Download The Hydrogen atom.

The Hydrogen atom.
In these lecture notes I will discuss the solution of the time-independent Schrödinger
equation for the hydrogen atom. The hydrogen atom is the simplest atom and can be
solved exactly. It is a useful model for all other atoms. The S.E. for the hydrogen atom
can be reduced to a ony-body problem in three dimensions. Even more essential the
problem has spherical symmetry, and it will therefore be advantageous to use spherical
coordinates to describe the solutions and tackle the problem. The angular part of the
problem shows up in many guises in physical chemistry and is not restricted at all to
finding atomic orbitals. We will use a very powerful way of finding solutions to this
problem that can be used in precisely the same way for such diverse problems as finding
the eigenfunctions of the rigid rotor, the description of spin eigenstates (singlets, triplets
etc.), or hyperfine splitting in atomic absorption spectra. As we will see later on, even
NMR and ESR spectra can be qualitatively understood in this manner. The method of
solving the angular problem involves working with operators and commutators and this
type of approach is used very often in the literature nowadays. It is useful to know, and
very elegant too, I may add - I can't resist telling you about it for the sheer beauty of it.
The second part of the Hydrogen atom problem involves the radial part of the
Schrödinger equation. I will also discuss this in some detail, but it has far less general
applicability than the angular part.
As always in quantum theory we analyse the classical problem first in order to derive the
quantum Hamiltonian. The classical energy of a proton and an electron would consist of
the kinetic energy + Coulomb interaction
E=
pp2
2 mp
+
pe 2
e2
−
r r
2me 4πε 0 re − rp
(6.1)
where the subscript p indicates the proton while e refers to the electron. This problem of
two-particles interacting through a central potential (depending only on the distance
between the particles) occurs often and as discussed nicely in for example Metiu, any
1
two-body problem in which the potential energy depends only on the distance between
the two bodies can be separated in a center of mass problem and a one-body problem that
involves the reduced mass. Let me give a brief derivation of the Hamiltonian, or the
classical energy for such a system.
General reduction of two-particle problem with central potential into center of mass
problem plus a one-body problem.
Assume we have coordinates
dr1
= m1r&1
dt
dr
r2 , p 2 = m2 2 = m2r&2
dt
r1 , p1 = m1
(C.1)
and define the center of mass coordinate and momentum
m1r1 + m2r2
, M = m1 + m2
m1 + m2
& cm = m r& + m r&
=MR
R cm =
P cm
1 1
(C.2)
2 2
and the relative coordinate
r = r2 − r1 , µ =
m1m2
, p = µ (r&2 − r&1 )
m1 + m2
(C.3)
Then we can show that the kinetic energy can be written in terms of these new
coordinates as
P cm ⋅ P cm p ⋅ p
+
=
2M
2µ
mm
1
(m1r&1 + m2r&2 ) ⋅ (m1r&1 + m2r&2 ) + 1 2 (r&2 − r&1 ) ⋅ (r&2 − r&1 )
2M
2M
1
(m12 r&12 + m2 2 r&2 2 + m1m2r&12 + m1m2r&2 2 )
=
2(m1 + m2 )
T=
=
(C.4)
1
(m1r&12 (m1 + m2 ) + m2 r&2 2 (m1 + m2 ))
2(m1 + m2 )
1
= (m1r&12 + m2 r&2 2 )
2
2
Therefore the classical energy expression consists of a kinetic energy of free particle
motion for the center of mass, plus an energy expression in the relative coordinate:
E=

( P cm ) 2  p 2
+
+ V (r ) 
2M
 2µ

(C.5)
Let us now continue the discussion for the Hydrogen atom along these lines.
General outline of solution for Hydrogen atom.
If we define
r
r
r mp rp + mere
Rc =
; M = mp + me
mp + me
(6.2a)
mm
mp
r r r
≈ me
r = re − rp ; µ = e p = me
me + mp
me + mp
(6.2b)
the energy becomes
2
r
PC
p2
e2
E=
+
−
; r= r
2 M 2µ 4πε 0r
(6.3)
The classical energy can be immediately translated into the quantum mechanical
Hamiltonian by making the substitutions px → − ih
∂
, etc, and realizing that
∂x
p 2 = px2 + p 2y + pz2 . This leads to the quantum mechanical Hamiltonian
2
∇2
e2
2 ∇C
2
$
−h
−
H total = − h
2M
2 µ 4πε 0r
where ∇ 2 =
(6.4)
∂2
∂2
∂2
(it is pronounced "nabla squared"). We are interested in
+
+
∂x 2 ∂y 2 ∂z 2
finding the eigenstates of this hamiltonian. The center of mass motion separates easily if
r
r
we try the solution Φ( Rc )Ψ( r ) , which leads to
r
2
∇2
r
1
e2
2 ∇ C Φ( Rc )
2
r +
−h
−
h
−
(6.5)
[
]Ψ( r ) = Etotal
r
2µ 4πε 0r
2 MΦ( Rc ) Ψ( r )
Each of the terms on the left hand side must equal a constant. The total energy is given by
the sum of the translational energy of the center of mass and the internal energy of the
3
hydrogen atom. One possible solution for the center of mass wave function is a plane
r r
r
h2 k 2
ik ⋅ Rc
wave Φ( RC ) = e
with energy ECM =
. It is infinitely degenerate though (every
2M
r
direction of k with the same length corresponds to the same energy) and this translational
energy can take on any positive value. It is not quantized. It would be quantized if we put
the system in a box, and this is what is used in statistical mechanics to discuss
r
translational energy. Henceforth we will only consider the internal wave function Ψ( r )
which is associated with the Hamiltonian
∇2
∇2
e2
e2
2
2
$
−
≈ −h
−
H = −h
2 µ 4πε 0r
2 me 4πε 0r
(6.6)
The latter Hamiltonian corresponds to a fixed proton in the origin. We could have started
from this formulation and we would have made only a very small error. Our problem
H$ Ψ( x , y , z ) = EΨ( x , y , z )
becomes
Because the potential only depends on the distance to the origin it is convenient to us
spherical coordinates. This will allow us to separate variables, and investigate various
problems separately. Defining
x = r sin θ cos ϕ ; y = r sin θ sin ϕ ; z = r cos θ ;
0 ≤ r < ∞ ; 0 ≤ θ ≤ π ; 0 ≤ ϕ ≤ 2π ;
(6.7)
with the corresponding inverse transformation
c
r = x2 + y2 + z2
h
1/ 2
c
y
; ϕ = arctan( ) ; θ = arccos( z x 2 + y 2 + z 2
x
h
−1/ 2
)
(6.8)
we can write the Schrödinger equation
e2
h2 2
H$ Ψ( r ,θ , ϕ ) = ( −
)Ψ( r ,θ , ϕ ) = EΨ( r ,θ , ϕ )
∇ −
2µ
4πε 0r
(6.9)
In order to discuss this problem further we need to obtain ∇2 in spherical coordinates.
This can be done by straightforward but very tedious manipulation using essentially the
chain rule. For example
∂f ( r , θ , ϕ ) ∂f ∂r ∂f ∂θ ∂f ∂ϕ
=
* +
*
+
*
∂x
∂r ∂x ∂θ ∂x ∂ϕ ∂x
for any f ( r ,θ , ϕ ) (chain rule).
∂
∂r ∂ ∂θ ∂ ∂ϕ ∂
+
+
→
∂x ∂r ∂x ∂θ ∂x ∂ϕ
∂x
4
∂r ∂ 2
1
x
= ( x + y 2 + z 2 )1/ 2 = ( x 2 + y 2 + z 2 )−1/ 2 * 2 x = = sin θ cos ϕ
r
∂x ∂x
2
So the procedure would be: "Take derivatives, and express (eventually) everything in
spherical coordinates." Very tedious! Let us work out one example explicitly, as it
involves a famous operator:
∂
∂
∂r
∂ϕ ∂
∂r ∂
∂θ
∂θ ∂
∂ϕ
L$z = ih( x − y ) = ih( x − y ) + ih( x
)
−y )
+ ih( x
−y
∂y
∂x
∂y
∂x ∂r
∂y
∂x ∂θ
∂y
∂x ∂ϕ
x
x
x
∂r
∂r xy yx
−y
=
−
=0
∂y
∂x r
r
∂θ
∂θ
−y
= mess *( xy − yx ) = 0
∂y
∂x
∂ϕ
∂ϕ
x
−y
) =1
−y
=x 2
− y( 2
2
∂y
∂x
x +y
x + y2
∂
. We have used this operator before for the particle on the ring!
and hence L$z = ih
∂ϕ
h2 2
Similar manipulations yield for −
∇ :
2µ
−
h2 2
h2 1 ∂ 2 ∂
L$2
∇ =−
+
(
r
)
∂r 2µr 2
2µ
2µ r 2 ∂r
(6.10)
where L$2 = L$2x + L$2y + L$2z is precisely the square of the angular momentum operator
1 ∂
1 ∂2
∂
L$2 = − h 2 [
(sin θ
)+ 2
]
sin θ ∂θ
∂θ sin θ ∂ϕ 2
(6.11)
Please note that h 2 is part of the definition of the operator L$2 . The operator L$2 is also
encountered when discussing the rigid rotor model for a diatomic molecule. It shows up
very frequently in quantum mechanics, and below we will discuss the eigenfunctions of
this operator, which are functions of the angular coordinates only. For now let us just
assume that such eigenfunctions exist and let us denote them ga (θ , ϕ ) , where
L$2 ga (θ , ϕ ) = h 2aga (θ , ϕ ) ,
(6.12)
such that the eigenvalue is h 2a . Assuming this, we try solutions for the Schrödinger
equation of the form
Ψ( r ,θ , ϕ ) = f ( r )* ga (θ , ϕ )
(6.13)
5
and substituting this form in the Schrödinger equation
$ ( r ) g (θ , ϕ ) = Ef ( r ) g (θ , ϕ )
Hf
a
a
(6.14)
we obtain
−
e2
h 2 1 ∂ 2 ∂f
h 2a
r
(
)
+
(
−
) f ( r ) = Ef ( r )
2 µ r 2 ∂r
2µr 2 4πε 0r
∂r
(6.15)
This is a one-dimensional differential equation for f ( r ) , that depends on the eigenvalue
a of the angular part of the wave function. We are hence left with two subproblems:
Problem a: Find the precise eigenvalues and eigenfunctions of the L$2 operator (the socalled angular problem).
Problem b: Find solutions of the radial equation for fixed value of a .
Intermezzo: Working with commutators.
In the following we will make extensive use of commutation relations. The following
rules come in very handy:
1.
cA$ , B$ = A$ , cB$ = c A$ , B$
In the above c is a number, but it could also be an operator that commutes with both A$
and B$ ! If an operator commutes with everything else we can always simply put it in
front of everything.
2.
$ $ , C$ = A$ B$ , C$ + A$ , C$ B$
AB
proof:
$ $ $ − CAB
$ $ $ = ( ABC
$ $ $ − ACB
$ $ $ ) + ( ACB
$ $ $ − CAB
$ $ $ ) = A$ B$ , C$ + A$ , C$ B$
ABC
6
This rule allows us to express unknown commutators in terms of elementary
commutators and operator products. It is often a very quick way to simplify matters.
Examples abound below!
3.
A$ , B$ + C$ = A$ ( B$ + C$ ) − ( B$ + C$ ) A$ = A$ , B$ + A$ , C$
Solution of angular problem using operator algebra.
We will show that the angular eigenvalue problem can be solved using only the
commutation relations between the angular momentum operators. So everything follows
from the following 'definitions' :
L$x , L$ y = ihL$z
L$z , L$x = ihL$ y
L$ y , L$z = ihL$x
(A.1)
L$2 = L$2x + L$2y + L$2z
In fact you derived these anticommutation relations yourself (MS problem 4-17) from the
r r
definition of L$ = r ∧ p , and the commutation relations rk , pl = ihδ kl , where k , l label the
three cartesian directions in space (x,y,z). Using the above mentioned rules this is very
easy:
$ $ z − zp
$$ y , zp
$$ x − xp
$$ z
L$x , L$ y = yp
$ $ x p$ z , z$ − yx p$ z , p$ z − z 2 p$ y , p$ x + xp
$$ y z$, p$ z = ih( xp
$$ y − yp
$ $ x ) = ihL$z
= yp
Please note how we moved commuting operators in front for immediate simplifications!
We will also use the following auxiliary operators:
L$+ = L$x + iL$ y
L$− = L$x − iL$ y
(A.2)
Using the commutation relations you can derive:
7
L$2 , L$x = L$2 , L$ y = L$2 , L$z = 0
(A.3)
L$2 = L$− L$+ + L$2z + hLz
= L$ L$ + L$2 − hL
(A.4)
L$2 , L$+ = L$2 , L$− = 0
(A.5)
L$z , L$+ = hL$+
(A.6)
L$z , L$− = −hL$−
(A.7)
+
−
z
z
examples:
L$2 , L$x = L$2x , L$x + L$2y , L$x + L$2z , L$x
= 0 + L$ y L$ y , L$x + L$ y , L$x L$ y + L$z L$z , L$x + L$z , L$x L$z
(A.8)
= − ih( L$ y L$z + L$z L$ y ) + ih( L$z L$ y + L$ y L$z ) = 0
L$z , L$+ = L$z , L$x + iL$ y = L$z , L$x + i L$z , L$ y
= ihLy + i( − ihLx ) = h( L$x + iL$ y ) = hL$+
(A.9)
Let us now continue and derive the eigenvalues of L$2 just by using the commutation
relations. In fact we will use that since L$2 , L$z = 0 these operators must have common
eigenfunctions. We could have used just as easily L$x or L$ y , but is is the standard
convention to use the pair L$2 and L$z . Let us call these common eigenfucntions ga ,b (θ , ϕ ) ,
where a is the eigenvalue of L$2 while b is the eigenvalue of Lz . They are arbitrary (real)
numbers at the moment.
L$2 ga ,b = aga ,b
(A.10)
L$z ga ,b = bga ,b
(A.11)
L$2 = L2x + L2y + L2z
(A.12)
However because
8
we immediately deduce that
a ≥ b2 ≥ 0
(A.13)
Next, if ga ,b is a common eigenfunction of L$2 , L$z then L$+ ga ,b , a new function, is also an
eigenfunction of both these operators. Proof:
L$2 ( L$+ ga ,b ) = L$+ L$2 ga ,b = a( L$+ ga ,b )
(A.15)
L$z ( L$+ ga ,b ) = L$+ L$z ga ,b + hL$+ ga ,b = ( b + h )( L$+ ga ,b )
(A.16)
hence L$+ ga ,b is an eigenfunction with eigenvalues ( a , b + h ) , or L$+ ga ,b is zero..... It is seen
that acting with L+ keeps the eigenvalue of L$2 the same but it increases the eigenvalue of
L$z by the amount h . This clearly indicates that the eigenvalue a of L$2 is degenerate: in
general there is more than one eigenfunction with the same eigenvalue!
Similarly
L$2 ( L$− ga ,b ) = L$− L$2 ga ,b = a( L$− ga ,b )
(A.17)
L$z ( L$− ga ,b ) = L$− L$z ga ,b − hL$− ga ,b = ( b − h )( L$− ga ,b )
(A.18)
L$− ga ,b is a common eigenfunction of L$2 , L$z with eigenvalues ( a , b − h ) (or it is zero.....).
L$+ and L$− are called ladder operators. They define eigenfunctions having adjacent
eigenvalue of L$z (shift b by ± h) but they leave the eigenvalue of L$2 unchanged.
However the ladder operators cannot act indefinitely, since b2 ≤ a . Let us call the
maximum eigenvalue bmax = lh . This means that the next higher function generated by L$+
has to vanish!
L$+ ga ,lh = 0
(A.19)
In the sequel I will suppress h in the subscribt: ga ,lh → ga ,l . It will turn out that the natural
unit of angular momentum is h, and the formulas take a simpler form if we write
bmax = lh . Since l is arbitrary, this is not a limitation, just a convenience. Interestingly
enough we can immediately find the eigenvalue of L$2 if bmax = lh . We use the specific
form for L$2 that acts with L$+ first (see Eqn. A.4), because we know L$+ ga ,l = 0.
9
L$2 ga ,l = ( L$− L$+ + L$2z + hL$z ) ga ,l = ( h 2l 2 + h 2l ) ga ,l = h 2l ( l + 1) ga ,l
(A.20)
Similarly acting by L$− the laddering down process must end. Let us call −kh the
minimum value, such that L$− ga , − k = 0 . The corresponding eigenvalue of L$2 is given by
(now we use the form of L$2 in which L$− acts first (eqn. A.4)):
L$2 ga , − k = ( L$+ L$− + L$2z − hL$z ) ga , − k = ( h 2 k 2 + h 2 k ) ga ,l = h 2 k ( k + 1) ga ,l
(A.21)
Since the eigenvalue a is the same we must have k = l .
We can summarize the above result by saying we get the complete set of eigenvalues:
a = l ( l + 1)h 2 , eigenvalues of L$2 , l ≥ 0
(A.22)
Corresponding values of L$z : mh, m = − l ,− l + 1,....., l − 1, l
(A.23)
And we can use as the defining equations for the (unnormalized) eigenfunctions:
L$+ gl ,l = 0; gl ,m−1 = L$− gl ,m
(unnormalized, laddering down)
(A.24)
or
L$− gl , − l = 0; gl ,m+1 = L$+ gl ,m (unnormalized, laddering up)
(A.25)
What are allowed values of l , (which up to now was only restricted to be ≥ 0 )? By
raising −l by 1 each time we must end up at +l and this means that there are only two
types of possibilities
A. l is integer
B. l is half integer.
The first possibility can describe functions gl ,m (θ , ϕ ) (see below). With the second type
of eigenvalue we cannot associate a well defined eigenfunction in the angular coordinates
however. Still, they turn out to have a physical meaning. They turn up when we describe
10
the spin of particles! Each value of l describes a set of eigenfunctions
gl ,m , m = − l ,− l + 1,..., l − 1, l . They are said to form a multiplet of dimension 2l + 1.
In the table below I have listed the lowest types of multiplets. With each value of l we
have 2l + 1 values of m, that range from − l ,− l + 1,...., l − 1, l , as shown
L$2 ( h 2 )
L$z ( h )
l ( l +1)
m
l=0
0
0
l =1
2
l=2
l =3
degeneracy L$2
spatial
spin
"name"
"name"
1
s
singlet
-1,0,1
3
p
triplet
6
-2,-1,0,1,2
5
d
quintet
12
-3,-2,-1,0,1,2,3
7
f
septet
l=
1
2
3
4
1 1
− ,
2 2
2
-
doublet
l=
3
2
15
4
3 1 1 3
− ,− , ,
2 2 2 2
4
-
quartet
l=
5
2
35
4
5 3 1 1 3 5
− , − ,− , , ,
2 2 2 2 2 2
6
-
sextet
At this point we have shown the general structure of the solutions, which we derived
using only the commutation relations between the operators. The first four relations in
this section is the only thing we needed, and all of the rest follows or can be derived.
Presently, in quantum mechanics the commutation relations are taken as the definition of
angular momentum. This is for example why spin is considered angular momentum: the
spin operators simply satisfy the same commutation relations!
11
If we assume the standard definition for angular momentum we can do a little more and
also derive the corresponding eigenfunctions in spherical coordinates. The operators
L$z , L$+ , L$− can be expressed in spherical coordinates, just like we did for Lz . They would
take the form:
∂
L$z = ih
∂ϕ
cos θ ∂
∂
L$+ = heiϕ [ + i
]
sin θ ∂ϕ
∂θ
cosθ ∂
∂
L$− = he − iϕ [ − i
]
sin θ ∂ϕ
∂θ
(A.26)
It is easy to find solutions that are eigenfuctions of L$z
− ih
Boundary condition:
∂
f (ϕ ) = mhf (ϕ ) → f (ϕ ) = eimϕ
∂ϕ
(A.27)
f (ϕ + 2π ) = f (ϕ ) → m is integer (only integer values allowed for
m!) This is the reason that the half-integer (spin) functions cannot be expressed in θ , ϕ
coordinates. They would not be single-valued functions in 3d-space!
Next we can solve for the θ -part. In spherical coordinates the eigenfunctions gl ,m are
conventionally denoted as Yl m (θ , ϕ ) . They are called the spherical harmonics. The ϕ dependent part of these functions is determined above, and the Yl m (θ , ϕ ) can be written as
Yl m (θ , ϕ ) = Pl m (θ )eimϕ ,
(A.28)
where the Pl m (θ ) are so-called associated Legendre polynomials. They can be easily
generated using the ladder operators. If we take the function with m = l it has to satisfy
bg
L+Yl l (θ , ϕ ) = 0 ; Yl m (θ , ϕ ) = Pl l θ eilϕ
(A.29)
And using the spherical coordinate form for the L$+ operator we find
eiϕ eilϕ [
∂
cos θ l
−l
]Pl (θ ) = 0
sin θ
∂θ
(A.30)
It is easily verified that the solution is
Pl l (θ ) = (sin θ )l = sin l θ
as
12
l sin l −1 θ cos θ − l
cos θ l
sin θ = 0
sin θ
b g
The highest m-valued function in a multiplet, Yl l θ , ϕ hence has the simple form
Yl l (θ , ϕ ) = sin l θeilϕ .
(A.31)
All of the other functions in the multiplet can be found by acting with L$− . There is one
further simplification in that we only need to generate functions up to m = 0 . One can
show that
Yl m (θ , ϕ ) = Pl m (θ )eimϕ
Hence the θ -part is the same for +m and − m . This follows from the form of L$2
1 ∂
∂
1
∂2
Lˆ2 = h 2 [
(sin θ
)+ 2
]
sin θ ∂θ
∂θ sin θ ∂ϕ 2
and L$2 Pl m (θ , ϕ )eimϕ and L$2 Pl − m (θ )e − imϕ yields the same differential equation for Pl m (θ )
(
1 ∂
∂
m2
(sin θ
) − 2 ) Pl m (θ ) = l (l + 1) Pl m (θ )
sin θ ∂θ
∂θ sin θ
Let us look at the non-trivial example of l = 2 (d-functions).
Y22 (θ , ϕ ) = sin 2 θe2 iϕ
(general formula)
L$− → Y21 (θ , ϕ ) ~ sin θ cosθeiϕ
L$− → Y20 (θ , ϕ ) ~ ( − sin 2 θ + cos2 θ + cos2 θ ) = 3 cos2 θ − 1
L$− → Y2−1 (θ , ϕ ) ~ sin θ cosθe − iϕ
L$− → Y2−2 (θ , ϕ ) ~ ( − sin 2 θ + cos2 θ − cos2 θ )e−2 iϕ ~ sin 2 θe−2 iϕ
cos θ 2
L$− → ( 2 sin θ cos θ − 2
sin θ ) = 0
sin θ
13
It is seen that the form of the functions is generated quite easily. The normalization
factors are less important, although even they can be obtained very generally from the
commutation relations!
To solve the Schrödinger equation for the Hydrogen atom we actually only require
angular eigenfunctions of L$2 , not of L$z . We develop the above formalism because of its
elegance and generality. In practice it is easier to think of real eigenfunctions. We can
therefore make linear combinations of degenerate eigenfunctions of L$2 . In particular we
can combine Pl m (θ )eimϕ and Pl m (θ )e − imϕ into Pl m (θ )cos( mϕ ) and Pl m (θ )sin( mϕ ) . For
the above d-functions this leads to the familiar cartesian forms of the d-orbitals
sin 2 θ cos 2ϕ = sin 2 θ (cos2 ϕ − sin 2 ϕ ) ~
sin 2 θ sin 2ϕ = 2 sin 2 θ sin ϕ cos ϕ ~
x2 − y2
r2
xy
r2
xz
r2
xy
sin θ cosθ sin ϕ ~ 2
r
sin θ cosθ cos ϕ ~
3 cos2 − 1 ~
3z 2 − 1
r2
All angular functions can easily be generated this way: Start from sin l θeilϕ , act with L$−
sequentially, and combine e ± imϕ into cos( mϕ ), sin( mϕ ) . Voila!
14
The radial equation for the Hydrogen atom and its solutions.
Above we discussed the angular part of the equations that determine the atomic orbitals
for the Hydrogen atom in great detail. Here we will discuss how the full set of solutions
can be obtained. The Hamiltonian in spherical coordinates was given by
h2 1 ∂ 2 ∂
L$2
e2
H$ = −
(
r
)
+
−
2µ r 2 ∂r
∂r 2µr 2 4πε 0r
(r.1)
And we try the function Ψ( r ,θ , ϕ ) = f ( r )Yl m (θ , ϕ ) →
[−
h 2 1 ∂ 2 ∂f
( l ( l + 1)h 2
e2
r
(
)
+
(
−
) f ( r )]Yl m (θ , ϕ ) = Ef ( r )Yl m (θ , ϕ )
2µ r 2 ∂r
2µr 2
4πε 0r
∂r
Let us simplify the notation somewhat and multiply through with
(r.2)
2µ
, and define
h2
2µe2
mee2
2
2 µE
2
≈
= , where a0 is the Bohr radius. This
=
ε
.
Let
us
also
use
2
2
2
4πε 0h
4πε 0h
a0
h
yields the radial equation
−
l ( l + 1)
1 ∂ 2 ∂f
2
(r
)+
f (r ) −
f ( r ) = εf ( r )
2
2
r ∂r
r
a0 r
∂r
(r.3)
or better yet,
2 ∂f ∂2 f l ( l + 1)
2
f (r ) −
f ( r ) = εf ( r )
−
− 2 +
2
r ∂r ∂r
r
a0 r
(r.4)
One more substitution to make. Try f ( r ) = p( r )e−αr , where p( r ) will be a polynomial in
r , p( r ) = a + br + cr 2 +..., hence
∂f ∂p −αr
e − αp( r )e −αr
=
∂r ∂r
2
∂ f
∂2 p
∂p
=
[
− 2α
+ α 2 p( r )]e −αr
2
2
∂r
∂r
∂r
e−αr is multiplied throughout, so we can cancel this. We end up with an equation for p( r )
−
2 dp 2α
2
d2p
dp
l ( l + 1)
p( r ) − 2 + 2α
p( r ) −
p( r ) = εp( r ) (r.5)
+
− α 2 p( r ) +
2
r dr
r
dr
dr
r
a0 r
15
repeated for convenience:
2 dp 2α
2
d2p
dp l ( l + 1)
p( r ) − 2 + 2α
p( r ) −
p( r ) − α 2 p( r ) = εp( r ) (r.5)
−
+
+
2
r dr
r
dr
dr
r
a0 r
Let us first examine some of the lower degree equations before discussing the general
solution. Rember that l = 0 for s-orbitals, l = 1 would yield p − orbitals, and so forth.
Let me just list some solutions:
l = 0, p( r ) = 1:
dp d 2 p
=
= 0 , substitute in (r.5)
dr dr 2
2α
2
−
− α 2 = ε ∀r
r a0 r
1
→ ε = −α 2 ; α = ;
a0
(r.6)
and therefore
f (r ) = e
− r / a0
h2
;E = −
2
2µa0
(r.7)
Another example:
l = 0, p( r ) = r − c →
d2 p
dp
= 1, 2 = 0
dr
dr
2 2α
2
− +
( r − c ) + 2α −
( r − c ) − α 2 ( r − c ) = ε ( r − c ) ∀r
r
r
a0 r
(r.8)
in such an equation the terms must match for each power in r, hence
−α2 = ε
2αr
+ 2α − 2r / ra0 = 0
order unity:
r
1
order : − 2 − 2αc + 2c / a0 = 0
r
order r:
(r.9)
and we obtain the solutions:
ε = −α 2 , α = 1 / 2a0 , c = 2a0
(r.10)
or
f ( r ) = ( r − 2a0 )e − r / 2 a0 ; E = −
1 h2
h2
=
−
2 µ ( 2a0 ) 2
4 2µa0 2
(r.11)
16
Let us take the simplest example of a p function
l = 1, p( r ) = r ,
d2 p
dp
= 1, 2 = 0 :
dr
dr
2
2 2
− + 2α + 2α + − − α 2r = εr
r
r a0
α = 1 / 2a0 , ε = −1 / 4a02
f ( r ) = re− r / 2 a0 , E = −
1 h2
4 2µa02
General characteristics:
Suppose highest power in polynomial is r m . Then, substituting in (r.5):
Terms of order m : −α 2 = ε (always!)
Terms of order m − 1: 2α + 2 mα −
2
2
= 0 → 2( m + 1)α =
a0
a0
1
h2
1
α=
,E = −
, m = 0 , 1, 2,...
2
2µa0 ( m + 1)2
( m + 1)a0
Independent of l ! Usually we put n = m +1 and call it the principle quantum number:
r n −1e− r / na0 , E = −
h2 1
2mea02 n 2
( n −1: highest power in r )
Call s the smallest exponent in p( r ) , p( r ) = r s (1 + ar + br 2 +...) .
In equation (r.5) we will then obtain terms starting from r s− 2 :
−2s − s( s − 1) + l ( l + 1) = 0 → s = l
b g
The first radial solution corresponding to Yl m θ , ϕ (no radial nodes) starts with r l !
Depends on quantum number l.
17
Summary of general solutions:
Ψn ,l ,m ( r ,θ , ϕ ) = pn ,l ( r )e − r / na0 Yl ,m (θ , ϕ );
E( n) = −
h2
2µa02 n 2
where pn ,l ( r ) = ar n −1 + br n − 2 +....+ r l is a polynomial in r having n − l − 1 radial nodes.
Possible energy levels and their degeneracies:
n = 1, 2, 3,....
l = 0,1, 2, 3,..., n − 1
m = − l ,− l + 1,...., l − 1, l
n −1
1
(
2
l
+
1
)
=
n
+
2
l = n + 2 ⋅ n( n − 1) = n 2
∑
∑
2
l =0
l =1
n −1
Degeneracy En :
We note that for the hydrogen atom the energy only depends on the principal quantum
number n . Hence the 2 s,2 p orbitals are degenerate as are 3s,3 p,3d and so forth. This is
only true for one-electron atoms (H, Ne7+, etc), but not for many-electron atoms in
general.
18