1. 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 2/2mL = h2/(8mL2)

2. 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

3. P(x) =|(x)|2 is probability/unit length
Problem
L01 Mar 20/02
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 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

4. 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)

6. Solution
Total probability of finding the particle between x=a and x=b is

7. Solution (cont’d) Since cos(2y)= cos2(y)-sin2(y)=1-2sin2(y)
we have sin2(y) = (1/2)(1 - cos(2y))

8. Solution(cont’d) For a=0, b=L we have L/L - (1/2)[sin(2)-sin(0)] = 1
(a) a=0, b=L/ /3 - (1/2)[sin(2/3)-sin(0)]= .195
(b) a=L/3, b=2L/ /3 - (1/2)[sin(4/3)-sin(2/3)]= .61
(c) a=2L/3, b=L /3 - (1/2)[sin(2)-sin(4/3)]= .195

9. Infinite Potential Well 
Nodes at ends => standing waves (x)=0 outside

10. Electron in a Finite Well
U0 <  L
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
Quantization
of energy

11. No longer a node
at x=0 and x=L
Electron has small
probability of
penetrating the walls
Area under each
curve is 1
oscillation
Exponential decay

12. Harmonic Oscillator Potential
P(x) 0 everywhere