PHYS 3313 – Section 001
Lecture #18
Wednesday, Nov. 13, 2013
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
Intrinsic Spin
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
1
Announcements
• Reminder: homework #6
– CH6 end of chapter problems: 34, 39, 46, 62 and 65
– Due on Monday, Nov. 18, in class
• Reading assignments
– CH7.6 and the entire CH8
• Colloquium at 4pm in SH101
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
2
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
3
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
4
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
¶y
¶R
= fg
¶r
¶r
¶y
¶f
= Rg
¶q
¶q
¶2y
¶2 g
= Rf 2
¶f 2
¶f
• Substituting them into the polar coord. Schrodinger Eq.
fg ¶ æ 2 ¶R ö
Rg ¶ æ
¶f ö
Rf ¶2 g 2 m
+ 2 ( E - V ) Rgf = 0
çè r
÷ø + 2
çè sin q ÷ø + 2 2
2
2
r ¶r
¶r
r sin q ¶q
¶q
r sin q ¶f
• Multiply both sides by r2 sin2 θ / Rfg
sin 2 q ¶ æ 2 ¶R ö sin q ¶ æ
¶f ö 1 ¶2 g 2 m 2 2
+ 2 r sin q ( E - V ) = 0
çè r
÷ø +
çè sin q ÷ø +
2
R ¶r
¶r
f ¶q
¶q
g ¶f
Reorganize
sin 2 q ¶ æ 2 ¶R ö 2 m 2 2
sinq ¶ æ
¶f ö 1 ¶2 g
çè r
÷ø - 2 r sin q ( E - V ) çè sinq ÷ø =
R ¶r
¶r
f ¶q
¶q
g ¶f 2
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
5
Solution of the Schrödinger Equation
• eiml f 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 (f ) = g (f + 2p )
e0 = e2p iml
g (f = 0 ) = g (f = 2p )
• mℓ must be zero or an integer (positive or negative) for this to work
• Now, set the remaining equation equal to −mℓ2 and divide either
side with sin2 and rearrange them as
1 ¶ æ 2 ¶R ö 2 mr 2
ml2
1
¶ æ
¶f ö
çè r
÷ø + 2 ( E - V ) = 2 çè sin q ÷ø
R ¶r depends
¶r
q andf sin
q ¶the
q right ¶side
q of
• Everything
on r on the leftsin
side
θ on
the equation.
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
6
Solution of the Schrödinger Equation
• Set each side of the equation equal to constant ℓ(ℓ + 1).
– Radial Equation
2
ù
1 ¶ æ 2 ¶R ö 2 mr 2
1 d æ 2 dR ö 2 m é
l ( l + 1) ú R = 0
çr
÷ + 2 ( E - V ) = l ( l + 1) Þ 2 çè r
÷ + 2 êE - V R ¶r è ¶r ø
r dr
dr ø
2
m
ë
û
– Angular Equation
sin2 q f sinq ¶q è ¶q ø
ç sinq ÷ = l (l +1) Þ
ml2
1 ¶ æ ¶f ö
1 d æ
df ö é
ml2 ù
ç sin q ÷ø + êl ( l + 1) - 2 ú f = 0
sinq dq è
dq
sin q û
ë
• Schrödinger equation has been separated into three ordinary
second-order differential equations, each containing only one
variable.
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
7
Solution of the Radial Equation
• The radial equation is called the associated Laguerre
equation, and the solutions R that satisfies 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 æ 2 dR ö 2 m
çè r
÷ø + 2 [ E - V ] R = 0
2
r dr
dr
• The derivative of
yields two terms, and we obtain
dR
r
dr
2
Wednesday, Nov. 13,
2013
d 2 R 2 dR 2 m æ
e2 ö
+
+ 2 çE +
R=0
2
÷
dr
r dr
4pe 0 r ø
è
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
8
Solution of the Radial Equation
-r a
• Let’s try a solution R = Ae 0 where A is a normalization constant,
and a0 is a constant with the dimension of length.
• Take derivatives of R, we obtain.
æ 1 2 m ö æ 2 me2
2ö1
çè a 2 + 2 E ÷ø + çè 4pe 2 - a ÷ø 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.
4pe 0 2
a0 =
Bohr’s radius
2
me
• Set the first parentheses equal to zero and solve for E.
E =-
2
2 ma
2
0
= -E0 = -13.6eV
• Both equal to the Bohr’s results
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
Ground state energy
of the hydrogen atom
9
Principal Quantum Number n
• The principal quantum number, n, results from the
solution of R(r) in the separate Schrodinger Eq. since
R(r) includes the potential energy V(r).
The result for this quantized energy is
2
mæ e ö 1
E0
En = - ç
=- 2
2
÷
2 è 4pe 0 ø n
n
2
• The negative sign of the energy E indicates that the
electron and proton are bound together.
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
10
Quantum Numbers
• The full solution of the radial equation requires an
introduction of a quantum number, n, which is a non-zero
positive integer.
• The three quantum numbers:
– n
– ℓ
– mℓ
Principal quantum number
Orbital angular momentum quantum number
Magnetic quantum number
• The boundary conditions put restrictions on these
– n = 1, 2, 3, 4, . . .
– ℓ = 0, 1, 2, 3, . . . , n − 1
– mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ
• The predicted energy level is En = - E0
n2
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
(n>0)
Integer
(ℓ < n) Integer
(|mℓ| ≤ ℓ) Integer
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Ex 7.3: Quantum Numbers & Degeneracy
What are the possible quantum numbers for the state n=4 in
atomic hydrogen? How many degenerate states are there?
n
4
4
4
4
ℓ
0
1
2
3
mℓ
0
-1, 0, +1
-2, -1, 0, +1, +2
-3, -2, -1, 0, +1, +2, +3
The energy of a atomic hydrogen state is determined only by the
primary quantum number, thus, all these quantum states,
1+3+5+7 = 16, are in the same energy state.
Thus, there are 16 degenerate states for the state n=4.
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
12
Hydrogen Atom Radial Wave Functions
• The radial solution is specified by the values of n and ℓ
• First few radial wave functions Rnℓ
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
13
Solution of the Angular and
Azimuthal Equations
iml f
-iml f
• The solutions for azimuthal eq. are e or e
• Solutions to the angular and azimuthal
equations are linked because both have mℓ
• Group these solutions together into functions
Y (q , f ) = f (q ) g (f )
---- spherical harmonics
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
14
Normalized Spherical Harmonics
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
15
Ex 7.1: Spherical Harmonic Function
Show that the spherical harmonic function Y11(  ) satisfies the angular
Schrodinger equation.
1 3
Y11 (q ,f ) = sinq eif = Asinq
2 2p
Inserting l = 1 and ml = 1 into the angular Schrodinger equation, we obtain
1 d æ
dY11 ö é
1 ù
1 d æ
dY11 ö æ
1 ö
ç sinq
÷ + 1(1+1) - 2 ú Y11 =
ç sinq
÷ + ç 2 - 2 ÷ø Y11
sinq dq è
dq ø êë
sin q û
sinq dq è
dq ø è
sin q
A d æ
d sinq ö
1 ö
A d
1 ö
æ
æ
=
( sinq cosq ) + A çè 2 - 2 ÷ø sinq
ç sinq
÷ + A çè 2 - 2 ÷ø sinq =
sinq dq è
dq ø
sin q
sinq dq
sin q
=
A d æ1
1 ö
A
1 ö
ö
æ
æ
sin
2
q
+
A
2
sin
q
=
cos2
q
+
A
2
sinq
çè
÷ø
çè
çè
2 ÷
2 ÷
ø
ø
sinq dq 2
sin q
sinq
sin q
A
1 ö
A
1 ö
æ
æ
2
=
(1- 2sin q ) + A çè 2 - sin2 q ÷ø sinq = sinq - 2Asinq + A çè 2 - sin2 q ÷ø sinq = 0
sinq
Wednesday, Nov. 13,
2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
16
Solution of the Angular and Azimuthal Equations
• The radial wave function R and the spherical
harmonics Y determine the probability density for
the various quantum states.
• Thus the total wave function  (r,  ) depends on
n, ℓ, and mℓ. The wave function can be written as
y nlm ( r,q ,f ) = Rnl ( r )Ylm (q ,f )
l
Wednesday, Nov. 13,
2013
l
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
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