24
4. Orbital Angular Momentum of Hydrogen Atom
NOTES:
Review
The potential energy of the electron in the Hydrogen atom is simply the Coulomb potential energy
associated with the orbiting electron (charge -e) and stationary proton (charge e):
V (r )   ke 2 r
(4.1)
We substitute this potential energy into the three-dimensional time-independent Schroedinger equation
and solve the resulting differential equation for the eigenfunction. Recall that this solution was very
tedious and involved. Let’s briefly summarize what we did in the solution.
First, we converted from rectangular (x,y,z) to spherical polar coordinates (r,. The resulting
Schroedinger equation is:
  2

 2m

1   2 
1
 

1
2 
r

sin


 V ( r )   ( r ,  ,  )  E ( r ,  ,  )




 2 r  r 
2
2
2
2 
  r sin   
r sin   
r

(4.2)
We then assumed that the eigenfunctions were separable into a product of a radial function and two
angular functions:
 ( r , ,  )  R ( r )  ( )  (  )
(4.3)
In solving the resulting equation, we found three quantum numbers: n, l, ml that are associated with the
different functions in (4.3). The radial function R depends on n, depends on ml, and  depends on
both l and ml. We went on to call the product of the two angular functions the spherical harmonics, Y ,
and thus wrote that the eigenfunctions can be labeled as:
 nlm (r , , )  Rn (r )Yl m (, )
(4.4)
What do these three quantum numbers quantize? Recall that the principal quantum number n
quantizes the energy according to
E n  13.6 n 2 eV .
(4.5)


The azimuthal quantum number l quantizes the magnitude of the orbital angular momentum according to
L  l (l  1) 
(4.6)
and the magnetic quantum number ml quantizes one component of the orbital angular momentum (we
arbitrarily chose the z component) according to
25
Lz  ml 
(4.7)
We also found that these quantum numbers were all integers that could only have certain values. For a
given n,
l  0, 1, 2,..., n -1
ml  l ,  l  1 ,...,0,..., l  1, l
(4.8)
The Angular Momentum Operators
We are going to concentrate on orbital angular momentum and show that its magnitude and z
component are indeed quantized as described above in (4.6) to (4.8) using an operator approach. As
with the simple harmonic oscillator, we are going to get this information without knowing the explicit
forms of the wavefunctions! In fact, you will see many similarities between this analysis and the analysis
for the s.h.o..
Let’s begin by establishing the operators that correspond to orbital angular momentum and its

components. For an electron that is some position r from the proton and which is orbiting the proton

with some tangential momentum p , the orbital angular momentum of the electron is
  
Lrp
(4.9)
The components of the angular momentum are found by using the definition of the cross product:
Lx  yp z  zp y
L y  zp x  xp z
(4.10)
Lz  xp y  yp x
The square of the magnitude of the angular momentum is simply
L2  L2x  L2y  L2z
(4.11)

We can view the components of L and its magnitude squared as operators by simply replacing the
position and momentum components by their operator equivalents. Thus, the corresponding operators
are
 

L x  i y  z 
y 
 z

 
L y  i z  x 
 x
z 
 

L z  i x  y 
x 
 y
L2  L2  L2  L2
x
y
z
(4.12)
26
Because of the spherical symmetry of the potential energy, it is convenient to use spherical polar
coordinates. Converting the operators to spherical polar coordinates gives the somewhat messy but
convenient (wait and see) expressions of



L x  i sin 
 cot  cos  

 




L y  i  cos 
 cot  sin  

 


L z  i

(4.13)
 

1 2 
2
2 1

L   
 sin   
  sin 2   2 
 sin   
The Schroedinger equation is expresses as usual as
H   E 
(4.14)
Comparing this equation to (4.2), we note that the Hamiltonian in spherical polar coordinates is the
expression inside the curly brackets on the left side of (4.2). In comparing this expression to the
expression for L2 in (4.13), notice that we can write the Hamiltonian as
 2
H 
2m
1 2
 1   2  
 r 2 r  r r    V (r )  2mr 2 L


(4.15)
Why is this important? Let’s assume that the eigenfunction is separable into a product of a radial
function and a function involving the two angles. Perhaps we can symbolize this by writing
  R , 
(4.16)
Now the Hamiltonian acts on the product of R and ,  and gives a number (the energy eigenvalue)
times the product. There are three parts to the Hamiltonian as expressed in (4.15). The first two parts
are purely radial operators: they only will operate on R . The third term involves a radial operation and
the operator L2 . Look at the expression for L2 in (4.13): it is a pure angle operator! In fact, it is the
only part of the Hamiltonian that operates on angle. In order for the total Hamiltonian to operate on the
total  and give a number times  then, L2 must operate on the angular part of  to give a number
times the angular part. That is,
L2 ,   c , 
(4.17)
Now multiplying the angular functions on both sides of this equation by the radial function won’t change
a thing since L2 doesn’t do anything to the radial function. So, we can write
27
L2 R ,   L2   c 
(4.18)
This equation states that the eigenfunctions  are not only eigenfunctions of the Hamiltonian, but they
are also eigenfunctions of L2 ! This is extremely beneficial to know since these eigenfunctions describe
the actual electron and any eigenvalues that we find associated with these eigenfunctions can be related
to physical quantities of the electron that can be measured. For example, the eigenfunction c in (4.18)
will be the value of the square of the orbital angular momentum for the electron described by the
eigenfunction Let’s try to find this c. Along the way, we will also find the allowed values for the z
component of angular momentum.

Before proceeding, let’s introduce some convenient notation. First, since the angular momentum
components and L2 only operate on angles, we only have to be concerned about the angular part of 
which we wrote as ,  . Let’s introduce two labels, l and m, that represent these two angular
functions. Hmmm....Are these going to be the same l and ml that we discussed above? Well, yes. Why
do it? Because it is a nice visual link to what the angular momentum operators do to the product of the
angular functions, as we will see later. (Note that we have dropped the subscript l on m. This is purely a
time-saving measure.) Thus, we can rewrite (4.18) as
L2 l , m  c l , m
(4.19)
Do we know anything at this moment about c? Yes, we do know that it must be real and positive since
the value of the square of the orbital angular momentum must be real and positive. (We also know that
it is real since an operator corresponding to a dyanmic variable has real eigenvlaues.) It also has units of
 2 . (Check this for yourself.) Let’s now proceed and see if we can find the exact value of c.
Commutation Relations
It will be useful to first evaluate several commutation relations involving the angular momentum
operators. For the individual component operators, you can show (Problem [4.1]) that
 L L   L L  L L  iL
 L L   L L  L L  iL
 L L   L L  L L  iL
x,
y
x
y
y
x
z
y,
z
y
z
z
y
x
z,
x
z
x
x
z
y
(4.20)
* The fact that no component commutes with another component means that you cannot simultaneously
measure all three components. This is a general theorem in quantum mechanics. If two operators do not
commute, then you cannot simultaneously measure the corresponding dynamic variables. Recall that you
showed that position and momentum do not commute in Problem [2.1]. This means that you cannot
simultaneously measure position and momentum. This, of course, is just what Heisenberg says in his
Uncertainty Principle!
While no component commutes with any other component, you can show that each component
commutes with L2 :
 L , L    L , L    L , L   0
2
2
x
2
y
z
28
(4.21)
* This means that you can measure simultaneously the magnitude of the angular momentum and one of
the components. We will choose this one component to be the z component. (Recall that the experiment
itself must determine a definite z axis by, for instance, establishing a magnetic field in the z direction.)
Another important piece of information is obtained by the commutation relations in (4.21). A lemma
in quantum mechanics states that if two operators commute, then an eigenfunction of one operator is
also an eigenfunction of another operator.
To show this, consider two operators A and B that commute so that
   BA
   0 . Let  be an eigenfunction of A with eigenvalue a so that
A , B  AB
 
A   a  . Now first operate on  with B and then operate on this vector with A
  with
   . But since the operators commute, you can replace AB
to form A B   AB






  so that A B   BA
    Ba
   a B  . Now this last equation can only be
BA
consistent with A   a  if B   b  so that we get
   bA   ba  . Dividing the last two equated expressions by b
A B   Ab


gives A   a  .
Thus, we can conclude that the eigenfunctions of L2 , which are the energy eigenfunctions of the
Hamiltonian, are also the eigenfunctions of L z ! Thus we expect that we can write
L z l , m  d l , m
(4.22)
Also, we assume that since L z only involves the angle , the label m will be useful in labeling the
eigenvalues d. The remaining label l will then be used to label the eigenvalues c of L2 in some yet
undetermined way.
Can we find this d which gives the value of the z component of the orbital angular momentum? Let’s
see. (Our list of eigenvalues to find is growing....we have to find c and d now!)
Raising and Lowering Operators
To find c and d, it is useful to introduce two new operators defined as
L   L x  iL y
L   L x  iL y
With the previous commutation relations in (4.20) & (4.21), you can show that
(4.23)
29
 L , L   L
 L , L   L
 L , L   2L
 L , L    L , L   0
z

z




z
2

2

(4.24)

These will be useful later as you will see. We can also derive some other useful relations involving the
raising and lowering operators. For example,
L  L   ( L x  iL y )( L x  iL y )

 i L , L 
 L2x  L2y  i L x L y  L y L x
 L2x  L2y
x

y
 L2x  L2y  L z
L  L   L2  L2z  L z
(4.25)
Notice that we make use of the first commutation relation in (4.20) and the last equation in (4.12) in
deriving (4.25). You can also show in a similar manner (Problem [4.2]) that
L  L   L2  L2z  L z
(4.26)
Let’s now go back and examine (4.22). What can we say about that eigenvalue d? It is the value of
the z component of orbital angular momentum. It must be real and it also has units of  . (Check this for
yourself!) So let’s write d as
d  m
(4.27)
We know that this m must be real but that is all we know at this point. We can write (4.22) as
L z l , m  m l , m
(4.28)
Let’s see what happens if we operate on the eigenfunction with L  first, and then operate on this
vector with L :
z
 


L z L  l , m  L  L z  L  l , m
(4.29)
Notice that we have used the first commutation relation in (4.24) to write L z L  as L  L z  L  . Using
(4.28), this equation becomes

L  L
 

l , m   (m  1) L l , m 
L z L  l , m  L  m  L  l , m
z


(4.30)
30


Aha! This is an eigenvalue equation! The vector L  l , m is a new eigenfunction of L z with an
eigenvalue that is one  more than the value of the z component of angular momentum. Because the
eigenvalue is increased, we call L  the raising operator.
Let’s see what happens if we now operate on the eigenfunction with L  first, and then operate on this
vector with L :
z
 


L z L  l , m  L  L z  L  l , m
(4.31)
Notice that we have used the second commutation relation in (4.24) to write L z L  as L  L z  L  .
Using (4.28), this equation becomes

L  L
 

l , m   (m  1) L l , m 
L z L  l , m  L  m  L  l , m
z

(4.32)


Aha, again! This is also an eigenvalue equation! The vector L  l , m
 is a new eigenfunction of L
z
with an eigenvalue that is one  less than the value of the z component of angular momentum. Because
the eigenvalue is decreased, we call L  the lowering operator.
What happens if we keep applying the raising operator to l , m ? We keep getting new vectors back
which are eigenfunctions of L with increasing eigenvalues. Eventually, we will get a value of Lz that is
z
so large that the value of L2 - Lz2 becomes negative. But wait....this quantity cannot be negative since L2
- Lz2 = Lx2 + Ly2 and the right hand side of this expression is always positive. Therefore, there must be
some maximum value of m, let’s call it mmax , such that
L  l , mmax  0
(4.33)
Let’s operate on this eigenfunction with L2  L2z . By (4.26), this is equivalent to operating with
L L  L so


z
Lˆ2  Lˆ2z l, mmax  Lˆ Lˆ  Lˆ z l, mmax
(4.34)
Using (4.33) and (4.28), this becomes
2
Lˆ2 l , mmax  mmax
 2 l , mmax  0  mmax  2 l , mmax
L2 l , mmax  mmax mmax  1 2 l , mmax
(4.35)
31
Aha! This is an eigenvalue equation that resembles the one we stated in (4.19)! That eigenvalue we
called c is mmax(mmax+1)  2 . Since we said that l is a convenient label for the eigenvalues of L2 , let’s let
mmax=l and assume that this equation is valid for all values of m , so that the eigenvalue equation for the
square of the orbital angular momentum is
where
L2 l , m  ll  1 2 l , m
(4.36)
mmax  l
(4.37)
We have found, then, that the eigenvalues c can be written as ll  1 2 .
The next question is: What happens if we keep applying the lowering operator to l , m ? We keep
getting new vectors back which are eigenfunctions of L with decreasing eigenvalues. Eventually, we
z
will get a value of Lz that is so large and negative that the value of L2 - Lz2 again becomes negative. As
we said before, though, this quantity cannot be negative since L2 - Lz2 = Lx2 + Ly2 and the right hand side
of this expression is always positive. Therefore, there must be some largest negative value of m, let’s
call it mmin, such that
L  l , mmin  0
(4.38)
Let’s operate on this eigenfunction with L2  L2z . By (4.25), this is equivalent to operating with
L L  L so


z
 L
2



 L2z l , mmin  L  L   L z l , mmin
(4.39)
Using (4.38) and (4.28), this becomes
2
Lˆ2 l , mmin  mmin
 2 l , mmin  0  mmin  2 l , mmin
L2 l , mmin  mmin mmin  1 2 l , mmin
(4.40)
This is an eigenvalue equation for L2 with an eigenvalue of mmin mmin  1 2 . But wait a minute! We
assumed for any m that the eigenvalue is ll  1 2 so we must have
ll  1 2  mmin mmin  1 2
This equation can be solved for l by writing it as
l 2  l  mmin  mmin  1  0
and using the quadratic equation:
(4.41)
32
l
 1  1  4mmin (mmin  1)
2

2
 1  1  4mmin  4mmin
2

 1  1  2mmin
2

l  mmin or l = mmin  1
Now we already said that l is the maximum value of m so we discard the second root. Thus, we find that
the most negative value of m is
mmin  l
(4.42)
The values of l and m
So far we have established some relationships between these labels l and m and the corresponding
eigenvalues of L2 and L z , but we still don’t know the actual values of the eigenvalues! We are in a
position now, though, to find the allowed values.
We know that the minimum (most negative) value of m is -l and that the maximum (most positive)
value of m is +l. We also know that the value of m changes by the integer one based on our analysis
with the raising and lowering operators. So we can write that,
m = -l, -l+1, ..., l-1, l
(4.43)
l - k = -l where k is an integer
(4.44)
2l = an integer
(4.45)
This means that
so
Aha! The only way this can be true is if:
1) l = integer
or
(4.46)
2) l 
1
 (odd integer )
2
Which is it? We have to rely on experimental observations to tell us! For the orbital angular momentum
of an atom, we find that the first condition is true: l takes on integer values.
We have done it! We have found the allowed values of the square of the orbital angular momentum (and
therefore the allowed values of the orbital angular momentum) and the z component of the orbital
angular momentum. These results were previously summarized at the beginning of this chapter in (4.6)
through (4.8).
33
Detailed Results of the Raising and Lowering Operations
Before we conclude, it is useful to see exactly what the raising and lowering operator do to the energy
eigenstates. This information can be used to find quantities such as the average value of the x
component of the orbital angular momentum without knowing the explicit forms of the eigenfunctions.
Since the raising operator increases the value of m by one and the lowering operator decreases the
value of m by one, we are tempted to write
L  l , m  cm1 l , m  1
(4.47)
L  l , m  cm1 l , m  1
(4.48)
Let’s find these two constants. We first find cm+1.
Let’s begin by taking the complex conjugates of both sides of (4.47). Using the fact that the raising
and lowering operators are complex conjugates of one another, we obtain
l , m L   cm1 * l , m  1
(4.49)
Let’s now evaluate the scalar product of L  L  using the information in (4.47) and (4.49):
l , m L  L  l , m  l , m  1 cm1 * cm1 l , m  1
 cm1
2
l, m  1 l, m  1
We require that the eigenstates be normalized so that
2
cm1  l , m L  L  l , m
(4.50)
Now let’s make use of (4.26) to replace L  L  :
2
cm1  l , m L2  L2z  L z l , m
(4.51)
We know what every operator does inside the bracket! Let’s operate with them to find
cm1  l , m l (l  1) 2  m2  2  m l , m
2
  2 l (l  1)  m(m  1) l , m l , m
so
cm1   l (l  1)  m(m  1)
and
(4.52)
34
L  l , m   l (l  1)  m(m  1) l , m  1
(4.53)
The derivation of cm-1 is similar to the above derivation. Here, we take the complex conjugates of both
sides of (4.48), form the scalar product involving L  L  , and make use of (4.25) to find
L  l , m   l (l  1)  m(m  1) l , m  1
(4.54)
35
PROBLEMS:
[4.1]
Using the expressions in (4.10), show that L x , L y  iL z


[4.2]
Show that L  L   L2  L2z  L z . This is equation (4.26). Follow a similar sequence of steps as used in
the derivation of (4.25).
[4.3]
Show that L  l , m   l (l  1)  m(m  1) l , m  1 . This is equation (4.54). Follow the sequence of
steps outlined in the text.
[4.4]
(a) By following the steps below, show that the average value of the x component of the orbital angular
momentum of an electron in any state is zero.
1
1) Show that you can write L x  L   L  .
2
2) Now evaluate the average value of L x for the state l , m . Make use of your knowledge of
how the raising and lowering operators operate on an eigenfunction (equations (4.53) &
(4.54)) , and the fact that the eigenstates are orthonormal.


(b) Using a similar sequence of steps, show that the average value of the y component of the orbital
angular momentum of an electron in any state is zero.
(c) Give a geometrical explanation of these results.