The 3D Harmonic Oscillator
The last problem in HW#9 involves the solutions to the 3D Harmonic Oscillator.
Gasciorowicz asks us to calculate the rate for the “ 2 p  1s ” transition, so the first
problem is to figure out what he means. Harmonic oscillator states in 1D are usually
labeled by the quantum number “n”, with “n=0” being the ground state [since
En  (n  12 )  ]. But in this problem, 1s means the ground state and 2p means the l  1
component of the first excited state, named in analogy to the hydrogen atom
wavefunctions where n=1 corresponds to the ground state.
In any case, this gives us a good opportunity to review what we learned in Ph234.
We discussed the 3D SHO there, and one of the homework problems (Shankar 12.6.11)
in HW#9 was to derive the wavefunctions. The student should review “The Isotropic
Oscillator” in Shankar (pages 351-2) and the solution to 12.6.11, which I have added to
the “Examples” on the Ph235 website. [Note that is Shankar’s notation, the ground state
has n=0 so Shankar if Shankar has written this problem, he would have asked for the
transition probability from the “1p” state to the “0s” state. I will use Shankar’s notation
below.]
But this problem gives us a good opportunity to review the solutions of
spherically symmetric potentials that we derived in Ph234, so that is what I will try to do
here.
For the general problem of a spherically symmetric potential V ( r ) , it is clearly
best to use spherical coordinates. Since  H , Lz  =0 [because a rotation about z does not
affect V(r)], we can always find (with separation of variables) eigenstates of the form
(1.1)
  r, ,   REl (r )Ylm  ,  
To find Rnl  r  , first write the Hamiltonian in spherical coordinates:
 2 2
H
  V (r )
2m
 21 d  2 d 
1
 
 
1
2 

r

sin



  V (r )




2m  r 2 dr  dr  r 2 sin   
  r 2 sin 2   2 
By using separation of variables, or by comparing (1.2) to the equation for L2 in
spherical coordinates [Shankar 12.5.36, p. 335], this can be written as
 2  1 d  2 d 
L2
H
r

 V (r )



2m  r 2 dr  dr   2mr 2
(1.2)
(1.3)
2
l  l  1
 2  1 d  2 d 

 V (r )

r
 
2m  r 2 dr  dr  
2mr 2
Now substitute REl  r   U El  r  / r and use the energy eigenvalue equation to obtain the
radial equation:
2
2
d 2U El
l (l  1)


U El  V (r )U El  EElU El
(1.4)
2
2m dr
2mr 2
So far, this development is the same any central potential. To find the form of U,
we need the asymptotic behavior of (1.4), and this depends on V ( r ) . In the case of a 3D
harmonic oscillator potential,
V (r )  12 m02 r 2
(1.5)
This term clearly dominates as r   , and in that limit equation (1.4) reduces to
2
d 2U El 1
(1.6)

 m02 r 2U El  0
2m dr 2
2
which has the asymptotic solution
 m0 
U El      El (  ) exp    
where   
(1.7)
 r


where El (  ) is a polynomial in  of finite order. Writiing (1.4) in terms of  gives
1/ 2
1
2
2
d 2U El    l (l  1)
2E

U El      2U El     
U 
2
2
d

 El
Substituting in (1.7) gives the equation for El (  ) :
 2E
   2    
 
1 
l  l  1 
  0
2 
(1.8)
(1.9)
Now go back to (1.4) and look at the limit as r  0 . In this limit the equation
becomes
2
2
d 2U El   
l (l  1)
(1.10)


U El   
2
2m d 
2mr 2
which tell us that the lowest power of  in     is l  1 . Since     must be a finite
polynomial, this means
N
      l 1  c j  j
(1.11)
j 0
Substituting this into (1.9) gives both the relation between N and En and also the
recursion relation for the c j ’s:
En    N  l  32 
and
(1.12)
En  ( j  l  32 ) 
 2 
c j 2  c j 
(1.13)

    j  l  3 j  l  2   l  l  1
Note that since cN 1  0 by definition of N, the recursion relation implies that c j  0 for
all odd j’s. Consequently N is even,