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# Rectangular potential barrier of length L Tunneling effect

### 1. Wave functions within the related regions:

```The Schrödinger time-independent one-dimension equation is:

- (ℏ 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 /ℏ2.

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)/  (ℏ 2)]ψ2(x) = 0

Equation of the form:
∂2ψ2(x)/∂x2 - α2 ψ2(x) = 0
where α2 = 2m(Vo - E)/  (ℏ 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) /(ℏ 2)]ψ2(x) = 0

This equation is the form:
∂2ψ2(x)/∂x2  + [k2]ψ2(x) = 0
with k2 = 2mE /ℏ2

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 /ℏ2.

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)/  (ℏ 2, and k2 = 2mE /ℏ2, 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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