Lehman CollegeScattering in one dimension ave Equation for Electrons ! E " = ! #" = ! j ! (free-particle) (free-particle) ave Equation ! !" p x " = ! k" = j !! t "! x p2 E = 2m!j

Quantum Mechanics

Luis A. Anchordoqui

Department of Physics and AstronomyLehman College, City University of New York

Lesson VMarch 5, 2019

L. A. Anchordoqui (CUNY) Quantum Mechanics 3-5-2019 1 / 35

Table of Contents

1 Scattering in one dimensionStep potentialPotential barrier and tunnelingThe ins and outs of tunneling


Scattering in one dimension

Schrodinger: A Wave Equation for Electrons

∂Eψ = !ωψ = −j!

(free-particle)

(free-particle)

..The Free-Particle Schrodinger Wave Equation !

∂!xψ =ψ p kψ = j!

∂tψ

∂x

p2

E =2m

−j!∂

∂tψ = − !2

2m

∂2ψ

∂x2

Erwin Schrödinger (1887–1961)

Image in the Public Domain

�

Schrodinger Equation and Energy Conservation

The Schrodinger Wave Equation

The quantity | |2 dx is interpreted as the probability that the particle can be

found at a particular point x (within interval dx)

0 x L n=3

P (x) = |ψ|2dx

2!Eψ(x) = −

|ψ|2

2m

∂2ψ(x)+ V (x)ψ(x)

∂x2

© Dr. Akira Tonomura, Hitachi, Ltd., Japan. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.

�

¹





0 x L n=3

P (x) = |ψ|2dx

2!Eψ(x) = −

|ψ|2

2m

∂2ψ(x)+ V (x)ψ(x)

∂x2


�

¹





0 x L n=3

P (x) = |ψ|2dx

2!Eψ(x) = −

|ψ|2

2m

∂2ψ(x)+ V (x)ψ(x)

∂x2


�

¹

LAST CLASS WE SAW

Notation ) i = j


Scattering in one dimension Step potential

Quantum Intuition

V(x) ={

0 for x < 0V0 for x ≥ 0

(1)252 Chapter 6 The Schrödinger Equation

ψ(x )

Energy

E

0 x

0I II

I II

V(x ) = V0

V(x) = 0

x

(a)

(b)

Figure 6-22 (a) A potential step. Particles are incident on the step from the left toward theright, each with total energy E ! V0 . (b) The wavelength of the incident wave (Region I) isshorter than that of the transmitted wave (Region II). Since k2 " k1 , however,the transmission coefficient T " 1.

ƒC ƒ 2 ! ƒA ƒ 2;

The general solutions are

Region I

6-63

Region II

6-64

Specializing these solutions to our situation where we are assuming the incident beamof particles to be moving from left to right, we see that the first term in Equation 6-63represents that beam since multiplying by the time part of yields aplane wave (i.e., a beam of free particles) moving to the right. The second term,

represents particles moving to the left in Region I. In Equation 6-64, D # 0since that term represents particles incident on the potential step from the right andthere are none. Thus, we have that the constant A is known or at least obtainable (de-termined by normalization of in terms of the density of particles in the beam asexplained above) and the constants B and C are yet to be found. We find them by ap-plying the continuity condition on and at x # 0, i.e., by requiring that

and . Continuity of at x # 0 yields

or

6-65a

Continuity of at x # 0 gives

6-65bk1A $ k1B # k2C

d%>dx A & B # C

%I(0) # A & B # %II(0) # C

%d%I(0)>dx # d%II(0)>dx%I(0) # %II(0)d%(x)>dx%(x)

Aeik1x

Be$ik2x,

'(x, t), ei(t,Aeik1x

(x ! 0) %II(x) # Ceik2x & De$ik2x

(x " 0) %I(x) # Aeik1x & Be$ik1x

254 Chapter 6 The Schrödinger Equation

ψ(x )

Energy

E

0 x

0

V(x) = V0

V(x) = 0

x

(a)

(b)

Figure 6-24 (a) A potentialstep. Particles are incidenton the step from the leftmoving toward the right,each with total energyE ! V0 . (b) The wavetransmitted into region II isa decreasing exponential.However, the value of R inthis case is 1 and no netenergy is transmitted.

1.0

0.6

0.8

0.4

0.2

00 321

Top of step

5

R, T

4

R

T

E/V0

Figure 6-25 Reflectioncoefficient R and transmissioncoefficient T for a potentialstep V0 high versus energy E(in units of V0).

Now let us consider the case shown in Figure 6-24a, where E ! V0 . Classically,we expect all particles to be reflected at x " 0; however, we note that k2 in Equation6-64 is now an imaginary number since E ! V0 . Thus,

6-71

is a real exponential function where (We choose the positiveroot so that as ) This means that the numerator and denominator ofthe right side of Equation 6-66 are complex conjugates of each other; hence

and R " 1 and T " 0. Figure 6-25 is a graph of both R and T versus en-ergy for a potential step. In agreement with the classical prediction, all of the particles(waves) are reflected back into Region I. However, another interesting result of our so-lution of Schrödinger’s equation is that the particle waves do not all reflect at x " 0.

ƒB ƒ 2 " ƒA ƒ 2

xS # .$II S 0% " 22m(V0 & E)>U.

$II(x) " Ceik2x " Ce&%x



A Simple Potential Step

Region 1 Region 2

CASE I : Eo > V

In Region 1: In Region 2:

ψA = Ae−jk1x ψC = Ce−jk1x

ψB = Be−jk1x

E = Eo

E = 0x

x = 0

2!Eoψ = −

2m

∂2ψ

∂x2

2!(Eo − V ) ψ = − ∂2ψ

2m

2k2 mEo1 =

∂x2

!2

k2 2m (Eo2 =

− V )

!2

V

10




CASE I : Eo > V

is continuous: is continuous:

ψ1 = Ae−jk1x + Bejk1x ψ2 = Ce−jk2x

ψ1(0) = ψ2(0) A + B = C

∂

∂xψ(0) =

∂

∂xψ2(0) A − B =

k2

ψ

∂ψC

k1∂x

Region 1 Region 2


ψB = Be−jk1x

E = Eo

E = 0

x = 0x

V

11




CASE I : Eo > V

B

A=

1 − k2/k1

1 + k2/k1

=k1 − k2

k1 + k2

C

A=

2

1 + k2/k1

=2k1

k1 + k2

A + B = C

A − B =k2

k1C

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

ψC = Ce−jk1x

E = Eo

E = 0

x = 0x

V

12



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)|2

< p > < v > = !k/m

J = ρv = q |ψ|2 (!k/m)

14




CASE I : Eo > V

JtransmittedTransmission = T =

Jincident=

JC

JA=

|ψC |2(!k2/m)

|ψA|2(!k1/m)=

!!!!C

A

!!!!2

k2

JreflectedReflection = R =

k1

JB=

Jincident JA=

|ψB |2(!k1/m)

|ψA|2(!k1/m)=

!!!!B

A

!!!!2

B 1=

− k2/k1

A

C

1 + k2/k1

2=

A 1 + k2/k1

Region 1 Region 2


ψB = Be−jk1x

E = Eo

E = 0

x = 0x

V

15




CASE I : Eo > V

1

T

R

T + R = 1

1

BReflection = R =

!!!!2

A

!!! =

!!k1 − k2

! !!!2

k1 + k2

!!

Transmission = T = 1

!

− R4k1k2

=|k1 + k2|2

k2

k1=

"1 − V

EoEo = V Eo = ∞

Region 1 Region 2


ψB = Be−jk1x

E = Eo

E = 0

x = 0x

V

16




CASE II : Eo < V

In Region 1: In Region 2:

2!Eoψ = − ∂2ψ

2m ∂x2

2!(Eo − V ) ψ = − ∂2ψ

2m

2k2 mEo1 =

∂x2

!2

κ2 2m (Eo=

− V )

!2

ψC = Ce−κx

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

E = Eo

E = 0

x = 0x

V

20



A Simple

ψ2 = Ce−κx

Potential Step

is continuous: is continuous:

CASE II : Eo < V

ψ1 = Ae−jk1x + Bejk1x

ψ1(0) = ψ2(0) A + B = C

∂

∂xψ(0) =

∂

∂xψ2(0)

ψ

∂ψ κA

∂x− B = −j

ψC = Ce−κx

Ck1

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

E = Eo

E = 0

x = 0x

V

21




CASE II : Eo < V

B

Total reflection � Transmission must be zero

1 + jκ/k1=

A

C

1 − jκ/k1

2=

A 1 − jκ/k1

R =

!!B!!A

!!!!2

= 1 T = 0

A + B = C

A − B = −jκ

ψC = Ce−κx

Ck1

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

E = Eo

E = 0

x = 0x

V

22



KEY TAKEAWAYS

Region 1 Region 2

CASE II : Eo < V


CASE I : Eo > V

Region 1 Region 2

BReflection = R =

!!!!A

!!!!2

=

!!!!k1 − k2

2

k1 + k2

!!!

4 k

!

Transmission = T = 1 − 1k2R =

|k1 + k2|2

PARTIAL REFLECTION

TOTAL REFLECTION

24


Scattering in one dimension Potential barrier and tunneling

Quantum Tunneling Through a Thin Potential Barrier

Total Reflection at Boundary

Frustrated Total Reflection (Tunneling)

R = 1 T = 0

T = 0

23



CASE II : Eo < V

Region 1 Region 2 Region 3

In Regions 1 and 3: In Region 2:

A Rectangular Potential Step

for Eo < V :

κxψA = Ae−jk1x ψC = Ce−

ψB = Bejk1x

ψF = Fe−jk1x

ψD = Deκx

VE = Eo

E = 0 −a a0

2!Eoψ = −

2m

∂2ψ

∂x2

(Eo − V )ψ = − !2

2m

∂2ψ

2k2 mEo1 =

∂x2

!2

κ2 =2m(V − Eo)

!2

T =

!!!!F

A

!!!!2

=1

1 + 14

V 2

Eo(V −Eo) sinh2(2κa)

25



A Rectangular Potential Step

for Eo < V :

T =

!!F!!A

!!!!2

=1

1 + 14

V 2

Eo(V −Eo) sinh2(2κa)

T =

!!!!F

A

!!!!2

≈ 1

1 + 14

V 2

Eo(V −Eo)

e−4κa

sinh2(2κa) ="e2κa − e−2κa

#2 ≈ e−4κa

UE

26



0 L

V0

x

Eo

metal metal

air gap

L = 2a

Example: Barrier Tunneling

Question: What will T be if we double the width of the gap?

•  Let�s consider a tunneling problem:

An electron with a total energy of Eo= 6 eV approaches a potential barrier with a height of V0 = 12 eV. If the width of the barrier is L = 0.18 nm, what is the probability that the electron will tunnel through the barrier?

T =

!!!!F

A

!!!!2

≈ 16Eo(V − Eo)

V 2e−2κL

κ =

"2me

!2(V − Eo) = 2π

"2me

h2(V − Eo) = 2π

"6eV

1.505eV-nm2 ≈ 12.6 nm−1

T = 4e−2(12.6 nm−1)(0.18 nm) = 4(0.011) = 4.4%

30



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

Eo

Although the amplitude of the wave is smaller after the barrier, no energy is lost in the tunneling process

31


Scattering in one dimension The ins and outs of tunneling

4/2/17, 7:27 PMRectangular Potential Barrier - Tedious Derivations

Page 1 of 7http://tediousderivations.blogspot.com/2013/08/rectangular-potential-barrier.html

Saturday, August 31, 2013

Rectangular Potential Barrier

Rectangular potential barriers, also called square potential barriers, are formed by energy potentials which create wall-like barricades for particles. Essentially, a potential barrier is a potential step except the energy potential returns tozero at some finite positive -position, , and remains zero beyond that point. Here, we'll derive the wave function of aparticle facing a rectangular potential barrier, then find the transmission and reflection coe!icients of the particle uponencountering the barrier.

Schrodinger Equations

Key to solving for the wave function of a particle hitting a potential barrier is finding the Schrodinger equations whichdescribe the system. First, define the energy potential, , of the system as this:

Writing the wave function of the particle as for , for , and for , theSchrodinger equations for , , and are respectively:

This can be simplified, considering the wavenumbers, and , of the wave function for inside and outside the barrierrespectively. Since and , this can be said of the wave function of a particlewith .

Notice, however that if , is imaginary and thus no longer an observable. By convention therefore, , definedby , is used instead for . The di!erential equations defining the wave function of aparticle with insu!icient energy are thus:

Author

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x a

V (x)

V (x) =⎧⎩⎨

0 ,,V0

0 ,

x < 00 < x < a

x > a

(1)

(x)ψ1 x < 0 (x)ψ2 0 < x < a (x)ψ3 x > ax < 0 0 < x < a x > a

E (x)ψ1

E (x)ψ2

E (x)ψ3

= − (x)ℏ2

2m

d2

dx2ψ1

= − (x) + (x)ℏ2

2m

d2

dx2ψ2 V0 ψ2

= − (x)ℏ2

2m

d2

dx2ψ3

(2)

(3)

(4)

k1 k2

= 2mE/k12 ℏ2 = 2m(E − )/k2

2 V0 ℏ2

E ≥ V0

0

0

0

= (x) + (x)d2

dx2ψ1 k1

2ψ1

= (x) + (x)d2

dx2ψ2 k2

2ψ2

= (x) + (x)d2

dx2ψ3 k1

2ψ3

(5)

(6)

(7)

E < V0 k2 κ= 2m( − E)/κ2 V0 ℏ2 E < V0





Saturday, August 31, 2013

Rectangular Potential Barrier

Rectangular potential barriers, also called square potential barriers, are formed by energy potentials which create wall-like barricades for particles. Essentially, a potential barrier is a potential step except the energy potential returns tozero at some finite positive -position, , and remains zero beyond that point. Here, we'll derive the wave function of aparticle facing a rectangular potential barrier, then find the transmission and reflection coe!icients of the particle uponencountering the barrier.

Schrodinger Equations

Key to solving for the wave function of a particle hitting a potential barrier is finding the Schrodinger equations whichdescribe the system. First, define the energy potential, , of the system as this:

Writing the wave function of the particle as for , for , and for , theSchrodinger equations for , , and are respectively:

This can be simplified, considering the wavenumbers, and , of the wave function for inside and outside the barrierrespectively. Since and , this can be said of the wave function of a particlewith .

Notice, however that if , is imaginary and thus no longer an observable. By convention therefore, , definedby , is used instead for . The di!erential equations defining the wave function of aparticle with insu!icient energy are thus:

Author

Vincent Chen is the guy responsible for the weird andwonderful things you see on this blog. He enjoysstudying math and learning neat stu!, then bloggingabout it in ad hoc posts.[...] Read More

Archive

▼ 2013 (18)► October (1)► September (5)▼ August (4)

Rectangular Potential BarrierHyperbolic FunctionsPotential StepProbability Current

► July (8)

Subscribe

Atom RSS

Tedious DerivationsMathematics, Quantum Mechanics, & Various Other Quirks

HOME BLOG AUTHOR search this site

x a

V (x)

V (x) =⎧⎩⎨

0 ,,V0

0 ,

x < 00 < x < a

x > a

(1)

(x)ψ1 x < 0 (x)ψ2 0 < x < a (x)ψ3 x > ax < 0 0 < x < a x > a

E (x)ψ1

E (x)ψ2

E (x)ψ3

= − (x)ℏ2

2m

d2

dx2ψ1

= − (x) + (x)ℏ2

2m

d2

dx2ψ2 V0 ψ2

= − (x)ℏ2

2m

d2

dx2ψ3

(2)

(3)

(4)

k1 k2

= 2mE/k12 ℏ2 = 2m(E − )/k2

2 V0 ℏ2

E ≥ V0

0

0

0

= (x) + (x)d2

dx2ψ1 k1

2ψ1

= (x) + (x)d2

dx2ψ2 k2

2ψ2

= (x) + (x)d2

dx2ψ3 k1

2ψ3

(5)

(6)

(7)

E < V0 k2 κ= 2m( − E)/κ2 V0 ℏ2 E < V0


Scattering in one dimension The ins and outs of tunneling4/2/17, 7:27 PMRectangular Potential Barrier - Tedious Derivations


If There Is Su!icient Energy

For , to find the wave function of the particle, equations , , and must be solved. These arehomogeneous second-order linear di!erential equations and have the following general solutions:

where , , , , , and are constants and and are the two solutions to the equation while and are the two solutions to the equation .

Notice, considering Euler's formula, that , , and represent waves travelling in the positivedirection while , , and represent waves travelling in the negative direction. Since reflection bythe barrier is conceivable, it is possible to have wave components travelling in the negative direction for , butthere is no reason to have waves doing so for . Thus, .

To solve for and in relation to , impose these four boundary conditions to ensure that the wave function is asmooth curve as and as :

0

0

0

= (x) + (x)d2

dx2ψ1 k1

2ψ1

= (x) − (x)d2

dx2ψ2 κ2 ψ2

= (x) + (x)d2

dx2ψ3 k1

2ψ3

(8)

(9)

(10)

E ≥ V0 (5) (6) (7)

(x)ψ1

(x)ψ2

(x)ψ3

= A + Be xrA e xrB

= C + De xrC e xrD

= F + Ge xrF e xrG

(11)(12)(13)

A B C D F G =rA rF =rB rG

+ = 0r2 k12 rC rD + = 0r2 k2

2

(x)ψ1

(x)ψ2

(x)ψ3

= A + Bei xk1 e−i xk1

= C + Dei xk2 e−i xk2

= F + Gei xk1 e−i xk1

(14)(15)(16)

Aei xk1 Cei xk2 F ei xk1

Be−i xk1 De−i xk2 Ge−i xk1

x < ax > a G = 0

(x)ψ1

(x)ψ2

(x)ψ3



= F ei xk1

(17)(18)(19)

B F Ax → 0 x → a

(x)limx→0−

ψ1

(x)limx→0−

d

dxψ1

(x)limx→a−

ψ2

(x)limx→a−

d

dxψ2

= (x)limx→0+

ψ2

= (x)limx→0+

d

dxψ2

= (x)limx→a+

ψ3

= (x)limx→a+

d

dxψ3

A + B

i A − i Bk1 k1

C + Dei ak2 e−i ak2

i C − i Dk2 ei ak2 k2 e−i ak2

= C + D

= i C − i Dk2 k2

= F ei ak1

= i Fk1 ei ak1

(20)(21)(22)(23)

A + Bk1 k1

A − Bk1 k1

C + Dk2 ei ak2 k2 e−i ak2

C − Dk2 ei ak2 k2 e−i ak2

= C + Dk1 k1

= C − Dk2 k2

= Fk2 ei ak1

= Fk1 ei ak1

(24)(25)(26)(27)










0

0

0

= (x) + (x)d2

dx2ψ1 k1

2ψ1

= (x) − (x)d2

dx2ψ2 κ2 ψ2

= (x) + (x)d2

dx2ψ3 k1

2ψ3

(8)

(9)

(10)

E ≥ V0 (5) (6) (7)

(x)ψ1

(x)ψ2

(x)ψ3

= A + Be xrA e xrB

= C + De xrC e xrD

= F + Ge xrF e xrG

(11)(12)(13)


+ = 0r2 k12 rC rD + = 0r2 k2

2

(x)ψ1

(x)ψ2

(x)ψ3




(14)(15)(16)



x < ax > a G = 0

(x)ψ1

(x)ψ2

(x)ψ3



= F ei xk1

(17)(18)(19)

B F Ax → 0 x → a

(x)limx→0−

ψ1

(x)limx→0−

d

dxψ1

(x)limx→a−

ψ2

(x)limx→a−

d

dxψ2

= (x)limx→0+

ψ2

= (x)limx→0+

d

dxψ2

= (x)limx→a+

ψ3

= (x)limx→a+

d

dxψ3

A + B

i A − i Bk1 k1



= C + D

= i C − i Dk2 k2

= F ei ak1

= i Fk1 ei ak1

(20)(21)(22)(23)

A + Bk1 k1

A − Bk1 k1



= C + Dk1 k1

= C − Dk2 k2

= Fk2 ei ak1

= Fk1 ei ak1

(24)(25)(26)(27)










0

0

0

= (x) + (x)d2

dx2ψ1 k1

2ψ1

= (x) − (x)d2

dx2ψ2 κ2 ψ2

= (x) + (x)d2

dx2ψ3 k1

2ψ3

(8)

(9)

(10)

E ≥ V0 (5) (6) (7)

(x)ψ1

(x)ψ2

(x)ψ3

= A + Be xrA e xrB

= C + De xrC e xrD

= F + Ge xrF e xrG

(11)(12)(13)


+ = 0r2 k12 rC rD + = 0r2 k2

2

(x)ψ1

(x)ψ2

(x)ψ3




(14)(15)(16)



x < ax > a G = 0

(x)ψ1

(x)ψ2

(x)ψ3



= F ei xk1

(17)(18)(19)

B F Ax → 0 x → a

(x)limx→0−

ψ1

(x)limx→0−

d

dxψ1

(x)limx→a−

ψ2

(x)limx→a−

d

dxψ2

= (x)limx→0+

ψ2

= (x)limx→0+

d

dxψ2

= (x)limx→a+

ψ3

= (x)limx→a+

d

dxψ3

A + B

i A − i Bk1 k1



= C + D

= i C − i Dk2 k2

= F ei ak1

= i Fk1 ei ak1

(20)(21)(22)(23)

A + Bk1 k1

A − Bk1 k1



= C + Dk1 k1

= C − Dk2 k2

= Fk2 ei ak1

= Fk1 ei ak1

(24)(25)(26)(27)





To solve for in relation to , consider equations , , and .

Using Euler's formula to expand and , the following can be derived:



Comparing equations and , is solved for in relation to .

Considering equations and alongside equation , and can also be solved for in relation to , butsince only , , and are needed to calculate the reflection and transmission coe!icients, the derivations of and

are omitted here. In order to find the reflection and transmission coe!icients, the wave function must be first writtenin terms of its incident, reflected, and transmitted components, , , and respectively.

The reflection and transmission coe!icients, and respectively, are defined as follows:

2 Ak1

2 Bk1

2 Ck2 ei ak2

2 Dk2 e−i ak2

= ( + )C + ( − )Dk1 k2 k1 k2

= ( − )C + ( + )Dk1 k2 k1 k2

= ( + )Fk1 k2 ei ak1

= ( − )Fk2 k1 ei ak1

(28)(29)(30)(31)

F A (28) (30) (31)

2 A = F − Fk1( +k1 k2)2

2k2ei( − )ak1 k2

( −k1 k2)2

2k2ei( + )ak1 k2 (32)

4 A = ( + F − ( − Fk1k2e−i ak1 k1 k2)2 e−i ak2 k1 k2)2 ei ak2 (33)

e−i ak2 ei ak2

4 A = (−2i sin a + 4 cos a − 2i sin a) Fk1k2e−i ak1 k12 k2 k1k2 k2 k2

2 k2 (34)

F = 2 Ak1k2e−i ak1

2 cos a − i ( + ) sin ak1k2 k2 k12 k2

2 k2

(35)

B A (29) (30) (31)

2 B = F − Fk1−k1

2 k22

2k2ei( − )ak1 k2

−k12 k2

2

2k2ei( + )ak1 k2 (36)

= ( − ) F4 Bk1k2e−i ak1

−k12 k2

2e−i ak2 ei ak2 (37)

e−i ak2 ei ak2

= F2 Bk1k2e−i ak1

−i ( − ) sin ak12 k2

2 k2

(38)

(35) (38) B A

B =−i ( − ) sin( a)Ak1

2 k22 k2

2 cos a − i ( + ) sin ak1k2 k2 k12 k2

2 k2

(39)

(30) (31) (35) C D AA B F C

D(x)ψi (x)ψr (x)ψt

(x)ψi

(x)ψr

(x)ψt

= Aei xk1

= Be−i xk1

= F ei xk1

(40)(41)(42)

R T













2 Ak1

2 Bk1

2 Ck2 ei ak2

2 Dk2 e−i ak2

= ( + )C + ( − )Dk1 k2 k1 k2

= ( − )C + ( + )Dk1 k2 k1 k2

= ( + )Fk1 k2 ei ak1

= ( − )Fk2 k1 ei ak1

(28)(29)(30)(31)

F A (28) (30) (31)

2 A = F − Fk1( +k1 k2)2

2k2ei( − )ak1 k2

( −k1 k2)2

2k2ei( + )ak1 k2 (32)


e−i ak2 ei ak2


2 k2 (34)


2 cos a − i ( + ) sin ak1k2 k2 k12 k2

2 k2

(35)

B A (29) (30) (31)

2 B = F − Fk1−k1

2 k22

2k2ei( − )ak1 k2

−k12 k2

2

2k2ei( + )ak1 k2 (36)

= ( − ) F4 Bk1k2e−i ak1

−k12 k2


e−i ak2 ei ak2

= F2 Bk1k2e−i ak1

−i ( − ) sin ak12 k2

2 k2

(38)

(35) (38) B A

B =−i ( − ) sin( a)Ak1

2 k22 k2

2 cos a − i ( + ) sin ak1k2 k2 k12 k2

2 k2

(39)

(30) (31) (35) C D AA B F C


(x)ψi

(x)ψr

(x)ψt

= Aei xk1

= Be−i xk1

= F ei xk1

(40)(41)(42)

R T













2 Ak1

2 Bk1

2 Ck2 ei ak2

2 Dk2 e−i ak2

= ( + )C + ( − )Dk1 k2 k1 k2

= ( − )C + ( + )Dk1 k2 k1 k2

= ( + )Fk1 k2 ei ak1

= ( − )Fk2 k1 ei ak1

(28)(29)(30)(31)

F A (28) (30) (31)

2 A = F − Fk1( +k1 k2)2

2k2ei( − )ak1 k2

( −k1 k2)2

2k2ei( + )ak1 k2 (32)


e−i ak2 ei ak2


2 k2 (34)


2 cos a − i ( + ) sin ak1k2 k2 k12 k2

2 k2

(35)

B A (29) (30) (31)

2 B = F − Fk1−k1

2 k22

2k2ei( − )ak1 k2

−k12 k2

2

2k2ei( + )ak1 k2 (36)

= ( − ) F4 Bk1k2e−i ak1

−k12 k2


e−i ak2 ei ak2

= F2 Bk1k2e−i ak1

−i ( − ) sin ak12 k2

2 k2

(38)

(35) (38) B A

B =−i ( − ) sin( a)Ak1

2 k22 k2

2 cos a − i ( + ) sin ak1k2 k2 k12 k2

2 k2

(39)

(30) (31) (35) C D AA B F C


(x)ψi

(x)ψr

(x)ψt

= Aei xk1

= Be−i xk1

= F ei xk1

(40)(41)(42)

R T


Scattering in one dimension The ins and outs of tunneling 4/2/17, 7:27 PMRectangular Potential Barrier - Tedious Derivations


where , , and are the incident, reflected, and transmitted probability currents respectively.

Applying the solutions for and found in equations and respectively gives:

Interestingly contrary to classical mechanics, quantum mechanics suggests that the particle may actually be reflectedby the potential barrier, despite having a total energy of equal or greater value than .

If There Is Insu!icient Energy

For , equations , , and must be solved to find , , and . To do this, follow themethodology employed in the previous section, "If There Is Su!icient Energy". The solutions of equations , ,and are identical to those of , , and respectively save for the use of in the place of .

R

T

= − jr

ji

= jt

ji

(43)

(44)

ji jr jt

R

T

= −(x) (x) − (x) (x)ψr

ddx

ψr∗ ψr

∗ ddx

ψr

(x) (x) − (x) (x)ψid

dxψi

∗ ψi∗ d

dxψi

=(x) (x) − (x) (x)ψt

ddx

ψt∗ ψt

∗ ddx

ψt

(x) (x) − (x) (x)ψid

dxψi

∗ ψi∗ d

dxψi

(45)

(46)

R

T

=|B|2

|A|2

=|F |2

|A|2

(47)

(48)

B F (39) (35)

R

T

=a( − )k1

2 k22 2 sin2 k2

4 a + ak12k2

2 cos2 k2 ( + )k12 k2

2 2 sin2 k2

= 4k12k2

2

4 a + ak12k2

2 cos2 k2 ( + )k12 k2

2 2 sin2 k2

(49)

(50)

R

T

=a( − )k1

2 k22 2 sin2 k2

4 + ak12k2

2 ( − )k12 k2

2 2 sin2 k2

= 4k12k2

2

4 + ak12k2

2 ( − )k12 k2

2 2 sin2 k2

(51)

(52)

R

T

= + 1⎡⎣ 4k1

2k22

a( − )k12

k22 2 sin2 k2

⎤⎦

−1

= + 1⎡⎣

a( − )k12

k22 2 sin2 k2

4k12k2

2

⎤⎦

−1

(53)

(54)

V0

E < V0 (8) (9) (10) (x)ψ1 (x)ψ2 (x)ψ3(8) (9)

(10) (5) (6) (7) iκ k2

(x)ψ1

(x)ψ2

(x)ψ3


= C + De−κx eκx

= F ei xk1

(55)(56)(57)










R

T

= − jr

ji

= jt

ji

(43)

(44)

ji jr jt

R

T

= −(x) (x) − (x) (x)ψr

ddx

ψr∗ ψr

∗ ddx

ψr

(x) (x) − (x) (x)ψid

dxψi

∗ ψi∗ d

dxψi

=(x) (x) − (x) (x)ψt

ddx

ψt∗ ψt

∗ ddx

ψt

(x) (x) − (x) (x)ψid

dxψi

∗ ψi∗ d

dxψi

(45)

(46)

R

T

=|B|2

|A|2

=|F |2

|A|2

(47)

(48)

B F (39) (35)

R

T

=a( − )k1

2 k22 2 sin2 k2

4 a + ak12k2

2 cos2 k2 ( + )k12 k2

2 2 sin2 k2

= 4k12k2

2

4 a + ak12k2

2 cos2 k2 ( + )k12 k2

2 2 sin2 k2

(49)

(50)

R

T

=a( − )k1

2 k22 2 sin2 k2

4 + ak12k2

2 ( − )k12 k2

2 2 sin2 k2

= 4k12k2

2

4 + ak12k2

2 ( − )k12 k2

2 2 sin2 k2

(51)

(52)

R

T

= + 1⎡⎣ 4k1

2k22

a( − )k12

k22 2 sin2 k2

⎤⎦

−1

= + 1⎡⎣

a( − )k12

k22 2 sin2 k2

4k12k2

2

⎤⎦

−1

(53)

(54)

V0

E < V0 (8) (9) (10) (x)ψ1 (x)ψ2 (x)ψ3(8) (9)

(10) (5) (6) (7) iκ k2

(x)ψ1

(x)ψ2

(x)ψ3


= C + De−κx eκx

= F ei xk1

(55)(56)(57)










R

T

= − jr

ji

= jt

ji

(43)

(44)

ji jr jt

R

T

= −(x) (x) − (x) (x)ψr

ddx

ψr∗ ψr

∗ ddx

ψr

(x) (x) − (x) (x)ψid

dxψi

∗ ψi∗ d

dxψi

=(x) (x) − (x) (x)ψt

ddx

ψt∗ ψt

∗ ddx

ψt

(x) (x) − (x) (x)ψid

dxψi

∗ ψi∗ d

dxψi

(45)

(46)

R

T

=|B|2

|A|2

=|F |2

|A|2

(47)

(48)

B F (39) (35)

R

T

=a( − )k1

2 k22 2 sin2 k2

4 a + ak12k2

2 cos2 k2 ( + )k12 k2

2 2 sin2 k2

= 4k12k2

2

4 a + ak12k2

2 cos2 k2 ( + )k12 k2

2 2 sin2 k2

(49)

(50)

R

T

=a( − )k1

2 k22 2 sin2 k2

4 + ak12k2

2 ( − )k12 k2

2 2 sin2 k2

= 4k12k2

2

4 + ak12k2

2 ( − )k12 k2

2 2 sin2 k2

(51)

(52)

R

T

= + 1⎡⎣ 4k1

2k22

a( − )k12

k22 2 sin2 k2

⎤⎦

−1

= + 1⎡⎣

a( − )k12

k22 2 sin2 k2

4k12k2

2

⎤⎦

−1

(53)

(54)

V0

E < V0 (8) (9) (10) (x)ψ1 (x)ψ2 (x)ψ3(8) (9)

(10) (5) (6) (7) iκ k2

(x)ψ1

(x)ψ2

(x)ψ3


= C + De−κx eκx

= F ei xk1

(55)(56)(57)




Applying the same boundary conditions as in the previous section and manipulating algebra in the same manner, it canalso be found that:


(In case you are unfamiliar with hyperbolic functions, is the hyperbolic sine function and is the hyperbolic cosine function.) To solve for in relation to , consider equations ,

, and .


As in the previous section, the wave function written in terms of its incident, reflected, and transmitted components is:

Furthermore, the reflection and transmission coe!icients, derivable using the same method as in the previous section,are again given by:

2i Ak1

2i Bk1

2κCe−κa

2κDeκa

= (i − κ)C + (i + κ)Dk1 k1

= (i + κ)C + (i − κ)Dk1 k1

= (κ − i )Fk1 ei ak1

= (i + κ)Fk1 ei ak1

(58)(59)(60)(61)

F A (58) (60) (61)

2i A = − F + Fk1(i − κk1 )2

2κe(i +κ)ak1

(i + κk1 )2

2κe(i −κ)ak1 (62)

4i κ A = −(i − κ F + (i + κ Fk1 e−i ak1 k1 )2 eκa k1 )2 e−κa (63)

4i κ A = [( − ) ( − ) + 2i κ ( + )] Fk1 e−i ak1 k12 κ2 eκa e−κa k1 eκa e−κa (64)

F = 2i κ Ak1 e−i ak1

( − ) sinh κa + 2i κ cosh κak12 κ2 k1

(65)

sinh u = ( − )/2eu e−u

cosh u = ( + )/2eu e−u B A (59)(60) (61)

2i B = F − Fk1+k1

2 κ2

2κe(i +κ)ak1

+k12 κ2

2κe(i −κ)ak1 (66)

= F − F4i κ Bk1 e−i ak1

+k12 κ2

eκa e−κa (67)

= F2i κ Bk1 e−i ak1

( + ) sinh κak12 κ2

(68)

(65) (68) B A

B =( + ) sinh(κa)Ak1

2 κ2


(69)

(x)ψi

(x)ψr

(x)ψt

= Aei xk1

= Be−i xk1

= F ei xk1

(70)(71)(72)








, and .




2i Ak1

2i Bk1

2κCe−κa

2κDeκa

= (i − κ)C + (i + κ)Dk1 k1

= (i + κ)C + (i − κ)Dk1 k1



(58)(59)(60)(61)

F A (58) (60) (61)

2i A = − F + Fk1(i − κk1 )2

2κe(i +κ)ak1

(i + κk1 )2

2κe(i −κ)ak1 (62)





(65)


cosh u = ( + )/2eu e−u B A (59)(60) (61)

2i B = F − Fk1+k1

2 κ2

2κe(i +κ)ak1

+k12 κ2

2κe(i −κ)ak1 (66)


+k12 κ2

eκa e−κa (67)



(68)

(65) (68) B A


2 κ2


(69)

(x)ψi

(x)ψr

(x)ψt

= Aei xk1

= Be−i xk1

= F ei xk1

(70)(71)(72)








, and .




2i Ak1

2i Bk1

2κCe−κa

2κDeκa

= (i − κ)C + (i + κ)Dk1 k1

= (i + κ)C + (i − κ)Dk1 k1



(58)(59)(60)(61)

F A (58) (60) (61)

2i A = − F + Fk1(i − κk1 )2

2κe(i +κ)ak1

(i + κk1 )2

2κe(i −κ)ak1 (62)





(65)


cosh u = ( + )/2eu e−u B A (59)(60) (61)

2i B = F − Fk1+k1

2 κ2

2κe(i +κ)ak1

+k12 κ2

2κe(i −κ)ak1 (66)


+k12 κ2

eκa e−κa (67)



(68)

(65) (68) B A


2 κ2


(69)

(x)ψi

(x)ψr

(x)ψt

= Aei xk1

= Be−i xk1

= F ei xk1

(70)(71)(72)




Contrary to classical expectations which would suggest that the particle has zero probability of travelling beyond , quantum mechanics asserts that the particle has a non-zero probability of tunneling through the rectangular

potential barrier, despite having a total energy less than . This phenomenon marks a major di!erence betweenquantum and classical mechanics.

Labels: energy potentials, Euler's formula, hyperbolic functions, probability current, quantum mechanics, rectangularpotential barrier, reflection and transmission coe!icients, Schrodinger equation, tunneling

7 comments:

1. Dhiraj UpadhyayMarch 20, 2016 at 1:52 AM

Thank you very much.

ReplyDelete

2. UnknownOctober 8, 2016 at 7:15 PM

This is, truly, a most tedious derivation.

ReplyDelete

3. UnknownOctober 12, 2016 at 9:18 AM

Thank you so much...I couldn't get this awesome derivation anywhere else..!!!!

ReplyDelete


Please, did u wrote this website in Latex ?

R

T

=|B|2

|A|2

=|F |2

|A|2

(73)

(74)

R

T

=κa( + )k1

2 κ2 2sinh2

κa + 4 κa( − )k12 κ2 2sinh2 k1

2κ2cosh2

= 4k12κ2

κa + 4 κa( − )k12 κ2 2sinh2 k1

2κ2cosh2

(75)

(76)

R

T

=κa( + )k1

2 κ2 2sinh2

κa + 4( + )k12 κ2 2sinh2 k1

2κ2

= 4k12κ2

κa + 4( + )k12 κ2 2sinh2 k1

2κ2

(77)

(78)

R

T

= + 1⎡⎣ 4k1

2κ2

κa( + )k12 κ2 2sinh2

⎤⎦

−1

= + 1⎡⎣


4k12κ2

⎤⎦

−1

(79)

(80)

x = 0V0





Contrary to classical expectations which would suggest that the particle has zero probability of travelling beyond , quantum mechanics asserts that the particle has a non-zero probability of tunneling through the rectangular

potential barrier, despite having a total energy less than . This phenomenon marks a major di!erence betweenquantum and classical mechanics.

Labels: energy potentials, Euler's formula, hyperbolic functions, probability current, quantum mechanics, rectangularpotential barrier, reflection and transmission coe!icients, Schrodinger equation, tunneling

7 comments:

1. Dhiraj UpadhyayMarch 20, 2016 at 1:52 AM

Thank you very much.

ReplyDelete


This is, truly, a most tedious derivation.

ReplyDelete

3. UnknownOctober 12, 2016 at 9:18 AM

Thank you so much...I couldn't get this awesome derivation anywhere else..!!!!

ReplyDelete


Please, did u wrote this website in Latex ?

R

T

=|B|2

|A|2

=|F |2

|A|2

(73)

(74)

R

T

=κa( + )k1

2 κ2 2sinh2

κa + 4 κa( − )k12 κ2 2sinh2 k1

2κ2cosh2

= 4k12κ2

κa + 4 κa( − )k12 κ2 2sinh2 k1

2κ2cosh2

(75)

(76)

R

T

=κa( + )k1

2 κ2 2sinh2

κa + 4( + )k12 κ2 2sinh2 k1

2κ2

= 4k12κ2

κa + 4( + )k12 κ2 2sinh2 k1

2κ2

(77)

(78)

R

T

= + 1⎡⎣ 4k1

2κ2


⎤⎦

−1

= + 1⎡⎣


4k12κ2

⎤⎦

−1

(79)

(80)

x = 0V0


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Lehman CollegeScattering in one dimension ave Equation for Electrons ! E " = ! #" = ! j ! (free-particle) (free-particle) ave Equation ! !" p x " = ! k" = j !! t "! x p2 E = 2m!j