Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
Lecture 9 (October 17, 2008):
(version 1.0; 20-Oct-2008::23:00)
Variational Methods and the ground state energy
[The first part of this lecture was spent reviewing the “Free Electron Gas”, and
applications, from the last lecture. That material is included in the preceding notes.]
Calculating a good upper limit on the ground state:
The first step in solving many problems in quantum mechanics is to find the
energy eigenfunctions and eigenvalues of the Hamiltonian H , and the natural place to
start is with the ground state (which is by definition the lowest-energy eigenstate). It is
easy to prove (done in class, and in Shankar 16.1) that for any wavefunction 
H


 H
 EG

(9.1)
where H is the Hamiltonian. If  is normalized (usually a good first step) then the
denominator in the middle term is 1. If we have chosen a good trial wavefunction, which
has the characteristics and boundary conditions of the ground state wf, then we would
expect H  to be very close to EG ; if we have chosen the ground state itself, then the
equality holds. But even if our guess is not very good, we will get a rigorous upper limit
on EG (which can be quite useful) which is usually reasonably close to EG . The
reasons for this are explained in Shankar at the top of page 433.
As an example of this, we considered the ground state of the infinite square well
potential (ISW) and worked through this in class. First we made a good, well-motivated
choice for our trial wavefunction, and got a result (here) very close to the exact solution.
Then we tried a not-so-good trial wf, giving a poorer limit (here), but one that was still
the right order of magnitude.
Variational methods:
We can try to optimize our trial wf by writing it as a function of some parameter
 . Then the resulting upper limit that we obtain from (9.1) will be a function of  . By
finding the value of  that minimizes the limit, we can both optimize our result and also
get a better understanding of how the wavefunction should be parameterized.
The example that Shankar uses (top of page 431) is the hydrogen atom. We
already know that the ground state wavefunction is  G   100 (where the subscripts are n,
l, and m) , and this solution has the form  G  A exp(r / a0 ) . If we choose as our trial wf
 (r )  A exp( r ) then our limit will be the exact solution ( EG  1.00 Ry ), given by
  1/ a0 . Instead, Shankar chooses the form  (r )  A exp   r 2  , and obtains a limit
of 0.85 Ry . I think it is worthwhile to work through this example in detail, to review
and get practice in doing radial integrals – that is done here.
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Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
The ground state of the Helium atom:
The Hamiltonian for the helium atom is
2
 1 1  e2
H 
12   22   Z0    

2m
 r1 r2  r12
where Z0  2
(9.2)
We know that the ground state has two electrons in the 1s  subshell, so
   100  r1  100  r2 
(9.3)
[Note: We have been looking at systems where the ground state has a symmetric spin
state and a totally antisymmetric spatial state. Helium provides a perfect example of this
in its excited states    a (r1 ) b (r1 ) . But if the two spatial states are the same, as they
clearly are in the ground state, then the 2-particle system must have a symmetric spatial
wavefunction (and the spin wave function must be the antisymmetric singlet).]
To find the ground state energy, first consider the He  ion (1 electron). This is
like the hydrogen atom but with Z0  2 . Then scaling from hydrogen, we get:
2
2
1
2  e 
EH 1    mc     13.6 eV
2
 c
(9.4)
2 2


Z
e
1
EHe 1    mc 2   0   4 EH 1  54.4 eV
2
 c 
Note that the radius of the orbit is reduced by a factor of two, so
a0
Z 0e2
1
(9.5)

and
 108.8 eV, K  54.4 eV
r
2
r
When we add a second electron, we might (very naively) expect it to have the same
energy, in which case we would predict
2
 Z 0e2 
1
2
EHe1    mc  2 
(9.6)
  108.8 eV
2
 c 
But the measured value is
EHe1 (exp)  79.0 eV
(9.7)
The reason for the discrepancy is obvious: we have neglected the e-e interaction term in
(9.2). We would expect the mean ee separation to be of order a0 / 2 (=e-nucleus
separation) or maybe twice this (if the electrons are on opposite sides of the nucleus).
The former would give a correction E  54.4 eV and the latter would give
E  27.2 eV . Using these estimates, we would guess that the true energy should be
(9.8)
EHe1  108.8 eV+  40.8  13.6 eV  68.0  13.6 eV
This is reasonably close to the measured value, but is not very precise.
How can we get a better and more quantitative estimate? There is a “classical
ansatz” that I have worked out, that gives a reasonable value, but I’m not sure how it
relates to the true QM picture. If you’re interested, the argument is given here; maybe we
will come back to it later. But for now, let’s pursue a variational approach. We know
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Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
that each electron “shields” the other from the full nuclear charge of Z 0 e . If each
electron sees an effective coulomb charge of Z, then the hydrogen-like wavefunction of
each should be
1/ 2
 Z3 
 100   3  e  Zr / a0
(9.9)
  a0 
(You should verify the normalization.) This is not an exact solution, since the effective
potential will not be proportional to 1/r, but this is a reasonable form to use in a
variational calculation. We will find the value of Z that minimizes the energy, using the
exact Hamiltonian of (9.2), and that will give us a rigorous upper limit on the ground
state energy.
This would be an exact eigenfunction of the Hamiltonian if we were to drop the
ee interaction term and replace Z 0 with Z , so we will rewrite the Hamiltonian as
H  HA  HB
 2
(9.10)
1   
1  e 2 
2
2
2 1
2 1
 




Ze


Z

Z
e



  

 
1
2
0
 2m
 r1 r2   
 r1 r2  r12 
where H A is the first term in brackets   , and H B is the second. Our variational trial
solution is
 Z 3   Z  r1  r2  / a0
(9.11)
   3 e
  a0 
This is an eigenfunctions of H A , with eigenvalue
2
 Ze 2 
1
2
E1 A    mc 2  
  2  2 Z EH 1
2
c


(9.12)
We need to evaluate  H B  . The first terms give an expectation value
 1
1 
Z
2
E1B1   Z  Z 0  e 2 

   Z  Z 0  e  2   4  Z  Z 0  Z  EH 1 (9.13)
r2 
a0
 r1
where a0 , as always, is the Bohr radius of the hydrogen atom. In the last step we used
the relation e2 / a0  2  EH 1 .
The only remaining term is
Vee
 Z3 
 3 
  a0 
2
2
 Z  r1  r2  / a0  e   Z  r1  r2  / a0 3
e
d r1d 3r2
 e

r
 12 
2
 r 2 dr d cos  r 2 dr d cos  d d 
 Z3  2
2 Z  r1  r2  / a0
1 2
2
2
1
2
 1 1

  3  e    e
2
2



a
r

r

2
r
r
cos

 0
1
2
1 2
2


(9.14)
2
 r 2 dr r 2 dr d cos 
 Z3 
2 Z r  r / a
2
  3   8 2 e 2    e  1 2  0  1 1 2 2
2
2


a
 0
 r1  r2  2r1r2 cos  2




where we have done the integrals over 1 , 2 , and cos 1 . The integral over cos 2 gives
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Ph235
Autumn 2008
F. Merritt
Oct 17, 2008

  1
1
d cos  2
 r 2  r 2  2r r cos  



  r 2  r 2  2r r cos   r1r2  1 2 1 2 2  1
1 
1
2
1 2
2 
1 
2
2

 r1  r2    r1  r2  


r1r2 
1

1 
r1r2 
 r1  r2 
2

 r1  r2 
2


(9.15)
2
 1

 r r 2r2  r if r2  r1 
12

1


 1 2r  2 if r  r 
1
2
 r1r2 1 r2

Using these expressions, we can now integrate over r2 :
Vee
2
r

 Z3 
2
2
 1

  3   8 2 e 2   r12 dr   e 2 Z  r1  r2  / a0 r22 dr2   e 2 Z  r1  r2  / a0 r22 dr2 
1
r1
r2
  a0 
r1
 0

2

 r1 2 Zr / a 2 2

 Z3 
2
2 Zr1 / a0 2
2 2
  3   8 e   e
r1 dr   e 2 0 r2 dr2   e 2 Zr2 / a0 r22 dr2 
1
r1
r2
  a0 
r1
 0

2
2

 2  a 3 2 Zr1 / a0  x 2

 Z3 
 a0 
2 Zr1 / a0 2
2 2
x
0
  3   8 e   e
r1 dr  

e
x
dx

2
e
xdx



 
0
1 r
 2Z  2 Zr1 / a0
  a0 
 1  2Z 

(9.16)
2

 2  a0 3   2 Zr1 / a0  x 2
 Z3 
 2Z    x
2 Zr1 / a0 2
2 2
  3   8 e   e
r1 dr  
e
x
dx

r
e
xdx



 
1


1 r
  a0 
 a0  2 Zr1 / a0

 1  2Z    0
The integrals over x can be integrated by parts, or you can use tables. If we call the
expression in brackets ... above I, then integrating gives:
2 Zr1 / a0



 2 Zr1   x
e   x  1 
I   e  x   x 2  2 x  2  



2 Zr1 / a0
0
 a0 




  2  e  x  x 2  2 x  2    xe  x  x  1
x  2 Zr1 / a0
(9.17)

 2 Zr1   
 2  e 2 Zr1 / a0 
  2 
a
 0   

Substituting this back into (9.16), we are left only with the integral over r1 :
2
Vee
 Z3 
  3   8 2 e2  e2 Zr1 / a0
  a0 
2

Z 
a 
  3   2e 2   0 
Z
 a0 
3
3

 2  a0 3  
 
2 Zr1 / a0  2 Zr1
 2   r12 dr
 

  2  e
 r1  2Z   
 a0
 
e2 Zr1 / a0

 
2 Zr / a  2 Zr1
 2   r1dr
2  e 1 0 
 a0
 

(9.18)
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Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
Changing variable gives
 2Zr1
 

e 2 Zr1 / a0 2  e 2 Zr1 / a0 
 2   r1dr1

 a0
 
2
5
 Z 3   2e2   a0 
x
x
 3  
  e 2  e  x  2 xdx
 a0   4   Z  0
2
Vee
 Z3 
a 
  3   2e2   0 
Z
 a0 
3


2
(9.19)
 Z 3   2e2   a0   5  Z e 2 5 5  1 e 2 
5
 3  
 Z 
   Z  EH 1
    
4
 a0   4   Z   4  a0 2 4 4  2 a0 
5
Summing equations (9.12), (9.13), and (9.19) gives
5 

H   2 Z 2  4 Z  Z  Z 0   Z  EH 1
(9.20)
4 

which agrees with Griffiths 7.32 . For Z  Z0  2 , this gives an expectation value of
74.83 eV , which is not bad as an upper limit (experiment = 78.98 eV ). But a better
limit is obtained by minimizing (9.20) with respect to Z. This gives
27
Z
 1.688  H  77.45 eV
(9.21)
16
This is a rather good result, less than 2% above the experimental value.
5