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
for a system described by the Hamiltonian , but you are
unable to solve the time-independent equation
g
E
..
schr o dinger H
What should you do
?
7.1 Theory
Theorem: Theorem:
H H
E
g
≡ Ψ
Ψ ≤
is any normalized function
That is, the expectation value of in the presumably in correct state is certain to overestimate the ground-state
energy. Of course, if just happens to be one of the excited states, then obviously exceeds .
H
Ψ
H
g
E Ψ
Proof Proof
Since the unknown eigenfunctions of form a complete set ,we can express as a linear combination of them:
H
Ψ
, with
n n
n n
n n
c H
E Ψ =
Ψ Ψ =
Ψ
And is normalized,
Ψ
2
1
m m
n n
m n
m n m
n n
m n
n
c c
c c c
= Ψ Ψ = Ψ
Ψ
= Ψ Ψ
=
2 m
m n
n m
n
m n n
m n
n n
m n
n
H c
H c
c E c E c
= Ψ
Ψ
= Ψ Ψ
=
Meanwhile,
Since
n g
E E
≤
,we get
g n
n g
E c
E H
= ≥
2
Which is what we were trying to prove.
To find the ground-state energy Aim
Processes
Examples Examples
Processes
Step 1. Select a trial wave function
Step 2. Calculate in this state
Step 3. Minimize the
Step 4. Take as the appropriate ground-state energy
H
H
Ψ
min
H
Example 1.
2 2
2 2
2
1 d
H m
x
ω
= − +
To find the ground-state energy for the one-dimensional harmonic oscillator:
ω
2 1
=
g
E
Of course, we already know the exact answer see chapter
2:
2
2 2
H m
x m dx
ω
= − +
Pick a Gaussian function as our trial state
2
bx
Ae x
−
= Ψ
where is a constant and is determined by normalization:
b
A
b A
dx e
A
bx
2 1
2 2
2
2
π
= =
∞ ∞
− −
4 1
2 =
π
b A
The mean value of is
H
2 2
2
2 2
2 2
2 2
2 2
2 2
1 2
2
= 2
8
bx bx
bx
d H
A e
e dx
m A
e x dx
m dx
b m
m b
ω
ω
∞ ∞
− −
− −∞
−∞
= − +
+
We can get the tightest bound through minimizing with
H
respect to :
b
8 2
2 2
2
= −
= b
m m
H db
d
ω
2
ω
m b
=
Putting this back into ,we find
H
ω
2 1
min
= H
Example 2.
To look for the ground state energy of the delta-function potential:
2 2
2
2 d
H x
m dx
αδ
= − −
Again ,we already know the exact answer see chapter 2:
2
g
b E
α π
= −
2m dx
The mean value of is
H
We also Pick a Gaussian function as our trial state:
2
bx
Ae x
−
= Ψ
2
d a
H =
− =
2 2
2
α
m b
=
Minimizing it,
2 2
2
2 2
2 2
2 2
2
2 2
= 2
bx bx
bx
d H
A e
e dx
A e
x dx m
dx b
b m
α δ
α π
∞ ∞
− −
− −∞
−∞
= − −
−
2 2
d a
H db
m b
π
= −
=
4
2
π α
m b
=
2 2
min
m H
α π
= −
which is indeed somewhat higher than
g
E
,since
2
π So
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:
Figure 7.1: “triangular” trial wave function for the infinite square well
2 a
a x
x Ψ
3 2
2 2
2 2
2
1 12
a a
a
a A
x dx a
x dx A
= +
− =
2 3
A a
a =
Where is determined by normalization:
A
In this case
The derivative of this step function is a delta function see
problem 2.24b
2 2
2
a x
A a
x A
x A
dx d
− +
− −
= Ψ
δ δ
δ
2 2
2 2
2
12 0 2
2 A
A a
a = −
Ψ − Ψ
+ Ψ =
= and hence
2
2 2
2 A
a H
x x
x a
x dx m
δ δ
δ
= − −
− +
− Ψ
2
0 2 2
2 2
2 a
m m
ma = −
Ψ − Ψ
+ Ψ =
=
The exact ground state is see chapter 2, so
the theorem works
2 2
2
2
g
E ma
π
=
2
12
π
write down a trial wave function
The variational principle is very powerful and easy to use
Advantages
Conclusions
Calculate
tweak the parameters to get the lowest possible value
H
Even if has no relation to the true wave function, one
often gets miraculously accurate values for
Ψ
g
E
Limitations
It applies only to the ground state
You never know for sure how close you are to the target 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
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