Chapter 5 - THE CLASSICAL HYDROGEN ATOM
Problems with solutions
5.1 Using SI units, find the period Tn of an electron in a hydrogen Bohr orbit of principal
quantum number n. Show that this period is consistent with Kepler's third law. Show that is the
same as Tc the period deduced from the correspondence principal. Hint: Eliminate the 's so the
constants match. Show that the result reduces to T  2 n3 in atomic units.
Solution:
 1  e2
 1  e2
 2 
2
2

From electrostatics F  
 2  me n rn   n  



3
 4 0  rn
 4 0  me rn  Tn 
 4 0  me r 3/ 2   2 n3  4 0  me  a 3/ 2
Solving for Tn we have Tn  2
n
0
e2
e2


Note that this result can also be obtained as follows.
2 rn
2 n 2 a0
4
Tn 
 2
 2 0  2 n3 a0
vn
e
e / 4 0 n


use a0   4 0  
2


2
me e 2


me e 2 a0
so that
4 0
 m e2 a
4 0
3
Tn  2  2 n  e 0
e
 4 0

4 0 me 3/ 2
a0
 a0  2 n3
e2

Note that in atomic units everything in Tn is unity except 2 n3 .
Now find the frequency from the correspondence principle according to which the frequency of
the motion should be the same as the difference between adjacent energies divided by . Since
2
2
me c 2  2

1  me c  
2
En  
 E 
n   c 
2
3
2
n
n
Tc
where the subscript c means "correspondence". But, for adjacent levels n  1 so we have
2 n3
Tc 
. To compare this expression with that for Tn we must replace .
 mec 2  2
One way to do this is to write  in terms of the Bohr radius, see Table 2, Chapter 1.
2


2 n3   me c  
1
1
3 2  me 
T

a
so
a0  






c
   2 n a0  
2  0
  me c 
a0  me c 
 
 mec    
But, we must replace  as above so
4 0
4 0 me
Tc  2 n3a0 2 me
 2 n3a03/ 2
 Tn
2
me e a0
e2
5.2 The (classical) Lenz vector (in a.u.) is
A  p  L  rˆ
where r̂ is the unit vector in the r direction. For a general central potential find a general
expression for the time derivative of A, i.e. A and show that A is a constant of the motion for a
Coulomb potential.
Solution to Chapter 5 problems
page 1
Chapter 5 - THE CLASSICAL HYDROGEN ATOM
Problems with solutions
Solution:
d
A   p  L   rˆ
dt
First work on the term
d
 p  L.
dt
d
 p  L  p  L  p  L  p  L because L  0 for any central potential.
dt
Now use Newton's 2nd law for an arbitrary central potential V(r).
We have
d
dV
 r  dV
 p  L   dV rˆ  L  1 dV r  L .
p  rˆ
  
which gives
dt
dr
r dr
dr
 r  dr
But L  r  p  r  mr which is L  r  r in atomic units so
d
 p  L   1 dV r  L   1 dV r  r  r    1 dV r r  r   rr 2
dt
r dr
r dr
r dr
1 d
r  r   rr so
Now r  r 
2 dt
d
 p  L   1 dV rrr  rr 2
dt
r dr
1 dV 3  r
r 

r  2 r 
r dr  r
r
dV d  r 
 r2
  
dr dt  r 
dV d
 r2
 rˆ
dr dt
dV 
 r2
rˆ
dr
d
dV   
dV  
rˆ  rˆ  1  r 2
Then A   p  L  rˆ  r 2
rˆ
dt
dr
dr 

1
dV
1
 2
For the Coulomb or Kepler potential we have V r    in atomic units. Therefore
r
dr
r

and A  0 .




5.3 a) Derive the equation of a Keplerian ellipse in terms of the Lenz vector A and show that the
eccentricity is   A .
b) Show that
2
n
Note that neither  nor n are "quantum numbers" because, classically, they are continuously
variable.  is the classical angular momentum and n is a measure of the energy. Recall that, in
atomic units E  1/  2n 2  .
  1
c) By averaging over a period show that the electric dipole moment of a Keplerian hydrogen
3
atom is  p   2 n 2 1 
2
/ n2 .
Solution to Chapter 5 problems
page 2
Chapter 5 - THE CLASSICAL HYDROGEN ATOM
Problems with solutions
d) For what value of the  is  p   0 ? What is special about these orbits?
e) Find the positions (or position) of the maxima (or maximum) in the radial probability density
for hydrogen atoms having  n 1, the maximum angular momentum. What is special about
these states? It may be helpful to recall that
dp
 r   2r 
1  2r 
q


Rn r   exp        L2
in
atomic
units
and
that
L


Lq    .

n 
p
d p
 n  n 
 n 
Solution:
a)
The Lenz vector is A  p  L  ar .
Now take the dot product with r.
A  r  p  L  r  a r  r Now permute the cross/dot product as follows: Ar cos  r  p  L  r
2
 1   cos  which is the equation
r
of an ellipse with A =  provided  < 1 and  is measured from pericenter.
2
 1   cos 
b) For a Coulomb/gravitational potential the equation of the orbit is of the form
r
or, since L =  Ar cos   2  r . Solving for r we have
where   1  2E 2 (atomic units). But E  
c)
1
2
in au so that   1  2
2n
n
p  r in atomic units.
By symmetery <y> = 0. Also <x> = <rcos> (ignore the minus sign).
1 
1
x   r cos d but, inserting the equation of the orbit and conservation of angular
0


momentum,   r 2 where A has been replaced by  we have
1 
r2
5 
cos
5 d   1 
cos
x 
r
cos


d


d


 
d

3 
2
3 
3
3

2n

2n
2n d   2  1   cos 2
1   cos 
Integrating we get

5  1  d
sin 
1
x 
 

3 
2
2n  2  d 1   1  cos    1   2


5
2n 3
Now, at the upper limit, tan
 1 d
 
 2  d
1



1 

2 
 1  

   / 2 .

2
 1  
2
tan 1

d
   1  cos 

1    tan


2
1   
2

Similarly, for the -limit. Therefore, the term in | | =
 and we have
Solution to Chapter 5 problems
page 3
Chapter 5 - THE CLASSICAL HYDROGEN ATOM
Problems with solutions
  5  1  3 
5  1  d 
2
2
x 
    5      

3  
3/ 2
2
2 5/ 2
2n  2  d  1   
 n  2  2  1   
Leave the  in the numerator for now, but replace it in the denominator.

2
  1 2  1  2
n

5/ 2
5
5  3  
3
 5  x  3  5
  n 2  so finally
n
n 2  5 2
n
 
 3
p    n2 1  2
n
 2
2
d) p  0 when  = n, that is, the maximum angular momentum. This corresponds to a circular
orbit.
 2r 
 2r 
e) For  = n – 1 the associated Laguerre polynomial is L2nnn111    L22nn11   When the
 n 
 n 
lower index and the upper one are the same, the polynomial is a constant as may be seen from
dp
L pp   
L p   because the Laguerre polynomial Lp is a pth order polynomial. Therefore,
d p
the radial wave function for  n 1 is given by
 r
Rn ,n 1 r   r n 1 exp    . The radial probability density, P(r), is given by
 n
2
 2r 
Pr   r 2 Rn ,n 1 r   r 2 n  exp    . The peak in the probability distribution can be found by
 n 
differentiating and setting equal to zero. We get rmax=n2. There is only one maximum in the
probability density so this is the equivalent of a Bohr orbit. Notice that this is the same result
that one obtains from the Bohr model of the atom, i.e. that rmax  n 2 a o . These states obviously
correspond to "circular" states.
5.4 This problem should show you why the conservation of the (classical) Lenz vector, A,
implies closed orbits for the Kepler problem. Use A in a.u. so that A  p  L  rˆ
a) Express A in terms of r and p alone, i.e. eliminate L. No cross products.
b) Show that –


1 
1 
  rmin  2E 

A  rmax  2 E 
rmax 
rmin 


where rmax and rmin are the maximum and minimum values of r, i. e. apocenter and pericenter.
c) Show that A is parallel to rmin and antiparallel to rmax .
d) From the answer to c) it is clear that for a circular orbit A = 0. Prove this mathematically
from the equations that you derived in b).
Solution:
a) L  r  p
so that
Solution to Chapter 5 problems
page 4
Chapter 5 - THE CLASSICAL HYDROGEN ATOM
Problems with solutions
A  p  r  p   rˆ
 rp 2  p p  r   rˆ
b) We eliminate p2 from this expression using the expression for the total energy.
p2 1
2
E


p 2  2E 
2 r
r
r
and letting rˆ  we have
r
1

A  r  2 E    p p  r 
r

Now, at rmin or rmax then p  r and pr  0, i.e. –
Then, letting r  rmin or rmax gives the desired result.


1 
1 
  rmin  2 E 
 and noting that the total energy is given by
c) Using A  rmax  2 E 
rmax 
rmin 


1
E
where a is the semi-major axis and is such that rmin  a  rmax we see that the quantity
2a
in ( ) is positive for rmin and negative for rmax thus establishing the directions.
d) The only way that A could be parallel and anti-parallel to two opposite vectors is if A = 0.


1 
1 
  rmin  2 E 

To see this mathematically, we use either of A  rmax  2 E 
rmax 
rmin 


where, for a circular orbit, rmax  rmin  R  a.
Since E = –1/2a, A = 0 in both equations. Note that this is also consistent with A being the
eccentricity.
Solution to Chapter 5 problems
page 5