ENEE313 - Homework 2

Solutions

1. Problem 2.1

Using the method of separation of variables solve the Schrodinger Wave Equation for a free electron in one dimension. Recall that for the free electron V=0, so that,

−ℏ𝟐

𝟐𝒎

𝛛𝟐𝚿(𝒙, 𝒕)

𝛛𝒙𝟐= −

ℏ

𝒋

𝛛𝚿(𝒙, 𝒕)

𝛛𝒕

Let Ψ(𝑥, 𝑡) = ψ(𝑥)ϕ(𝑡) From SWE,

−ℏ2

2𝑚

∂2(ψ(𝑥)ϕ(𝑡))

∂𝑥2= −

ℏ

𝑗

∂(ψ(𝑥)ϕ(𝑡))

∂𝑡

−ℏ2

2𝑚ϕ(𝑡)

∂2ψ(𝑥)

∂𝑥2= −

ℏ

𝑗ψ(𝑥)

∂ϕ(𝑡)

∂𝑡

By dividing from [ψ(𝑥)ϕ(𝑡)]

−ℏ2

2𝑚

1

ψ(𝑥)

∂2ψ(𝑥)

∂𝑥2= −

ℏ

𝑗

1

ϕ(𝑡)

∂ϕ(𝑡)

∂𝑡

Since we have separated variables, we can use 𝑑

𝑑𝑥 and

𝑑

𝑑𝑡 instead of partial derivations. As both LHS and RHS have

independent variables, to this equation to be satisfied, both sides need to be equal to a constant. Let that constant to be 𝐸. LHS:

−ℏ2

2𝑚

1

ψ(𝑥)

𝑑2ψ(𝑥)

𝑑𝑥2= 𝐸

𝑑2ψ(𝑥)

𝑑𝑥2= −

2𝑚𝐸

ℏ2ψ(𝑥)

Let 𝑘 = √2𝑚𝐸

ℏ2

𝑑2ψ(𝑥)

𝑑𝑥2= −𝑘2ψ(𝑥)

Solution for this differential equation is,

ψ(𝑥) = 𝐴𝑒𝑗𝑘𝑥 + 𝐵𝑒−𝑗𝑘𝑥 Here 𝐴 and 𝐵 are constants and need to solved using boundary conditions. RHS:

−ℏ

𝑗

1

ϕ(𝑡)

𝑑ϕ(𝑡)

𝑑𝑡= 𝐸

𝑑ϕ(𝑡)

𝑑𝑡= −

𝑗𝐸

ℏϕ(𝑡)

Solution for this differential equation is,

ϕ(𝑡) = 𝐶𝑒−𝑗𝐸ℏ

𝑡

Here, 𝐸

ℏ= ω

∴ ϕ(𝑡) = 𝐶𝑒−𝑗ω𝑡

Combining solutions,

Ψ(𝑥, 𝑡) = ψ(𝑥)ϕ(𝑡) = (𝐴𝑒𝑗𝑘𝑥 + 𝐵𝑒−𝑗𝑘𝑥) ⋅ 𝐶𝑒−𝑗ω𝑡 = 𝐷𝑒𝑗(𝑘𝑥−ω𝑡) + 𝐹𝑒−𝑗(𝑘𝑥+ω𝑡)

Here 𝐷 = 𝐴 ⋅ 𝐶 and 𝐹 = 𝐵 ⋅ 𝐶

(a) Assume the electron is moving in the positive 𝒙 direction and show that the solution is that of a plane wave. Show also that for the solution to exist the following relation between 𝑬, our separation constant, and 𝒌 must

hold: 𝑬 =ℏ𝟐𝒌𝟐

𝟐𝒎, where 𝑬 is the total energy of the electron. (Note that 𝒌 is the wavevector and 𝒌 =

𝟐𝛑

𝛌 where

𝛌 the electron wavelength). If electron is moving to positive 𝑥 direction, 𝑥 must be increase with 𝑡. In the above solution, we have two waves.

1. 𝐷𝑒𝑗(𝑘𝑥−ω𝑡) ∶ Here, to make the phase constant, 𝑥 should be increased with 𝑡, that is the wave is moving to positive 𝑥 direction.

2. 𝐹𝑒−𝑗(𝑘𝑥+ω𝑡) ∶ But in this case, to make the phase constant, 𝑥 should be decreased with 𝑡, that is the wave is moving to negative 𝑥 direction.

Therefore, if the electron is moving to positive 𝑥 direction,

Ψ(𝑥, 𝑡) = 𝐷𝑒𝑗(𝑘𝑥−ω𝑡) From the definition of 𝑘,

𝑘 = √2𝑚𝐸

ℏ2

𝑘2 =2𝑚𝐸

ℏ2

𝐸 =ℏ2𝑘2

2𝑚

(b) Starting from the classical expression for kinetic energy, 𝑬 =𝟏

𝟐𝒎𝒗𝟐, show that the momentum of the free

electron can be written as 𝒉𝒌. Since the electron is free, we assume the entire energy of the electron to be kinetic energy.

𝐸 =1

2𝑚𝑣2 =

1

2𝑚𝑝2

∴𝑝2

2𝑚=

ℏ2𝑘2

2𝑚

𝑝2 = ℏ2𝑘2

𝑝 = ℏ𝑘

(c) We know the momentum operator is given by ℏ

𝒋

𝛛

𝛛𝒙. Show that the average (or expected) momentum is given

by ℏ𝒌

⟨𝑝⟩ = 𝑙𝑖𝑚𝐿→∞

∫ ψ*(𝑥)ℏ𝑗

∂∂𝑥

ψ(𝑥)𝑑𝑥𝐿

−𝐿

∫ ψ*(𝑥)ψ(𝑥)𝑑𝑥𝐿

−𝐿

⟨𝑝⟩ = limL→∞

∫ 𝐷𝑒−𝑗(𝑘𝑥−ω𝑡) ℏ𝑗

∂∂𝑥

𝐷𝑒𝑗(𝑘𝑥−ω𝑡)𝐿

−𝐿𝑑𝑥

∫ 𝐷2𝐿

−𝐿𝑑𝑥

⟨𝑝⟩ =ℏ

𝑗⋅ 𝑗𝑘 𝑙𝑖𝑚

𝐿→∞

∫ 1𝑑𝑥𝐿

−𝐿

∫ 1𝑑𝑥𝐿

−𝐿

⟨𝑝⟩ = ℏ𝑘

2. Problem 2.2 (Use your code to solve parts of the problem)

If an electron is in an infinite potential well of length 𝟏. 𝟎 𝒏𝒎 :

(a) Is the electron bound or unbound?

Since the potential walls are infinite, energy of the electron should be less than that. Therefore, the electron is bound.

(b) Calculate the values of the first three energy levels that the electron can have in 𝑱𝒐𝒖𝒍𝒆𝒔 and 𝒆𝑽.

Outside the well potential is infinite which makes the wave-function,

Ψ = 0

Inside the well, from SWE,

−ℏ2

2𝑚

𝑑2ψ(𝑥)

𝑑𝑥2+ 0 ⋅ ψ(𝑠) = 𝐸 ⋅ ψ(𝑥)

𝑑2ψ(𝑥)

𝑑𝑥2= −

2𝑚𝐸

ℏ2ψ(𝑥)

Let 𝑘 = √2𝑚𝐸

ℏ2

∴𝑑2ψ(𝑥)

𝑑𝑥2= −𝑘2ψ(𝑥)

Solution for this second order differential equation,

ψ(𝑥) = 𝐴𝑠𝑖𝑛(𝑘𝑥) + 𝐵𝑐𝑜𝑠(𝑘𝑥)

Let us apply boundary conditions,

i. ψ(0) = 0

𝐴 ⋅ 𝑠𝑖𝑛(0) + 𝐵 ⋅ 𝑐𝑜𝑠(0) = 0

0 + 𝐵 = 0

𝐵 = 0

ψ(𝑥) = 𝐴𝑠𝑖𝑛(𝑘𝑥)

ii. 𝜓(𝐿) = 0

𝐴 ⋅ 𝑠𝑖𝑛(𝑘𝐿) = 0

For non-trivial solution, 𝐴 ≠ 0

∴ 𝑠𝑖𝑛(𝑘𝐿) = 0

𝑘𝑛𝐿 = 𝑛π

𝑘𝑛 =𝑛π

𝐿

∴ ψ(𝑥) = 𝐴𝑠𝑖𝑛 (𝑛π

𝐿𝑥)

From the definition of the 𝑘,

√2𝑚𝐸𝑛

ℏ2=

𝑛π

𝐿

𝐸𝑛 =𝑛2π2ℏ2

2𝑚𝐿2

𝐸1 =π2ℏ2

2𝑚𝐿2=

(1.055 × 10−34𝐽𝑠)2π2

2 × 9.109 × 10−31𝑘𝑔 × (10−9𝑚)2

𝐸1 = 6.02 × 10−20𝐽 = 0.375𝑒𝑉

𝐸2 = 4𝐸1 = 2.41 × 10−19𝐽 = 1.5𝑒𝑉

𝐸3 = 9𝐸1 = 5.42 × 10−19𝐽 = 3.375𝑒𝑉

(c) Solve the Schrodinger equation to get the wave-function for the first three energy levels.

To find the value of A, we need to normalize the wave.

∫ ψ*(𝑥)ψ(𝑥)𝑑𝑥∞

−∞

= 1

∫ 𝐴2𝐿

0

𝑠𝑖𝑛2 (𝑛π

𝐿𝑥) 𝑑𝑥 = 1

∫ 𝑠𝑖𝑛2 (𝑛π

𝐿𝑥) 𝑑𝑥

𝐿

0

=1

𝐴2

𝐿

2=

1

𝐴2

𝐴 = √2

𝐿

Therefore,

ψ1(𝑥) = √2

𝐿𝑠𝑖𝑛 (

π

𝐿𝑥)

ψ2(𝑥) = √2

𝐿𝑠𝑖𝑛 (

2π

𝐿𝑥)

ψ3(𝑥) = √2

𝐿𝑠𝑖𝑛 (

3π

𝐿𝑥)

(d) Plot the wave-function for the first three levels.

• First energy level:

• Second energy level:

• Third energy level:

(𝑚−

1 2)

(𝑚−

1 2)

(𝑚−

1 2)

(e) Plot the square magnitude of the wave-functions for the first three levels. Comment on the what each of these plots tell us with respect to finding the probability at a specific location in the well, for each energy level.

• First energy level:

• Second energy level:

• Third energy level:

Highest probability of finding electron in each case is at the peeks of the PDFs.

𝑚−

1

𝑚−

1

𝑚−

1

(f) Calculate the probability of finding the particle in the region (𝟎 < 𝒙 <𝑳

𝟒) if the electron is in the 𝒏𝒕𝒉 energy

level.

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = ∫ ψ*(𝑥)ψ(𝑥)𝑑𝑥

𝐿4

0

= ∫2

𝐿𝑠𝑖𝑛2 (

𝑛π

𝐿𝑥) 𝑑𝑥

𝐿4

0

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =2

𝐿∫ 𝑠𝑖𝑛2 (

𝑛π

𝐿𝑥) 𝑑𝑥

𝐿4

0

=2

𝐿[𝐿

8−

𝐿𝑠𝑖𝑛 (𝑛π2

)

4𝑛π] =

1

4−

𝑠𝑖𝑛 (𝑛π2

)

2𝑛π

(g) Calculate the expectation value for position and momentum for a particle in the 𝒏𝒕𝒉 energy state of the well. Do your answers make physical sense? Compare this to what you would expect classically.

For the expectation value of the position,

⟨𝑥⟩ = ∫ ψ*(𝑥)𝐿

0

⋅ 𝑥 ⋅ ψ(𝑥)𝑑𝑥

For the sake of argument let’s take n=1.

⟨𝑥⟩ =2

𝐿∫ 𝑥

𝐿

0

⋅ 𝑠𝑖𝑛2 (π

𝐿𝑥) 𝑑𝑥 =

2

𝐿⋅

𝐿2

4=

𝐿

2

Similarly, we can calculate ⟨𝑥⟩ for other n values as well. For the expectation value of the momentum,

⟨𝑝⟩ = ∫ ψ*(𝑠)ℏ

𝑗

∂

∂𝑥ψ(𝑥)𝑑𝑥

𝐿

0

⟨𝑝⟩ = ∫2

𝐿

𝐿

0

𝑠𝑖𝑛 (𝑛π

𝐿𝑥)

ℏ

𝑗

∂

∂𝑥𝑠𝑖𝑛 (

𝑛π

𝐿𝑥)

⟨𝑝⟩ =𝑛πℏ

𝑗𝑙2∫ 𝑠𝑖𝑛 (

2𝑛π

𝐿𝑥) 𝑑𝑥

𝐿

0

= 0

• Since the probability distribution is symmetric around 𝐿

2 , expected position being that makes sense.

• A particle inside the well would go back and forth without any special bias to any direction. Therefore, expected momentum being 0 is valid.

3. Problem 2.4 (Use your code to solve parts of the problem)

Lets say you are designing an optical communication device to be used in Fiber Optic Services (FIOS) and you want to

use a higher frequency light for your optical communication.

(a) If you want to increase the frequency of the light that the electrons in the well could absorb, what could you

do to the potential energy well? Explain qualitatively in a sentence or two.

When frequency of light is increased, if electrons need to absorb them, the energy gap between the energy levels should

be higher. Since energy of a specific level in the potential well is inversely proportional to the length (or width) of the

potential well, it needs to be decreased.

(b) If your well in 𝟏. 𝟎 𝒏𝒎 in length, and can be approximated to have infinite potential on the sides, and you are

using the first three energy levels for your optical system, what three wavelengths of light will you use for your

optical communications?

Transition Wavelength (𝒏𝒎)

1 to 2 (or 2 to 1) 1099.1

1 to 3 (or 3 to 1) 412.16

2 to 3 (or 3 to 2) 659.46

4. Problem 2.5

A new kind of digital electronics is being developed. In this new technology, an electron is trapped in a 2-dimensional

potential energy well and the different quantum states of the system are being used as different logic levels. Assume

the well has infinitely high barriers on all four sides. Inside the well 𝑽 = 𝟎. Also, the well has length of 𝟎. 𝟓𝒏𝒎 and

width of 𝟏. 𝟎𝒏𝒎.

(a) Solve Schrodinger equation to get the wave function of the well. (Hint, this is a 2-D problem, so use separation

of variables: 𝚿(𝒙, 𝒚) = 𝑿(𝒙)𝒀(𝒚), and then solve two directions independently.)

Outside the well, since the potential is infinite,

Ψ(𝑥, 𝑦) = 0

Inside the well, using Schrodinger equation,

−ℏ2

2𝑚(

∂2

∂𝑥2+

∂2

∂𝑦2) Ψ(𝑥, 𝑦) = 𝐸Ψ(𝑥, 𝑦)

Let, Ψ(𝑥, 𝑦) = 𝑋(𝑥)𝑌(𝑦)

−ℏ2

2𝑚(

∂2

∂𝑥2+

∂2

∂𝑦2) (X(x)Y(y)) = 𝐸(X(x)Y(y))

−ℏ2

2𝑚[𝑌(𝑦)

∂2𝑋(𝑥)

∂𝑥2+ 𝑋(𝑥)

∂2𝑌(𝑦)

∂𝑦2] = 𝐸𝑋(𝑥)𝑌(𝑦)

[−ℏ2

2𝑚⋅

1

𝑋(𝑥)

∂2𝑋(𝑥)

∂𝑥2] + [−

ℏ2

2𝑚⋅

1

𝑌(𝑦)

∂2𝑌(𝑦)

∂𝑦2] = 𝐸

Let us define two terms,

𝐸𝑥 = −ℏ2

2𝑚⋅

1

𝑋(𝑥)

∂2𝑋(𝑥)

∂𝑥2

𝐸𝑦 = −ℏ2

2𝑚⋅

1

𝑌(𝑦)

∂2𝑌(𝑦)

∂𝑦2

𝐸𝑥 + 𝐸𝑦 = 𝐸

These are again similar to 1-D infinite potential well equations. Using that solution,

𝑋(𝑥) = √2

𝐿𝑥𝑠𝑖𝑛 (

𝑛𝑥π

𝐿𝑥𝑥) ; 𝐸𝑥 =

𝑛𝑥2ℏ2π2

2𝑚𝐿𝑥2

𝑌(𝑦) = √2

𝐿𝑦𝑠𝑖𝑛 (

𝑛𝑦π

𝐿𝑦𝑦) ; 𝐸𝑦 =

𝑛𝑦2 ℏ2π2

2𝑚𝐿𝑦2

Therefore,

Ψ(𝑥, 𝑦) =2

√𝐿𝑥𝐿𝑦

⋅ 𝑠𝑖𝑛 (𝑛𝑥π

𝐿𝑥𝑥) 𝑠𝑖𝑛 (

𝑛𝑦π

𝐿𝑦𝑦)

(b) Derive an expression for the allowed energy levels.

Allowed energy levels 𝐸 = 𝐸𝑥 + 𝐸𝑦

𝐸 =𝑛𝑥

2ℏ2π2

2𝑚𝐿𝑥2 +

𝑛𝑦2ℏ2π2

2𝑚𝐿𝑦2 =

ℏ2π2

2𝑚[(

𝑛𝑥

𝐿𝑥)

2

+ (𝑛𝑦

𝐿𝑦)

2

]

(c) Calculate the energies of the first 3 levels in 𝑱𝒐𝒖𝒍𝒆𝒔 and 𝒆𝑽.

i. 𝑛𝑥 = 1, 𝑛𝑦 = 1

[(𝑛𝑥

𝐿𝑥)

2

+ (𝑛𝑦

𝐿𝑦)

2

] = [(1

0.5 × 10−9)

2

+ (1

10−9)

2

] = 5 × 1018

ii. 𝑛𝑥 = 1, 𝑛𝑦 = 2

[(𝑛𝑥

𝐿𝑥)

2

+ (𝑛𝑦

𝐿𝑦)

2

] = [(1

0.5 × 10−9)

2

+ (2

10−9)

2

] = 8 × 1018

iii. 𝑛𝑥 = 2, 𝑛𝑦 = 1

[(𝑛𝑥

𝐿𝑥)

2

+ (𝑛𝑦

𝐿𝑦)

2

] = [(2

0.5 × 10−9)

2

+ (1

10−9)

2

] = 17 × 1018

iv. 𝑛𝑥 = 2, 𝑛𝑦 = 2

[(𝑛𝑥

𝐿𝑥)

2

+ (𝑛𝑦

𝐿𝑦)

2

] = [(2

0.5 × 10−9)

2

+ (2

10−9)

2

] = 20 × 1018

v. 𝑛𝑥 = 1, 𝑛𝑦 = 3

[(𝑛𝑥

𝐿𝑥)

2

+ (𝑛𝑦

𝐿𝑦)

2

] = [(1

0.5 × 10−9)

2

+ (3

10−9)

2

] = 13 × 1018

Therefore, lowest energy levels are (1,1), (1,2) and (1,3). Corresponding energies are,

𝐸1,1 = 3.015 × 10−19𝐽 = 1.882𝑒𝑉

𝐸1,2 = 4.824 × 10−19𝐽 = 3.011𝑒𝑉

𝐸1,3 = 7.839 × 10−19𝐽 = 4.893𝑒𝑉

5. Problem 2.6

Consulting a 1-D quantum well with finite barriers of height 𝑽 on either side of the well:

(a) Write down the time independent SWE for each of the three regions of the system.

For region I:

−ℏ2

2𝑚

𝑑2ψ𝐼(𝑥)

𝑑𝑥2+ 𝑉 ⋅ ψ𝐼(𝑥) = 𝐸 ⋅ ψ𝐼(𝑥)

For region II:

−ℏ2

2𝑚

𝑑2ψ𝐼𝐼(𝑥)

𝑑𝑥2= 𝐸 ⋅ ψ𝐼𝐼(𝑥)

For region III:

−ℏ2

2𝑚

𝑑2ψ𝐼𝐼𝐼(𝑥)

𝑑𝑥2+ 𝑉 ⋅ ψ𝐼𝐼𝐼(𝑥) = 𝐸 ⋅ ψ𝐼𝐼𝐼(𝑥)

(b) Solve the SWE for each of the three regions for the bound electrons, you do not have to obtain values for the

unknown coefficients, except those that would keep the wave-function from going to infinity at 𝒙 = ±∞.

For bound electrons, 𝑉 > 𝐸,

For region II:

Since the potential distribution is symmetric around 𝑥 = 0, the wave function also should be symmetric or anti-

symmetric around that point. Therefore, we have two solutions in this region.

Symmetric solution,

ψ𝐼𝐼,𝑆(𝑥) = 𝐴𝑐𝑜𝑠(𝑘𝐼𝐼𝑥)

Anti-symmetric solution,

ψ𝐼𝐼,𝐴(𝑥) = 𝐴𝑠𝑖𝑛(𝑘𝐼𝐼𝑥)

Here,

𝑘𝐼𝐼 =2𝑚𝐸

ℏ2

For region I:

Since the solution should decay when 𝑥 goes to −∞, therefore,

ψ𝐼(𝑥) = ±𝐶𝑒𝑥𝑝(𝑘𝑥)

Here + is for symmetric solution and – is for anti-symmetric solution.

Here,

𝑘 =2𝑚(𝑉 − 𝐸)

ℏ2

For region III:

Again, the solution should decay when 𝑥 goes to ∞.

ψ𝐼𝐼𝐼(𝑥) = 𝐶𝑒𝑥𝑝(−𝑘𝑥)

(c) Sketch what you expect the wave-function to look like. Remember the continuity requirements for the wave-

function.

The wave functions for a 2nm, 5eV finite potential energy well are drawn for n=1, 2, 3. These were calculated using the

expressions in Appendix A of the text, and then using Matlab and Newton’s method to get the energy eigenvalues. Once

the energies were obtained. We are able to solve for the coefficients and plot the exact wave function. You were not

required to do this for the HW, just to sketch the functions. A key thing to keep in mind is that for the finite well, the wave

functions goes beyond the walls of the well (vertical lines). This can not happen according to classical physics since the

energy of the particle is less than the energy of the height of the well wall (5eV). Also, try to notice the units for the wave

function. For the 1-D well, the are (1/length)1/2.

(d) Apply the boundary conditions and set up the explicit set of equations that you need to solve to determine the

coefficient.

Since the potential barrier is finite, wave function should be continuous and continually differentiable at the boundary.

For symmetric solutions:

At 𝑥 =𝐿

2

ψ𝐼𝐼 (𝐿

2) = ψ𝐼𝐼𝐼 (

𝐿

2)

𝐴𝑐𝑜𝑠 (𝑘𝐼𝐼

𝐿

2) = 𝐶𝑒𝑥𝑝 (−𝑘

𝐿

2)

𝑑ψ𝐼𝐼(𝑥)

𝑑𝑥|𝑥=

𝐿2

=𝑑ψ𝐼𝐼𝐼(𝑥)

𝑑𝑥|𝑥=

𝐿2

𝐴𝑘𝐼𝐼𝑠𝑖𝑛 (𝑘𝐼𝐼

𝐿

2) = 𝐶𝑘𝑒𝑥𝑝 (−𝑘

𝐿

2)

For anti-symmetric solutions:

ψ𝐼𝐼 (𝐿

2) = ψ𝐼𝐼𝐼 (

𝐿

2)

𝐴𝑠𝑖𝑛 (𝑘𝐼𝐼

𝐿

2) = 𝐶𝑒𝑥𝑝 (−𝑘

𝐿

2)

𝑑ψ𝐼𝐼(𝑥)

𝑑𝑥|𝑥=

𝐿2

=𝑑ψ𝐼𝐼𝐼(𝑥)

𝑑𝑥|𝑥=

𝐿2

𝐴𝑘𝐼𝐼𝑐𝑜𝑠 (𝑘𝐼𝐼

𝐿

2) = −𝐶𝑘𝑒𝑥𝑝 (−𝑘

𝐿

2)

(e) What physically does it mean that the wave-function decays exponentially on either side of the quantum well.

Compare this result to the infinite well case above and explain what the difference means physically.

• If wave function does not decay exponentially on either side of the quantum well, the pdf also will go to infinity.

• Since it exponentially decays, even though it penetrates the voltage barrier, we can say that the electron (or the

quantum particle) can be found within the finite space.

• In the infinite potential well case, an electron cannot penetrate the infinite potential wall and is trapped inside

the well.

• Most importantly, in classical physics, a particle (or an object) cannot pass through a potential barrier if its energy

is lower than the barrier. However, in this case, we can find that the probability of finding electron is not zero on

the other side of the barrier.

6. Problem 2.7

A flash memory stick (also known as a travel-drive) contains billions of transistors for storing charge. Each charge-storing

transistor gives rise to a `bit', with a 1 or 0 depending on whether or not the charge is stored. The way charge is stored

is by having electrons tunnel through an oxide barrier to charge a capacitor. When there is no voltage applied to the

transistor, the barrier is too high for significant tunneling and thus charge storage does not take place. When a voltage

is applied to the transistor, the tunneling barrier is lowered enough so that tunneling and charging and thus storage of

`bits' takes place.

(a) If the rectangular barrier is 1nm thick and 𝟒. 𝟐 𝒆𝑽 high, write down the time-independent SWE for the three regions of interest. Take the potential energy on either side of the barrier to be 𝟎 𝒆𝑽.

Let the energy of the electron be 𝐸. For region I:

−ℏ2

2𝑚

𝑑2ψ𝐼(𝑥)

𝑑𝑥2+ 0 ⋅ ψ𝐼(𝑥) = 𝐸 ⋅ ψ𝐼(𝑥)

−ℏ2

2𝑚

𝑑2ψ𝐼(𝑥)

𝑑𝑥2= 𝐸ψ𝐼(𝑥)

For region II: Let, 𝑉0 = 4.2𝑒𝑉

−ℏ2

2𝑚

𝑑2ψ𝐼𝐼(𝑥)

𝑑𝑥2+ 𝑉0 ⋅ ψ𝐼𝐼(𝑥) = 𝐸 ⋅ ψ𝐼𝐼(𝑥)

−ℏ2

2𝑚

𝑑2ψ𝐼𝐼(𝑥)

𝑑𝑥2+ 𝑉0ψ𝐼𝐼(𝑥) = 𝐸ψ𝐼𝐼(𝑥)

For region III: This is again similar to the region I,

−ℏ2

2𝑚

𝑑2ψ𝐼𝐼𝐼(𝑥)

𝑑𝑥2+ 0 ⋅ ψ𝐼𝐼𝐼(𝑥) = 𝐸 ⋅ ψ𝐼𝐼𝐼(𝑥)

−ℏ2

2𝑚

𝑑2ψ𝐼𝐼𝐼(𝑥)

𝑑𝑥2= 𝐸ψ𝐼𝐼𝐼(𝑥)

(b) Solve the SWE to obtain the wave-function in each of the three regions. Take the electron energy to be 1.1eV , which is obviously lower than the barrier energy. You do not have to obtain values for the constant coefficient in front of the exponents.

For region I: From SWE,

𝑑2ψ𝐼(𝑥)

𝑑𝑥2= −𝑘𝐼

2ψ𝐼(𝑥)

Here, 𝑘𝐼 = √(2𝑚𝐸

ℏ2 )

Therefore,

ψ𝐼(𝑥) = 𝐴 ⋅ 𝑒𝑗𝑘𝐼𝑥 + 𝐵 ⋅ 𝑒−𝑗𝑘𝐼𝑥 For region II: Because 𝑉0 > 𝐸, from SWE,

𝑑2ψ𝐼𝐼(𝑥)

𝑑𝑥2= 𝑘𝐼𝐼

2 ψ𝐼𝐼(𝑥)

Here, 𝑘𝐼𝐼 = √2𝑚(𝑉0−𝐸)

ℏ2

Therefore,

ψ𝐼𝐼(𝑥) = 𝐶 ⋅ 𝑒𝑘𝐼𝐼𝑥 + 𝐷 ⋅ 𝑒−𝑘𝐼𝐼𝑥 For region III: Similar to region I,

ψ𝐼𝐼𝐼(𝑥) = 𝐹 ⋅ 𝑒𝑗𝑘𝐼𝑥 + 𝐺 ⋅ 𝑒−𝑗𝑘𝐼𝑥 However, the ψ𝐼𝐼𝐼(𝑥) represents an electron that passed through the barrier, and therefore it should move to +𝑥 direction. Therefore, we can conclude that, 𝐺 = 0.

ψ𝐼𝐼𝐼(𝑥) = 𝐹 ⋅ 𝑒𝑗𝑘𝐼𝑥

(c) Now lower the barrier to 1.2eV by applying a bias of 3.0V to the device. How has the solution to the SWE changed? (The electron energy is still 1.1eV ). Describe quantitatively, by referring to your solutions above, the difference in the barrier penetration and thus charging between cases in parts (b) and (c).

From thin barrier approximation, (textbook equation 2.86) Transmission coefficient can be found as follows:

𝑇 =|𝐹|2

|𝐴|2≈ 𝑒

−2𝐿ℏ √2𝑚(𝑉𝐵−𝐸)

For 𝑉𝐵 = 𝑉0 = 4.2𝑒𝑉

𝑇1 = 𝑒−

2×1×10−9𝑚1.055×10−34𝐽𝑠

√2×9.109×10−31𝑘𝑔×4.967×10−19𝐽= 1.473 × 10−8

For 𝑉𝐵 = 𝑉0 = 1.2𝑒𝑉

𝑇2 = 𝑒−

2×1×10−9𝑚1.055×10−34𝐽𝑠

√2×9.109×10−31𝑘𝑔×1.602×10−20𝐽= 0.0392

As we can see here, once the 𝑉𝐵 is lowered, T is increased drastically. Therefore, charging in the capacitor should be very high compared to the first case.

7. Problem 2.10

For the infinite well potential configuration shown in Figure 2.4:

(a) Confirm that the solutions given by Equation 2.42, do indeed work when substituted back into the

Schrodinger Wave Equation. Inside the potential well SWE,

−ℏ2

2𝑚

𝑑2ψ(𝑥)

𝑑𝑥2+ 𝑉 ⋅ ψ(𝑥) = 𝐸 ⋅ ψ(𝑥)

Substitute,

ψ(𝑥) = √2

𝐿𝑠𝑖𝑛 (

𝑛π

𝐿𝑥)

LHS,

𝐿𝐻𝑆 = −ℏ2

2𝑚

𝑑2 (√2𝐿 𝑠𝑖𝑛 (

𝑛π𝐿 𝑥))

𝑑𝑥2+ 0 = −

ℏ2

2𝑚⋅ (−

𝑛2π2

𝐿2√

2

𝐿𝑠𝑖𝑛 (

𝑛π

𝐿𝑥)) =

𝑛2π2ℏ2

2𝑚𝐿2ψ(𝑥)

Form the definition,

𝑘 = √2𝑚𝐸

ℏ2=

𝑛π

𝐿

∴ 𝐸 =𝑛2π2ℏ2

2𝑚𝐿2

From LHS, 𝐿𝐻𝑆 = 𝐸 ⋅ ψ(𝑥) = 𝑅𝐻𝑆

(b) Show that the different solutions are orthonormal to each other, i.e. show that Equation 2.44 equals 0 for

𝒏 ≠ 𝒎 and 1 for 𝒏 = 𝒎. By substituting the wave function in equation 2.44 of the textbook,

δ𝑛𝑚 = ∫ √2

𝐿

𝐿

0

𝑠𝑖𝑛 (𝑛π

𝐿𝑥) ⋅ √

2

𝐿𝑠𝑖𝑛 (

𝑚π

𝐿𝑥) 𝑑𝑥 =

1

𝐿∫ 2𝑠𝑖𝑛 (

𝑛π

𝐿𝑥) 𝑠𝑖𝑛 (

𝑚π

𝐿𝑥)

𝐿

0

𝑑𝑥

∴ δ𝑛𝑚 =1

𝐿[∫ 𝑐𝑜𝑠 (

(𝑛 − 𝑚)π

𝐿𝑥) 𝑑𝑥

𝐿

0

− ∫ 𝑐𝑜𝑠 ((𝑛 + 𝑚)π

𝐿𝑥) 𝑑𝑥

𝐿

0

]

When 𝑛 = 𝑚,

δ𝑛𝑚 =1

𝐿[∫ 𝑐𝑜𝑠(0)𝑑𝑥

𝐿

0

− ∫ 𝑐𝑜𝑠 (2𝑛π

𝐿𝑥) 𝑑𝑥

𝐿

0

] = 1

When 𝑛 ≠ 𝑚,

δ𝑛𝑚 =1

2[𝑠𝑖𝑛((𝑛 − 𝑚)𝑥) − 𝑠𝑖𝑛((𝑛 + 𝑚)𝑥)] = 𝑐𝑜𝑠(𝑛π) ⋅ 𝑠𝑖𝑛(−𝑚π) = 0

Therefore, solutions of the wave function are orthonormal to each other.