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Infinite (and finite) square well potentials
Announcements:
• There will be a homework set put up this afternoon and
due Wednesday after break.
• No class on Friday.
• HW1-8 and Exam 1 grades are now in CULearn. Please
check to make sure they are correct.
In case you are curious, the average
scores (of those turned in) are below:
EX1
HW1 HW2 HW3 HW4 HW5 HW6 HW7 HW8
Avg: 75.0 44.2 43.7 42.2 35.6 32.7 39.2 39.2 37.7
EX1 is out of 100, HW4 is out of 40 and the rest are out of 50
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
1
Some wave function rules
y(x) and dy(x)/dx must be continuous
These requirements are used to match boundary conditions.

|y(x)|2
 y ( x) dx  1
2
must be properly normalized

This is necessary to be able to interpret
|y(x)|2 as the probability density
y ( x)  0 as x  
This is required to be able to normalize y(x)
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
2
Infinite square well (particle in a box) solution
y ( x)  B sin( kx)
2 2
2 2 2
n

k
n
 
k

and
which
gives
us
an
energy
of
E


n
2
a
After applying boundary conditions we found
2m
2ma
Things to notice:
Energy
Energies are quantized.
Minimum energy E1 is not zero.
Consistent with uncertainty principle.
x is between 0 and a so Dx~a/2.
Since DxDp≥ħ/2, must be uncertainty
in p. But if E=0 then p=0 so Dp=0,
violating the uncertainty principle.
When a is large, energy levels get
closer so energy becomes more like
continuum (like classical result).
http://www.colorado.edu/physics/phys2170/
16E1
n=4
9E1
n=3
4E1
n=2
E1
0
n=1
Physics 2170 – Spring 2009
V=0
a
3
x
Finishing the infinite square well
We need to normalize y(x). That is, make sure that

 y ( x)
2
dx  1

For the region x<0 and x>a the probability
|y(x)|2 is zero so we just need to ensure that
a
 y (x) dx  1
2
0
Putting in y ( x)  B sin( kx) and doing the integral we find B  2/a
Therefore y ( x)  2/a sin( kx)
But we also know k  n / a
 
So we can write the solution as y ( x)  2 sin nx
a
a
 
Adding in the time
( x, t)  y ( x) (t )  2 sin nx e iEt/
a
a
dependence:
Is it still normalized?  *(t)  (t) is not a function of x so can pull
out of the integral and find *(t)(t)  eiEt/eiEt/  1
So the time dependent piece is already normalized.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
4
Clicker question 1
2 2
En  n2   2  n2 E1
2ma
Set frequency to DA
2/a
2/a
a/3
2a/3
a
2/a
a/3 2a/3
a
a/ 2
a
2/a
a/ 2
a
2/a
2/a
a
a
a
Am I more likely to find the particle close to a/2 in the n=2 or n=3 state?
A. n=2 state
B. n=3 state
C. No difference
For n=2 state:
|y 2 (a/2) |2  0
For n=3 state:
|y 3(a/2) |2  2/a
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
5
Be careful to understand everything we plot…
Energy
V(x)
Total energy
E (n=3)
E (n=2)
V=0 eV
E (n=1)
0
a
y(x)
0
Sometimes plot three
things on the same graph!
x
Potential Energy V(x)
Total Energy E
Wave Function y(x)
http://phet.colorado.edu/simulations/sims.php?sim=Quantum_Bound_States
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
6
Comparing classical and quantum results
 
( x, t )  2 sin nx e iEnt / 
a
a
2 2

  n2 E
En  n
1
2
2
2ma
Note: time dependence depends on energy
a
Classical physics
Particle can have any energy
Lowest kinetic energy is 0
(particle is at rest)
Quantum physics
Particle can only have
particular energies (quantized)
Lowest energy state in box has
kinetic energy (zero point motion)
How small would a box need to be for E1 to be 4.7 eV?
 (6.58 1016 eVs)(3 108 m/s )
n


a

 2.8 1010 m
2mEn
2(0.511106 eV)(4.7 eV)
About the size of an atom so our model wouldn’t work anyway
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
7
Reading Quiz 1
Set frequency to DA
Please answer this question on your own.
No discussion until after.
Q. Classically forbidden regions are where…
A. a particle’s total energy is less than its kinetic energy
B. a particle’s total energy is greater than its kinetic energy
C. a particle’s total energy is less than its potential energy
D. a particle’s total energy is greater than its potential energy
E. None of the above.
This would imply that the kinetic energy is
negative which is forbidden (at least classically).
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
8
Motivation for a finite square well
Infinite square well approximation
assumes that electrons never get
V(x) out of the well so V(0)=V(a)=∞
and y(0)=y(a)=0.
Energy
V(x)
0
A more accurate potential
function V(x) gives a chance
of the electron being outside
0
a
x
What if the particle
energy is higher?
E
What about two wires very close together?
These scenarios require the more accurate potential
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
9
Need to solve the finite square well
Energy
V(x)
wire
0
0
Need to solve TISE:
4.7 eV
 2 d 2y ( x)

 V ( x)y ( x)  Ey ( x)
2
2m dx
x < 0: V(x) = 4.7 eV
Work function
x > a: V(x) = 4.7 eV
0 < x < a: V(x) =0
x
a
This will be used to understand quantum tunneling which
provides the basis for understanding
Radioactive decay
Scanning Tunneling Microscope which is used to study surfaces
Binding of molecules
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
10
Analyzing the finite square well
Energy
V(x)
 2 d 2y ( x)
4.7 eV TISE:  2m dx 2  V ( x)y ( x)  Ey ( x)
wire
We rewrite the TISE as
Eparticle
0
0
Region I
a
Region II
d 2y ( x) 2m
 2 V ( x)  E y ( x)
2
dx

x
Region III
Consider three regions
In Region II: total energy E > potential energy V so V − E < 0
2
2
m
d
2


V
(
x
)

E
Replace 2
with −k to get y (2x)  k 2 y ( x) (k is real)

dx
Same as infinite square well so sin(kx) and cos(kx) or eikx and e-ikx
Region II: y II (x)  A cos(kx)  B sin( kx)
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
11