PHYS 3313 – Section 001
Lecture #17
Wednesday, Nov. 7, 2012
Dr. Jaehoon Yu
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Solutions for Schrodinger Equation for
Hydrogen Atom
Quantum Numbers
Principal Quantum Number
Orbital Angular Momentum Quantum
Number
Magnetic Quantum Number
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
1
Announcements
• Research team member list have been updated on the web!
– Remember the deadline for your research paper is Monday, Nov.
26!!
• Reminder: homework #6
– CH6 end of chapter problems: 34, 39, 46, 62 and 65
– Due on Monday, Nov. 12, in class
• Reading assignments
– CH7.6 and the entire CH8
• Colloquium today
– At 4pm, Wednesday, Nov. 7, in SH101
– Dr. Nick White of Goddard Space Center, NASA
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
2
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
3
Group – Research Topic Association
Research Group
Research Topic
Presentation Date
1
2
3
4
5
6
7
8
9
10
6
5
7
2
1
9
10
4
3
8
12/5-4
12/5-5
12/5-1
12/3-2
12/3-3
12/3-5
12/3-1
12/5-3
12/5-2
12/3-4
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
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Solution of the Schrödinger Equation for Hydrogen
• Substitute  into the polar Schrodinger equation and separate the
resulting equation into three equations: R(r), f(θ), and g( ).
Separation of Variables
• The derivatives in Schrodinger eq. can be written as

R
 fg
r
r

f
 Rg


2
2 g
 Rf 2
 2

• Substituting them into the polar coord. Schrodinger Eq.
fg   2 R 
Rg  
f 
Rf
2 g 2 
 2 E  V Rgf  0
 r
  2
 sin   2 2
2
2
r r
r
r sin 

r sin  
h
• Multiply both sides by r2 sin2 θ / Rfg
sin 2    2 R  sin  
f  1 2 g 2  2 2
 2 r sin  E  V   0
 r
 
 sin  
2
R r
r
f 

g 
h
Reorganize
sin2    2 R  2  2 2
sin  
f  1 2 g

 r
  2 r sin  E  V  
 sin  
R r
r
h
f 

g  2
Monday, Nov. 5, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
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Solution of the Schrödinger Equation
• Only r and θ appear on the left-hand side and only 
appears on the right-hand side of the equation
• The left-hand side of the equation cannot change as 
changes.
• The right-hand side cannot change with either r or θ.
• Each side needs to be equal to a constant for the equation to
be true in all cases. Set the constant −mℓ2 equal to the righthand side of the reorganized equation
d2g
2

m
-------- azimuthal equation
l g
2
d
• It is convenient to choose a solution to be eiml  .
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
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Solution of the Schrödinger Equation
• eiml  satisfies the previous equation for any value of mℓ.
• The solution be single valued in order to have a valid solution for
any  , which requires g    g   2 
e0  e2 iml
g   0   g   2 
• mℓ must be zero or an integer (positive or negative) for this to work
• If the sign were positive, the solution would not be normalized.
• Now, set the remaining equation equal to −mℓ2 and divide either
side with sin2 and rearrange it.
1   2 R  2 r 2
ml2
1
 
f 
 r
  2 E  V   2 
 sin 
R r
r
h
sin  f sin 

• Everything depends on r on the left side and θ on the right side of
theWednesday,
equation.
Nov. 7, 2012
PHYS 3313-001, Fall 2012
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Dr. Jaehoon Yu
Solution of the Schrödinger Equation
• Set each side of the equation equal to constant ℓ(ℓ + 1).
– Radial Equation

1 d  2 dR  2  
h2
1   2 R  2 r 2
E V 
l l  1 R  0

r
  2 E  V   l l  1  2  r
r dr
dr  h2 
2
R r  r 
h

– Angular Equation
sin2  f sin  

 sin   l l  1 

ml2
1
 
f 
1 d 
df  
ml2 
 sin    l l  1  2  f  0
sin  d 
d
sin  

• Schrödinger equation has been separated into three
ordinary second-order differential equations, each
containing only one variable.
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
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Solution of the Radial Equation
• The radial equation is called the associated Laguerre
equation, and the solutions R that satisfy the appropriate
boundary conditions are called associated Laguerre
functions.
• Assume the ground state has ℓ = 0, and this requires mℓ = 0.
We obtain
1 d  2 dR  2 
 r
  2 E  V R  0
2
r dr
dr
h
dR
• The derivative of r
yields two terms, we obtain
dr
2
d 2 R 2 dR 2  
e2 

 2 E 
R0
2

dr
r dr h 
4 0 r 
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
9
Solution of the Radial Equation
• Let’s try a solution R  Aer a0 where A is a normalization constant,a0
is a constant with the dimension of length.
• Take derivatives of R, we obtain.
 1 2    2  e2
21
 a 2  h2 E    4 h2  a  r  0
0
0
0
• To satisfy this equation for any r, each of the two expressions in
parentheses must be zero.
• Set the second parentheses equal to zero and solve for a0.
4 0 h2
a0 
Bohr’s radius
 e2
• Set the first parentheses equal to zero and solve for E.
h2
E 
 E0  13.6eV
2
2 a0
• Both equal to the Bohr’s results
Wednesday, Nov. 7, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
Ground state energy
of the hydrogen atom
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