Chapter 3
The Hydrogen Atom
Application of the Schrödinger Equation to the
Hydrogen Atom

Solution of the Schrödinger Equation for Hydrogen

Quantum Numbers

Magnetic Effects on Atomic Spectra – The Normal
Zeeman Effect

Intrinsic Spin

Energy Levels and Electron Probabilities

Dr. H. Hanafy
 Application of the Schrödinger Equation to the
Hydrogen Atom
 The approximation of the potential energy of the electron-proton
system is electrostatic:
 Rewrite the three-dimensional time-independent Schrödinger
Equation.
For Hydrogen-like atoms (He+ or Li++)
 Replace e2 with Ze2 (Z is the atomic number).
 Use appropriate reduced mass μ.
• The potential (central force) V(r) depends on the
distance r between the proton and electron.
Transform to
spherical polar
coordinates because
of the radial
symmetry.
Insert the Coulomb
potential into the
transformed
Schrödinger
equation.
Equation 7.3
Dr. H. Hanafy
• The wave function ψ is a function of r, θ, φ .
Equation is separable.
Solution may be a product of three functions.
Equation
7.4
• We can separate Equation 7.3 into three separate
differential equations, each depending on one
coordinate: r, θ, or φ .
Dr. H. Hanafy
 Solution of the Schrödinger Equation for
Hydrogen
• Substitute Eq (7.4) into Eq (7.3) and separate the
resulting equation into three equations: R(r), f(θ),
and g( ).
Separation of Variables
• The derivatives from Eq (7.4)
• Substitute them into Eq (7.3)
• Multiply both sides by r2 sin2 θ / Rfg
Equation 7.7
Dr. H. Hanafy
Solution of the Schrödinger Equation
• Only r and θ appear on the left side and only φ appears on
the right side of Eq (7.7)
• The left side of the equation cannot change as changes.
• The right side cannot change with either r or θ.
• Each side needs to be equal to a constant for the equation
to be true.
Set the constant −mℓ2 equal to the right side of Eq (7.7)
Equation
7.8
-------- azimuthal equation
• It is convenient to choose a solution to be
Dr. H. Hanafy
.
Solution of the Schrödinger Equation
•
satisfies Eq (7.8) for any value of mℓ.
• The solution must be single valued in order to have
a valid solution for any , which is
• mℓ to be zero or an integer (positive or negative)
for this to be true.
• If Eq (7.8) were positive, the solution would not be
normalized.
Dr. H. Hanafy
Solution of the Schrödinger Equation
• Set the left side of Eq (7.7) equal to −mℓ2 and
rearrange it.
• Everything depends on r on the left side and θ on
the right side of the equation.
Dr. H. Hanafy
Solution of the Schrödinger Equation
• Set each side of Eq (7.9) equal to constant ℓ(ℓ + 1).
----Radial equation
Equation
7.10
----Angular equation
Equation
7.11
• Schrödinger equation has been separated into three
ordinary second-order differential equations [Eq (7.8),
(7.10), and (7.11)], each containing only one variable.
Dr. H. Hanafy
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.
Eq (7.10) becomes
• The derivative of
yields two terms and V is the
Coulomb potentials.
Equation 7.13
Write those terms:
Dr. H. Hanafy
Solution of the Radial Equation
• Try a solution
A is a normalized constant.
a0 is a constant with the dimension of length.
Take derivatives of R and insert them into Eq (7.13).
Equation 7.14
• To satisfy Eq (7.14) for any r is for each of the two
expressions in parentheses to be zero.
Set the second parentheses equal to zero and solve for
a0.
Set the first parentheses equal to zero and solve for E.
Both equal to the Bohr result.
Dr. H. Hanafy
Quantum Numbers
• The appropriate boundary conditions to Eq (7.10)
and (7.11) leads to the following restrictions on
the quantum numbers ℓ and mℓ:
– ℓ = 0, 1, 2, 3, . . .
– mℓ = −ℓ, −ℓ + 1, . . . , −2, −1, 0, 1, 2, . ℓ . , ℓ − 1, ℓ
– |mℓ| ≤ ℓ and ℓ < 0.
• The predicted energy level is
n = 1,2,3…..
n>l
Dr. H. Hanafy
Hydrogen Atom Radial Wave Functions
• First few radial wave functions Rnℓ
• Subscripts on R specify the values of n and ℓ.
Dr. H. Hanafy
Solution of the Angular and Azimuthal
Equations
• The solutions for Eq (7.8) are
.
• Solutions to the angular and azimuthal equations are
linked because both have mℓ.
• Group these solutions together into functions.
---- spherical harmonics
Dr. H. Hanafy
Normalized Spherical Harmonics
Dr. H. Hanafy
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. The total wave
function
depends on n, ℓ, and mℓ.
The wave function becomes
Dr. H. Hanafy
7.3: Quantum Numbers
The three quantum numbers:
– n
– ℓ
– mℓ
Principal quantum number
Orbital angular momentum quantum number
Magnetic quantum number
The boundary conditions:
– n = 1, 2, 3, 4, . . .
Integer
– ℓ = 0, 1, 2, 3, . . . , n − 1
Integer
– mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ Integer
The restrictions for quantum numbers:
– n>0
– ℓ<n
– |mℓ| ≤ ℓ
Dr. H. Hanafy
Principal Quantum Number n
• It results from the solution of R(r) in Eq (7.4) because
R(r) includes the potential energy V(r).
The result for this quantized energy is
• The negative means the energy E indicates that the
electron and proton are bound together.
Dr. H. Hanafy
Radial probability density
Radial probability density
r 2 Rnl2 (r )
<r>
(au)
Orbital
n l
1s
1 0
1.5
2s
2 0
6.0
2p
2 1
5.0
3s
3 0
13.5
n2
ao
*
r   d 3r  nlm
(r )r nlm (r )

  dr r Rnl  r 
3
0
2
Dr. H. Hanafy
Shapes of the spherical harmonics
Y11
Y00
Y10
z
y
x
Re[Y11 ]
l  1, m  0
l  0, m  0
Y00 
Y10 
1
4
3
cos 
4
l  1, m  1
Y11  
3
sin  exp(i )
8
(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)
Dr. H. Hanafy
Shapes of spherical harmonics (2)
Y22
Y21
Y20
Re[Y22 ]
Re[Y21 ]
z
y
x
l  2, m  0
Y20 
l  2, m  2
Y22 
15 2
sin  exp(2i )
32
5
(3cos 2   1)
16
l  2, m  1
Y21  
15
sin  cos  exp(i )
8
(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)
Dr. H. Hanafy
Orbital Angular Momentum Quantum
Number ℓ
• It is associated with the R(r) and f(θ) parts of the
wave function.
• Classically, the orbital angular momentum
with L = mvorbitalr.
• ℓ is related to L by
.
• In an ℓ = 0 state,
.
It disagrees with Bohr’s semiclassical “planetary”
model of electrons orbiting a nucleus L = nħ.
Dr. H. Hanafy
Orbital Angular Momentum Quantum
Number ℓ
• A certain energy level is degenerate with respect to ℓ
when the energy is independent of ℓ.
• Use letter names for the various ℓ values.
– ℓ=
5...
– Letter =
h...
0
1
2
3
4
s
p
d
f
g
• Atomic states are referred to by their n and ℓ.
• A state with n = 2 and ℓ = 1 is called a 2p state.
• The boundary conditions require n > ℓ.
Dr. H. Hanafy
8.2: Total Angular Momentum
Orbital angular momentum
Spin angular momentum
Total angular momentum
L, Lz, S, SzJ and Jz are quantized.
24
Total Angular Momentum
• If j and mj are quantum numbers for the single electron (hydrogen
atom).
• Quantization of the magnitudes.
• The total angular momentum quantum number for the single electron
can only have the values
25
Spin-Orbit Coupling
• An effect of the spins of the electron and the orbital angular momentum
interaction is called spin-orbit coupling.
• The dipole potential energy
.
• The spin magnetic moment .
•
•
.
is the magnetic field due to the proton.
where cos a is the angle between
.
26
Total Angular Momentum
No external magnetic field:
• Only Jz can be known because the uncertainty principle forbids Jx or Jy
from being known at the same time as Jz.
27
Total Angular Momentum
With an internal magnetic field:
•
will precess about
.
28
Total Angular Momentum
• Now the selection rules for a single-electron atom become
– Δn = anything
– Δmj = 0, ±1
Δℓ = ±1
Δj = 0, ±1
• Hydrogen energy-level diagram for n = 2 and n = 3 with the spin-orbit
splitting.
29
Many-Electron Atoms
Hund’s rules:
1) The total spin angular momentum S should be maximized to the extent
possible without violating the Pauli exclusion principle.
2) Insofar as rule 1 is not violated, L should also be maximized.
3) For atoms having subshells less than half full, J should be minimized.
•
For labeled two-electron atom
•
There are LS coupling and jj coupling to combine four angular momenta J.
30
LS Coupling
• This is used for most atoms when the magnetic field is weak.
• If two electrons are single subshell, S = 0 or 1 depending on whether the
spins are antiparallel or parallel.
•
•
•
•
For given L, there are 2S + 1 values of J.
For L > S, J goes from L − S to L + S.
For L < S, there are fewer than 2S + 1 possible J values.
The value of 2S + 1 is the multiplicity of the state.
31
•
•
•
LS Coupling
The notation for a single-electron atom becomes
n2S+1 LJ
The letters and numbers are called spectroscopic symbols.
There are singlet states (S = 0) and triplet states (S = 1) for two electrons.
32
LS Coupling
• There are separated energy
levels according to whether
they are S = 0 or 1.
• Allowed transitions must have
ΔS = 0.
• No allowed (forbidden)
transitions are possible
between singlet and triplet
states with much lower
probability.
33
LS Coupling
• The allowed transitions for the LS coupling scheme are
– ΔL = ±1
– ΔJ = 0, ±1
ΔS = 0
(J = 0 → J = 0 is forbidden)
• A magnesium atom excited to the 3s3p triplet state has no lower
triplet state to which it can decay.
• It is called metastable, because it lives for such a long time on the
atomic scale.
34
jj Coupling
• It is for the heavier elements, where the nuclear charge causes the spinorbit interactions to be as strong as the force between the individual and
.
35
8.3: Anomalous Zeeman Effect
• More than three closely spaced optical lines were observed.
• The interaction that splits the energy levels in an external magnetic field
is caused by
interaction.
• The magnetic moment depends on
Orbital contribution
and
Spin magnetic moment
• The 2J + 1 degeneracy for a given total angular momentum state J is
removed by the effect of the
.
• If the
is small compared to internal magnetic field, then and
precess about while precesses slowly about
.
36
•
Anomalous
Zeeman
Effect
The total magnetic moment is
μB is the Bohr magneton and
it is called the Landé g factor.
•
•
•
The magnetic total angular momentum numbers mJ from −J to J in integral steps.
splits each state J into 2J + 1 equally spaced levels separated ΔE = V.
For photon transitions between energy levels
ΔmJ = ±1, 0 but
is forbidden when ΔJ = 0.
37
The vector model
This is a useful semi-classical model of the quantum results.
Imagine L precesses around the z-axis. Hence the magnitude of L and
the z-component Lz are constant while the x and y components can take a
range of values and average to zero, just like the quantum eigenfunctions.
A given quantum number l determines the
magnitude of the vector L via
L  l (l  1)
2
z
2
L  l (l  1)
The z-component can have the 2l+1 values
corresponding to
Lz  m , l  m  l
In the vector model this means that only particular special
angles between the angular momentum vector and the zaxis are allowed
Dr. H. Hanafy
L
θ
Selection Rules
• We can use the wave functions to calculate
transition probabilities for the electron to change
from one state to another.
Allowed transitions:
• Electrons absorbing or emitting photons to change
states when Δℓ = ±1.
Forbidden transitions:
• Other transitions possible but occur with much
smaller probabilities when Δℓ ≠ ±1.
Dr. H. Hanafy