Zero-point Energy

• Minimum energy corresponds to n=1

• n=0 => n(x)=0 for all x

• => P(x)=0 => no electron in the well

• zero-point energy

• never at rest!

• Uncertainty principle: if x=L/, then px > ħ/(x) = ħ /L => E > (px

)2/2m = 2 ħ 2/2mL2 = h2/(8mL2)

Problem • An electron is trapped in an infinite well which is

250 pm wide and is in its ground state. How much energy must it absorb to jump up to the state with n=4?

• Solution: standing waves => n(/2)=L

• E=p2/2m= h2/2m 2 = n2h2/8mL2

• E4-E1= (h2/8mL2)(42 -1) =15(6.63x10-34)2/[8(9.11x10-31)(250x10-12)2=90.3 eV

Problem • An electron is trapped in an infinite well that is 100 pm

wide and is in its ground state. What is the probability that you can detect the electron in an interval of width x=5.0 pm centered at x= (a) 25 pm (b)50 pm (c)90 pm ?

• P(x) =|(x)|2 is probability/unit length• P(x)dx = probability that electron is

located in interval dx at x• L=100 x 10-12 m (x)=(2/L)1/2 sin(x/L)• P(x) x = (2/L)sin2(x/L) x =(1/10) sin2(x/L)• a) P(x) x =.1 sin2(/4) = .05• b) P(x) x =.1 sin2(/2) = .1• c) P(x) x =.1 sin2(.9) = .0095

Problem • A particle is confined to an infinite potential

well of width L. If the particle is in its ground state, what is the probability that it will be found between (a) x=0 and x=L/3 (b) x=L/3 and x=2L/3 (c) x=2L/3 and x=L?

• If x is not small, we must integrate!

(a)(b) (c)

1

2

0

2

2

1

( ) sin

( ) sin

( )

xL

x

L

P xL

x

L

P x dxL

FHGIKJ

FHGIKJ

zP a x b

L

x

Ldx

a

b

( ) sin FHGIKJz2 2

Solution• Total probability of finding the particle

between x=a and x=b is

Let yx

Lx

Ly

dxLdy

Limits x a yLa

x b yLb

:

2 2

L

x

Ldx

a

bz FHGIKJsin

2 2

L

Ly dy

a

L

b

L

sin bgz

Solution (cont’d)• Since cos(2y)= cos2(y)-sin2(y)=1-2sin2(y)

• we have sin2(y) = (1/2)(1 - cos(2y))

11 2

z cos( )y dya

L

b

Lb g y y

a

L

b

L

a

L

b

L

1

2

2sin( )

LNM

OQP

b a

L

b

L

a

L

1

2

2 2

sin sin

Solution(cont’d)

• For a=0, b=L we have L/L - (1/2)[sin(2)-sin(0)] = 1

• (a) a=0, b=L/31/3 - (1/2)[sin(2/3)-sin(0)]= .195

• (b) a=L/3, b=2L/3 1/3 - (1/2)[sin(4/3)-sin(2/3)]= .61

• (c) a=2L/3, b=L 1/3 - (1/2)[sin(2)-sin(4/3)]= .195

LNM

OQP

b a

L

b

L

a

L

1

2

2 2

sin sin

Nodes at ends => standing waves (x)=0 outside

Infinite Potential Well

Electron in a Finite Well

• Only 3 levels with E < U0

• states with energies E > U0 are not confined

• all energies for E > U0 are allowed since the electron has enough kinetic energy to escape to infinity

U0 <

0 L

Quantizationof energy

No longer a nodeat x=0 and x=L

Electron has smallprobability of penetrating the walls

Area under eachcurve is 1

oscillation Exponential decay

Harmonic Oscillator Potential

2

0 0( ) axx A e P(x) 0 everywhere