Download Degeneracy

Review: “Particle in rigid 3D box”
Today:
• Rest of the 3D rigid box
• The Hydrogen atom
−
h2  ∂ 2 ∂ 2 ∂ 2 

ψ ( x, y, z) + V ( x, y, z)ψ ( x, y, z) = Eψ ( x, y, z)
+
+
2m  ∂x2 ∂y 2 ∂z 2 
V(x,y,z) = 0 inside; ∞ outside.
Final Exam is on Wednesday, Dec. 14,
1:30pm-4pm in G1B20 (our classroom).
Allowed materials: One letter paper-sized,
handwritten cheat-sheet and a calculator. (No
cell-phones, PCs, textbooks, BBQs etc.)
Separation of variables approach:
ψ(x,y,z) = X(x)Y(y)Z(z)
Solutions:
c
a
b
π 2h2
π
k = n , Ex = n2
, n = 1,2,3....
a
2ma 2
X ( x) = A ⋅ sinkx x ,
and similar for Y(y) and Z(z).
Exam format: Similar to Exam 1- 3 (all
bubble; emphasis on understanding
concepts)
2D box: Square of the wave function for nx=ny=1
And the total energy is:
E = E x + E y + Ez
Or for a cube (a=b=c):
E = E0 (n x2 + n 2y + nz2 ) , E0 =
100%
0%
ny
nx
Sometimes, there are several solutions with the exact
same energy. Such solutions are called ‘degenerate’.
2ma 2
2D box: Square of the wave function of selected
excited states
‘Percent’ relative
to maximum
Degeneracy
π 2h 2
Q1:
The ground state energy of the 2D box of size L x L is 2E0, where
E0 = π2ħ2/(2mL2) is the ground state energy of a 1D box of size L.
y
L
E=E0(nx2+ny2)
E=
E0(nx2+ny2+nz2)
L
x
What is the energy of the 1st excited state of this 2D box?
a)3E0
b)4E0
c) 5E0
d)8E0
Degeneracy of 1 means “non-degenerate”
Q2:
Imagine a 3D cubic box of sides L x L x L. What is
the degeneracy of the ground state and the first
excited state?
E = E (n 2 + n 2 + n 2 )
0
x
y
z
–
– Why? Scattering showed hard core.
– Problem: electrons should spiral into nucleus in ~10-11 sec.
–
–
–
–
+
• Bohr – fixed energy levels
L
1, 1
3, 1
1, 3
3, 3
0, 3
– Why? Known that negative charges can be removed from atom.
–
– Problem: Rutherford showed nucleus is hard core.
• Rutherford – Solar System
Degeneracy of ground state
Degeneracy of 1st excited state
a)
b)
c)
d)
e)
Review Models of the Atom
• Thomson – Plum Pudding
– Why? Fits to spectral lines of Hydrogen
– Problem: No reason for fixed energy levels
Structure doesn’t fit physics
L
+
• deBroglie – electron standing waves
+
– Why? Explains fixed energy levels
– Problem: still only works for Hydrogen.
L
• Schrodinger – quantum wave functions
– Why? Explains everything!
– Problem: None (except that it’s hard to understand)
What is Schrodinger’s model of
the hydrogen atom?
Schrodinger’s Solutions for Hydrogen
If you understand the 1D quantum well (and how we
went from 1D to 3D), then it is easy for you to
understand Schrödinger's hydrogen atom!
Electron is cloud of probability whose wave
function Ψ(x,y,z,t) is the solution to the
Schrodinger equation:
−
h2  ∂ 2 ∂ 2 ∂ 2 

Ψ( x, y, z, t )
+
+
2m  ∂x2 ∂y 2 ∂z 2 
∂
+ V ( x, y, z)Ψ( x, y, z, t ) = ih Ψ( x, y, z, t )
∂t
# of protons in nucleus
Coulomb constant
where:
2
V ( x, y , z ) = −
V
All that’s really different is the potential V(r) ∝ 1/r
(Coulomb)
r
2
Zke
Zke
=− 2
r
( x + y 2 + z 2 )1/ 2
OK. Mathematically it is a bit ‘involved,’ For now,
focus on the concept and let someone else do the
math for you!
(Coulomb potential; no time dependence)
Can get rid of time dependence and simplify:
Equation in 3D, looking for Ψ(x,y,z,t):
−
h2  ∂ 2 ∂ 2 ∂ 2 

Ψ( x, y, z, t)
+
+
2m  ∂x2 ∂y 2 ∂z 2 
∂
+ V ( x, y, z)Ψ( x, y, z, t ) = ih Ψ( x, y, z, t)
∂t
Quick note on vector derivatives
Laplacian (or ‘Laplace operator’) in cartesian coordinates:
∇2ψ ≡
∂ 2ψ ∂ 2ψ ∂ 2ψ
+
+
∂x 2 ∂y 2 ∂z 2
Same thing! Just different coordinates.
Since V(x,y,z) not function of time:
Ψ ( x, y, z , t ) = ψ ( x, y, z )e − iEt / h
Laplacian in spherical coordinates:
Eψ ( x, y, z )e − iEt / h
Time independent Schrodinger Equation (analogous to 1D):
h2  ∂ 2 ∂ 2 ∂ 2 
−  2 + 2 + 2 ψ ( x, y, z) + V ( x, y, z)ψ ( x, y, z) = Eψ ( x, y, z)
2m  ∂x ∂y ∂z 
∇2ψ =
∂ 
∂ψ 
1 ∂  2 ∂ψ 
1
1
∂ 2ψ
r
+ 2
 sin θ
+ 2 2
2
r ∂r  ∂r  r sin θ ∂θ 
∂θ  r sin θ ∂φ 2
3D Schrödinger with Laplacian (coordinate free):
−
r
r
r
h2 2 r
∇ ψ ( r ) + V ( r )ψ ( r ) = Eψ ( r )
2m
Since potential is spherically
symmetric ( V ( r ) = − Zke 2 / r ),
easier to solve w/ spherical coords:
Q3:
z
Watch out:
Physicist’s
convention!
Time-independent Schrodinger
eqn. in spherical coordinates:
r
φ
V(r)
In 1D (electron in a wire):
we got quantization from applying
boundary conditions in terms of x.
θ
y
−
h 2  1 ∂  2 ∂ψ 

r
2m  r 2 ∂r  ∂r 
+
1
∂ 
∂ψ 
1
∂ 2ψ 
sin
+
+V ( r )ψ = Eψ
θ


∂θ  r 2 sin 2 θ ∂φ 2 
r 2 sin θ ∂θ 
x
(x,y,z) =
(rsinθcosφ, rsinθsinφ, rcosθ)
Technique for solving: “Separation of Variables”
r
In 3D: have 3 degrees of freedom.
Boundary conditions in terms of r,θ,φ
What are the boundary conditions on the wave function ψ in r ?
A. ψ must go to 0 at r=0
B. ψ must go to 0 at r=infinity
C. ψ at r=infinity must equal to ψ at r=0, but value can be non-zero.
D. A and B
E. A, B, and C
(similar to particle in a 3D box, but now in spherical coordinates)
ψ (r ,θ , φ ) = R(r ) f (θ ) g (φ )
Ψ (r , θ , φ , t ) = R (r ) f (θ ) g (φ ) e − iEt / h
Q4:
r
x
φ
y
ψ nlm (r ,θ , φ ) = Rnl (r ) f lm (θ ) g m (φ )
Shape of ψ depends on n, l ,m. Each (n,l,m) gives unique ψ
n=1, 2, 3 … = Principle Quantum Number
2p
a. 1
b. 3
c. 4
d. 6
e. 9
l=0, 1, 2, 3 …= Angular Momentum Quantum Number
(restricted to 0, 1, 2 … n-1)
=s, p, d, f
n=2
m = ... -1, 0, 1.. = z-component of Angular Momentum
l=1
(restricted to –l to l)
m= -1,0,1
The Hydrogen atom
3D-Schrödinger with V(r) = -ke2/r
h2 2
ke 2
∇ ψ ( r,θ , φ ) −
ψ ( r, θ , φ ) = Eψ ( r, θ , φ )
2m
r
2
with: ∇ ψ =
θ
In 3D, now have 3 degrees of freedom:
Boundary conditions in terms of r,θ,φ
Have 3 quantum numbers (n, l, m)
How many quantum numbers are there in 3D?
In other words, how many numbers do you need to
specify unique wave function? And why?
−
z
In 1D (electron in a wire):
Have 1 quantum number (n)
In 1D (electron in a wire):
Have 1 quantum number (n).
Need to specify value of n to know
what state electron is in.
nπx −iEnt / h
2
Ψ( x, t ) = ψ ( x )e −iEnt / h =
sin(
)e
L
L
In 3D, now have 3 degrees of freedom (ignore spin):
3 sets of boundary conditions in terms of r,θ,φ
1 ∂  2 ∂ψ 
1
1
∂ 2ψ
∂ 
∂ψ 
sin
+
+
θ
r




∂θ  r 2 sin 2 θ ∂φ 2
r 2 ∂r  ∂r  r 2 sin θ ∂θ 
Comparing H atom & Infinite Square Well
Infinite Square Well: (1D)
• V(x) = 0 if 0<x<L
∞ otherwise
∞
ψ (r ,θ , φ ) = R(r ) f (θ ) g (φ )
Once we have ψ(r,θ,ϕ) we also know the time-dependent solution:
Ψ (r , θ , φ , t ) = R (r ) f (θ ) g (φ ) e −iEt / h
Done!
r
∞
0
L
x
• Energy eigenstates:
n 2π 2 h 2
En =
Use separation of variables:
H Atom: (3D)
• V(r) = -ke2/r
2mL2
• Wave functions:
ψ n ( x) = L2 sin( nLπx )
Ψ n ( x, t ) = ψ n( x )e − iEnt / h
• Energy eigenstates:
mk 2 e 4
2h 2 n 2
• Wave functions:
En = −
ψ nlm (r ,θ , φ ) = Rnl (r ) f lm (θ ) g m (φ )
Ψ nlm (r , θ , φ , t ) = ψ nlm (r , θ , φ )e − iEnt / h
: generalized Laguerre polynomials
: spherical harmonic functions
Answers to clicker questions
Q1: C; nx=1, ny=2 or nx=2 ny=1s
Q2: C
Ground state = 1,1,1 : E1 = 3E0
1st excited state: 2,1,1 1,2,1 1,1,2 : all same E2 = 6 E0
Q3: B
ψ must be ‘normalizable’, so needs to go to zero at r∞
(Also makes sense physically: not probable to find electron there)
NOTE: ψ(0)=const. is sufficient! Probability of finding it around r~0 is proportional to: 4πr2|ψ|2.
Q4: B