Lectures 3-4: One-electron atoms
o
Schrödinger equation for one-electron atom.
o
Solving the Schrödinger equation.
o
Wavefunctions and eigenvalues.
o Atomic orbitals.
o
See Chapter 7 of Eisberg & Resnick.
  (x, y,z,t)

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The Schrödinger equation
o
One-electron atom is simplest bound system in nature.
o Consists of positive and negative particles moving in 3D Coulomb potential:
V  V (x, y,z) 
o
Ze 2
4 0 x 2  y 2  z 2
Z =1 for atomic hydrogen, Z =2 for ionized helium, etc.

o Electron in orbit about proton treated using reduced mass:

o
mM
m M
Total energy of system is therefore,
 KE  PE  E
1 2
( px  py2  pz2 )  V (x, y,z)  E
2
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
The Schrödinger equation
o
Using the Equivalence Principle, the classical dynamical quantities can be replaced with their
associated differential operators:
px  i
E i



, py  i
, pz  i
x
y
z

t
o
  2
 2  2 

Substituting, we obtain the operator equation:   2  2  2  V (x, y,z)  i
2 x
y z 
t
o
Assuming electron can be described by a wavefunction of form,   (x, y,z,t)
2

 2 (x,y,z,t)  2 (x,y,z,t)
 2 (x,y,z,t) 
(x,y,z,t)




V(x,y,z
)(x,y,z
,t)

i
can write 


2 
x 2
y 2
z 2
t

2

    V  i
2
t
2
or

2

where,


2 2 2
  2 2 2
x y z
2
is the Laplacian operator.
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The Schrödinger equation
o
Since V(x,y,z) does not depend on time, (x, y,z,t)   (x, y,z)eiEt / is a solution to the
Schrödinger equation and the eigenfunction  (x, y,z) is a solution of the time-independent
Schrödinger equation:

o
2
2


 2 (x, y,z)  V (x, y,z)  E (x, y,z)
As V = V(r), convenient to use spherical polar coordinates.


where
2 
o
2
2
 2 (r,,  )  V (r) (r,,  )  E (r,,  )
(1)
1  2
1


1
2
(r
)
(sin

)
r 2 r r r 2 sin  
 r 2 sin 2   2
Can now use separation of variables to split the partial
differential
equation into a set of ordinary differential equations.

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Separation of the Schrödinger equation
(r,,)  R(r)()()
o
Assuming the eigenfunction is separable:
o
Using the Laplacian, and substituting (2) and (1):

2
(2)
1   2 R 
1
 
R 
1
 2 R
  2 r

sin 

 V (r)R  ER
2 r r 
r  r 2 sin   
  r 2 sin 2   2 
o
Carrying out the differentiations,


   2 dR  R  
d 
R d 2
  2
r

sin 

 V (r)R  ER
2  r r  dr  r 2 sin   
d  r 2 sin 2  d 2 
2
o
Note total derivatives now used, as R is a function of r alone, etc.
o
Now multiply through by 2r 2 sin 2  /R 2 and taking transpose,
1 d 2
sin 2  d  2 dR  sin  d 
d  2 2 2

r sin [E  V (r)]
r

sin 

2
 d
R dr  dr   d 
d  2

(3)
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Separation of the Schrödinger equation
o As the LHS of Eqn 3 does nor depend on r or  and RHS does not depend on  their
common value cannot depend on any of these variables.
1 d 2
d 2
2
 ml  2  ml2
o Setting the LHS of Eqn 3 to a constant:
2
 d
d
and RHS becomes

(4)
1 d  2 dR 
1
d 
d  2 2
ml2

r [E  V (r)]   2
r

sin 

R dr  dr  sin  d 
d  2
sin 

1 d  2 dR  2r 2
ml2
1
d 
d 

r
 2 [E  V (r)]  2 
sin 

R dr  dr 
sin  sin  d 
d 
o Both sides must equal a constant, which we choose as l(l+1):

1 d 
d  ml2

 l(l  1)
sin 

sin  d 
d  sin 2 
1 d  2 dR  2r 2
R
r
 2 [E  V (r)]R  l(l  1) 2
2
r dr  dr 
r
(5)
(6)
o We have now separated the time-independent Schrödinger equation into three
ordinary differential equations, which each only depend on one of  (4),  (5) and R(6).

.
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Summary of separation of Schrödinger equation
o
Express electron wavefunction as product of three functions: (r,,)  R(r)()()
o As V ≠ V(t), attempt to solve time-independent Schrodinger equation.

o
Separate into three ordinary differential equations for R(r),( ) and ( ).
o
Eqn. 4 for () only has acceptable solutionsfor certain value
of ml.

o
Using these values for ml in Eqn. 5, () only has acceptable values for certain
values of l.
o
With these values for l in Eqn. 6, R(r) only has acceptable solutions for certain
values of En.
o
Schrödinger equation produces three quantum numbers!
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Azimuthal solutions (())
( )  e iml 
o
A particular solution of (4) is
o
As the einegfunctions must be single valued, i.e.,  =>

e iml 0  e iml 2
and using Euler’s formula,
1 cosml 2  isin ml 2
o
This is only satisfied if ml
= 0, ±1, ±2, ...
o
Therefore, acceptable solutions to
(4) only exist when ml can only have certain integer
values, i.e. it is a quantum number.
o
ml is called the magnetic quantum number in spectroscopy.
o
Called magnetic quantum number because plays role when atom interacts with magnetic
fields.
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Polar solutions (())
o
Making change of variables (z = rcos, Eqn. 5 transformed into an associated Legendre
equation:
2

d 
(7)
m l 
2 d 
(1- z ) +l(l +1)  = 0

dz 
dz  
1- z 2 
o
Solutions to Eqn. 7 are of form
lml ( )  sin |ml | Fl|ml | (cos  )

where Fl|m | (cos  ) are associated Legendre polynomial functions.
l

o
 remains finite when


 = 0, 1, 2, 3, ...
ml = -l, -l+1, .., 0, .., l-1, l
Can write the associated Legendre functions using quantum number subscripts:
00 = 1
10 = cos
 20 = 1-3cos2

2±2 = 1-cos2
1±1 = (1-cos2)1/2
2±1 = (1-cos2)1/2cos
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Spherical harmonic solutions
o
Customary to multiply () and () to form so called spherical harmonic functions
which can be written as:
m
Y ( , ) = 
m
( ) m ( )
i.e., product of trigonometric and polynomial

functions.
o
First few spherical harmonics are:
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Y00= 1
Y10= cos
Y1±1= (1-cos2)1/2 e±i
Y 20= 1-3cos2 Y2±1= (1-cos2)1/2cos e±i
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Radial solutions (R( r ))
o
What is the ground state of hydrogen (Z=1)? Assuming that the ground state has n = 1, l = 0
Eqn. 6 can be written
1 d  2 dR  2 
e 2 
r
 E 
R  0
r 2 dr  dr  2 
40 r 
o
Taking the derivative
d 2 R 2 dR 2 
e 2 

 2 E 
R  0
dr 2 r dr
40 r 


o
(7)
Try solution R  Aer / a, where A and a0 are constants. Sub into Eqn. 7:
0

o
o

To satisfy this Eqn. for any r, both expressions in brackets must equal zero. Setting the second
expression to zero =>
40 2
a0 
e 2
Setting first term to zero =>
Same as Bohr’s results
2
E 
 13.6 eV
2


2a0
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Radial solutions (R( r ))
o
Radial wave equation
d 2 R 2 dR 2 
e 2 

 2 E 
R  0
dr 2 r dr
40 r 

has many solutions, one for each positive integer of n.

o
Solutions are of the form (see Appendix N of Eisberg & Resnick):
l


Zr 
Zr
Zr / na0
Rnl (r)  e
  Gnl  
a0  a0 
where a0 is the Bohr radius. Bound-state solutions are only acceptable if
Z 2e 4
En  
(4 0 ) 2 2 2 n 2

Z2
 13.6 2
n
eV
where n is the principal quantum number, defined by n = l +1, l +2, l +3, …

o
En only depends on n: all l states for a given n are degenerate (i.e. have the same energy).
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Radial solutions (R( r ))
o
Gnl(Zr/a0) are called associated Laguerre polynomials, which depend on n and l.
o
Several resultant radial wavefunctions (Rnl( r )) for the hydrogen atom are given below
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Radial solutions (R( r ))
o
The radial probability function Pnl(r ), is the probability that the electron is found between r
and r + dr:
Pnl (r)dr 

2
Rnl (r)Ylml (,  ) r 2 dV
 4 r 2 Rnl r 2 dr
2
o
Some representative radial probability functions are given at right:
o
Some points to note:
o The r2 factor makes the radial probability density
vanish at the origin, even for l = 0 states.
o For each state (given n and l), there are n - l - 1
nodes in the distribution.
o The distribution for states with l = 0, have n maxima,
which increase in amplitude with distance from origin.
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Radial solutions (R( r ))
o
Radial probability distributions for an electron in several of the low energy orbitals of
hydrogen.
o
The abscissa is the radius
in units of a0.
s orbitals
p orbitals
d orbitals
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Hydrogen eigenfunctions
o
Eigenfunctions for the state described by the quantum numbers (n, l, ml) are therefore of form:
 nlm (r,,  )  Rnl (r) lm ( )m ( )
l
l
l
and depend on quantum numbers:

n = 1, 2, 3, …
l = 0, 1, 2, …, n-1
ml = -l, -l+1, …, 0, …, l-1, l
o
Energy of state on dependent on n:
En  
o
13.6Z 2
n2
Usually more than one state has same
energy, i.e., are degenerate.

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Born interpretation of the wavefunction
o
Principle of QM: the wavefunction contains all the dynamical information about the system it
describes.
o
Born interpretation of the wavefunction: The probability (P(x,t)) of finding a particle at a
position between x and x+dx is proportional to |(x,t)|2dx:
P(x,t) = *(x,t) (x,t) = |(x,t)|2
o
P(x,t) is the probability density.
o
Immediately implies that sign of wavefunction has no
direct physical significance.
(x,t)
P(x,t)
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Born interpretation of the wavefunction
o
In H-atom, ground state orbital has the same sign everywhere => sign of orbital must be all
positive or all negative.
o
Other orbitals vary in sign. Where orbital changes sign,  = 0 (called a node) => probability
of finding electron is zero.
o
Consider first excited state of hydrogen: sign of
wavefunction is insignificant (P = 2 = (-)2).
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Born interpretation of the wavefunction
o
Next excited state of H-atom is asymmetric about origin. Wavefunction has opposite sign on
opposite sides of nucleus.
o
The square of the wavefunction is identical on
opposite sides, representing equal distribution
of electron density on both side of nucleus.
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Atomic orbitals
o
Quantum mechanical equivalent of orbits in Bohr model.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
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s orbitals
0
-0.2
0.2
o
Named from “sharp” spectroscopic lines.
o
l = 0, ml = 0
0.2
0
-0.2
o
n,0,m = Rn,0 (r ) Y0,m (, )
Y0,0 
-0.2
0
0.2
1
4
o
Angular solution:
o
Value of Y0,0 is constant over sphere.
o

1  Z  Zr / a


  e
For n = 0, l = 0, ml = 0 => 1s orbital 1,0,0
 a0 
o
The probability density is
3/2
0
Pnlml (r,,  ) nlml (r,,  )*nlml (r,,  )
1 Z 
 P1,0,0 (r,,  )    e2Zr / a 0
 a0 
3

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p orbitals
o
Named from “principal” spectroscopic lines.
o
l = 1, ml = -1, 0, +1 (n must therefore be >1)
o
n,1,m = Rn1 (r ) Y1,m (, )
o
Angular solution:
o
A node passes through the nucleus and separates the two lobes
of each orbital.

Dark/light areas denote opposite sign of the wavefunction.
o
Y1,0 
3
cos
4
 3 1/ 2
 2,1,0 (r,,  )  R2,1 (r)  cos  2 pz
4 
o
Three p-orbitals denoted px, py , pz

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d orbitals
o
Named from “diffuse” spectroscopic lines.
2
1
o
l = 2, ml = -2, -1, 0, +1, +2 (n must therefore be
>2)
0
-1
o
-2
-1
n,2,m = Rn1 (r ) Y2,m (, )
Y2,0 
Angular solution:
o
There are five d-orbitals, denoted
o
0
-0.5
0.5
0
0.5
1
1
5 1
(3cos2  1)
4 2
o

-1
-0.5
dz 2 , dxz,dyz ,dxy ,dx 2 y 2
m = 0 is z2. Two orbitals of m = -1 and +1 are xz
and yz. Two orbitals with m = -2 and +2 are
designated xyand x2-y2.
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Quantum numbers and spectroscopic notation
o Angular momentum quantum number:
o l = 0 (s subshell)
o l = 1 (p subshell)
o l = 2 (d subshell)
o l = 3 (f subshell)
o …
o
Principal quantum number:
o n = 1 (K shell)
o n = 2 (L shell)
o n = 3 (M shell)
o …
o
If n = 1 and l = 0 = > the state is designated 1s. n = 3, l = 2 => 3d state.
o
Three quantum numbers arise because time-independent Schrödinger equation contains three
independent variables, one for each space coordinate.
o
The eigenvalues of the one-electron atom depend only on n, by the eigenfunctions depend on
n, l and ml, since they are the product of Rnl(r ), lml () and ml().
o
For given n, there are generally several values of l and ml => degenerate eigenfunctions.
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Orbital transitions for hydrogen
o
Transition between different energy levels of
the hydrogenic atom must follow the
following selection rules:
l = ±1
m = 0, ±1
o
A Grotrian diagram or a term diagram shows
the allowed transitions.
o
The thicker the line at right, the more probable
and hence more intense the transitions.
o The intensity of emission/absorption
lines could not be explained via Bohr
model.
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Schrödinger vs. Bohr models
o
Schrodinger’s QM treatment had a number of advantages over semi-classical
Bohr model:
1. Probability density orbitals do not violate the Heisenberg Uncertainty
Principle.
1. Orbital angular momentum correctly accounted for.
1. Electron spin can be properly treaded.
1. Electron transition rates can be explained.
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