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Quantum Mechanics

Luis A. Anchordoqui

Department of Physics and Astronomy

Lehman College, City University of New York

Lesson V

March 5, 2019

Table of Contents

1 Scattering in one dimension

Step potential

Potential barrier and tunneling

The ins and outs of tunneling

Scattering in one dimension Step potential Quantum Intuition V(x) = { 0 for x < 0 V

for x ≥ 0 (1) quation ψ(x ) Energy E 0 x 0 I II I II V(x ) =V (^0) V(x ) = 0 x (a) (b) 6-22 ( a ) A potential step. Particles are incident on the step from the left toward the ach with total energy E! V 0. ( b ) The wavelength of the incident wave (Region I) is than that of the transmitted wave (Region II). Since k 2 " k 1 , however, smission coefficient T " 1. ƒCƒ

ƒAƒ

254 Chapter 6 The Schrödinger Equation

ψ(x ) Energy E 0 x 0 V(x ) =V 0 V(x ) = 0 x (a) (b)

Figure 6-24 ( a ) A potential

step. Particles are incident on the step from the left moving toward the right, each with total energy E! V 0

. ( b ) The wave

transmitted into region II is a decreasing exponential. However, the value of R in this case is 1 and no net energy is transmitted.

Now let us consider the case shown in Figure 6-24 a , where E! V

0

. Clas

we expect all particles to be reflected at x " 0; however, we note that k

2

in E

Scattering in one dimension Step potential

A Simple

Potential Step

Region 1 Region 2

CASE I : E

o

> V

In Region 1:

In Region 2:

ψ A = Ae

−jk 1 x

ψ C = Ce

−jk 1 x

ψ B = Be

−jk 1 x

E = E

o

E = 0

x

x = 0

ℏ E (^) o ψ = − 2 m

∂

ψ

∂x

ℏ (E o − V ) ψ = −

∂

ψ

2 m

2 k

mEo

1

∂x

ℏ

k

2 m (Eo

2

− V )

ℏ

V

10

Scattering in one dimension Step potential

A Simple

Potential Step

CASE I : E

o

> V

B

A

=

1 − k 2 /k 1

1 + k 2 /k (^1)

=

k 1 − k 2

k 1 + k (^2)

C

A

=

2

1 + k 2 /k 1

=

2 k 1

k 1 + k (^2)

A + B = C

A − B =

k (^2)

k (^1)

C

Region 1 Region 2

A

= Ae

−jk 1 x

ψ B = Be

−jk 1 x

C

= Ce

−jk 1 x

E = E

o

E = 0

x = 0

x

V

12

Scattering in one dimension Step potential

Quantum Electron Currents

Given an electron of mass

that is located in space with charge density

and moving with momentum corresponding to

… then the current density for a single electron is given by

m

ρ = q |ψ(x)|

< p > < v >^ =^ ℏk/m

J = ρv = q |ψ|

(ℏk/m)

Scattering in one dimension Step potential

A Simple

Potential Step

CASE I : E

o

> V

T

R

T + R = 1

B Reflection = R =

∣ ∣ ∣ ∣

A

∣ ∣ ∣ =

∣ ∣ k 1 − k (^2)

∣

∣ ∣

∣ 2

k 1 + k (^2)

∣ ∣

Transmission = T = 1

∣

− R

4 k 1 k (^2)

|k 1 + k 2 |

k (^2)

k (^1)

V

E (^) o = V Eo = ∞ E (^) o

Region 1 Region 2

ψ A = Ae

−jk 1 x

ψ C = Ce

−jk 1 x

ψ B = Be

−jk 1 x

E = Eo

E = 0

x = 0

x

V

16

Scattering in one dimension Step potential

A Simple

Potential Step

CASE II : E

o

< V

In Region 1:

In Region 2:

ℏ E (^) o ψ = −

∂

ψ

2 m ∂x

ℏ (E (^) o − V ) ψ = −

∂

ψ

2 m

2 k

mEo

1

∂x

ℏ

κ

2 m (Eo

− V )

ℏ

ψ C = Ce

−κx

Region 1 Region 2

ψ A = Ae

−jk 1 x

ψ B = Be

−jk 1 x

E = Eo

E = 0

x = 0

x

V

20

Scattering in one dimension Step potential

A Simple

Potential Step

CASE II : E

o

< V

B

Total reflection  Transmission must be zero

1 + jκ/k (^1)

A

C

1 − jκ/k (^1)

2

A 1 − jκ/k 1

R =

∣ ∣ B ∣ ∣ A

∣ ∣ ∣ ∣

= 1 T^ = 0

A + B = C

A − B = −j

κ

ψ C = Ce

−κx

C

k (^1)

Region 1 Region 2

ψ A = Ae

−jk 1 x

ψ B = Be

−jk 1 x

E = Eo

E = 0

x = 0

x

V

22

Scattering in one dimension Step potential

KEY TAKEAWAYS

Region 1 Region 2

CASE II : Eo < V

A Simple Potential Step

CASE I : E

o

> V

Region 1 Region 2

B

Ref lection = R =

A

2

=

k 1 − k (^2)

2

k 1 + k (^2)

4 k

T ransmission = T = 1 −

1 k^2 R = |k 1 + k 2 |

2

PARTIAL REFLECTION

TOTAL REFLECTION

24

Scattering in one dimension Potential barrier and tunneling

CASE II : E

o

< V

Region 1 (^) Region 2 Region 3

In Regions 1 and 3:

In Region 2:

A Rectangular

Potential Step

for E

o

< V :

κx

ψ A = Ae

−jk 1 x

ψ C = Ce

B

= Be

jk 1 x

ψ F = F e

−jk 1 x

ψ D = De

κx

V

E = Eo

E = 0

−a 0 a

ℏ E (^) o ψ = − 2 m

∂

ψ

∂x

(E (^) o − V )ψ = −

ℏ

2 m

∂

ψ

2 k

mEo

1

∂x

ℏ

κ

=

2 m(V − E (^) o )

ℏ

T =

∣ ∣ ∣ ∣

F

A

∣ ∣ ∣ ∣

=

1

1 +

V 2

E (^) o (V −E (^) o )

sinh

(2κa)

25

Scattering in one dimension Potential barrier and tunneling

A Rectangular

Potential Step

for E

o

< V :

T =

∣ ∣ F ∣ ∣ A

∣ ∣ ∣ ∣

=

1

1 +

V

2

E (^) o (V −E (^) o )

sinh

(2κa)

T =

∣ ∣ ∣ ∣

F

A

∣ ∣ ∣ ∣

≈

1

1 +

V

2

E (^) o (V −E (^) o )

e

− 4 κa

sinh

(2κa) =

[

e

2 κa − e

− 2 κa

] 2 ≈ e

− 4 κa

E U

26

Scattering in one dimension Potential barrier and tunneling

Multiple Choice Questions

Consider a particle tunneling through a barrier:

1. Which of the following will increase the

likelihood of tunneling?

a. decrease the height of the barrier

b. decrease the width of the barrier

c. decrease the mass of the particle

2. What is the energy of the particles that have successfully
escaped?

a. < initial energy

b. = initial energy

c. > initial energy

0 L

V

x

E

o

Although the amplitude of the wave is smaller after the barrier, no

energy is lost in the tunneling process

31

encountering the barrier. Scattering in one dimension The ins and outs of tunneling

Schrodinger Equations

Key to solving for the wave function of a particle hitting a potential barrier is finding the Schrodinger equations which

describe the system. First, define the energy potential, , of the system as this:

Writing the wave function of the particle as for , for , and for , the

Schrodinger equations for , , and are respectively:

This can be simplified, considering the wavenumbers, and , of the wave function for inside and outside the barrier

respectively. Since and , this can be said of the wave function of a particle

with.

V ( x )

V ( x ) =

V 0 ,

x < 0

0 < x < a

x > a

ψ 1 ( x ) x < 0 ψ 2 ( x ) 0 < x < a ψ 3 ( x ) x > a

x < 0 0 < x < a x > a

E ψ 1 ( x )

E ψ ( x )

2

E ψ 3 ( x )

= − ( x )

2

2 m

d

2

dx

2

= − ( x ) + ( x )

2

2 m

d

2

dx

2

ψ 2 V 0 ψ 2

= − ( x )

2

2 m

d

2

dx

2

k 1 k 2

k^ =^2 mE /

1

2

2

k =^2 m ( E^ −^ )/

2

2

V 0 ℏ

2

E ≥ V

L. A. Anchordoqui (CUNY) Quantum Mechanics 3-5-2019 20 / 35