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© The scientific sentence. 2010

Rectangular potential barrier of length L
Tunneling effect




1. Wave functions within the related regions:

The Schrödinger time-independent one-dimension equation is: 
  
- ( 2/2m)∂2ψ(x)/∂x2 + V(x) ψ(x) = E ψ(x)    (1.1)


We have three regions:
1. Where the particle is free:  -∞ < x < 0 , V(x) = 0 : ψ1(x)
2. Where the particle encouters the potential barrier Vo: 0 < x< L : ψ2(x)
3. Where the particle is free:  L < x < + ∞, V(x) = 0 : ψ3(x)

Region 1:

The particle is free, the solution of the Schrodinger equation is then of the form: ψ1(x) = C exp{ikx} With k2 = 2mE /2. This wave function ψ1(x) contains two parts: incident and reflected from the poteitial: ψ1(x) = ψ1I(x) + ψ1R(- x) Therefore: ψ1(x) = C1I exp{ikx} + C1R exp{- ikx}

Region 2:

2.1: Within the barrier: Within the second region, inside the barrier: E < Vo, the Schrödinger time-independant equation is the following: 2ψ2(x)/∂x2 + [2m(E - Vo)/ ( 2)]ψ2(x) = 0 Equation of the form: ∂2ψ2(x)/∂x2 - α2 ψ2(x) = 0 where α2 = 2m(Vo - E)/ ( 2) This equation has the solution of the form: ψ2(x) = C2I exp{αx} + C2R exp{- αx} The wave function is exponentially decaying within the barrier. 2.2: Above the barrier: Within the second region, above the barrier: with E > Vo, the Schrödinger time-independant equation is the equation of a free particle with the total energie is: E - Vo: ∂2ψ2(x)/∂x2 + [2m(E - Vo) /( 2)]ψ2(x) = 0 This equation is the form: ∂2ψ2(x)/∂x2 + [k22(x) = 0 with k2 = 2mE /2 This equation has the solution of the traveling wave of the form: ψ2(x)= C2I cos(kx) + C2R sin (kx) The wave function oscillates with wave vector k

Region 3:

The particle is again free, the solution of the Schrodinger equation is then of the form: ψ1(x) = C exp{ikx} With k2 = 2mE /2. This wave function ψ3(x) is transmitted: Therefore: ψ3(x) = C3T exp{+ ikx} The wave function oscillates, decays, and oscillates agian. That is the Tunneling effect.

2. Transmission and reflection coefficients: E <= Vo

The wave functions:
- Incident (at left): ψ1(x) = C1I exp{ikx} + C1R exp{- ikx}
- Within the barrier: ψ2(x) = C2I exp{αx} + C2R exp{- αx}
- Transmitted (at right): ψ3(x) = C3T exp{+ ikx}

must be continue at the walls of the potential barrier: 
- At x = 0: ψ1(0) = ψ2(0), and 
- At x = L: ψ2(L) = ψ3(L),
That is:

-- At x = 0:
ψ1(0) = C1I + C1R (at left) =
= ψ2(2) = C2I + C2R  (within the barrier)

-- At x = L:
ψ2(L) = C2I exp{αL} + C2R exp{- αL}  (within the barrier) 
= ψ3(x) = C3T exp{+ ikL (at right)


as well as their derivatives:
- Incident (at left): dψ1(x)/dx = ik C1I exp{ikx} - ik C1R exp{- ikx}
- Within the barrier: dψ2(x)/dx = α C2I exp{αx} - αC2R exp{- αx}
- Transmitted (at right): ψ3(x)= ik C3T exp{+ ikx}

-- At x = 0:
ik C1I - ik C1R (at left) =
α C2I - αC2R (within the barrier)

-- At x = L:
 α C2I exp{αL} - αC2R exp{- αL}  (within the barrier) 
= ik C3T exp{+ ikL} (at right)


We write the constants C1I = 1, C1R = R, C3T = T. 
Rearranging, yields:
1 + R  = C2I + C2R     (1)
C2I exp{αL} + C2R exp{- αL} = T exp{+ ikL}    (2)
ik  - ik R = α C2I - α C2R    (3)
α C2I exp{αL} - αC2R exp{- αL} =  ik T exp{+ ikL}   (4)

(1) and (3) give:
(α + ik)C2I + (ik - α)C2R = 2ik    (I)

(2) and (4) give:
C2I(ik - α) exp{αL} + C2R(ik + α) exp{- αL} = 0    (II)

(I) and (II) give:
2ik (α +ik) = C2I [(α + ik)2 - (α - ik)2  exp{2αL}]
Therefore:
with S = [(α + ik)2 - (α - ik)2  exp{2αL}], we have:
C2I = 2ik (α + ik) / S
From (II), the expression of C2R is:
C2R = (α - ik)exp {2αL} C2I / (α + ik) 
= 2ik (α - ik) exp{2αL} / S


From (2), we get:
T = exp {- ikL}[2ik (α + ik) exp{αL} + 2ik (α - ik)exp{αL}]/S = 
 = 4iαk exp {- ikL} exp{αL}/S
 
 
T = 4iαk exp {- ikL} exp{αL}/[(α + ik)2 - (α - ik)2  exp{2αL}]


T = 4iαk exp {- ikL} /[(α + ik)2 exp{- αL} - (α - ik)2  exp{αL}]
The denominator is developed as: 

(α + ik)2 exp{- αL} - (α - ik)2  exp{αL} = 
(α2 - k2)(exp{- αL} - exp{αL}) + 2iαk (exp{- αL} + exp{αL})
 
The transmission coefficient τ = |T|2
 
τ = |4iαk exp {- ikL}| /|(α2 - k2)(exp{- αL} - exp{αL}) + 2iαk (exp{- αL} + exp{αL})| 

 We have:
N = |4iαk exp {- ikL}| = 16 α2k2, and 
D = |(α2 - k2)(exp{- αL} - exp{αL}) + 2iαk (exp{- αL} + exp{αL})| = 

[(α2 - k2)(exp{- αL} - exp{αL})]2
+ [2αk (exp{- αL} + exp{αL})]2 = 

[(α2 - k2)2 (exp{- 2αL} - exp{2αL} - 2)]
+ [4α2k2 (exp{- 2αL} + exp{2αL} + 2)]2
=
[(α2 - k2)2 + 4α2k2][exp{- 2αL} + exp{2αL}]
- 2 (α2 - k2)2 + 8α2k2

We have:
[(α2 - k2)2 + 4α2k2][exp{- 2αL} + exp{2αL}] = 
[(α2 + k2)2][exp{- 2αL} + exp{2αL}],
and 
- 2 (α2 - k2)2 + 8α2k2 = - 2 α4 -2 k4 + 4α2k2) + 8α2k2 = 
- 2α4 -2k4 + 12α2k2) = - 2 α4 -2 k4 - 4α2k2 + 16α2k2) =
- 2 (α2 + k2)2 + 16α2k2)
Then:
D = [(α2 + k2)2][exp{- 2αL} + exp{2αL}] - 2 (α2 + k2)2 + 16α2k2 = 
[(α2 + k2)2][exp{- 2αL} + exp{2αL} - 2] + 16α2k2 

With [exp{- 2αL} + exp{2αL} - 2] = [exp{- αL} - exp{αL}]2 
= sinh2(αL), we have:

D = 16α2k2  + [(α2 + k2)2] sinh2(αL)
 
Therfore:
 τ = N/D = 16 α2k2/16α2k2  + [(α2 + k2)2] sinh2(αL) 
 
 τ = 16 α2k2/16α2k2  + [(α2 + k2)2] sinh2(αL) 
 
  
τ = 1/1 + {[(α2 + k2)2] sinh2(αL)}/16 α2k2

With α2 = 2m(Vo - E)/  ( 2, and k2 = 2mE /2, we have:
 
Transmission: 
τ = 1/{1 + [(Vo2 sinh2(αL))/(4E(Vo - E))]}
Reflection: 
ρ = 1 - τ


Remarks:
- We have no transmission at E = Vo.
- At E > Vo, the wave function in the region 2 
continues to oscillate. Here the α is an imaginary number = ik. 
Therefore sinh(αL) becomes sin(kL). And even in this case, we 
have transmission and reflexion effects:
 
τ = 1/{1 + [(Vo2 sin2(kL))/(4E(E - Vo))]}
ρ = 1 - τ




  


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