Physics 3 for Electrical EngineeringPhysics 3 for Electrical Engineering

Ben Gurion University of the Negevwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter

Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin

Week 6. Quantum mechanics – probability current • 1D scattering and tunneling • simple 1D potentials • parity • general 1D potential • 2D and 3D square wells and degeneracy Sources: Merzbacher (2nd edition) Chap. 6;Merzbacher (3rd edition) Chap. 6; Tipler and Llewellyn, Chap. 6 Sects. 1-3,6;7פרקים בפיסיקה מודרנית, יחידה

Probability current

When we solved the finite square well potential for bound

states (E < V0 ), we threw away the outside term ek′|x| because it is not normalizable.

Recall

x

V(x)

V0

0 L/2−L/2

outside outsideinside

. /||2' , /2 0 EVmkmEk

Probability current

But for free states (E > V0 ) the solutions outside

are linear combinations of eik′x and e−ik′x ; and they are not normalizable either. Should we throw them away?

x

V(x)

V0

0 L/2−L/2

outside outsideinside

, 2/ || , )(ψ2

)(ψ )(2

22

0 Lxxdx

d

mxVE

Probability current

We know why the wave eik′x is not normalizable: its momentum is with zero uncertainty (Δp = 0), so the uncertainty Δx in its position must be infinite – the particle can be anywhere along the x-axis with equal probability.

x

V(x)

V0

0 L/2−L/2

outside outsideinside

'kp

'kp

Probability current

We know why the wave eik′x is not normalizable: its momentum is with zero uncertainty (Δp = 0), so the uncertainty Δx in its position must be infinite – the particle can be anywhere along the x-axis with equal probability. Should we throw away the

wave eik′x?

x

V(x)

V0

0 L/2−L/2

outside outsideinside

Probability current

If we ask what fraction of an incoming beam ψ(x) = Aeikx is reflected and what fraction is transmitted, then the normalization doesn’t matter.

Probability current

If we ask what fraction of an incoming beam ψ(x) = Aeikx is reflected and what fraction is transmitted, then the normalization doesn’t matter.

The flux density is ρv where ρ = |ψ(x)|2 = |A|2 and v = .mk /

Probability current

Consider the time derivative of the probability density ρ(x,t):

where we have applied Schrödinger’s equation and its complex conjugate. (What happened to the potential term?)

, ),(),(),(),(2

),(),(),(),(2

),(),(),(),(

)],(),([),(

**

2

2**

2

2

**

*

txx

txtxx

txxm

i

txx

txtxx

txm

i

txt

txtxt

tx

txtxt

txt

Probability current

We found

which implies the 1D “continuity equation”

if we define

, ),(),(),(),(2

),( **

txx

txtxx

txxm

itx

t

, ),(),( txJx

txt x

. ),(),(),(),(2

),( **

txx

txtxx

txm

itxJ x

Probability current

Let’s check: If Ψ(x,t) = Aeikx-iωt , then

which is just what we found from our earlier calculation.

, )( 2

),(),(),(),(2

),(

2*

**

Am

kikikAA

m

i

txx

txtxx

txm

itxJ x

So we could look for an outside solution of the form

in which the incoming flux /m splits into a reflected flux

/m and a transmitted flux /m .

x

V(x)

V0

0 L/2−L/2

outside outsideinside

, 2/ ,

, 2/ , )(ψ'

''

LxFe

LxBeAexxik

xikxik

xikFe 'xikxik BeAe '' ikxCe ikxDe

2' Ak

2' Bk 2

' Fk

1D scattering and tunneling

But let’s consider a simpler scattering problem:

x

V(x)

V0

0

1D scattering and tunneling

But let’s consider a simpler scattering problem:

We take E > V0 and

in which the incoming flux /m splits into a reflected flux

/m and a transmitted flux /m . The reflection

probability is R = and the transmission probability is

T =

x

V(x)

V0

0

, 0 ,

, 0 , )(ψ'

xCe

xBeAexxik

ikxikx

xikCe 'ikxikx BeAe

2Ak

2Bk 2

' Ck

. /'/2

kkAC

2/ AB

Continuity of ψ at x = 0: A + B = C .Continuity of dψ/dx at x = 0: ik(A – B) = ik′C .

Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′) and B/A = (k–k′)/(k +k′) .

x

V(x)

V0

0

xikCe 'ikxikx BeAe

Continuity of ψ at x = 0: A + B = C .Continuity of dψ/dx at x = 0: ik(A – B) = ik′C .

Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′) and B/A = (k–k′)/(k +k′) . Does R + T = 1?

x

V(x)

V0

0

xikCe 'ikxikx BeAe

Continuity of ψ at x = 0: A + B = C .Continuity of dψ/dx at x = 0: ik(A – B) = ik′C .

Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′) and B/A = (k–k′)/(k +k′) . Does R + T = 1?

x

V(x)

V0

0

xikCe 'ikxikx BeAe

TA

C

k

k

kk

kk

kk

kk

A

BR

1

'1

)'(

'41

)'(

)'(2

22

22

Isn’t something very odd – “quantumly” odd – going on here?

x

V(x)

V0

0

Isn’t something very odd – “quantumly” odd – going on here?

A classical particle with E > V0 would never reflect back from the potential step…but here there is reflection unless k = k′ , i.e. unless V0 = 0.

x

V(x)

V0

0

Isn’t something very odd – “quantumly” odd – going on here?

A classical particle with E > V0 would never reflect back from the potential step…but here there is reflection unless k = k′ , i.e. unless V0 = 0.

In fact, even if we replace V0 by –V0 so that k >> k′, there is still reflection at x = 0!

x

V(x)

−V0

0

The case E < V0 :

Now R= 1, T = 0: no transmission. How deep do particles penetrate the potential step?

x

V(x)

V0

0

. 0 ,

, 0 , )(ψ'

xCe

xBeAexxk

ikxikx

xkCe 'ikxikx BeAe

0.5

1.0

Summary of scattering from the potential step:

Snapshots of the probability density, for an incident wave packet:

Another scattering problem: the “square” potential barrier:

x

V(x)

V0

0 L

Another scattering problem: the “square” potential barrier:

As before,

x

V(x)

V0

0

; L ,

; 0 ,

; 0 , )(ψ''

xFe

LxDeCe

xBeAex

ikx

xkxk

ikxikx

ikxikx BeAe L

ikxFexkCe ' xkDe '

. /||2' , /2 0 EVmkmEk

Boundary conditions (ψ and dψ/dx continuous) at x = 0:

; '

;

DCkBAik

DCBA

Hence

. '

1 '

12 ; '

1 '

12 Dik

kC

ik

kBD

ik

kC

ik

kA

x

V(x)

V0

ikxikx BeAe ikxFexkCe ' xkDe '0 L

Boundary conditions (ψ and dψ/dx continuous) at x = L:

; '

; ''

''

ikLLkLk

ikLLkLk

ikFeDeCek

FeDeCe

Hence

x

V(x)

V0

ikxikx BeAe ikxFexkCe ' xkDe '

. '

12 ; '

12 '' ikLLkikLLk Fek

ikDeFe

k

ikCe

0 L

If we combine these equations we can show

'sinh

'

'

2'cosh Lk

k

k

k

kiLk

ikLe

A

F

Hence

x

V(x)

V0

ikxikx BeAe ikxFexkCe ' xkDe '

'sinh'

'

4

1'cosh

1

22

22

Lk

k

k

k

kLk

A

FT

0 L

For k′L >> 1, both cosh k′L and sinh k′L approach

and the expression for T simplifies:

x

V(x)

V0

ikxikx BeAe ikxFexkCe ' xkDe '

2'//'

16 '22

kkkk

e

A

FT

Lk

2'Lke

0 L

Simulation of a Gaussian wave packet with kinetic energy 500 eV, incident on a square potential barrier with height 600 eV and thickness 25 pm. About 17% of the wave packet tunnels through the barrier.

To see the probability density:

http://www.youtube.com/watch?v=4-PO-RHQsFA&NR=1

To see the real (black) and imaginary (red) components of the wave function:

http://www.youtube.com/watch?v=_3wFXHwRP4s

Three applications of quantum tunneling:

1. Light crosses a “total internal reflection barrier”.

2. The NH3 molecule:

Distance of N atom from plane of 3 H atoms, at various energy levels Ei

Three applications of quantum tunneling:

1. Light crosses a “total internal reflection barrier”.

2. The NH3 molecule:

Distance of N atom from plane of 3 H atoms, at various energy levels Ei

3. α–decay of a U238 nucleus:

Simple 1D potentials – summary

Simple 1D potentials – summary

Parity

When the potential term V(x) of the 1D Schrödinger equation is symmetric, i.e. V(x) = V(−x) , then the Schrödinger equation itself is symmetric and the solutions must have the same physical symmetry. But what is physical, ψ(x) or |ψ(x)|2 ?

Parity

When the potential term V(x) of the 1D Schrödinger equation is symmetric, i.e. V(x) = V(−x) , then the Schrödinger equation itself is symmetric and the solutions must have the same physical symmetry. But what is physical, ψ(x) or |ψ(x)|2 ?

Only |ψ(x)|2 (and the probability current) are physical. Hence ψ(x) can be a solution whether ψ(−x) = ψ(x) or ψ(−x) = −ψ(x). But we give these cases different names: in the first case ψ(x) has even parity; in the second ψ(x) has odd parity.

Parity

When the potential term V(x) of the 1D Schrödinger equation is symmetric, i.e. V(x) = V(−x) , then the Schrödinger equation itself is symmetric and the solutions must have the same physical symmetry. But what is physical, ψ(x) or |ψ(x)|2 ?

Only |ψ(x)|2 (and the probability current) are physical. Hence ψ(x) can be a solution whether ψ(−x) = ψ(x) or ψ(−x) = −ψ(x). But we give these cases different names: in the first case ψ(x) has even parity; in the second ψ(x) has odd parity.

If two solutions are degenerate (have the same energy) then a linear combination of them need not have definite parity. But if the energies are nondegenerate (one solution for each energy) then the solutions will have definite parity (even or odd).

General 1D potential

Now let’s consider a 1D potential without any special symmetry:

What can we say about the solutions of the time-independent Schrödinger equation?

V(x)

x

General 1D potential

If ψ(x) > 0 and E > V(x) , then ψ(x) will turn down until it becomes negative. If ψ(x) < 0 and E > V(x), then ψ will turn up until it becomes positive. The larger E − V(x) , the more up-and-down curves in ψ(x) and the more nodes (zeros) in ψ(x).

ψ )(2ψ

22

2ExV

m

dx

d

E

V(x)

x

General 1D potential

Hence the lowest-energy solution will have no node.

ψ )(2ψ

22

2ExV

m

dx

d

E0

ψ(x) V(x)

x

General 1D potential

The next-lowest-energy solution will have one node.

ψ )(2ψ

22

2ExV

m

dx

d

E1

ψ(x) V(x)

x

General 1D potential

As the energy increases, so does the number of nodes.

ψ )(2ψ

22

2ExV

m

dx

d

E5

ψ(x) V(x)

x

),(ψ ),(2

),(ψ2

2

2

22yxyxV

yxmyxE

2D and 3D square wells and degeneracy

The time-independent Schrödinger equation for a particle in two dimensions:

Suppose V(x,y) is an infinite square well potential, vanishing for |x| < Lx/2 and |y| < Ly/2, and infinite otherwise. What are the five lowest energy levels? Are the energies nondegenerate? If not, what are their degeneracies? How do we normalize the solutions in 2D? What are the normalized solutions? Does it matter whether or not Lx and Ly are equal?

What are the corresponding questions and answers in 3D?

The time-independent Schrödinger equation for a particle in three dimensions is

where r = (x,y,z). How do we define the probability current J(r,t) in 3D so that the continuity condition,

holds?

)(ψ )(2

)(ψ 22

rrr

V

mE

, 0 ),(),(

ttt

rJr