The variational principle

The variational principleQuantum mechanics 2 - Lecture 5

Igor Lukacevic

UJJS, Dept. of Physics, Osijek

November 8, 2012

Igor Lukacevic The variational principle

Contents

1 Theory

2 The ground state of helium

3 The linear variational problem

4 Literature

Theory

Contents

1 Theory

4 Literature

Theory

What is a problem we would like to solve?

To find approximate solutions of eigenvalue problem

Oφ(x) = ωφ(x)

Theory

To find approximate solutions of eigenvalue problem

Oφ(x) = ωφ(x)

A question

Can you remember any eigenvalue problems?

Theory

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

Theory

Theorem - the variational principle

Given any normalized function ψ (that satisfies the appropriate boundaryconditions), then the expectation value of the Hamiltonian represents an upperbound to the exact ground state energy

〈ψ|H|ψ〉 ≥ E0 .

Theory

〈ψ|H|ψ〉 ≥ E0 .

A question

What if ψ is a ground state w.f.?

Theory

〈ψ|H|ψ〉 ≥ E0 .

A question

What if ψ is a ground state w.f.?

〈ψ|H|ψ〉 = E0

Theory

ψ are normalized ⇒ 〈ψ|ψ〉 = 1

Theory

On the other hand, (unknown) ψ form a complete set ⇒ |ψ〉 =∑α cα|ψα〉

Theory

On the other hand, (unknown) ψα form a complete set ⇒ |ψ〉 =∑α cα|ψα〉

〈ψ|ψ〉 =⟨∑

cβψβ

∣∣∣∑α

cαψα⟩

=∑αβ

c∗βcα 〈ψβ |ψα〉︸︷︷︸δαβ

=∑α

|cα|2 = 1

Theory

〈ψ|ψ〉 =⟨∑

cβψβ

∣∣∣∑α

cαψα⟩

=∑αβ

=∑α

|cα|2 = 1

〈ψ|H|ψ〉 =⟨∑

cβψβ

∣∣∣H∣∣∣∑α

cαψα⟩

︸︷︷︸∑α cαH|ψα〉︸︷︷︸

Eα|ψα〉

=∑αβ

c∗βcαEα 〈ψβ |ψα〉︸︷︷︸δαβ

=∑α

Eα|cα|2

Theory

〈ψ|ψ〉 =⟨∑

cβψβ

∣∣∣∑α

cαψα⟩

=∑αβ

=∑α

|cα|2 = 1

〈ψ|H|ψ〉 =⟨∑

cβψβ

∣∣∣H∣∣∣∑α

cαψα⟩

︸︷︷︸∑α cαH|ψα〉︸︷︷︸

Eα|ψα〉

=∑αβ

c∗βcαEα 〈ψβ |ψα〉︸︷︷︸δαβ

=∑α

Eα|cα|2

But Eα ≥ E0 , ∀α, hence

〈ψ|H|ψ〉 ≥∑α

E0|cα|2 = E0∑α

|cα|2 = E0

Theory

Example: One-dimensional harmonic oscilator

a] Find the ground state energy and w.f. of one-dimensional harmonic oscilator:

H = − ~2

2m∆ +

2mω2x2 .

Theory

H = − ~2

2m∆ +

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

√2α

π from normalization condition (do it for HW)

Theory

H = − ~2

2m∆ +

2mω2x2 .

ψ(x) = Ae−αx2

α parameter

√2α

π from normalization condition

(ii) calculate 〈H〉 = 〈T 〉+ 〈V 〉

Theory

ψ(x) = Ae−αx2

α parameter

√2α

〈T 〉 =~2α2m

〈V 〉 =mω2

On how to solve these kind ofintegrals, see Ref. [5].

Theory

ψ(x) = Ae−αx2

α parameter

√2α

〈T 〉 =~2α2m

〈V 〉 =mω2

On how to solve these kind ofintegrals, see Ref. [5].

〈H〉 =~2α2m

Theory

(iii) minimize 〈H〉 wrt parameter α

dα〈H〉 = 0 =⇒ α =

Theory

dα〈H〉 = 0 =⇒ α =

(iv) insert back into 〈H〉 and ψ(x):

〈H〉min =1

ψmin(x) = 4

√mω

π~e−

mω2~ x2

Theory

dα〈H〉 = 0 =⇒ α =

(iv) insert back into 〈H〉 and ψ(x):

〈H〉min =1

ψmin(x) = 4

√mω

π~e−

mω2~ x2

exact ground state energy and w.f.

A question

Why did we get the exact energy and w.f.?

Theory

b] Do the same, but with trial function ψ(x) = Bxe−βx2

Theory

(i) normalization gives B =

√2√π

Theory

√2√π

)3/4(ii) calculate 〈H〉 =

2mβ +

Theory

√2√π

2mβ +

(iii) minimize 〈H〉 =⇒ β =mω

Theory

√2√π

2mβ +

2~(iv) get minimal values

〈H〉min =3

ψmin(x) =

√2√π

)3/4xe−

mω2~ x2

Theory

√2√π

2mβ +

2~(iv) get minimal values

〈H〉min =3

ψmin(x) =

√2√π

)3/4xe−

mω2~ x2

exact 1st excited stateenergy and w.f.

Theory

In conclusion...

ψb]trial (x) = Bxe−βx2

ψa]trial (x) = ψgs

exact(x) = Ae−αx2

b]trial (x)|ψgs

exact(x)⟩

Theory

In conclusion...

b]trial (x)|ψgs

exact(x)⟩

Also, 〈H〉b]min accounts for 1st excited state

Theory

In conclusion...

b]trial (x)|ψgs

exact(x)⟩

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.

The ground state of helium

Contents

1 Theory

4 Literature

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|~r1 −~r2|

A question

What does each of these terms mean?

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|~r1 −~r2|

A question

What does each of these terms mean?

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|~r1 −~r2|

kinetic energy of electrons 1 and 2

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|~r1 −~r2|

electrostatic attraction between the nucleusand electrons 1 and 2

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|r1 − r2|

electrostatic repulsion between the electrons1 and 2

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|~r1 −~r2|

Our mission to calculate the ground state energy Egs

E expgs = −78.975 eV

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|~r1 −~r2|

A question

Can you identify the troublesome term in H?

H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

r2− 1

|~r1 −~r2|

A question

Can you identify the troublesome term in H?

Vee =e2

4πε0

|~r1 −~r2|

For start, let us ignore Vee

A question

Can you “guess” what happens then with H, how ψ looks like and what’s theenergy?

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

Now, let us account for Vee :

ψ0 99K trial w.f. (is this justifiable?)

〈H〉 = 8E1 + 〈Vee〉

〈Vee〉 =

4πε0

)2 ∫e−4

r1+r2a

|~r1 −~r2|d~r1d~r2

〈H〉 = 8E1 + 〈Vee〉

〈Vee〉 =

4πε0

)2 ∫e−4

r1+r2a

|~r1 −~r2|d~r1d~r2

A question

What do you expect for 〈Vee〉 and why?

〈H〉 = 8E1 + 〈Vee〉

〈Vee〉 =

4πε0

)2 ∫e−4

r1+r2a

|~r1 −~r2|d~r1d~r2 = 34 eV

Calculate〈Vee〉 usingRefs. [2] and[5].

〈H〉 = 8E1 + 〈Vee〉

〈Vee〉 =

4πε0

)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%

Zeff effective nuclearcharge

Trial w.f.

ψ1(~r1,~r2) =Z 3

a3πe−Zeff

r1+r2a

Zeff effective nuclearcharge

Trial w.f.

ψ1(~r1,~r2) =Z3

a3πe−Zeff

r1+r2a

Zeff - variationalparameter

Let us rewrite the Hamiltonian:

H = − ~2

2m(∆1 + ∆2)− e2

4πε0

[Zeff − 2

Zeff − 2

r2− 1

|~r1 −~r2|

H = − ~2

2m(∆1 + ∆2)− e2

4πε0

[Zeff − 2

Zeff − 2

r2− 1

|~r1 −~r2|

=⇒ 〈H〉 =

[−2Z 2

eff +27

For calculation details, see Ref. [2].

H = − ~2

2m(∆1 + ∆2)− e2

4πε0

[Zeff − 2

Zeff − 2

r2− 1

|~r1 −~r2|

=⇒ 〈H〉 =

[−2Z 2

eff +27

Now minimizing 〈H〉 we getZmin

eff = 1.69

H = − ~2

2m(∆1 + ∆2)− e2

4πε0

[Zeff − 2

Zeff − 2

r2− 1

|~r1 −~r2|

=⇒ 〈H〉 =

[−2Z 2

eff +27

Now minimizing 〈H〉 we getZmin

eff = 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).

The linear variational problem

Contents

1 Theory

4 Literature

ψ normalized trial function depends on α1, α2 . . .

〈ψ|H|ψ〉 very complex function of α1, α2 . . .

Suppose

|ψ〉 =N∑

ci |ψi 〉 , 〈ψi |ψj〉 = δij

Suppose

|ψ〉 =N∑

=⇒ (H)ij = Hij = 〈ψi |H|ψj〉 matrix representation in basis {|ψi 〉}

Suppose

|ψ〉 =N∑

H hermitian{|ψi 〉} real

}⇒ H symmetric

Suppose

|ψ〉 =N∑

}⇒ H symmetric

ψ normalized ⇒∑

c2i = 1

Suppose

|ψ〉 =N∑

}⇒ H symmetric

c2i = 1

the expectation value depends on cij :

=⇒ 〈ψ|H|ψ〉 =∑

cijHij

c2i = 1

=⇒ 〈ψ|H|ψ〉 =∑

cijHij

Unfortunately,∂

∂ck〈ψ|H|ψ〉 = 0 , k = 1, 2, . . . ,N

is unsolvable for ck are mutually dependent.

c2i = 1

=⇒ 〈ψ|H|ψ〉 =∑

cijHij

Unfortunately,∂

∂ck〈ψ|H|ψ〉 = 0 , k = 1, 2, . . . ,N

Lagrange’s method of undetermined multipliers

L(c1, . . . , cN ,E) = 〈ψ|H|ψ〉 − E(〈ψ|ψ〉 − 1

cicjHij − E

c2i − 1

Unfortunately,∂

∂ck〈ψ|H|ψ〉 = 0 , k = 1, 2, . . . ,N

Lagrange’s method of undetermined multipliers

L(c1, . . . , cN ,E) = 〈ψ|H|ψ〉 − E(〈ψ|ψ〉 − 1

cicjHij − E

c2i − 1

〈ψ|H|ψ〉 and L are minimal for same ck

Let us now choose c1, c2, . . . , cN−1 as independent

⇒ cN is given by∑

c2i = 1

Then we have∂L∂ck

= 0 , k = 1, 2, . . . ,N − 1

but not necessarily∂L∂cN

c2i = 1

= 0 , k = 1, 2, . . . ,N − 1

But we still have undetermined multiplier E , so now we choose it so that

∂L∂ck

= 0 , k = 1, 2, . . . ,N − 1,N

c2i = 1

= 0 , k = 1, 2, . . . ,N − 1

∂L∂ck

= 0 , k = 1, 2, . . . ,N − 1,N

On the other hand

∂L∂ck

cjHkj +∑

ciHik − 2Eck

= 0 , k = 1, 2, . . . ,N − 1

∂L∂ck

= 0 , k = 1, 2, . . . ,N − 1,N

On the other hand

∂L∂ck

cjHkj +∑

ciHik︸︷︷︸equal, since Hij=Hji

−2Eck

So, ∑j

Hijcj − Eci = 0

Or in matrix formHc = Ec

A question

What represents this equation?

⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑

cαi cβi = δαβ

⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑

cαi cβi = δαβ

Eαβ = Eαδαβ , Ciα = cαi =⇒ HC = EC

⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑

cαi cβi = δαβ

Solving gives N orthonormal solutions

|ψα〉 =N∑

cαi |ψi 〉 , α = 0, 1, . . . ,N − 1

⇒ Hcα = Eαcα , α = 0, 1, . . . ,N − 1 , (cα)†cβ =∑

cαi cβi = δαβ

|ψα〉 =N∑

cαi |ψi 〉 , α = 0, 1, . . . ,N − 1

What about E ’s:〈ψβ |H|ψα〉 = Eαδαβ

HC = EC

|ψα〉 =N∑

cαi |ψi 〉 , α = 0, 1, . . . ,N − 1

For example,E0 = 〈ψ0|H|ψ0〉 ≥ E0

A question

What’s the meaning of other E ’s?

HC = EC

|ψα〉 =N∑

cαi |ψi 〉 , α = 0, 1, . . . ,N − 1

For example,E0 = 〈ψ0|H|ψ0〉 ≥ E0

A question

What’s the meaning of other E ’s? Eα ≥ Eα , α = 1, 2, . . .

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}.

Literature

Contents

1 Theory

4 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, Skolska 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. Bronstejn, K. A. Semendjajev, Matematicki prirucnik, Tehnickaknjiga, Zagreb, 1991.

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