Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
4.1 The wave function for a hydrogen atom is given by
  r , ,    r 2e  r / 3a0  3cos 2   1
a) What are the quantum numbers n, , m for this atom? No calculations allowed.
b) What is the most probable value of r for the electron in this state? What is significant about
the result?
Solution:
a) The n quantum numbers is 3 because the denominator in the exponential in the radial parts of
the wave functions is na0 . The function  3cos 2   1 is proportional to Y20  ,   . Note that
there is no  dependence which confirms that m  0 . Answer: n  3;  2; m  0
b) The most probable value of r is the position of the maximum in the probability distribution
2
r 2  . The max will be the same as the max of r so, taking the derivative with respect to r:
and setting it equal to zero:
 1 3

d
r r 2e r / 3a0  e r / 3a0  
r  3r 2   0  r0  32 a0
dr
 3a0

This is the third Bohr radius. It turns out that for the state of maximum  for a given n the most
probable value of r is always the Bohr radius.


4.2 The wave function for a hydrogen atom is given by
 r,  ,    R54 r Ym  ,    R53 r Y30  ,    R52 r Y21  ,  
For what values of  and m (= m) will  be an eigenfunction of the Hamiltonian? What is the
energy in electron volts? in Rydbergs? in au? in cm-1?
Solution:
Since the radial parts of the wave function all have n = 5,  r,  ,   is already an eigenfunction
of the Hamiltonian. Recall that the energy for a hydrogen atom doesn't depend on either  or m .
Therefore the only requirements on  or m are that they be consistent with n = 5 and, of course,
each other.
Since the radial part of the first term has  = 4, so too must the spherical harmonic.
Further, m must be such that –4 ≤ m ≤ +4.
The energy for n = 5 is:
13.6 eV
1
1 1
1.097373  10 5 cm 1
E5  


0
.
544
eV


Ry



au


 4389.492 cm 1
2
2
2
2
5
5
2 5
5
4.3 An electron is in the ground state of tritium, i.e a hydrogen atom with two neutrons the
symbol for which is 3H. A nuclear reaction instantaneously changes the nucleus to 3He which
retains the single electron from the tritium atom. That is, the product is a singly ionized 3He
which is a one-electron atom. Calculate the probability that the atomic electron remains in the
ground state of the 3He ion.
Solution:
We may expand the wave function for the "old" atom, the tritium, on the complete set of
eigenfunctions of the Hamiltonian for the "new" one-electron atom, the 3He+.

tritium
 100
  Anm  nHe
m
n 1 
m
Solution to Chapter 4 problems
page 1
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
But, we are only interested in the probability that the 3He is in the ground state so we need only
A100 because the desired probability is | A100 |2. Furthermore, the spherical harmonics will be the
same in the new and old wave functions. In fact, Y00  ,    1/ 4 so when we integrate the
square of Y00 over all angles we get 1. Therefore, we need only work with the radial parts of the
two wave functions. These wave functions are identical except that the tritium wave functions
have Z = 1 while the He wave functions have Z = 2.
He
tritium
 100
 A100
Formally, we take the inner product  100
Now we must perform the integral in coordinate space. The wave functions are
3
Z
 100
 Z 2
 Z
 2  exp  
 a0 
 a0

r 

so that
3
He
tritium
 100
 100


0
and A100 
2
 3r 
4 3 
22
r 2 drR10Z  2  r  R10Z 1  r   2 2  r 2 dr exp     8 3
0
a0
3
 a0 
64  8
 0.70
36
4.4 An electron in the Coulomb field of a proton is described by the normalized wave function

  r   16 4  100  r   3  211  r    210  r   10  211  r 

where the  r  are spherical (orbital) eigenfunctions of the hydrogen atom Hamiltonian.
Find:
a) The expectation value of the energy in a.u.
b) The expectation value of L̂2 . Keep the 's in the answer.
c) The expectation value of L̂ z .
Solution:
a)
1
1
2
Hˆ   Hˆ   2 42 E1  32 E2   1 E2  10 E2   4 E1  5E2 
6
9
1  1 1 
21
 1 1 
  4     2   5     2   a.u.   a.u.
9  2 1 
72
 2 2 
b)
1 2 
1
10 2
2
Lˆ2  2 0  32  1 2    1  1 2   10  1 2  2  2  9  1  10  

6 
6
9


c)
1
Lˆz  2 0  32  
6 
   1   0   10      
2
1
36
4.5 A hydrogen atom is in an eigenstate of L2 and Lz . Show that, although neither Lx nor L y




are well defined, their sum L2x  L2y is well defined. Find the value of L2x  L2y in terms of 
Solution to Chapter 4 problems
page 2
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
and m . Ignore spin.
Solution:
L
2
x
Therefore
L
2
x
 L2y  L2  L2z
 L2y  nm  L2  L2z  nm


   1  m   2 nm
2
which does indeed have a definite value.
4.6 Show that the integrals
z nnm 1m  n   1m z nm


0
0
  r 2 dr Rn r Rn  1 r rdr  Pm  P 1m   cos   sin d
and
znn m1m  n   1 m z n m


  r 2 Rn  r  Rn 
0
1
 r  rdr 0
P m   P
1 m
  cos  sin  d
are given in atomic units by
z nnm 1m 
3
n
2


 m2 n2  2
2  12  1
2

and
z nnm 1m 
3
n
2
  1

 m 2 n 2    1
2  32  1
2
2

Solution:
To evaluate the -integral use the following relation for associated Legendre functions:
Pm   cos 
and the orthogonality relation
  12  m 2 P   
2  m2
 
P
2  32  1 1 m
2  12  1 1 m


0
Pm  P 'm  sin d    ' . Evaluate the r-integral using
Solution to Chapter 4 problems
page 3
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
 2
Rn  r     
n
3/ 2
  n   1 !


 2n  n   
1/ 2
 2r 
 r  2 1  2r 
   exp     Ln  1  
 n 
 n
 n 
and the recursion relation for associated Laguerre functions
L  x   L 1  x   L 11  x 
which is applicable only to the associated Laguerre polynomials as defined in the radial wave
function given above, i.e. Definition 2. You will also need the orthogonality relation


0
 p e   Lqp  Lqp'  d 
By letting     1 obtain Rnln   1 
q  p ! 
q!
qq '
3
2
n n 2    1 .
2
Solution:
The -integral:
Replace Pm   cos  with the identity given. For z nnm 1m the only term that survives is the one
with Pm1   . Therefore


0

Pm  P 1m   cos   sin d   
0 



2  m2
P 1 m   P 1m  sin d

2  12  1

2  m2
2  12  1
For z nnm 1m the only term that survives is the one with Pm1   . Therefore


0

Pm  P 1m   cos  sin d   
0 


  12  m 2 P   P  sin d
2  32  1 1 m  1m
  12  m 2
2  32  1
The r-integral:
Solution to Chapter 4 problems
page 4
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions

Rnn  1   r 2 dr Rn r Rn  1 r rdr
0
 2  3 / 2  n    1! 1 / 2 
   
 
 n   2nn   !  
 2  3 / 2  n   !  1 / 2    2 r / n  2r  2  1 2  1  2r  2  1  2r  3
  Ln  1   Ln   r dr
  
  e
 n
 n
 n
 n   2nn    1!  0
Now let x  2r / n so dr 
n
dx . We get
2

Rnn  1   r 2 dr Rn r Rn  1 r rdr
0
1 n    1!n   
4 n    1!n   1 / 2
1/ 2



0
e  x x 2   2 L2n11  x L2n1 x dx
To use the orthogonality integral we must use the recursion relation to raise the upper index to
match the power of x, i.e. to 2  2 .
In the recursion relation L  x   L 1 x   L 11  x  we identify   n    1 and   2  1 in the
first associated Laguerre function under the integral.
We have then L2n11 x   L2n21 x   L2n2 2  x 
Now we have to do the same thing with the second Laguerre function.   n   and   2  1 .
Therefore,
L2n1  x   L2n  x   L2n 1  x 
Clearly the recursion relation must be applied twice more.

 

L2n1 x   L2n1 x   L2n11 x   L2n11 x   L2n1 2 x 
 L2n1 x   2 L2n11 x   L2n1 2 x 

 
 

 L2n2 x   L2n21 x   2 L2n21 x   L2n2 2 x   L2n2 2 x   L2n23 x 
2  1
n 
L
 x   3L
2  1
n   1
 x   3L
2 2
n   2
x   L
2 2
n   3
x 
Now, in order for the integral to be non-zero we must have both associated Laguerre polynomials
the same. Therefore, the only terms in the product that will be non-zero are
Solution to Chapter 4 problems
page 5
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions




 3 L2n21  x   3 L2n2 2 x 
2
2
We have then


0
0
and using


0








e  x x 22 L2n11 x L2n1 x d  3 e  x x 22 L2n21 x  dx  3 e  x x 22 L2n22 x  dx


0
 p e   Lqp  Lqp'  d 

q  p ! 
q!
2
0
2
qq '



e  x x 2   2 L2n11  x L2n1 x d  3 e  x x 2   2 L2n21  x  dx  3 e  x x 2   2 L2n2 2  x  dx
0
2
0
2
 n    1  2  2 ! n    2  2  2 !
 3

n    1!
n    2! 

 n    1!
n   ! 
 3


 n    1! n    2 !
 n    1n   !
n   ! 
 3


 n    1n    2 ! n    2 !
n   !  n    1  1
 3
n    2!  n    1 
n   !
2n
 3
n    2! n    1
n   ! n
 6
n    1!
We have then
R
n   1
n
1 n    1!n   

4 n    1!n   1 / 2
1/ 2
1 n    1!n   
4 n    1!n   1 / 2
1/ 2

1 n    1!n   

4 n    1!n   1 / 2
1/ 2

3 n   
n   n
2 n   1 / 2

3
n n2  2
2


0
e  x x 2   2 L2n11  x L2n1 x dx

n   ! 
 6 n    1! n 


 n    1!n    
n
6
n    1! 

1/ 2
If we let     1 we obtain Rnn 1
Solution to Chapter 4 problems
page 6
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
Rnln   1 
3
2
n n 2    1
2
We have then
z
n  1m
nm


z
n  1m
nm
2  m2
3
 n n2  2
2  12  1 2
3
n
2


 m2 n2  2
2  12  1
2

2

  1  m 2 3
2

 n n 2    1
2  32  1 2
2
2
3   1  m 2 n 2    1 
 n
2  32  1
2
4.7 A collection of hydrogen atoms is in a parabolic eigenstate  given by
1
1
200 
210
2
2
where the kets on the right are the spherical eigenkets nm .
 
a) What is the energy eigenvalue, in atomic units, corresponding to  ?
b) What are the possible sets of parabolic quantum numbers that characterize  ? Virtually no
computation is required. Merely list the combinations that are possible from inspection of the
wave function given.
Solution:
a) Since n = 2 the eigenvalue is:
1
1
1
E2   2  
  au
2
2n
22
8
b) The parabolic quantum numbers n, n1, n2, m are related by:
n  n1  n2  m  1
Since n = 2 and m = 0 (from the wave function above), we must have
n1  n2  2  1  1
Since n1 and n2 can take on only integral values, the possible combinations for parabolic quantum
numbers are:
n1n2 m  100
or
n1n2 m  010
Solution to Chapter 4 problems
page 7
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
4.8 The normalized energy eigenfunctions for the hydrogen atom in parabolic coordinates are
given by
m 3/ 2
exp  im 
n1 ! n2 !
m /2 m
m
 n1 n2 m  , ,   
exp         Ln1  m   Ln2  m  
3/ 2
3/ 2
 n  n1  m   n2  m 
where    2E and the L's are associated Laguerre polynomials.
a) Find the normalized hydrogen atom energy eigenfunctions in parabolic coordinates for n = 2,
n1 = 1, n2 = 0 = m in terms of the hydrogen atom energy eigenfunctions in spherical coordinates.
b) - h) What are the probabilities of measuring the following:
b) the energy of the ground state.
c) the energy of the first excited state.
d) total angular momentum zero
e) total angular momentum 2 .
f) z-component of angular momentum zero.
g) z-component of angular momentum .
h) Find the normalized ground state eigenfunction in parabolic coordinates in terms of the
spherical eigenfunctions. What is noteworthy about this eigenfunction and why?
Solution:
1
1! 0 !
0
 1

a)  100  , ,   
 3/ 2 3/ 2  3/ 2 exp             L1    L0  
2 1! 0!
 2

but L0 x   1 and L1 x   1  x so that
1
 1

  3/ 2 exp          1   
2
 2

The defining relations for parabolic coordinates are
  r  z  r 1  cos 
 100  , ,   
  r  z  r 1  cos 
    tan 1
y
x
from which we find
x   cos 
y   sin 
1
   
2
1
r     
2
 rz
rz
z
 1  1
Also,    2 E   2  2   which in this case is  = 1/2 because n =2.
 2n  n
We have then
Solution to Chapter 4 problems
page 8
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
 100  , ,   

Now Y00  ,   
1 1
 
2  2 
3/ 2

 r   1
exp     1     r  z  
 2   2


1
1
 r   1
exp     1    r 1  cos   
2 2 2
 2   2

1
and Y10  ,   
1
3
 r
exp    and
cos while R20 r  
4
2
 2
4
1
 r
R21 r 
r exp    so that we can rewrite u100 in a more transparent way.
2 6
 2
1 1
1
 r  r  1 1
 r
 100 
exp   1     r  exp    cos
4 2
4
 2  2  2 2
 2
1
1
1 1

Y00  ,   R20  r   2 6 R21  r  
Y10  ,  
2
2 3
2
1
1
or  100 
R20  r  Y00  ,   
R21  r  Y10  ,  
2
2
b) The energy depends only on the radial part of the wavefunction, Rn r  . Since the only
principal quantum number that appears in  100 is n = 2, the probability of measuring E1, the
ground state energy is zero.
c) n = 2 corresponds to the first excited state energy so that the probability of measuring E2 is
unity.
d) The total angular momentum is given by   1 where  appears in the spherical
harmonic Ym  ,   . Therefore, the probability of measuring  = 0 is given by the square of the
2
1
 1 
coefficient of the spherical harmonic that contains  = 0, i.e. 
 
2
 2
2
1
 1 
e) Probability of  =1:  
 
2
2

f) Since both spherical harmonics in  100 have m = 0 the probability of measuring m = 0 is unity.
A better way of saying this is that m is a good quantum number in both spherical and parabolic
coordinates and m = 0 for this parabolic eigenstate.
g) In view of f) the answer is zero.
h) Parabolic quantum numbers for the ground state are: n = 1, n1 = n2 = m = 0. From the
expression given for the wavefunction in parabolic coordinates we have
1
 1

 000  , ,   
exp       

 2

1
 1 

exp   r 

 2 
To convert this to spherical coordinates we use Equations (5.1)
  r  z  r 1  cos  ;   r  z  r 1  cos  ;   
Adding the first two together we get     2r so in spherical coordinates this wave function is
Solution to Chapter 4 problems
page 9
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
 1 
exp   r  which is 100  r, ,  . Thus, the ground state parabolic wave function and the

 2 
ground state spherical eigenfunctions are the same. The reason these wave functions are the
same is that the ground state is non-degenerate.
1
4.9 A hydrogen atom is in an eigenstate that is characterized by the parabolic quantum numbers
n1 1 ; n2  0  m . The wave function in spherical coordinates in atomic units is:
3/ 2


1 1
1
–r / 2
  – 1  2 r 1  cos    e
2  2 
That is, you have been spared the trouble of writing the wave function in parabolic coordinates
and converting to spherical coordinates. Some selected spherical harmonics and radial hydrogen
atom wave functions are:
1
3
3
Y00 
;Y10 
cos  ;Y11 
sin e i
4
8
4
1
1
R10  2e – r ; R20 
1 – 12 r e – r / 2 ; R21 
re – r / 2
2
2 6
a) If a measurement of the energy of the atom in this state is made what are the possible values of
the energy that could be measured? Give your answer in atomic units. What probabilities are
associated with each value?
b) If a measurement of the total orbital angular momentum of the atom in this state is made what
are the possible values that could be measured? What probabilities are associated with each
possible angular momentum?
c) If a measurement of the z-component of the angular momentum of the atom in this state is
made what are the possible values that could be measured? What probabilities are associated
with each possible z-component ?
Solution:
a) Since n1 1 ; n2  0  m we have n  n1  n2  m  1 1  0  0  1  2 . Therefore, the
eigenstate is characterized by n = 2 and the only possible result of the measurement is
1
1
E2  –
 – au
2
22
8
b) This is the only part that requires any calculation. We must cast u100 u110 in terms of the
spherical harmonics and the radial wave functions. From the spherical harmonics and radial
wave functions given, we have:
1 – 12 r  e – r / 2  2 R20 ; re – r / 2  2 6 R21
un1n2m  u100 




1  4 Y00 ; cos  
u100 
4
Y10
3
1 
4 
1
Y10 
 – 2 R20 4 Y00  2 2 6 R21
3
4  

8
1
– R20Y00  R21Y10 
4 
2
From the last form we see that  = 1 and 0 are possible. Since the expansion coefficients are the
same, these values are equally probable.

– R20Y00  R21Y10  
Solution to Chapter 4 problems
page 10
Chapter 4 - THE QUANTUM MECHANICAL HYDROGEN
Problems with solutions
c) The system is in an eigenstate of Lz, i.e m = 0. Therefore, the only possible value of the zcompontent of angular momentum that could be measured is 0. This is also clear from the last
form of the wave function since both spherical harmonics have m = 0.
4.10 Show that the expectation value of the electric dipole moment of an atom having N
electrons in a state of well-defined parity vanishes. Use atomic units. Suppose a hydrogen atom
is in a spherical eigenstate. What is the dipole moment? How about a parabolic eigenstate?
Solution:
The dipole moment is given by
 N 
pˆ    rq 
 q1 
so the expectation value of the dipole moment is
 N 
pˆ     rq  
 q 1 
where, of course, the integral on the rhs is over all space.
Since rq  xq iˆ  yq ˆj  zq kˆ the dipole operator has definite parity – odd. If the state vector 
has definite parity as specified in the statement of the problem, then the integrand on the rhs
vanishes because the product  * is necessarily even.
Spherical eigenstates have definite parity so a hydrogen atom in a spherical eigenstate has no
dipole moment. Parabolic eigenstates do not necessarily have definite parity so a hydrogen atom
in such an eigenstate can have a permanent dipole moment. Of course, those states for which the
spherical and parabolic eigenfunctions are the same cannot have dipole moments.
Solution to Chapter 4 problems
page 11