量子力学
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
CHAPTER 7
The variational principle
7.1
Theory
7.2
The ground state of helium
7.3
The hydrogen molecule ion
Suppose you want to calculate the ground-state energy Eg
for a system described by the Hamiltonian H , but you are
..
unable to solve the (time-independent) schr o dinger equation
What should you do
?
7.1 Theory
Theorem:
is any normalized function
Eg ≤ Ψ H Ψ ≡ H
That is, the expectation value of H in the (presumably
in correct) state Ψ is certain to overestimate the ground-state
energy. Of course, if Ψ just happens to be one of the excited
states, then obviously H exceeds E .
g
Proof
Since the (unknown) eigenfunctions of H form a complete
set ,we can express Ψ as a linear combination of them:
Ψ = ∑ cn Ψ n
H Ψ n = En Ψ n
, with
n
And Ψ is normalized,
1= Ψ Ψ =
∑c
m
Ψm
m
∑c Ψ
n
n
n
= ∑∑ c m cn Ψ m Ψ n = ∑ cn
*
m
n
n
2
Meanwhile,
H =
∑c
m
Ψ m H ∑ cn Ψ n
m
n
= ∑∑ c m En cn Ψ m Ψ n = ∑ En cn
*
m
n
n
Since Eg ≤ En ,we get
H ≥ E g ∑ cn = E g
2
n
Which is what we were trying to prove.
2
Examples
Aim
To find the ground-state energy
Processes
Step 1.
Select a trial wave function Ψ
Step 2.
Calculate H
Step 3.
Minimize the H
Step 4.
Take H minas the appropriate ground-state energy
in this state
Of course, we already know
the exact answer (see chapter
1
2):
E g = hω
2
Example 1.
To find the ground-state energy for the one-dimensional
harmonic oscillator:
h2 d 2 1
2 2
H =−
+
m
ω
x
2
2m dx 2
Pick a Gaussian function as our trial state
Ψ ( x) = Ae
− bx 2
where b is a constant and A is determined by normalization:
1= A
2
∫
∞
e
−∞
− 2 bx
2
dx = A
2
π
2b
 2b 
A= 
π 
1
4
The mean value of H is
h2 2 ∞ −bx2 d 2 −bx2
1
2 ∞ −2bx2 2
2
H =−
A ∫ e
e
dx + mω A ∫ e x dx
2
−∞
−∞
2m
dx
2
h2b mω 2
=
+
2m 8b
( )
We can get the tightest bound through minimizing H with
respect to
b:
d
h 2 mω 2
H =
− 2 =0
db
2m 8b
Putting this back into H ,we find
1
H min = hω
2
mω
b=
2h
Again ,we already know the
exact answer (see chapter 2):
2b
Eg = −α
π
Example 2.
To look for the ground state energy of the delta-function potential:
h2 d 2
H =−
− αδ ( x)
2
2m dx
2m
We also Pick a Gaussian function as our trial state:
Ψ ( x) = Ae
The mean value of H is
− bx 2
h 2 2 ∞ − bx2 d 2 − bx2
2 ∞ −2 bx 2
H =−
A ∫ e
e
dx − α A ∫ e δ ( x)dx
2
−∞
−∞
2m
dx
h 2b
2b
=
−α
2m
π
(
)
Minimizing it,
d
h2
a
H =
−
=0
db
2m
2π b
2m2α 2
b=
πh 4
So
H
min
mα 2
=−
π h2
which is indeed somewhat higher than E g ,since π > 2
Example 3.
To find an upper bound on the ground-state energy of
the one-dimension infinite square well, using the “triangular”
trial wave function (figure 7.1):
Ψ (x)
a
a
2
Figure 7.1: “triangular” trial wave function
for the infinite square well
x
Where A is determined by normalization:
3
a
a
2
 a2 2

1 = A  ∫ x dx + ∫a (a − x) 2 dx  = A
12
2
 0

2
2 3
A=
a a
In this case
The derivative of this step function is a delta function (see
problem 2.24b)
dΨ
a ) + Aδ ( x − a )
=
A
δ
(
x
)
−
2
A
δ
(
x
−
2
dx 2
and hence
h2 A 
a ) + δ ( x − a ) Ψ ( x)dx
H =−
δ
(
x
)
2
δ
(
x
−
−
2

2m ∫ 
2 2
2
h2 A 
h
A
12
h
=−
Ψ (0) − 2Ψ (a ) + Ψ (a )  =
=
2


2m
2m
2ma 2
h2π 2
The exact ground state is Eg =
2 ( see chapter 2), so
2ma
the theorem works ( 12 > π 2 )
Conclusions
Advantages
The variational principle is very powerful and easy to use
write down a trial wave function
Calculate H
tweak the parameters to get the lowest possible value
Even if Ψ has no relation to the true wave function, one
often gets miraculously accurate values for Eg
Limitations
It applies only to the ground state
You never know for sure how close you are to the target
and all you can certain of is that you have got an upper
bound.
7.2
The ground state of helium
Our task:
To calculate the ground–state energy by using the
Variational Principle
Theoretically reproduce the value :
E g = −78.975 ev (experimental)
r r
r1 − r2
−e
r2
−e
r2
+2e
Figure 7.2:the helium atom
The Hamiltonian for the helium atom system (ignoring
fine structure and small correction) is
2
2 
h
e 2 2
1
H =−
∇12 + ∇ 2 2 −
+ − ur ur
2m
4πε 0  r1 r2 r1 − r2

Let
e2
1
V ee =
ur ur
4 π ε 0 r1 − r2
(
)




If we ignore the electron-electron repulsion Vee is, the
ground-state wave function is just
ur ur
ur
ur
8
Ψ0 r1, r2 ≡ Ψ100 r1 Ψ100 r2 = 3 e−2( r1 +r2 ) a
πa
where Ψ100 is hydrogen-like wave function with Z = 2 .
( )
()
( )
Consequently ,the energy that goes with this simplified
picture is 8 E1 = −109 ev (see Chapter 5 ).
In the following we will apply the variational principle ,
using the Ψ0 as the trial wave function. The eigenfunction of
Hamiltonian is:
H Ψ0 = ( 8E1 + Vee ) Ψ0
Thus
H = 8E1 + Vee
where
 e  8 
=
 3 
 4πε 0   π a 
2
Vee
2
∫
e
−4( r1 + r2 ) a
ur 3 ur
ur ur d r1d r2
r1 − r2
3
To get the above integral value conveniently, we do the
r
integral first and orient the r2 coordinate system so that the
r
r1 polar axis lies along (see Figure 7.3).
By the law of cosines,
ur ur
r1 − r2 = r12 + r2 2 − 2r1r2 cos θ 2
r
r2
and hence
e−4r2 a 3 ur
e−4r2 a
I 2 ≡ ∫ ur ur d r2 = ∫
r22 sin θ2 dr2 dθ2 dφ2
r1 − r2
r12 + r22 − 2r1r2 cos θ2
φ2 integral is trivial
( 2π );the φ1 integral is:
The
∫
sin θ 2
π
0
=
2
2
r1 + r2 − 2 r1 r2 cos θ 2
dθ 2
r12 + r2 2 − 2 r1 r2 cos θ 2 π
0
r1 r2
Figure 7.3: choice
r of coordinate for the
r2 integral
=
1
rr
12
(
2
2
−
+
r12 + r22 + 2rr
r
r
12
1
2 − 2rr
12
1
= ( r1 + r2 ) − r1 − r2 
rr
12
Thus
=
{
)
2 r1 , if r2 < r1
2 r2 , if r2 > r1
∞
 1 r1 −4r2 a 2

−4r2 a
I 2 = 4π  ∫ e
r2 dr2 + ∫ e
r2 dr2 
0
r
1
 r1

π a3 
 2r1  −4r1 a 
=
1 − 1 +  e


8r1  
a 

It follows that Vee is equal to
 e2   8    2r1  −4r1 a  −4r1 a
e
r1 sin θ1dr1dθ1dφ1

  3  ∫ 1 − 1 +  e

a 

 4πε 0   π a   
v
The angular integals are easy( 4π),and the r1 integal becomes
∫
∞
0
 −4r a 
2r 2  −8r a 
5a 2
−r +
 re
dr =
e
a 
128



Finally , then,
Vee
5  e2 
5
=

 = − E1 = 34ev
4a  4πε 0 
2
And therefore
H = −109ev + 34ev = −75ev
Not bad , but we
can do better!
Can we think of a more realistic trial function
than
Ψ
0
?
We try the product function
Z 3 − Z ( r1 + r2 ) a
r r
Ψ1 (r1 , r2 ) ≡ 3 e
πa
and treat Z as a variable rather than setting it equal to 2.
The idea is that as each electron shields the nuclear charge
seen by the other ,the effective Z is less than 2.
In the following, we will treat Z as a variational parameter,
picking the value that minimizes H .
Rewrite H in the following form:
2
2 
Z − 2) ( Z − 2)


(
e
Z
Z
e
1
h2
2
2

H =−
∇
+
∇
−
+
+
+
+
ur ur
(


1
2 )
2m
4πε 0  r1 r2  4πε 0  r1
r2
r1 − r2

The expectation value of H is evidently
2

 1
e
2
H = 2Z E1 + 2 ( Z − 2 ) 
+ Vee

 4πε 0  r
Here 1 r is the expectation value of 1 r in the (one-particle)
hydrogenic ground state Ψ100 (but with nuclear charge Z).




And according to Chapter 6, we know
1
a
=
r
Z
The expection value of Vee is the same as before ,except
that instead of Z = 2 ,we now want arbitrary Z—so we
multiply a by Z 2 :
Vee
5Z  e 2 
5Z
=
E1

=−
8a  4πε 0 
4
Putting all this together, we find
H =  2 Z 2 − 4 Z ( Z − 2 )( 5 4 ) Z  E1 =  −2 Z 2 + ( 27 4 ) Z  E1
The lowest upper bound occurs when H is minimized:
d
H =  −4 Z + ( 27 4 )  E1 = 0
dZ
from which it follows that
27
Z =
= 1 .6 9
16
Putting in this value for Z, we find
6
13
H =   E1 = −77.5ev
22
Much nearer to experimental value!
7.3
The hydrogen molecule ion
−e
r2
R
r1
R
r
2
Figure 7.4 :the hydrogen molecule ion, H 2 +.
The Hamiltonian for this system is
h2 2
e2  1 1 
∇ −
H =−
 + 
2m
4πε 0  r1 r2 
To construct the trial wave function , imagine that the ion is
formed by taking a hydrogen atom in its ground state
r
Ψ g (r ) =
1
πa 3
e
−r
a
and then bringing in proton from far away and nailing it down a
distance R away. If R is substantially greatly than the Bohr radius
a, the electron’s wave function probably isn’t changed very much.
But we would like to treat the two protons equally ,so that the electron
has the same probability of being associated with either one. So we
consider a trial function of the form
Ψ = A  Ψ g ( r1 ) + Ψ g ( r2 ) 
Normalize this trial function:
2
r
r
r
2 3r
2 
3
3 
1 = ∫ Ψ d r = A  ∫ Ψ g ( r1 ) d r + ∫ Ψ g ( r2 ) d r + 2 ∫ Ψ g ( r1 )Ψ g ( r2 ) d r 


2
3
r
= 2 A 1 + ∫ Ψ g ( r1 )Ψ g ( r2 ) d r 
2
Let
3
I ≡ Ψ g ( r1 ) Ψ g ( r2 )
1
=
π a3
∫e
− ( r1 + r2 )
a
r
d r
3
Picking coordinates so that the pronton 1 is at the origin and
proton 2 is on the z-axis at the point R (Figure 7.5),we have
r1 = r
r2 = r 2 + R 2 − 2rR cos θ
and therefore
1
−r
−
I = 3 ∫ e ae
πa
r 2 + R 2 − 2 rR cosθ
a
r 2 sin θ drdθ dφ
The φ integral is trivial ( 2π ).
To do the θ integral ,let
y ≡ r 2 + R 2 − 2rR cos θ
so that
d ( y 2 ) = 2 ydy = 2rR sin θ dθ
Figure 7.5: coordinates for the
calculation of I
Then
∫
π
0
e
− r 2 + R 2 − 2 rR cos θ a
1 r+R − y a
sin θ dθ =
e
ydy
∫
r
−
R
rR
a  −( r + R ) a
− r−R
=−
e
r
+
R
+
a
−
e
(
)
rR 
a
( r − R + a )
The r integral is now straightforward:
R
2  −R a ∞
−2r a
−R a
I = 2 −e ∫ ( r + R + a)e rdr + e ∫ ( R − r + a) rdr
0
0
a R 
∞
Ra
+e ∫ ( r − R + a) e−2r ardr 

R
Evaluating the integrals, we find
R 1 R 2

I = −e − R a 1 + ( ) + ( ) 
a 3 a 

In the terms of I , the normalization factor is
1
A =
2(1 + I )
2
Next we must calculate the expectation value of H in the
trial state Ψ. Noting that
 h2 2
e2 1 
∇ −
−
 Ψ g ( r1 ) = E1Ψ g ( r1 )
4πε 0 r1 
 2m
Where E1 = −13.6 ev is the ground-state energy of atomic
hydrogen and the same with r2 in place of
 h2 2
e2  1
H Ψ = A −
∇ −
 +
4πε 0  r1
 2m
 e2   1
= E1 Ψ − A 
  Ψ g ( r1 ) +
 4πε 0   r2
1
r2
r1 ,we have

   Ψ g ( r1 ) + Ψ g ( r2 ) 


1
Ψ g ( r2 ) 
r1

It follows that

 e2  
1
1
H = E1 − 2 A 
  Ψ g ( r1 ) Ψ g ( r1 ) + Ψ g ( r1 ) Ψ g ( r2 ) 
r2
r1
 4πε 0  

2
Calculate the two remaining quantities ,the so-called direct integral,
D ≡ a Ψ g ( r1 )
1
Ψ g ( r1 )
r2
and the exchange integral,
X ≡ a Ψ g ( r1 )
The results are
D =
1
Ψ g ( r2 )
r1
a 
a 
− 1 +  e−2R
R 
R
 a  −R a
X = 1 +  e
 R
a
0
 1
Putting all this together, and recalling that E1 = −  e

 4πε 0  2a
we conclude that
( )

(D + X ) 
H = 1 + 2
E1

(1 + I ) 

This is only the electron’s energy----there is also potential energy
associated with the proton-proton repulsion:
V pp =
e2
1
2a
=−
E1
4πε 0 R
R
Thus the total energy of the system, in units of − E1
and expressed as a function of x ≡ R a ,is less than
2  (1 − (2 / 3) x 2 )e− x + (1 + x)e−2 x 
F ( x) = −1 + 

2
−x
x  1 + (1 + x + (1/ 3) x )e

Figure 7.6:
plot of the function F (x ) ,showing existence of
a bound state.
Evidently bonding does occur, for there exists a region in which the graph
goes below -1,indicating that the energy is less than that of a neutral atom plus
a free proton (to wit, −13.6 ev ).The Equilibrium separation of the protons is
o
about 2.4 Bohr radii, or1.27 A .