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Finite Square Well: Potential

V(x) =

V0 x 0 region I0 0 < x < L region II

V0 x

L region III

cL.Frommhold. p.44/44

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Finite Square Well: Schrdinger Eq.

2

2m

d2

dx2+ V0 = E (regions I and III)

2

2m

d2

dx2= E (region II)

For E < V0, that is

+ k2 = 0 (region II)

2 = 0 (regions I and III)

with k2 = 2mE/2 and 2 = 2m(V0 E)/2.


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Valid Wave Functions(x, t), (x)

Remember: Not every solution

of the Schrdinger equation is a wave function

must be finite everywhere

must be single valued everywhere

and /x must be continuous

must be square integrablecL.Frommhold

. p.42/44

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Finite Square Well: Solutions

General (mathematical) solution

(x) = A ex + B ex regions I and III

(x) = C cos kx + D sin kx region II

Wave function (x)I(x) = A e

x in region I

II(x) = B cos kx + C sin kx in region II

III(x) = D ex in region III

Require continuity of and at x = 0 and x = L


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Finite Square Well: Wave Functions


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and Finite Square Well

Question:

Why lowerenergies forfinite well ?


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Bound States: Finite and

Well

Compare wave functions for n = 1 state


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Well



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Well



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Well



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Correspondence Principle (n = 20)

|

20

(x,

t)|

2

For n clas-sical bahavior ?

What isclassical

behavior ?

probability

= constto find particle at

x, dx


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Schrdinger Eqn. in 3 Dimensions

Time-dependent equation it = H(r, t)

with notation according to

H = 2

2m

2 + V(r)

2

2m2

=p2x +p

2y +p

2z

/2m

px = i x

, etc.

Time-independent equation H(r) = E (r)

with (r, t) = (r) exp(

Et/)


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Particle in 3-Dim. Infinite Box

V(x,y,z ) =

0 if 0 < x < Lx; 0 < y < Ly; 0 < z < Lz elsewhere

Schrdinger eq.:

2

2m

2(r) = E (r)

Ansatz:

(r) = u(x) v(y) w(z)


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3-D Infinite Box II

2

2m uv w + u vw + u v w = E u v w

Divide both sides by u v w

2

2mu

u

+v

v

+w

w = E = Eu + Ev + Ew

Separate variables to get

2

2m u = Eu u, with u(0) = u(Lx) = 0

22m v = Ev v, with v(0) = v(Lv) = 0

2

2m w = Ew w, with w(0) = w(Lw) = 0


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3-D Infinite Box III

Solution:

(r) = A sin kx,n x sin ky,m y sin kz, z

with kx,n = n/Lx, ky,m = m/Ly, kz, = /Lz,

A =

8/(Lx Ly Lz) and

Emn =22

2m n2

L2x+

m2

L2y+

2

L2zIn three dimensions, we have three quantum numbers

,m,n = 1, 2, 3


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Harmonic Oscillator Classical

F = m a becomes x = m d2x/dt2

(Note that this is the same DEq. as y

+2

y = 0)

Solution: x(t) = A cos t + B sin t, with =/m = classical oscillator frequency.

Potential energy P E = 12x2 (parabolic well)

Kinetic energy KE =12mx

2

Total energy E = P E+ KE is a continuum,somewhere between 0 and

J.


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Harmonic Oscillator Quantal

Schrdinger equation 2

2m d

2

dx2 + 12x2

= E

Solution n(x) = m1

4 12nn!

Hn() e2/2

with = m/x; n = 0, 1, 2 ; = /m; andHn(x) = Hermite polynomials; H0 = 1, H1 = 2x, . . .Energy En = (n + 12) quantized !


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HO Results I


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HO Results II

Wave functions

Note: a n = 0 state exists(groundstate)

Also shown are the

n = 1, 2, 3 state wave functions


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HO Some Basic Facts

Re-appearance of the classical oscillator frequency

transitions n n 1; En =

Quantization of energy En nRemember freezing of vibrational states ?

Zero point energy E0 = 12 = 0 (Heisenberg !)Even and odd wave functions:even n

even n; odd n

odd n

Corresponence principle as n (next viewgraph)


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HO Correspondence Principle

Classical harmonic oscillator: x(t) = A cos t;

velocity x = A sin t = A2

x2

Time spent between x and x + dx: dt = dx/|x|

Classical probability of finding particle at dx:

dP= dtT /2

=2

T

dxA2 x2 =

1

dxA2 x2

Compare with QM (n = 10):cL.Frommhold

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Correspondence Principle (n = 100)

. p.23/44

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Short Summary

Definition: On =

n(x) O n(x) dx

Hn = En ; H2n = E2n ; En = H2n H2n = 0En is sharp, no uncertainty; a constant of motion

Infinite square well: xn = L/2x2n = L

2

3 1 3

2n22 (x)2 =

L2

12 1 6

n22pn = 0 p2n = 2 k2n (p)n = kn

x p

(L/

12) n/L = n/

12 > /2


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Square Barriers Classical

Kinetic particle energyexceeds barrier height

KE > V0Classical result:

100% transmission0% reflection

Kinetic particle energyless than barrier height

KE < V0Classical result:



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Square Wells and Free States

Kinetic energy > 0Same case as KE

exceeds barrier height

when V0 < 0

Classical result:



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Wave Picture of Square Barrier

Expect1. an incident wave2. a transmitted wave

3. a reflected wave

Divide up space into

Region I: < x 0 Region II: 0 < x < L Region III: L x <

and solve the Schrdinger equationscL.Frommhold

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Free Particle Schrdinger Eq.

For V0 = const, the Schrdinger eq. may be written

2

2m

d2

dx2 + V0

(x) = E (x) or

k2 = 0

with the (positive !) constant k2 given by

k2 =

(E V0) /(2m2) if E > V0 (upper sign)

(V0

E) /(2m2) if E < V0 (lower sign)

Wave functions oscillatory (+ sign) or evanescent

( sign)cL.Frommhold

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Difference of a Sign

For V0 = const, the Schrdinger equation is of theform k2 = 0, with solutions

+(x) = A ekx + B ekx or = C cos kx + D sin kx

(x) = F ex + G ex or = H cosh x + I sinh x

For the + sign we get oscillatory solutionsso that k2 = (E V0)/2m2 and k = 2/

For the sign we get exponentials

so that k2 = 2 = (V0 E)/2m2 and is thereciprocal penetration depth, d = 1/

The sign of the constant k2 is important !


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Beams as Propagating Waves

Remember: (x, t) = (x) exp(E t/).

If (x) = exp(kx), (x, t) = exp (kx t) is awave traveling in the positive x-direction

If (x) = exp(kx), (x, t) = exp (kx + t) is awave traveling in the negative x-direction


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Square Barriers

Region I: I(x) = A exp(kx) + B exp(kx)

= incident + reflected wave

Region II, if E > V0: II(x) = Cexp(kx) + D exp(kx)

Region II, if E < V0: II(x) = Cexp(x) + D exp(x)

Region III: III(x) = F exp(kx)no backward wave exists !

k2 = 2mE/2

(k)2 = 2m(E V0)/22 = 2m(V0 E)/2


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Reflection and Transmission

Continuity of (x) at x = 0 and x = L

Continuity of d/dx at x = 0 and x = L

This gives 4 linear Equations for the 5 unknownsA,B,C,D, and F that must be solved,

which leaves a scaling parameter free.

Reflection and transmission coefficients:R =

|I(reflected)|2

|I(incident)|2=

BB

A

A

T =|III(transmitted)|2

|I(incident)

|2

=FF

AA


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Transmission and Reflection Coeffs.

Make and continuous at x = 0 and x = L !

Compute transmission coefficientT =

F F

A A

Compute reflection coefficientR =

B B

A

ANote thatT + R = 1


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Wave Functions(x) fromy(x)

Square Barrier, if E > V0: continuity of and atboundaries, x = 0 and x = L

Region I: (x) = A exp kx + B exp

kx

(incident and reflected wave)

Region II: (x) = Cexp kx + D exp kx

Region III: (x) = F exp kx (no reflected wave)with 2k2 = 2mE and 2k2 = 2m(E V0).

Solving for A, B, and F, with E > V0, we get

T =F F

A A=

1

1 +V20sin2 kL

4E(EV0)

and T + R = 1


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Square Barriers withE > V0

Note: the sign of V0 does not matter

100% transmission if k2L = n

anti-reflection coating of optical lensescL.Frommhold

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(x) fromy(x), The Other Case

Square Barrier, if E < V0: continuity of and atboundaries, x = 0 and x = L

Region I: (x) = A exp kx + B exp

kx

(incident and reflected wave)

Region II: (x) = Cexp x + D exp x

Region III: (x) = F exp kx (no reflected wave)with 2k2 = 2mE and 22 = 2m(V0 E).

Solving for A, B, and F, we get

T =F F

A A=

1

1 +V20sinh2 L

4E(V0E)

and T + R = 1


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Square Barrier withE < V0

Note that (sinh x)2

> 0Tunneling is unheard of in classical physics !But it is well known in wave optics


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Alpha-Decay

Radioactivity: -decay,e.g., 241Am 237Np +or generally

AZ

A4(Z

2) +

strong force overcomesCoulomb repulsion;

binding the nucleons

and particles.But, if E() > 0, the particle will tunnel rays !(nuclei stable if E() < 0)


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Energy from Nuclear Fusion

D + D 3

He + n + 3.27 MeVD + D T + H + 4.05 MeV,D + T

4He + n + 17.6 MeV

D + 3He 4He + H + 18.3 MeV, etc.

But strong Coulomb repulsion tends to oppose fusion:

Need high temperatures ( 107 K) and tunneling


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Molecular Physics: NH3

nitrogen wavefunctions: tunneling!


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Tunneling of e.m. Waves

Total Reflection

an evanescent (nonpropagating)wave exists

Pick up transmitted wave,employing a second prism:

tunneling of e.m. waves


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Field Emission

Field emission:tunneling

V(x) = W e ExScanningtunneling

microscope:


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Square Wells/Barriers: Summarizing

Why do we study square wells/barriers?

Because math is simple when V = const.

Sinusoidal/exponential wave functionsKnow wave functions, know everything!

Simple models of complex systems

for intuitive understanding of quantum phenomena

Free and bound states:

Bound states: standing waves; energies quantized

Free particles: no energy quantizationBut angular momentum is always quantized,

as we will see.cL.Frommhold. p.3/44

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Barriers and Wells

Tunneling impossible in classical physicsradioactivity, nuclear fusion, tunneling microscopes,

semiconductor devices,. . .

Transmission and reflection properties

resemble closely those of e.m. radiation(but not predictions of classical mechanics)

Physical meaning of quantum number nn counts nodesenergies increase with number of nodes


R l d C d A i i

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Related Cases and Approximations

The piecewise constant potential function can be usedto approximate arbitrary potential functions:

Construct overall wave function with any desired precision !


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