Download The “classically forbidden regions” are where … a. a particle`s total

The “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
Answer is c.
Today’s class:
• 1 more thing in 1-d Quantum
harmonic oscillator
• Back to 3-d
• Pauli exclusion principle
One last example for the 1D
Schrödinger equation
In past classes we did:
• Free particle: V=0 everywhere
• Particle in rigid box: V=0 inside, V=∞ outside
• Particle in non-rigid box: V=0 ins., 0<V<∞ outside
Next:
• Harmonic oscillator: V(x) x2
One last 1D example:
The harmonic oscillator
Classical harmonic oscillator:
Mass ‘m’ experiences restoring force F = -kx
(x: displacement from equilibrium point).
In QM: don’t want to deal with forces. What can
we do?
Can derive corresponding PE function:
x
F = -kx
1 2
U ( x)   F ( x' )dx'  kx
2
0
x
Why is this important?
Vibration in molecules!! (Used for molecular detection)
x
1 2
U ( x)  kx
2
U(x)
Potential close to
the ground-state
looks very similar
to x2
x (Å)
Real time molecular detection
Real time molecular detection
Simple harmonic oscillator
All we have to do is to solve the Schrödinger equation
with a parabolic potential:
2 d 2 ( x ) 1 2

 kx  ( x )  E ( x )
2
2m dx
2
(… or a lengthy
calculation...)
(“Hermite polynomials”)
Energy levels
Simple harmonic oscillator (cont.)
Probability density
(“Where is the particle most likely to be found”)
Harmonic
oscillator
Iodine spectrum
Absorption (%)
Atmospheric greenhouse effect
Notice the CO2 absorption is saturated. The details of the
interactions are key to understanding the effects of more
CO2 in the atmosphere.
That’s all about the quantum harmonic oscillator
Up next:
Brief review of energy eigenstates
and superposition states.
Review: “Particle in rigid 3D box”
2  2 2 2 
 2  2  2  ( x, y, z )  V ( x, y, z ) ( x, y, z )  E ( x, y, z )

2m  x y z 
V(x,y,z) = 0 inside; ∞ outside.
Separation of variables approach:
(x,y,z) = X(x)Y(y)Z(z)
Solutions:
X ( x)  A  sin k x x ,
c
a
2 2



2
k  n , Ex  n
, n  1,2,3....
2
a
2ma
and similar for Y(y) and Z(z).
And the total energy is:
Or for a cube (a=b=c):
E  Ex  E y  Ez
E  E0 ( nx2  n 2y  nz2 ) , E0 
 2 2
2ma 2
b
Degeneracy
Sometimes, there are several solutions with the exact
same energy. Such solutions are called ‘degenerate’.
E = E0(nx2+ny2+nz2)
Degeneracy of 1 means “non-degenerate”
Pauli exclusion principle.
• Particles come in two types (there
are basically only two possibilities)
– This is a consequence of relativity
and quantum together.
The two types are Fermions and
Bosons (we’ll come back to this)
Enrico Fermi
Satyendra Nath Bose
Wolfgang Pauli
Pauli exclusion principle.
• No two identical fermions can
be put into the same quantum
state. They are, in a way,
antisocial.
Wolfgang Pauli
E1
Stay out
E0
Pauli exclusion principle.
• You can put them in
another energy state
(takes more energy)
• You can instead change
their internal quantum
numbers (if they have
them).
• Electrons do have 1
internal number that can
be +1/2 or -1/2 so 2 of
them can get into a state.
Wolfgang Pauli
E1
Stay out
E0
Electron Spin.
• The internal number that can be switched in
electrons is called spin.
• It is an angular momentum that can only
1
1
take on two quantized states + ℏ or − ℏ
2
2
• Two electrons can occupy any given state
(one with “up spin” and one with “down spin”
• This is different for different particles
(bosons love to party)
• Usually we use arrows
E0
not tiny faces.
Imagine a 3D cubic box of sides L x L x L. What is
the degeneracy and how many electrons can be
put into the a state with energy equal to the first
excited energy?
2
2
2
E  E0 ( n x  n y  n z )
Degeneracy of ground state
Number of electrons you can put in there.
L
a)
b)
c)
d)
1, 2
1,1
3, 6
3, 3
L
L
1st excited state: 2,1,1 1,2,1 1,1,2 : all same E2 = 6 E0
So 3 degeneracy. And each “state” can take +1/2 and -1/2
electrons so 6 total
What is the minimum energy required to put 4
electrons into the 3-d rigid box? We are given
4 electrons with zero energy to start with.
a)
b)
c)
d)
e)
12 E0
39 E0
18 E0
0 E0
24 E0
2 in the ground state and 2 in the next state.