Tutorial 3

Spring 2012 Tutorial 3 Quantum Electro optics 049052

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Tutorial 2:

WKB approximation:

Exercise 1:

Lets examine the passage of an electron trough an insulator:

In many devices there is a layer of a semiconductor or a metal attached to a layer of an

insulator (MOS capacitor of example).

The difference between the work function of the semiconductor/metal and the insulator is

denoted by , and is a constant electric field (created by the voltage differences between

two electrodes) which drops mainly on the insulator (there is a limit t the voltage drop on

the semiconductor). FE is the Fermi level, that beneath it all the states are occupied.

We will define the axes like in the figure, so we can see that the potential is a function of x:

FV x E e x

e is the electron charge.

We will calculate the transmission coefficient which is the result of electron tunneling trough

the barrier.


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You have seen in the lecture that the transmission coefficient in the WKB approximation is:

'2exp

b

a

T p dx

So we can calculate:

2p m E V x

In our case V is a function of x:

2 2F Fp m E e x E m e x

We will put this expression in the integral:

3 2'

0 0

3 2 3 2

2 2 2 2exp 2 exp

3

4 2 4 2exp 0 exp

3 3

eem

T m e x dx e xe

m me e

e e

And get:

3 24 2exp3

mT

e

This is the famous Fowler Nordheim formula.


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Exercise 2:

We will now calculate the transmission coefficient of a particle with mass m and energy E

trough a potential barrier:

2 21

2V x m x

0 And we do not have to assume that 1 :

We will start by calculating the expression for semi classical momentum:

2 2

2

0

2

0 0

12

2 21

m E m xp x Exx

x E x

Where we have defined:

0 0x m E

Lets see what happens in great distance, meaning 2 20 0x Ex E . We will expand the term

inside the square root in power series, and keep the first term only:

2

0

2 2 2

0 0 0 0

21

Exx x E

x E x x E x

For 0x the wave function in the WKB approximation for x far enough is:

0

' '

0

1exp

x

x

ix p x dx

x p

'

' ' '

2 2

0 0 0 00 0 00 0

1 1exp exp exp

x x xi x E i iE dx

x dx x dxx E x x E xx p x p

Now we can calculate:

1 1 12 22 40 0

0 2 2 2

0 0 0 00

2 211 1

Ex Exx xx

x E x x E xx p


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And integrate:

0

1 12 22 40

2 2

0 0 0 0 0

1 22

0

2

0 0

21 exp exp ln

2

21

iE E

Exx ix iE x

x E x x E x

Exx

x E x

0

11 2

4 2 2

2 2

0 0 0

exp exp2 2

iE E

ix x ix

x x x

Since we are looking at x the term 2

0

2

0

21

Ex

E x can be neglected because this term

appears not in the exponent. This will give us the asymptotic form for the incoming and

outgoing waves:

0 0

0

1 1

2 22 2

2 2

0 0 0 0

1

22

2

0 0

exp exp2 2

exp2

E Ei iE E

EiE

x ix x ixr x

x x x xx

x ixt x

x x

Where r and t are the reflection and transmission coefficients.

We can link r and t by using analytical continuation technique. We will take the variable

0x x to be complex.

0

ix ex

Now the wave function will take the form:

01 2 2

2 exp2

E ii

i Ei e

t e


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According to the figure when and the angle the coefficients must be equal.

Meaning that the incoming wave and outgoing wave must be the same for :

021

2 exp2

EiE

ir

Equating the expressions:

0 0

0 0

0

1 2 2 21

2 2

1 2 2 21

2 2

1

2 2 2

0

2

0

exp exp2 2

exp exp2 2

1exp

2 2 2

exp2

E i Ei ii E E

E i Ei ii E E

i

Ei

E

i e it e r

i e ir t e

i e i Er t i i

E

E it te e

E

0

E

Eite r

We know from conservation of the current flux that 2 2

1t r so we can get the

expression for t explicit:

0 0

22

21

E E

E Et ite T Te

Where 2

T t so we get:

02

1

1E E

Te

Notice that is the difference between energy levels if there was no minus sign in the

potential barrier. Also that the more energy the particle has the closer the transmission

coefficient is to 1.

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Tutorial 3