Upload
others
View
36
Download
0
Embed Size (px)
Citation preview
The variational principle
The variational principleQuantum mechanics 2 - Lecture 5
Igor Lukačević
UJJS, Dept. of Physics, Osijek
November 8, 2012
Igor Lukačević The variational principle
The variational principle
Contents
1 Theory
2 The ground state of helium
3 The linear variational problem
4 Literature
Igor Lukačević The variational principle
The variational principle
Theory
Contents
1 Theory
2 The ground state of helium
3 The linear variational problem
4 Literature
Igor Lukačević The variational principle
The variational principle
Theory
What is a problem we would like to solve?
To find approximate solutions of eigenvalue problem
Oφ(x) = ωφ(x)
Igor Lukačević The variational principle
The variational principle
Theory
What is a problem we would like to solve?
To find approximate solutions of eigenvalue problem
Oφ(x) = ωφ(x)
A question
Can you remember any eigenvalue problems?
Igor Lukačević The variational principle
The variational principle
Theory
What is a problem we would like to solve?
To find approximate solutions of Oφ(x) = ωφ(x).
A question
Can you remember any eigenvalue problems?
Hψα = Eαψα , α = 0, 1, . . .
whereE0 ≤ E1 ≤ E2 ≤ · · · ≤ Eα ≤ · · · , 〈ψα|ψβ〉 = δαβ
Igor Lukačević The variational principle
The variational principle
Theory
Theorem - the variational principle
Given any normalized function ψ̃ (that satisfies the appropriate boundaryconditions), then the expectation value of the Hamiltonian represents an upperbound to the exact ground state energy
〈ψ̃|H|ψ̃〉 ≥ E0 .
Igor Lukačević The variational principle
The variational principle
Theory
Theorem - the variational principle
Given any normalized function ψ̃ (that satisfies the appropriate boundaryconditions), then the expectation value of the Hamiltonian represents an upperbound to the exact ground state energy
〈ψ̃|H|ψ̃〉 ≥ E0 .
A question
What if ψ̃ is a ground state w.f.?
Igor Lukačević The variational principle
The variational principle
Theory
Theorem - the variational principle
Given any normalized function ψ̃ (that satisfies the appropriate boundaryconditions), then the expectation value of the Hamiltonian represents an upperbound to the exact ground state energy
〈ψ̃|H|ψ̃〉 ≥ E0 .
A question
What if ψ̃ is a ground state w.f.?
〈ψ̃|H|ψ̃〉 = E0
Igor Lukačević The variational principle
The variational principle
Theory
Proof
ψ̃ are normalized ⇒ 〈ψ̃|ψ̃〉 = 1
Igor Lukačević The variational principle
The variational principle
Theory
Proof
ψ̃ are normalized ⇒ 〈ψ̃|ψ̃〉 = 1On the other hand, (unknown) ψ form a complete set ⇒ |ψ̃〉 =
∑α cα|ψα〉
Igor Lukačević The variational principle
The variational principle
Theory
Proof
ψ̃ are normalized ⇒ 〈ψ̃|ψ̃〉 = 1On the other hand, (unknown) ψα form a complete set ⇒ |ψ̃〉 =
∑α cα|ψα〉
So,
〈ψ̃|ψ̃〉 =〈∑
β
cβψβ
∣∣∣∑α
cαψα〉
=∑αβ
c∗βcα 〈ψβ |ψα〉︸ ︷︷ ︸δαβ
=∑α
|cα|2 = 1
Igor Lukačević The variational principle
The variational principle
Theory
Proof
ψ̃ are normalized ⇒ 〈ψ̃|ψ̃〉 = 1On the other hand, (unknown) ψα form a complete set ⇒ |ψ̃〉 =
∑α cα|ψα〉
So,
〈ψ̃|ψ̃〉 =〈∑
β
cβψβ
∣∣∣∑α
cαψα〉
=∑αβ
c∗βcα 〈ψβ |ψα〉︸ ︷︷ ︸δαβ
=∑α
|cα|2 = 1
Now
〈ψ̃|H|ψ̃〉 =〈∑
β
cβψβ
∣∣∣H∣∣∣∑α
cαψα〉
︸ ︷︷ ︸∑α cαH|ψα〉︸ ︷︷ ︸
Eα|ψα〉
=∑αβ
c∗βcαEα 〈ψβ |ψα〉︸ ︷︷ ︸δαβ
=∑α
Eα|cα|2
Igor Lukačević The variational principle
The variational principle
Theory
Proof
ψ̃ are normalized ⇒ 〈ψ̃|ψ̃〉 = 1On the other hand, (unknown) ψα form a complete set ⇒ |ψ̃〉 =
∑α cα|ψα〉
So,
〈ψ̃|ψ̃〉 =〈∑
β
cβψβ
∣∣∣∑α
cαψα〉
=∑αβ
c∗βcα 〈ψβ |ψα〉︸ ︷︷ ︸δαβ
=∑α
|cα|2 = 1
Now
〈ψ̃|H|ψ̃〉 =〈∑
β
cβψβ
∣∣∣H∣∣∣∑α
cαψα〉
︸ ︷︷ ︸∑α cαH|ψα〉︸ ︷︷ ︸
Eα|ψα〉
=∑αβ
c∗βcαEα 〈ψβ |ψα〉︸ ︷︷ ︸δαβ
=∑α
Eα|cα|2
But Eα ≥ E0 , ∀α, hence
〈ψ̃|H|ψ̃〉 ≥∑α
E0|cα|2 = E0∑α
|cα|2 = E0
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
a] Find the ground state energy and w.f. of one-dimensional harmonic oscilator:
H = − ~2
2m∆ +
1
2mω2x2 .
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
a] Find the ground state energy and w.f. of one-dimensional harmonic oscilator:
H = − ~2
2m∆ +
1
2mω2x2 .
How to do this using the variational principle...
(i) pick a trial function which somehow resembles the exact ground state w.f.:
ψ(x) = Ae−αx2
α parameter
A =4
√2α
π from normalization condition (do it for HW)
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
a] Find the ground state energy and w.f. of one-dimensional harmonic oscilator:
H = − ~2
2m∆ +
1
2mω2x2 .
How to do this using the variational principle...
(i) pick a trial function which somehow resembles the exact ground state w.f.:
ψ(x) = Ae−αx2
α parameter
A =4
√2α
π from normalization condition
(ii) calculate 〈H〉 = 〈T 〉+ 〈V 〉
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
How to do this using the variational principle...
(i) pick a trial function which somehow resembles the exact ground state w.f.:
ψ(x) = Ae−αx2
α parameter
A =4
√2α
π from normalization condition
(ii) calculate 〈H〉 = 〈T 〉+ 〈V 〉
〈T 〉 = ~2α
2m
〈V 〉 = mω2
8α
On how to solve these kind ofintegrals, see Ref. [5].
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
How to do this using the variational principle...
(i) pick a trial function which somehow resembles the exact ground state w.f.:
ψ(x) = Ae−αx2
α parameter
A =4
√2α
π from normalization condition
(ii) calculate 〈H〉 = 〈T 〉+ 〈V 〉
〈T 〉 = ~2α
2m
〈V 〉 = mω2
8α
On how to solve these kind ofintegrals, see Ref. [5].
〈H〉 = ~2α
2m+
mω2
8α
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
(iii) minimize 〈H〉 wrt parameter α
d
dα〈H〉 = 0 =⇒ α = mω
2~
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
(iii) minimize 〈H〉 wrt parameter α
d
dα〈H〉 = 0 =⇒ α = mω
2~
(iv) insert back into 〈H〉 and ψ(x):
〈H〉min =1
2~ω
ψmin(x) = 4√
mω
π~e−
mω2~ x
2
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
(iii) minimize 〈H〉 wrt parameter α
d
dα〈H〉 = 0 =⇒ α = mω
2~
(iv) insert back into 〈H〉 and ψ(x):
〈H〉min =1
2~ω
ψmin(x) = 4√
mω
π~e−
mω2~ x
2
exact ground state energy and w.f.
A question
Why did we get the exact energy and w.f.?
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
b] Do the same, but with trial function ψ(x) = Bxe−βx2
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
b] Do the same, but with trial function ψ(x) = Bxe−βx2
(i) normalization gives B =
√2√π
(mω~
)3/4
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
b] Do the same, but with trial function ψ(x) = Bxe−βx2
(i) normalization gives B =
√2√π
(mω~
)3/4(ii) calculate 〈H〉 = 3~
3
2mβ +
3mω2
8
1
β
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
b] Do the same, but with trial function ψ(x) = Bxe−βx2
(i) normalization gives B =
√2√π
(mω~
)3/4(ii) calculate 〈H〉 = 3~
3
2mβ +
3mω2
8
1
β
(iii) minimize 〈H〉 =⇒ β = mω2~
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
b] Do the same, but with trial function ψ(x) = Bxe−βx2
(i) normalization gives B =
√2√π
(mω~
)3/4(ii) calculate 〈H〉 = 3~
3
2mβ +
3mω2
8
1
β
(iii) minimize 〈H〉 =⇒ β = mω2~
(iv) get minimal values
〈H〉min =3
2~ω
ψmin(x) =
√2√π
(mω~
)3/4xe−
mω2~ x
2
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
b] Do the same, but with trial function ψ(x) = Bxe−βx2
(i) normalization gives B =
√2√π
(mω~
)3/4(ii) calculate 〈H〉 = 3~
3
2mβ +
3mω2
8
1
β
(iii) minimize 〈H〉 =⇒ β = mω2~
(iv) get minimal values
〈H〉min =3
2~ω
ψmin(x) =
√2√π
(mω~
)3/4xe−
mω2~ x
2
exact 1st excited stateenergy and w.f.
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
In conclusion...
ψb]trial (x) = Bxe
−βx2
ψa]trial (x) = ψ
gsexact(x) = Ae
−αx2
}=⇒
〈ψ
b]trial (x)|ψ
gsexact(x)
〉= 0
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
In conclusion...
ψb]trial (x) = Bxe
−βx2
ψa]trial (x) = ψ
gsexact(x) = Ae
−αx2
}=⇒
〈ψ
b]trial (x)|ψ
gsexact(x)
〉= 0
Also, 〈H〉b]min accounts for 1st excited state
Igor Lukačević The variational principle
The variational principle
Theory
Example: One-dimensional harmonic oscilator
In conclusion...
ψb]trial (x) = Bxe
−βx2
ψa]trial (x) = ψ
gsexact(x) = Ae
−αx2
}=⇒
〈ψ
b]trial (x)|ψ
gsexact(x)
〉= 0
Also, 〈H〉b]min accounts for 1st excited state
Corollary
If 〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efes , where Efes is the energy of the 1st excitedstate.
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Contents
1 Theory
2 The ground state of helium
3 The linear variational problem
4 Literature
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|~r1 −~r2|
)
A question
What does each of these terms mean?
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|~r1 −~r2|
)
A question
What does each of these terms mean?
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|~r1 −~r2|
)
kinetic energy of electrons 1 and 2
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|~r1 −~r2|
)
electrostatic attraction between the nucleusand electrons 1 and 2
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|̃r1 − r̃2|
)
electrostatic repulsion between the electrons1 and 2
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|~r1 −~r2|
)
Our mission to calculate the ground state energy Egs
E expgs = −78.975 eV
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|~r1 −~r2|
)
A question
Can you identify the troublesome term in H?
Igor Lukačević The variational principle
The variational principle
The ground state of helium
H = − ~2
2m(∆1 + ∆2)
− e2
4π�0
(2
r1+
2
r2− 1|~r1 −~r2|
)
A question
Can you identify the troublesome term in H?
Vee =e2
4π�0
1
|~r1 −~r2|
Igor Lukačević The variational principle
The variational principle
The ground state of helium
For start, let us ignore Vee
A question
Can you “guess” what happens then with H, how ψ looks like and what’s theenergy?
Igor Lukačević The variational principle
The variational principle
The ground state of helium
For start, let us ignore Vee
A question
Can you “guess” what happens then with H, how ψ looks like and what’s theenergy?
H = H1 + H2
ψ0(~r1,~r2) = ψ100(~r1)ψ100(~r2) =23
a3πe−2
r1+r2a
E0 = 8E1 = −109 eV
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Now, let us account for Vee :
ψ0 99K trial w.f. (is this justifiable?)
〈H〉 = 8E1 + 〈Vee〉
〈Vee〉 =(
e2
4π�0
)(23
a3π
)2 ∫e−4
r1+r2a
|~r1 −~r2|d~r1d~r2
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Now, let us account for Vee :
ψ0 99K trial w.f. (is this justifiable?)
〈H〉 = 8E1 + 〈Vee〉
〈Vee〉 =(
e2
4π�0
)(23
a3π
)2 ∫e−4
r1+r2a
|~r1 −~r2|d~r1d~r2
A question
What do you expect for 〈Vee〉 and why?
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Now, let us account for Vee :
ψ0 99K trial w.f. (is this justifiable?)
〈H〉 = 8E1 + 〈Vee〉
〈Vee〉 =(
e2
4π�0
)(23
a3π
)2 ∫e−4
r1+r2a
|~r1 −~r2|d~r1d~r2 = 34 eV
HW
Calculate〈Vee〉 usingRefs. [2] and[5].
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Now, let us account for Vee :
ψ0 99K trial w.f. (is this justifiable?)
〈H〉 = 8E1 + 〈Vee〉
〈Vee〉 =(
e2
4π�0
)(23
a3π
)2 ∫e−4
r1+r2a
|~r1 −~r2|d~r1d~r2 = 34 eV
〈H〉 = −109 eV + 34 eV = −75 eV
E expgs = −79 eV
Rel. error 5.1%
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Zeff effective nuclearcharge
Trial w.f.
ψ1(~r1,~r2) =Z 3effa3π
e−Zeffr1+r2
a
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Zeff effective nuclearcharge
Trial w.f.
ψ1(~r1,~r2) =Z3effa3π
e−Zeffr1+r2
a
Zeff - variationalparameter
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Let us rewrite the Hamiltonian:
H = − ~2
2m(∆1 + ∆2)−
e2
4π�0
(Zeffr1
+Zeffr2
)+
e2
4π�0
[Zeff − 2
r1+
Zeff − 2r2
− 1|~r1 −~r2|
]
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Let us rewrite the Hamiltonian:
H = − ~2
2m(∆1 + ∆2)−
e2
4π�0
(Zeffr1
+Zeffr2
)+
e2
4π�0
[Zeff − 2
r1+
Zeff − 2r2
− 1|~r1 −~r2|
]
=⇒ 〈H〉 =[−2Z 2eff +
27
4Zeff
]E1
For calculation details, see Ref. [2].
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Let us rewrite the Hamiltonian:
H = − ~2
2m(∆1 + ∆2)−
e2
4π�0
(Zeffr1
+Zeffr2
)+
e2
4π�0
[Zeff − 2
r1+
Zeff − 2r2
− 1|~r1 −~r2|
]
=⇒ 〈H〉 =[−2Z 2eff +
27
4Zeff
]E1
Now minimizing 〈H〉 we getZmineff = 1.69
Igor Lukačević The variational principle
The variational principle
The ground state of helium
Let us rewrite the Hamiltonian:
H = − ~2
2m(∆1 + ∆2)−
e2
4π�0
(Zeffr1
+Zeffr2
)+
e2
4π�0
[Zeff − 2
r1+
Zeff − 2r2
− 1|~r1 −~r2|
]
=⇒ 〈H〉 =[−2Z 2eff +
27
4Zeff
]E1
Now minimizing 〈H〉 we getZmineff = 1.69
Which gives
〈H〉min = Emin = −77.5 eV ,Egs − Emin
Egs= 1.87%
Note:For more precise results see E. A. Hylleraas, Z. Phys. 65, 209 (1930) or C. L. Pekeris,
Phys. Rev. 115, 1216 (1959).
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Contents
1 Theory
2 The ground state of helium
3 The linear variational problem
4 Literature
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized trial function depends on α1, α2 . . .
〈ψ̃|H|ψ̃〉 very complex function of α1, α2 . . .
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized trial function depends on α1, α2 . . .
〈ψ̃|H|ψ̃〉 very complex function of α1, α2 . . .
Suppose
|ψ̃〉 =N∑
i=1
ci |ψi 〉 , 〈ψi |ψj〉 = δij
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized trial function depends on α1, α2 . . .
〈ψ̃|H|ψ̃〉 very complex function of α1, α2 . . .
Suppose
|ψ̃〉 =N∑
i=1
ci |ψi 〉 , 〈ψi |ψj〉 = δij
=⇒ (H)ij = Hij = 〈ψi |H|ψj〉 matrix representation in basis {|ψi 〉}
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized trial function depends on α1, α2 . . .
〈ψ̃|H|ψ̃〉 very complex function of α1, α2 . . .
Suppose
|ψ̃〉 =N∑
i=1
ci |ψi 〉 , 〈ψi |ψj〉 = δij
=⇒ (H)ij = Hij = 〈ψi |H|ψj〉 matrix representation in basis {|ψi 〉}
H hermitian{|ψi 〉} real
}⇒ H symmetric
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized trial function depends on α1, α2 . . .
〈ψ̃|H|ψ̃〉 very complex function of α1, α2 . . .
Suppose
|ψ̃〉 =N∑
i=1
ci |ψi 〉 , 〈ψi |ψj〉 = δij
=⇒ (H)ij = Hij = 〈ψi |H|ψj〉 matrix representation in basis {|ψi 〉}
H hermitian{|ψi 〉} real
}⇒ H symmetric
ψ̃ normalized ⇒∑
i
c2i = 1
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized trial function depends on α1, α2 . . .
〈ψ̃|H|ψ̃〉 very complex function of α1, α2 . . .
Suppose
|ψ̃〉 =N∑
i=1
ci |ψi 〉 , 〈ψi |ψj〉 = δij
=⇒ (H)ij = Hij = 〈ψi |H|ψj〉 matrix representation in basis {|ψi 〉}
H hermitian{|ψi 〉} real
}⇒ H symmetric
ψ̃ normalized ⇒∑
i
c2i = 1
the expectation value depends on cij :
=⇒ 〈ψ̃|H|ψ̃〉 =∑
ij
cijHij
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized ⇒∑
i
c2i = 1
the expectation value depends on cij :
=⇒ 〈ψ̃|H|ψ̃〉 =∑
ij
cijHij
Unfortunately,∂
∂ck〈ψ̃|H|ψ̃〉 = 0 , k = 1, 2, . . . ,N
is unsolvable for ck are mutually dependent.
Igor Lukačević The variational principle
The variational principle
The linear variational problem
ψ̃ normalized ⇒∑
i
c2i = 1
the expectation value depends on cij :
=⇒ 〈ψ̃|H|ψ̃〉 =∑
ij
cijHij
Unfortunately,∂
∂ck〈ψ̃|H|ψ̃〉 = 0 , k = 1, 2, . . . ,N
is unsolvable for ck are mutually dependent.
Lagrange’s method of undetermined multipliers
L(c1, . . . , cN ,E) = 〈ψ̃|H|ψ̃〉 − E(〈ψ̃|ψ̃〉 − 1
)=∑
ij
cicjHij − E
(∑i
c2i − 1
)
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Unfortunately,∂
∂ck〈ψ̃|H|ψ̃〉 = 0 , k = 1, 2, . . . ,N
is unsolvable for ck are mutually dependent.
Lagrange’s method of undetermined multipliers
L(c1, . . . , cN ,E) = 〈ψ̃|H|ψ̃〉 − E(〈ψ̃|ψ̃〉 − 1
)=∑
ij
cicjHij − E
(∑i
c2i − 1
)
〈ψ̃|H|ψ̃〉 and L are minimal for same ck
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Let us now choose c1, c2, . . . , cN−1 as independent
⇒ cN is given by∑
i
c2i = 1
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Let us now choose c1, c2, . . . , cN−1 as independent
⇒ cN is given by∑
i
c2i = 1
Then we have∂L∂ck
= 0 , k = 1, 2, . . . ,N − 1
but not necessarily∂L∂cN
= 0
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Let us now choose c1, c2, . . . , cN−1 as independent
⇒ cN is given by∑
i
c2i = 1
Then we have∂L∂ck
= 0 , k = 1, 2, . . . ,N − 1
but not necessarily∂L∂cN
= 0
But we still have undetermined multiplier E , so now we choose it so that
∂L∂ck
= 0 , k = 1, 2, . . . ,N − 1,N
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Let us now choose c1, c2, . . . , cN−1 as independent
⇒ cN is given by∑
i
c2i = 1
Then we have∂L∂ck
= 0 , k = 1, 2, . . . ,N − 1
but not necessarily∂L∂cN
= 0
But we still have undetermined multiplier E , so now we choose it so that
∂L∂ck
= 0 , k = 1, 2, . . . ,N − 1,N
On the other hand
∂L∂ck
=∑
j
cjHkj +∑
i
ciHik − 2Eck
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Then we have∂L∂ck
= 0 , k = 1, 2, . . . ,N − 1
but not necessarily∂L∂cN
= 0
But we still have undetermined multiplier E , so now we choose it so that
∂L∂ck
= 0 , k = 1, 2, . . . ,N − 1,N
On the other hand
∂L∂ck
=∑
j
cjHkj +∑
i
ciHik︸ ︷︷ ︸equal, since Hij=Hji
−2Eck
So, ∑j
Hijcj − Eci = 0
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Or in matrix formHc = Ec
A question
What represents this equation?
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Or in matrix formHc = Ec
⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑
i
cαi cβi = δαβ
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Or in matrix formHc = Ec
⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑
i
cαi cβi = δαβ
Eαβ = Eαδαβ , Ciα = cαi =⇒ HC = EC
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Or in matrix formHc = Ec
⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑
i
cαi cβi = δαβ
Eαβ = Eαδαβ , Ciα = cαi =⇒ HC = EC
Solving gives N orthonormal solutions
|ψ̃α〉 =N∑
i=1
cαi |ψi 〉 , α = 0, 1, . . . ,N − 1
Igor Lukačević The variational principle
The variational principle
The linear variational problem
Or in matrix formHc = Ec
⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑
i
cαi cβi = δαβ
Eαβ = Eαδαβ , Ciα = cαi =⇒ HC = EC
Solving gives N orthonormal solutions
|ψ̃α〉 =N∑
i=1
cαi |ψi 〉 , α = 0, 1, . . . ,N − 1
What about E ’s:〈ψ̃β |H|ψ̃α〉 = Eαδαβ
Igor Lukačević The variational principle
The variational principle
The linear variational problem
HC = EC
Solving gives N orthonormal solutions
|ψ̃α〉 =N∑
i=1
cαi |ψi 〉 , α = 0, 1, . . . ,N − 1
What about E ’s:〈ψ̃β |H|ψ̃α〉 = Eαδαβ
For example,E0 = 〈ψ̃0|H|ψ̃0〉 ≥ E0
A question
What’s the meaning of other E ’s?
Igor Lukačević The variational principle
The variational principle
The linear variational problem
HC = EC
Solving gives N orthonormal solutions
|ψ̃α〉 =N∑
i=1
cαi |ψi 〉 , α = 0, 1, . . . ,N − 1
What about E ’s:〈ψ̃β |H|ψ̃α〉 = Eαδαβ
For example,E0 = 〈ψ̃0|H|ψ̃0〉 ≥ E0
A question
What’s the meaning of other E ’s? Eα ≥ Eα , α = 1, 2, . . .
Igor Lukačević The variational principle
The variational principle
The linear variational problem
In conclusion
Solving the matrix eigenvalue problem
HC = EC ,
by diagonalization, is equivalent to the variational principle in a subspacespanned by {|ψi 〉 , i = 1, 2, . . . ,N}.
Igor Lukačević The variational principle
The variational principle
Literature
Contents
1 Theory
2 The ground state of helium
3 The linear variational problem
4 Literature
Igor Lukačević The variational principle
The variational principle
Literature
Literature
1 A. Szabo, N. Ostlund, Modern Quantum Chemistry, Introduction toAdvanced Electronic Structure theory, Dover Publications, New York,1996.
2 D. J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., PearsonEducation, Inc., Upper Saddle River, NJ, 2005.
3 I. Supek, Teorijska fizika i struktura materije, II. dio, Školska knjiga,Zagreb, 1989.
4 Y. Peleg, R. Pnini, E. Zaarur, Shaum’s Outline of Theory and Problems ofQuantum Mechanics, McGraw-Hill, 1998.
5 I. N. Bronštejn, K. A. Semendjajev, Matematički priručnik, Tehničkaknjiga, Zagreb, 1991.
Igor Lukačević The variational principle
TheoryThe ground state of heliumThe linear variational problemLiterature