ( x)e  iEt /
- solution of Schr. eq. for time independent case
 conditions on wavefunction  :
d
 and
must be contineous across the any boundary:
dx
lim  ( a   )  ( a   )   0 - should always be fulfilled
 
 d 

 d 
lim 



  0 - can be violated in the case of discontineous V(x)
 
 dx  x  a   dx  x a  
If we know  ( x, t ) then we know
x2
P=   *( x, t ) ( x, t ) dx - probability to find particle (wave package) between (x1,x2)



x1
 * dx  1 - normalization condition: the total probability should be 1
 f    * f ( x) dx - physical meaning of the observable f(x) is statistical average over all possible states


<p x > =  * pˆ x dx   *(-i
) dx  -i  *
dx  for momentum operator pˆ x

x

x
For stationary case, V(x), e iEt / time dependent part is cancelling and <p x > doesnot depend on t.

 *
j=
( *

) - particle current density
2mi
x
x
Free-particle in one dimension
Stationary Schrodinger eq. for free particle H  E with H=
p2
and V(x)=0 looks like
2m
d 2

 E and has solution:
2m dx 2
 (x)=A' sin(kx) + B' cos(kx) or equivalently
2
 (x)=A eikx + B e-ikx , where k 2  2mE /
2
 (x)=A ei(kx- t) + B e-i(kx+ t) - time-dependent solution with E= 
I . A ei(kx- t) - a wave traveling to x=+
II. B e -i(kx+ t) - a wave traveling to x=-
Boundary conditions are not valid in this case because both
cos 2 kx and sin 2 kx diverge at x=  .
In contrast we put at x=- a source (accelerator) which emits particles
with the rate I particles per second with momentum p= k in positive direction x=+.
Then B=0 because there is no particles travelling in opposite direction to x=-.
The particle current should coinside with initial current of the emitted particles:

 *
j=
( *

)
( A* e-i(kx- t) (ik )A ei(kx- t)  A ei(kx- t) ( ik )A* e-i(kx- t) ) 
2mi
x
x
2mi
k
j
2 | A |2 ik 
| A |2  I , so
2mi
m
A= mI / k .
The plane wave solution for free particle is
 (x,t)= mI / k ei(kx- t) with E=  and p= k.
Infinite potential cell
V(x)=
Stationary Schrodinger eq. for free particle in box
H  E with H=
V(x)=
, at x>a and x<0 
p
+V(x) and V(x)= 

2m
 0, at 0  x  a 
2
looks like
d 2
 E at 0<x<a and has solution:
2m dx 2
 (x)=A sin(kx) + B cos(kx) with continuity at x=0 and x=a :
 (0)=0 =>B=0,
 (a)=0 =>Asin(kL)=0, A  0 then
2

sin(ka)=0 with solutions at kL= n where n=1,2,3 ... and k 2  2mE /
V(x)=0
2
0
a
x
2 2
k2
 2
En 

n - here energy E n is quantized, only certain values are permitted
2m 2mL2
2 2
2  nx
 2
 nx

(x,t)=
sin
for
E
=
n and n=1,2,3... for a=L
A is found from normalization condition for wave function  (x)=A sin(
)
n
n
L
L
2mL2
a
2
a
  * dx  1
0

L
0
| A |2 sin 2
 nx
a
dx  1. Using cos2x  1  2sin 2 x what gives sin 2 x  (1  cos 2 x) / 2 one gets
a
1 a
n
1
n
| A |2 (1  cos 2
x)dx  | A |2 (a   cos 2
xdx) 

0
2 0
a
2
a
1
a a
n
n
| A |2
= | A |2 (a 
cos
2
xd
2
x
)

a
2
2 n 0
a
a
2
2 n
| A |2
1=
a , because  cos xdx  sin x |02 n  sin(2 n)  sin(0)  0 at n=1,2,3...
0
2
A= 2 / a .
The plane wave solution for free particle in the infinite box is
1
 n (x,t)=
2 2
2  nx
 2
sin
for E n =
n and n=1,2,3....
a
a
2ma 2
Potential Barier, E>V0 and (tunnelling effect) E<V0
E > V0
ik 1 x
-ik 1 x
ψ(x)
= A1 e
1
ik 2 x
ψ(x)
= A2 e
2
ik 1 x
+ B1 e
-ik 2 x
+ B2 e
a
0
-κ 2 x
ψ(x)
= Ce
2
ψ'' + k ( x) ψ = 0
2
ψ(
0)
= ψ(
0) , ψ2(a)
=
ψ(
a)
1
2
3
ψ'(
0)
= ψ'2( 0) , ψ'2( a)
= ψ'(
a)
1
3
whe re k(x)
=






a






∂ρ
= ∇ j = 0 - c ontinuity eq . where ρ( x, t)
=| ψ( x, t)
|2 -propability density,
∂t
h
j(x, t)
=
-( ∇ψ*) ψ - flux
ψ *(∇ψ)
2mi
For the I region x < 0, V(x)
=0
the solution is the sum of
div j +
incoming particle(free particle
solution) Ae ik1x
Be -ik1x :
and wave reflected from the barrier
ψ(
x)
= A1 eik1x + B1 e-ik1x with h 2k 12 = 2mE
1
j1 =
hk 1
| A 1 |2 - | B1 |2  = j1 - j2 .
m
For the II region at 0 < x < a,
for V(x)
= V < 0E
or
equations are ψ2 '' + k 2 2( x) ψ2 = 0
or
ψ 2 '' - κ 2 2( x) ψ 2 = 0
or
ψ(
x)
= Ceκ 2 x + De-κ 2 x
2
ψ(
x)
= A2 eik 2 x + B2 e-ik 2 x
2
solutions are
V(x)
= V > 0E
hk 2
hκ 2
currents are
j2 =
or
j2 = 2
Im  C * D 
| A 2 | 2 - | B 2 | 2 
m
m
Finally in t he III re gion(x > a) it is only outg oing particles
ik 3 x
ψ(
x)
= Fe
3
j3 =
with k 3 = k 1 =
hk 3
| F | 2 
m
2mE / h
2
+κ 2 x
+ De
-ik 1 x
ik 1 x
+ B1 e
V0
ψ(x)
= Fe
3
I
II
III
x
 p2

Stationary Schrodinger eq. 
+ V(x)  ψ = Eψ for step potential V0
 2m

 0, at x < 0 
 h2



2
∇ + V(x)  ψ = Eψ wit h V( x)
=  V0 , at a ≥ x ≥ 0 
 2m

 0, at x > 0 


is writen as

2m
k1 =
, at x < 0

h2


2m
=
(E2- V , ) E0> V
if
 k 2(x)
0

h
,
0at< x <

2m
iκ ( x)
=i
( V0 - E)
,if E < V0
 2
h2

2m

k3 =
, at x > a

h2

ψ(x)
= A1 e
1
III
II
I
ik 1 x
ψ(x)
= Fe
3
V0
E < V0
 0, at x < 0 


V(x)
= V , at0 a
≥ x ≥ 0
 0, at x > 0 


0
a
x
5 coefficients A,B, C,D,F can be defined from 4 boundary conditions,
it means that one - for example, A , is defined from somewhere else .
ψ(0)
= ψ(0),
ψ(a)
= ψ(a)
1
2
2
3
ψ'(0)
= ψ'2(0), ψ'2(a)
= ψ'(a)
1
3
System of linear equations
A+ B=
C+ D
(1)
A+ B=
C+ D
ik 1A - ik 1B = iκ 2 C + -iκ 2D
( 2)
ik 1A - ik 1B = -κ 2 C + κ 2D
iκ 2 a
-iκ 2 a
ik 1a
-κ 2 a
κ 2a
Ce
+ De
=
Fe (3)
Ce
+
De
=
Fe ik1a
iκ 2 Ce -κ 2 a + -iκ 2De κ 2 a = ik 1Fe ik1a ( 4) - κ2 Ce -κ 2 a + κ 2De κ 2 a = ik 1Fe ik1a
F =
sinhx =(e -x e /-x
)2
2Ae -ik1a( k 1κ 2 )
where
2
2
x
2k 1κ 2 coshκ 2 a - i(k -1 κ 2 )sinhκ 2 a
coshx =(e +
e /-x
)2
16k 12κ 2 2
| F |2
=
transmission coefficient for barrier V0 > E.
2
2
2
2
|A|
4κ 2k 1 +(k12 + κ2 2 )
sinh2 κ2 a
At κ 2 a ≪ 1,
T ≈ 1,( κ2 = 2m / h(V0 - E)
)
4k 12κ 2 2
-2κ 2 a
κ 2 a ≫ 1, T =
e
≪ 1, T becomes very sensitive to the
2
(k +
κ 2 2 )2
1
E of the particle due to the second exponential term.
It is especially important for life time of the nuce i in the case of α - de cay or
tunnelin g of α - particle through nuclear pote ncial.
the first term is due to the reflection losses at two boundaries x = 0 and x = a,
the decrea sing exponent d escribes the amplitude decay within the b arrier.
the first term is slowly varying with E and is usually neglected.
T=
The last result can be used for the case of arbitarary smooth potential V(x)
1/ 2
WKB approximation

-2
κ x
 2m

T e  2  at x  0  exp  2  dx  2 V ( x )  E   
Wentzel
– Kramers – Brillouin



For the case V0 < E κ 2 - > -ik 2 , sin h2 κ 2 a- > sin 2k 2 a and
at k 2 a = πn takes place T = 1 - resonance transm ission.
A.2.Density of the states and Three dimensional box
In the case of the well with infinite potential barrier
the solution of the Schrodinger eq.
are the standing waves vanishing at the ends of the well
corresponding to discrete
energy levels always > 0
because of uncertainty principle:
Δp˜1/L -> Δ(E=p2/2m)˜1/L2 .
For 3 dimensional well the energy
and the wavefunction are
or
where nx>0, ny>0, nz>0
The number of the states is proportional to the number of lattice points
dinstanced from each other by (π/L): (L/π)3
Or in the Fermi sphere it will be 1/8 of the sphere (because nxyz>0)
n(k0)=1/8 ּ4/3πk03 ּ(L/π)3=V/(2π)3ּ4/3 πk03 - for all k<k0 while for k0<k<k0+dk0:
dn(k0)=V/(2π)3ּ4πk02dk0, dn(p)=V/(2πћ)3ּ4πp2dp, dn(E)=V/(2π)3ּ4πk02dk0 where
ρ(k0)=dn(k0)/dk0=V/(2π)3ּ4πk02 is density of states for k0<k<k0+dk0
ρ(p)=dn(p)/dp=V/(2πћ)3ּ4πp2 is density of states for momenta p<p<p+dp:k=ћp
ρ(E)=ρ(p)dp(E)/dE=V/(2πћ)3ּ4πp2/v =V/(2πћ)3ּ4πmp - for E<E<E+dE, E=p2/2m,p=mv
ρ(E)=V/(2πћ)3ּp2/vdΩ for solid angle Ω, d3p=p2dpdΩ ¨
dE=p/mdp=vdp
For particles with spin spin multiplicity factor should be included:
for spin=1/2 according to the Pauli principle such factor is 2.
A.3 Perturbation theory and the Second Golden Rule
In perturbation theory the Hamiltonian at any time t may be written as
H(t)=H0+V(t), where H0 is unperturbed Hamiltonian and V(t) is small.
The solution for eigenfunctions of H starts by expanding in the terms
of the complete set of energy eigenfunctions |un> of H0 : H0 |un> =En |un>
|ψ(t)>=Σcn(t) |un> exp{-iEnt/ћ}, where En are the corresponding energies.
If |ψ(t)> is normalized to unity <ψ(t)|ψ(t)>=1 then |cn(t)|2 is probability that
at time t the system will be in the state |un>. Substituting it in Schrodinger eq :
iћ ∂cf(t)/∂t=ΣVfn(t) cn(t) exp{-iωfnt}, where matrix element
Vfn(t)= <uf|V(t)|un> and angular frequency ωfn=(Ef-En)/ ћ.
Initial state of the system |ui> at t=0 then cn(0)=δni and ci(t)=ci(0)+1/iћ∫0tVii(t’)dt’
Final state of the system f≠i
cf(t)=
1/iћ∫0tVfi(t’)ei ωfi t’dt’ general, for V(t)
For const V =0 and then V0
cf(t)= Vfi/(ћ ωfi )(1-ei ωfi t)
Probability for transition from state i to state f :
Pfi=|cf (t)|2 =4|Vfi|2/ћ2 [sin2(1/2 ωfi t)/ω2fi ]. At large t it is valid only if
Ћ | ωfi |= |Ef-En|<2πћ/t (uncertainty principle) then [sin2(1/2 ωfi t)/ω2fi ]->1/2 πћtδ(Ef-Ei).
Pfi =2πt/ћ |Vfi|2δ(Ef-Ei) – probability
dPfi /dt= 2π/ћ |Vfi|2δ(Ef-Ei) – transition probability per unit time – valid for discrete final states.
dTfi /dt= ∫ dPfi /dt ρ(Ef)dEf = 2π/ћ [|Vfi|2 ρ(Ef)]Ef=Ei– transition probability per unit time – for continious spectra
Dirac Delta Function
Appendix C
1911 Rutherford scattering
Only if
cos θ <0 or θ >90o
In Ernest Rutherford's laboratory, Hans Geiger and Ernest Marsden (a 20 yr old undergraduate student)
carried out experiments to study the scattering of alpha particles by thin metal foils.
In 1909 they observed that alpha particles from radioactive decays occasionally scatter at angles greater
than 90°, which is physically impossible unless they are scattering off something more massive than
themselves. This led Rutherford to deduce that the positive charge in an atom is
concentrated into a small compact nucleus.
During the period 1911-1913 in a table-top apparatus, they bombarded the foils with high energy alpha
particles and observed the number of scattered alpha particles as a function of angle.
Based on the Thomson model of the atom, all of the alpha particles should have been found within a small
fraction of a degree from the beam, but Geiger and Marsden found a few scattered alphas at angles over 140
degrees from the beam.
Rutherford's remark "It was quite the most incredible event that ever happened to me in my life. It was almost
as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The
scattering data was consistent with a small positive nucleus which repelled the incoming positively charged
alpha particles. Rutherford worked out a detailed formula for the scattering (Rutherford formula), which
matched the Geiger-Marsden data to high precision.
Appendix C 2. Rutherford scattering CM versus QM
Classical mechanics : Rutherford's model enables us to derive a formula for the angular distribution of scattered a-particles.
Basic Assumptions:
•
a) Scattering is due to Coulomb interaction between a-particle and positively charged atomic nucleus.
•
b) Target is thin enough to consider only single scattering (and no shadowing)
•
c) The nucleus is massive and fixed.
•
This simplifies the calculation, but it could be avoided by working in the centre of mass frame.
•
d) Scattering is elastic
Finite potential cell
Stationary Schrodinger eq. for free particle in box
I
V , at x>a and x<0 
p2
H  E with H=
+V(x) and V(x)=  0

2m
 0, at 0  x  a 
looks like
d 2
 V ( x)  E at 0<x<a and has solution:
2m dx 2
 (x)=A sin(kx) + B cos(kx) with continuity at x=0 and x=a :
 (0)=0 =>B=0,
 (a)=0 =>Asin(kL)=0, A  0 then
II
III
2

sin(ka)=0 with solutions at kL= n where n=1,2,3 ... and k 2  2mE /
2
2 2
k2
 2

n - here energy E n is quantized, only certain values are permitted
2m 2mL2
 nx
A is found from normalization condition for wave function  (x)=A sin(
)
a
En 
V ( x)  V0
V ( x)  0
V ( x)  V0
2
0
a
x
a
  * dx  1
0

L
0
| A |2 sin 2
 nx
a
dx  1. Using cos2x  1  2sin 2 x what gives sin 2 x  (1  cos 2 x) / 2 one gets
a
1 a
n
1
n
| A |2 (1  cos 2
x)dx  | A |2 (a   cos 2
xdx) 

0
2 0
a
2
a
1
a a
n
n
| A |2
= | A |2 (a 
cos
2
xd
2
x
)

a
2
2 n 0
a
a
2
2 n
| A |2
1=
a , because  cos xdx  sin x |02 n  sin(2 n)  sin(0)  0 at n=1,2,3...
0
2
A= 2 / a .
The plane wave solution for free particle in the infinite box is
1
 n (x,t)=
2 2
2  nx
 2
sin
for E n =
n and n=1,2,3....
a
a
2ma 2
Discrete energy for
the particle in the box
and continious energy
for free particle
5 coefficients A,B,C,D,F can be defined from 4 boundary conditions,
it means that one - for example, A, is defined from somewhere else.
ψ(0)
= ψ(0),
1
2
ψ(a)
= ψ(a)
2
3
ψ'(0)
= ψ'(0),
ψ'(a)
= ψ'(a)
1
2
2
3
A+ B= C+ D
(1)
ik 1A - ik 1B = -κ 2 C + κ 2D
(2)
Ce -κ 2 a + De κ 2 a = Fe ik1a (3)
-κ 2Ce -κ 2 a + κ 2De κ 2 a = ik 1Fe ik1a (4)
System of linear equ ations
- B+
C+
D +
0*F
= A
+ Bκ 2 / ik 1C + κ 2 / ik 1D +
0*F = A
e -κ 2a C +
e κ 2 aD e ik1aF = 0
- e C-κ 2+a
e κ 2 aD - ik 1 / κ 2 e ik1aF = 0
- B+
C+
D + 0*F = A
(1+ 2)
(3)
*(1κ 2 / ik 1 )
* eκ 2 a :
ik 1a
2a
(1- κ 2 / ik 1 )C +( 1+ κ2 / ik1 )
D
= 2A (1- κ 2 / ik 1 )
C +( 1- κ2 / ik1 )
e2κ 2 a D - (1- κ2 /ik 1e) κe
F =0
κ2a
(1+ 2)
-(3)
*(1κ 2 / ik 1 )
*e :
2κ 2 a
κ 2 a ik 1a
{(1+ κ 2 / ik 1 )
-(1- κ2 / ik1 )
e
}D +(1- κ2 / ik1 )
e e F = 2A (5)
(3)
+(4)
/
κ2 :
2e κ 2 aD -(1+ ik /1 κ 2 )eik1aF =
1
2
0 => D = (1 + ik 1/
κ 2 )e -κ 2 a e ik1 a F
1
ik 1a
2a
{(1+ κ 2 / ik 1 )
-(1- κ2 / ik1 )
e
}(1+ ik1 / κ2 )e-κ 2 a eik1a F +(1- κ2 / ik1e) κe
F = 2A (6)
2 -κ a
κ2a
a
2
{(1+ κ 2 / ik 1 )e -(1- κ2 / ik1 )e
}(1+ ik1 / κ2 )F + 2(1- κ2 / ik1 )e κ 2F
= 4Ae -ik1a(6)
4Ae -ik1a
F=
=
{(1+ κ 2 / ik 1 )e-κ 2 a -(1- κ2 / ik1 )eκ 2 a }(1+ ik1 / κ2 )
+ 2(1- κ2 / ik1 )e κ 2 a
4Ae -ik1a
=
=
(1+ κ 2 / ik 1 )(1+ ik1 / κ2 )e-κ 2 a +{2(1- κ2 / ik1 )
-(1- κ2 / ik1 )(1+ik 1 / κ 2 )}eκ 2 a
4Ae -ik1a
=
=
(2 + κ 2 / ik 1 + ik 1 / κ 2 )e-κ 2 a +(1- κ2 / ik1 )(1- ik1 / κ2 )e κ 2 a
4Ae -ik1a
=
=
(2ik 1κ 2 + κ 22 - k 12 )e-κ 2 a /( ik1 κ2 )
+{2ik1 κ2 - κ22 + k12 }eκ 2 a /( ik1 κ2 )
4Ae -ik1a(ik 1κ 2 )
=
=
( 2ik 1κ 2 + κ 22 - k 12 )e -κ 2+a{2ik 1κ 2 - κ 22 + k 12 }eκ 2 a
2κ 2 a
2Ae -ik1a(k 1κ 2 )
where sinhx =(e -x e /-x)2,coshx =(e + e x /)2 -x
2k 1κ 2 coshκ 2 a - i(k -12 κ 22 )sinhκ 2 a
2k 1κ 2
2k 1κ 2
| F |2
T=
=
*

| A |2
2k 1κ 2 coshκ 2 a - i(k12 - κ22 )sinhκ2 a
2k1 κ2 coshκ2 a + i(k -12 κ 22 )sinhκ2 a
2
2
4k 1 κ 2
= 2 2

2κ 2 a
a
2κ 2 a
4κ 2k (e
+ e2κ 2 a + 2)
/ 4 +(k41 -2k 21κ 2 2 + κ 4 2 )(e 2+κ 2e
- 2)
/4
1
4k 12κ 2 2
= 2 2
=
2κ 2 a
4κ 2k (e
+ e2κ 2 a - 2 + 4)
/ 4 +(k41 - 2k21 κ2 2 + κ4 2 )(e2κ 2 a + e2κ 2 a - 2)
/4
1
4k 12κ 2 2
= 2 2

4κ 2k 1 4 / 4 +(k 41 + 2k 21κ2 2 + κ 4 2 )(e2κ 2 a + e 2κ 2 a - 2)
/ 4
2
2
4k 1 κ 2
T 2 2
transmission coefficient for barrier V0  E.
2
4κ 2k 1 +(k 12 + κ 2 2 )
sinh2κ 2 a
=