Infinite Potential Well

Physical Electronics, Spring 2017 23

Infinite Potential Well

3.3 Infinite Potential Well: A Confined Electron

Consider the behavior of the electron when it is confined to a certain region,0 < x < a where V = 0 and KE < V (i.e., electron cannot escape).

IΨ(x)I2 = 0 outside 0 < x < a. Therefore,Schrödinger equation in 0 < x < a:

2

2 2

2 0d m Edxψ ψ+ =

x = 0 x = a0

E1

E3

E2

E4

n = 1

n = 2

n = 3

n = 4

Ener

gyof

electr

on

Energy levels in the well ψ(x) ∝ sin(nπx/a) Probability density ∝ |ψ(x)|2

ψ1

ψ2

ψ3

ψ4

0 a a0x

0 a x0

V(x)

V = 0

Electron

V = 8 V = 8

Electron in a one-dimensional infinite PE well. The energy of theelectron is quantized. Possible wavefunctions and the probabilitydistributions for the electron are shown.

8

( ) exp( ) exp( )x A jkx B jkxψ = + −

General solution :




Using boundary conditions ψ(0) = 0, we obtain B = -A : Therefore,

( ) [exp( ) exp( )] 2 sinx A jkx jkx Aj kxψ = − − =

To obtain the energy E as a function of k, substitute the above equation into Schrodinger equation:

2 22

2

22 (sin ) (2 sin ) 02

m kAjk kx E Aj kx Em

− + = → =

Since the electron only has kinetic energy, the total energy is then

2

2x

xpE KE p km

= = → = ± The momentum px may be in the +xdirection or the – direction, so theaverage momentum is zero, pav = 0.




First define k using the boundary condition ψ(a) = 0,

( )( ) 2 sin 0 1,2,3,...a Aj ka ka n nψ π= = → = =

( ) 2 sinnn xx Aj

aπψ → =

2 2 2 2

2 2

( )2 8n

n h nEma maπ

→ = =

All ψn for n = 1, 2, 3, …constitute the eigenfunctionof the system, and identifiesthe possible states for theelectron.

We know that the total probability of finding the electron in 0 < x < a is unity.

1 12 2 2

0

1 2( ) 1 ( ) sin2

x a

nx

n xx dx A x ja a a

πψ ψ=

=

= → = → = ∫

2

2

1 8)12(

manhEEE nn+

=−=∆ +

:Quantized energy. As a→∞, electron is free.


Example 3.7 Electron confined within atomic dimension


Consider an electron in an infinite potential well of size 0.1 nm (typical size of an atom). What is the ground energy of the electron? What is the energy required to put the electron at the third energy level? How can this energy be provided?

The electron is confined in an infinite potential well, so its energy is given by

( ) ( )( )( )

2 2342 218

1 22 31 9

6.6 10 16.025 10 37.6

8 8 9.1 10 0.1 10n

J sh nE E J or eVma kg m

−−

− −

× ⋅= → = = ×

× ×

The frequency of the electron associated with this energy is 18

16 15 134

6.025 10 5.71 10 / 9.092 101.055 10

E J rad s or sJ s

ω ν−

−−

×= = = × = ×

× ⋅

The energy difference between E3 and E1 levels is ( )23 1 1 1 13 8 300.8E E E E E eV− = ⋅ − = =

This energy can be provided by a photon of exactly that energy; no less, and no more. Since the photon energy is E = hν = hc / λ,

( )( )34 8

19

6.6 10 3.0 10 /4.12

300.8 1.6 10J s m shc nm

E eVλ

−

−

× ⋅ ×= = =

× ×


3.4 Heisenberg’s Uncertainty Principle

For an electron trapped in a one-dimensional infinite PE well, so the uncertainty in the position of the electron is a. The momentum of the electron is either px=ћk in the +x direction or –ћk in the –x direction. The uncertainty Δpx=2ћk.

( )( ) ( ) 2xx p a h

aπ ∆ ∆ = =

We cannot exactly and simultaneously know both the position and momentum of a particle along a given coordinate. Heisenberg’s Uncertainty Principle : Quantum nature of the universe, not the measurement accuracy issue.

Similarly, the uncertainty ΔE in the energy E (or angular frequency ω) and the time duration Δt during which it possesses the energy, exist.

xx p∆ ∆ ≥

1By analogy,

t 1 t

xx p x k

Eω

∆ ∆ ≥ → ∆ ∆ ≥

∆ ∆ ≥ → ∆ ∆ ≥

“The simultaneous measurement of two conjugate variables (such as themomentum and position or the energy and time for a moving particle) entailsa limitation on the precision (standard deviation) of each measurement.Namely: the more precise the measurement of position, the more imprecisethe measurement of momentum, and vice versa. In the most extreme case,absolute precision of one variable would entail absolute imprecisionregarding the other.”


3.5 Tunneling Phenomenon: Quantum Leak

V(x)

I II III

Vo

x=0 x=a

E<Vo ψΙ(x)ψΙΙ(x) ψΙΙΙ(x)Incident

Reflected Transmitted

A1

A2

x

(b)

(a)

AB

C

D

E

Startherefromrest

(a) The roller coaster released from A can at most make to C, but not to E. ItsPE at A is less than the PE at D. When the car is the bottom its energy istotally KE. CD is the energy barrier which prevents the car making to E. Inquantum theory, on the other hand, there is a chance that the car could tunnel(leak) through the potential energy barrier between C and E and emerge on theother side of the hill at E .(b) The wavefunction of the electron incident on a potential energy barrier (Vo).The incident and reflected waves interfere to give yI(x). There is no reflectedwave in region III. In region II the wavefunction decays with x because E < Vo.

Finite Potential Barrier

“Note that the only way theSchrödinger equation willhave the solution ψ(x) = 0 is if the PE is infinite.Therefore, within any zero or finite PE region, there will always be a solution ψ(x) and there always will be some probability of finding the electron.”



1 2

1 2

1 2

( ) exp( ) exp( )( ) exp( ) exp( )( ) exp( ) exp( )

I

III

II

x A jkx A jkxx C jkx C jkxx B x B x

ψψψ α α

= + −= + −= + −

Solving the Schrödinger equation in each region:

22

2mEk =

① k and α are positive numbers.② exp(jkx) and exp(－jkx) are the traveling wave to the direction of +x and -x, respectively.③ In region III, there is no reflected wave, so C2 = 0.

We must now apply the boundary conditions and the normalization condition to determine the various constants A1, A2, B1, B2, and C1. Then obtain ψⅠ(x), ψⅡ(x), and ψⅢ(x).

2 22III 01

2 2 21 0I

( ) 1 where 1 sinh ( ) 4 ( )( )

x VCT DA D a E V Ex

ψαψ

= = = =+ −

Transmission coefficient :

If αa >>1, then sinh(αa) ≈ (1/2) exp(αa). For this case, the above equation becomes

002

0

16 ( ) exp( 2 ) exp( 2 )E V ET a T aV

α α−= − = −

The reflection coefficient :

222

1

1AR TA

= = −

2

02 2

22

2

2 ( ) 0

0

d m V Edx

ddx

ψ ψ

ψ α ψ

− − =

→ − =

( )02

2

2m V Eα

−=



x

V(x)

Metal

ψ(x) Second MetalVacuum

Vo

(b)x

V(x)

Metal

ψ(x)Vacuum

Vo

E < Vo

(a)

Materialsurface

Probe ScanItunnel

x

Itunnel

Image of surface (schematic sketch)(c)

(a) The wavefunction decays exponentially as we move away from the surface because the PE outside the metal is Vo and the energy of the electron, E < Vo..(b) If we bring a second metal close to the first metal, then the wavefunction can penetrate into the second metal. The electron can tunnel from the first metal to the second. (c) The principle of the Scanning Tunneling Microscope. The tunneling current depends on exp(-αa) where a is the distance of the probe from the surface of the material and α is a constant.

Scanning Tunneling Microscope (STM)



STM image of a graphite surface where contours represent electron concentrations within the surface, and carbon rings are clearly visible. Two Angstrom scan.

STM image of Ni (100) surface

STM image of Pt (111) surface



Quantum Leak Used in Nonvolatile Memory : Floating-Gate Tunneling Oxide (FLOTOX) Technology



Example 3.12 Tunneling current through metal-to-metal contactsConsider we have a thin CuO layer between two Cu wires, and suppose that CuO acts as a square potential energy barrier of 10 eV. The thickness of CuO is 5 nm, and the kinetic energy of conduction electrons in Cu is about 7 eV. What will be the transmission coefficient? Do again in case of 1 nm.

( ) ( )( )( )( )

121 31 192

0 9 122 34

2 9.1 10 10 7 1.6 10 /28.9 10

1.05 10

kg eV eV J eVm V Em

J sα

− −−

−

× − ×− = = = × × ⋅

( )( )9 1 98.9 10 5 10 44.5 1a m mα − −= × × = >>

( )( )( )

39022

0

16 7 10 716 ( ) exp( 2 ) exp( 89.0) 7.4 1010

eV eV eVE V ET aV eV

α −−−= − = − ≅ ×

With a=1 nm :( )( )

( )8

2

16 7 10 7exp( 17.8) 6.2 10

10eV eV eV

TeV

−−= − ≅ ×

Notice that reducing the layer thickness by five times increases the transmission probability by 1031! We should note that when a voltage is applied across the two wires, the potential energy height is altered (PE=charge × voltage), which results in a large increase in the transmission probability and hence results in a current.


3.6 Potential Box: Three Quantum Numbers

y

z

x

0

a

c

b

V =∞

V =∞

V =∞

V =∞V=0

Electron confined in three dimensions by a three dimensional infinite "PEbox". Everywhere inside the box, V = 0, but outside, V = ∞. The electroncannot escape from the box. What is the energy and wavefunction of theelectron?

“To examine the properties of a particle confined to a region of space, we take a three-dimensional space: 1-D 3-D (three-dimensional potential well)”

What will be the behavior of the electron in this case?

0)(222

2

2

2

2

2

=−+∂∂

+∂∂

+∂∂ ψψψψ VEm

zyx



To solve the Schrödinger equation:

1. Use the technique known as a variable separation:

2. Then substitute this in 3-D Schrödinger equation to obtain three different differential equations, and derive the each solution separately.

3. Finally, the total wave function is then

( ) ( ) ( ) ( )x y zx, y, z x y zψ ψ ψ ψ=

( ) ( ) ( )( ) sin sin sinx y zx, y, z A k x k y k zψ =

4. Apply the boundary conditions at x=a, y=b, and z=c.

31 21 2 3, , where , , are quantum numbers.x y z

nn nk k k n n na b c

ππ π= = =

5. The eigenfunction of the electron is

=

czn

byn

axnAz y, x,nnn

πππψ 321 sinsinsin)(321



6. Derive “A” using the normalization condition :

1 2 3

322 2( ) 1n n n x, y, z dxdydz A

aψ = → =

∫∫∫7. The energy of the electron is then calculated by substituting eigenfunction into Schrödinger equation :

( )2

2 2 2( , , )2x y z x y zE E k k k k k k

m= = + +

1 2 3

22 2231 2

2 2 28n n nnn nhE

m a b c

→ = + +

In three dimensions, we have three quantum numbers, each one arising from boundary conditions along one of the coordinates.The lowest energy for the electron is equal to E111, not zero. E211, E121, E112 have the same energy, so the number of states that have the same energy is termed the degeneracy of that energy level. The second energy level E211 is thus three-fold degenerate.

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Infinite Potential Well