Semiconductor Devices

Quantum Tunneling ThroughA Barrier

Jasmine Robertssubmitted to: Dr. Wilfred Haensch, IBM

May 4th 2013

Abstract

The goal of this report was to examine the quantum mechanicalphenomenon of tunneling with respect to the Schrodinger Equation.A quantum mechanical wave is said to tunnel when it travels (prop-agates) through a classically forbidden region. In a more physical in-terpretation, it is when the energy (E) of the wave is lower than thepotential at a specific point V (x). [A point where V (x) is greater thanE is referred to as a classical turning point.] The method used to findthe one-dimensional transmission coefficient was the WKB (analytical)

approximation method. It was found to be e2σ

(1+ 14 e

−2σ)2. The probabil-

ity density was also calculated to sum over both final and inital statesstates.

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1 Schrodinger Equation in One-Dimension

The tunnel effect is based on complex valued wave functions where asolution begins with the one-dimensional stationary Schrodinger Equation.The wavefunction describes a wave of probability, the square of whose am-plitude is equal to the probability of finding a particle at position x and timet. Due to the one-dimensional nature of many semiconductor heterostruc-tures,the one-dimensional Schrodinger equation is sufficient for most appli-cations. The wavefunction describes a wave of probability, the square ofwhose amplitude is equal to the probability of finding a particle at positionx and time t. To derive the one-dimensional Schrodinger equation, examinethe total energy equation

p2

2m+ U(x) = Etotal (1.1)

Substituting dynamical variables to the wave-equation yields the one-dimensionaltime-dependent Schrodinger equation

−~2m

∂2

∂x2(1.2)

Using the notation from the Hamiltonia operator, the time-dependent Schrodingerequation can be written as

HΨ(x, t) = −~i

∂

∂tΨ(x, t) (1.3)

Since the Schr/”odinger equation is a partial differential equation, the equa-tion can be separated into a spatial and temporal part

Ψ(x, t) = ψ(x, t)f(t) (1.4)

Subsituted into the previous equation into Schrodinger’s yields:

1

Ψ(x)HΨ(x) =

i~f(t)

d

dtf(t) (1.5)

The left side of this equation depends solely on x, while the right side de-pends solely on t. x and t are independent variables and the equation isonly satisified when both are constant

i~f(t)

d

dtf(t) = const. (1.6)

Set the const. = Ef(t) = e

−iEt~ (1.7)

Inserting this into Eq (1.4) yields the time-dependent Schrodinger equation:

Ψ(x, t)f(t) = e−iEt

~ (1.8)

2

The time-independent Schrodinger equation is obtained by inserting Eq (1.8)into (1.2). Thus,

− ~2m

d2

dx2Ψ(x) + U(x)ψ(x) = Eψ(x) (1.9)

Using the Hamiltonian operator

Hψ(x) = Eψ(x). (1.10)

Since H is an operator and E is a real number, the Schrodinger equationtakes the form of an eigenvalue equation. The eigenfunctions ψn(x) andeigenvalues En are found by solving the Schrodinger equation. The eigen-values are discrete. The lowest eigenstate energy is ground state while alhigher energies are excited states.

2 WKB Approximation

The WKB (Wentzel, Kramers,and Broiullon) method is the most widelyused approximation for solving tunneling problems. It approximates solu-tions of linear partial differential equations with spatially varying coefficientsand is often applied to one-dimensional cases.

Suppose we want to find the approximate solution of the Schrodingerequation:

ψ′′(x) +Q(x)ψ(x) = 0 (2.1)

where the primes indicate derivatives with respect to x and

Q(x) =2m

~2(E − V (x)) (2.2)

First, ψ(x) is replaced by φ(x)

ψ(x) = exp[φ(x)], (2.3)

This is substitued into (2.1)

φ′′(x) + φ′2(x) +Q(x) = 0. (2.4)

The equation can be transformed into a Riccati equation by changing φ toy′; 1

φ′(x) = y(x) (2.5)

1A Riccati equation is any first-order ordinary differential equation that is quadraticin the unknown function. In other words, it is an equation of the form of

y′(x) = q0(x) + q1(x)y(x) + q2(x)y2(x)

where q0(x) 6= 0 and q2(x) 6= 0

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to obtainy′(x) + y2x+Q(x) = 0 (2.6)

In tunneling problems, y(x) is a complex function and is written as:

y(x) = α(x) + iβ(x) (2.7)

By substituting (2.7) in (2.6) and separating the real and imaginary parts,two first order differential equations are found:

α′ + α2 − β2 +Q(x) = 0, (2.8)

andβ′ + 2αβ = 0 (2.9)

Equation (2.9) can be integrated to yield∫α(x)dx = ln(

1√βx

+ C) (2.10)

From Eqs. (2.5), (2.7) and (2.10) φ and ψ can be found;

φ(x) = ln(1√β(x)

) + i

∫β(x)dx, (2.11)

and

ψ(x) = eφ(x) =1√β(x)

[A1cos(

∫β(x)dx) +A2sin(

∫β(x)dx)]. (2.12)

This expression for ψ(x) is the exact solution of Eq.(2.1). To find an approx-imate solution to (2.1), if α′ and α are ignored, then the algebraic equationfor the unknown function β:

β(x) =√Q(x) (2.13)

If Q(x) is positive from (2.12) and (2.13) then

ψ(x) =1

[Q(x)]14

exp[i(

∫ √Q(x)dx− θ)]. (2.14)

If Q(x) is negative, then

ψ(x) =1

[Q(x)]14

[A1exp(

∫ √−Q(x)dx) +A2exp(−

∫ √−Q(x)dx)]. (2.15)

Equations (2.8) and (2.9) show that this approximation is not valid at thosepoints where Q(x) is small or zero, but it is acceptable whenever Q(x) is

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large. IN order to estimate the error in this approximation, we go back toRicatti equation (2.6) and write y(x) as

y(x) = y0(x) + η(x), (2.16)

where y0(x) is the solution of the Ricatti equation in the WKB approxima-tion, i.e.

y0(x) = i√Q(x)− 1

4

Q′(x)

Q(x). (2.17)

From Eqs. (2.16), (2.17) and (2.6)

η′(x) + η2(x) + 2y0(x)η(x) +1

16[(lnQ(x))′]2 − 1

4(lnQ(x)))” = 0 (2.18)

Since η(x) is assumed to be a small function compared to y(x), η2(x) andη′(x) in (2.18) can be ignored. η(x) is defined as

η(x) =lnQ(x))”− 1

4 [(lnQ(x))′]2

8i√Q(x)

(2.19)

In the denominator, the y0 was replaced by its largest part i√Q(x).

Equation (2.19) shows that the WKB approximation is valid when the righthand side of (2.19) is small. In tunneling problems Q(x) changes sign, thisthere is a region wher Q(x) is positive (oscillatory solution) and a partwhere Q(x) is negative (exponential solution). Therefore, Q is continuousif it passes through zero.

When Q(x) = 0, or it is small, this method breaks down and we haveto find another way of determining the solution. Assume that the WKBmethod is not valid for all points on the x-axis between A and B, andbetween these points Q(x) changes sign. If Q(x) is a smoothly varyingfunction we can approximate it by a straight line, i.e.

Q(x) = a(x− x0), A ≤ x ≤ B (2.20)

The origin of the coordinate system so that it coincides with x0, the Schrodingerequation for the segment AB can be written as

ψ”(x) + axψ(x) = 0, A ≤ x ≤ B (2.21)

This differential equation can be intgrated and the solution is given as theBessel function. If

ψ(x) = xνJp(βxα) (2.22)

where Jp is the Bessel function of order p, then ψ(x) satisfies the differentialequation

ψ”(x) + (1− 2γ

x)ψ′(x) + (α2β2x2α−2 +

γ2 − α2p2

x2)ψ(x) = 0 (2.23)

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Now if we compare Eq. (2.21) with (2.23) we conclude that the the lattersimplifies to the former provided

α =3

2, γ =

1

2, p =

1

3and α2β2 = α (2.24)

If β =1 and this a = 94 . The solution of (2.21) in terms of the Bessel function,

ψ(x) =√x[A1J 1

3(x

32 ) +A2J−1

3(x

32 )]. (2.25)

Alternatively we can write ψ as a linear combination of f(x) and g(x). where

f(x) =√x[J 1

3(x

32 ) +A2J−1

3(x

23 ) (2.26)

g(x) =√x[−J 1

3(x

32 ) +A2J−1

3(x

23 ) (2.27)

From the asymptotic expansion of the Bessel function we can find asymptoticforms of f(x) and g(x)

f(x)→ 2

√2

πx−14 cos(

π

6)cos(x

32 − π

4), as x→∞ (2.28)

and

g(x)→ −2

√2

πx−14 sin(

π

6)sin(x

32 − π

4), as x→∞ (2.29)

To find the asymptotic form when x tends to −∞, we first change thevariable from xto t;

t2 = x3 (2.30)

In terms of this new variable:

f(−x) = t13K 1

3(it), (2.31)

and the asymptotic xpansion of f(−x) becomes

f(−x)→ t13

√2

πte−tsin(

π

3) =

√2

πx−

14 sin(

π

3)exp(−x

3

2) (2.32)

In the same way g(−x) we find

g(−x)→√

2

πx−

14 exp(

x3

2). (2.33)

With the aid of these relations we find the connection formula for the wavefunction from a region where Q(x) is negative to a prt of Q(x) is positive.

Q(x) =9

4x (2.34)

6

and ∫ x

0

√Q(x)dx = x

32 (2.35)

If Q(x) < 0, we take x = −z, then we have√−Q(x) =

3

2z

12 , (2.36)

and ∫ −x0

√−Q(x)dx = −z

32 (2.37)

Now we choose the point x = x0 at which Q(x0) = 0 as the referencepoint and for negative Q(x) we write ψ(x) as

ψ(x) =1

[−Q(x)]14

[A1exp(

∫ x

x0

√−Q(x)dx) +A′2exp[(

∫ x

x0

√Q(x)dx)] (2.38)

in a similar way

ψ(x) =1

[Q(x)]14

[A1exp(i

∫ x

x0

√−Q(x)dx) +A′2exp[(−i

∫ x

x0

√Q(x)dx)]

(2.39)The four constants A1, A2, A

′1 and A′2 are relaeed to eachother. For

joining ψ(x) on the two sides of x = x0, we usef(x) to find A1 and g(x)to find A2. By comparing Eqs. (3.32) and (3.28), and writing the latter interms of complex functions

A′1 = exp(−iπ4

)A1 (2.40)

andA′2 = exp(i

π

4)A1 (2.41)

In the same way writing (2.33) and (2.29) where the second equation is writ-ten in terms of a complex function, the following expressions are obtained:

A′1 =i

2exp(−iπ

4)A2, (2.42)

and

A2− =−i2exp(i

π

4)A2 (2.43)

Thus the relations between Ai’s and A′i’s are

A′1 = exp(−iπ4

[A1 +i

2A2] (2.44)

and

A′2 = exp(iπ

4[A1 −

i

2A2] (2.45)

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These can be simplified:

A′1 =1√2

[(1− i)A1 +1

2(1 + i)A2] (2.46)

and

A′2 =1√2

[(1 + i)A1 +1

2(1− i)A2] (2.47)

We have assumed that Q(x) goes from negative to positive. For cases whereQ(x) goes from postiive to negative values:

A′1 =1

12

√2

[1

2(1− i)A1 + (1 + i)A2] (2.48)

and

A′2 =1√2

[1

2(1 + i)A1 +

1

2(1− i)A2] (2.49)

For regions where ψ(x) is the sum of two exponentials, it is sufficient to takethe coefficents of the two exponentias as real quantities.For the region where ψ(x) is oscillatory we need real functions, and for thiswhen Q(x) is positive we write ψ(x) as

ψ(x) =C

[Q(x)]14

cos[

∫ x

x0

√Q(x)dx− θ], (2.50)

The two constants C and θ are substituted for A′1 and A′2 in (2.39). Bycomparing (2.48) and (2.50)

A′1 =C

2e−iθ, A′2 =

C

2eiθ (2.51)

We replace the two constants C and θ for A′1 and A′2 in terms of C and θ,we can use Eqs(2.48) to go fromto region of positive Q(x) to the region ofnegative Q(x) and find A1 and A2 in terms of C and θ

A1 =C

2cos(θ − π

4) (2.52)

andA2 = −Csin(θ − π

4) (2.53)

By solving for θ and C

tan(θ − π

4) = − A2

2A1(2.54)

and

C =√

4A21 +A2

2 (2.55)

Considering the tunneling of a particle of mass m and energy E through anarbitrary potential barrier V (x). Denoting the classical turning points by a

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and b, the approximaate solution of the Schrodinger equation can be writtenas

ψ1(x) =A1

(Q(x))14

exp[i

∫ x

a

√Q(x−)dx′]+

B1

(Q(x))14

exp[−i∫ x

a

√Q(x′)dx′], x < a

(2.56)

ψ2(x) =A2

(−Q(x))14

exp[

∫ x

a

√−Q(x−)dx′]+

B2

(−Q(x))14

exp[

∫ x

a

√−Q(x′)dx′], a < x < b

(2.57)

ψ1(x) =A3

(Q(x))14

exp[i

∫ x

b

√Q(x−)dx′]+

B3

(Q(x))14

exp[−i∫ x

b

√Q(x′)dx′], x > b

(2.58)Using the connection formula for joining the three wave functinos and theneliminating the constraints A2 and B2 from these formulae

[A1

B1] =

1

2[

2eσ + 12e−σ i(2eσ − 1

2e−σ)

−i(2eσ + 12e−σ) 2eσ + 1

2e−σ ][

A3

B3] (2.59)

where

σ =

∫ b

a

√−Q(x)dx. (2.60)

The transmission coefficient is defined by the relation |A3A1|2, i.e.,

T (E) = |A3

A1|2 =

e−2σ

(1 + 14e−2σ)2

(2.61)

3 One-Dimensional Transition Probability

3.1 Emitter/Collector States

Resonant tunneling structures (RTS) consist of two tunneling barriersand a quantum well layer between the barriers. Current flows from theone electrode, denoted as emitter, through the double barrier structure tothe receiving electrode, denoted as collector. The emitter and collector aredoped to provide charge carriers for transport. RT structures are usually n-type due to the larger quantum energies attainable with the lighter electronsas compared to the heavier holes. Each one is characterized by a completeset eigenfunctions ΨI and ΨF and the corresponding eigenenergies. (I andF are the initial and final states respectively). When the two systems arebrought into contact, say by tunneling barrier, one may approximate thewave function of the total system by expansion of the form

Ψ(t) =∑i

ai(t)Ψiexp(iEit

~) +

∑j

bj(t)Ψjexp(iEjt

~) (3.1)

9

The emitter states are denoted by the i indices and the collector statesare denoted by j. Equation (3.1) is a good approximation if the coupling isweak between the two systems and the barriers and relatively thick. If theelectron is assumed to start from state Ψ(0)I , one has initial conditions forthe coefficients a0(t = 0) = 1 and ai(0) = 0 = bj(0) = 0. This is now insertedinto (1.) If a weakly coupled system is assumed, a perturbation approachmay be taken which starts with the assumptions ai(t) ≈ 1, bj(t) ≈ 0 anddai/dt ≈ 0 for all times t. Thus Ψ(t) ≈ Ψi=0 ≡ ΨI . The time dependentSchrodinger equation is:∑

j

i~dbJdj

Ψjexp(−iEkt

~) = (H − EI)ΨIexp(−

iEIt

~) (3.2)

After multiplication with a state ΨF taken from all possible states in thecollector system, and repeating the same theory known from time-dependentperturbation theory, the transition probability between ΨI and ΨF becomesthe following expression:

|bF (t)|2

t=

2π

~|MIF |2δ(EF − EI) (3.3)

With a transition matrix element given by MIF = ΨF |H − EI |ΨI .To obtain the total transition probability, one has to sum over all initialand final states and to weight every term in the usual way by the combinedprobability to find the initial state occupied and the final state empty. Thisleads to the expression.

PIF =2π

~∑m,n

∑jI ,jF

|M2IF [fEI − fEF ]δ(EI − EF + eVb) (3.4)

The δ function ensure the conservation of energy. It is alread assumedthat a bias voltage is applied which drops completely across the tunnelingbarrier and leads to a relative shift in Fermi levels across the collector andemitter states.

3.2 Small Perturbation Theory

Let |f〉 be the final state that |j〉 transitions to. VIF ≡ 〈f |Vp |j〉 and ωjf ≡(Ej − EF )/~

i~aF =∑j

ajVjF ei(Ej−EF )/~ (3.5)

In this approximation, there are only two states, the initial excited andfinal states. aI(t) = 1∀t, aF (0) = 0while other a’s are always equal to zero.Integrating (3.5)

af = VIF1− eiωIF

hωIF(3.6)

10

P = |af |2 = |VIF |2 (3.7)

From the Heisenberg Uncertainty Principle, ∆E∆T ≥ ~2 . So assume the

energy of the excited state is distributed according to a distribution ρ(ω).Integrating to obtain the occupation probability:

P =

∫ ∞−∞

dω|V 2IF |~2

ρ(ω)sin2((ωIF − ω)t/2)

(sin2((ωIF − ω)/2)2(3.8)

the

sin2((ωIF − ω)t/2)

(sin2((ωIF − ω)/2)2(3.9)

narrows as t increases. The decay needs to be considered from the excitedstate to the final state, so large t

P =|VIF |2

~2ρ(ωIF )2t

∫ ∞−∞

dxsin2x

x2(3.10)

The integral evaluates to π, therefore

P =2π

~2|VIF |2ρ(ωIF )t (3.11)

Since EIF = ~ωIF →P =

2π

~|VIF |2ρ(EIF )t (3.12)

Schrodinger Time-Independent Probability Density

11

Probability Density of Transition Including Inital and Final States

4 References

Garg, Anupam. Tunnel Splittings for One-Dimensional Potential WellsRevisited. American Journal of Physics. 68 (5), 2000. 430-437.

Liboff, Richard L. Introductory Quantum Mechanics. San Francisco:Cornell University Press, 1990.

R.L.Liboff, Introductory Quantum Mechanics (Addison-Wesley,Reading, Massachusetts, 1992), 2nd ed., Chap. 7.E.

Merzbacher, Quantum Mechanics, (Wiley, New York, 1970), 2nd ed.,Chap. 6.

R.Eisberg and R.Resnick, Quantum Physics of Atoms, Molecules, Solids,Nuclei, and Particles(Wiley, New York, 1985), 2nd ed., Chap. 6.

S.Gasiorowicz,Quantum Physics, (Wiley, New York, 1974), Chap. 5.D.Park, Introduction to the Quantum Theory, (McGraw-Hill, New York,

1992), Chap. 4.L.I.Schiff, Quantum Mechanics, (McGraw-Hill, New York, 1968), Chap.

5.R.G.Winter,Quantum Physics, (Wadsworth, Belmont, California, 1979),

Chap. 2.R.L.Liboff, Introductory Quantum Mechanics (Addison-Wesley,

Reading, Massachusetts, 1992), 2nd ed., p. 257.

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