The variational principle - UNIOSfizika.unios.hr/~ilukacevic/dokumenti/materijali_za_studente/qm2/... · The variational principle The variational principle Quantum mechanics 2 -

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

Proof

ψ are normalized ⇒ 〈ψ|ψ〉 = 1



Theory

Proof


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



Theory

Proof


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

So,

〈ψ|ψ〉 =⟨∑

β

cβψβ

∣∣∣∑α

cαψα⟩

=∑αβ

c∗βcα 〈ψβ |ψα〉︸︷︷︸δαβ

=∑α

|cα|2 = 1



Theory

Proof



So,

〈ψ|ψ〉 =⟨∑

β

cβψβ

∣∣∣∑α

cαψα⟩

=∑αβ


=∑α

|cα|2 = 1

Now

〈ψ|H|ψ〉 =⟨∑

β

cβψβ

∣∣∣H∣∣∣∑α

cαψα⟩

︸︷︷︸∑α cαH|ψα〉︸︷︷︸

Eα|ψα〉

=∑αβ

c∗βcαEα 〈ψβ |ψα〉︸︷︷︸δαβ

=∑α

Eα|cα|2



Theory

Proof



So,

〈ψ|ψ〉 =⟨∑

β

cβψβ

∣∣∣∑α

cαψα⟩

=∑αβ


=∑α

|cα|2 = 1

Now

〈ψ|H|ψ〉 =⟨∑

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

〈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∆ +

1

2mω2x2 .



Theory



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)



Theory



H = − ~2

2m∆ +

1

2mω2x2 .



ψ(x) = Ae−αx2

α parameter

A =4

√2α

π from normalization condition

(ii) calculate 〈H〉 = 〈T 〉+ 〈V 〉



Theory




ψ(x) = Ae−αx2

α parameter

A =4

√2α



〈T 〉 =~2α2m

〈V 〉 =mω2

8α

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



Theory




ψ(x) = Ae−αx2

α parameter

A =4

√2α



〈T 〉 =~2α2m

〈V 〉 =mω2

8α

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

〈H〉 =~2α2m

+mω2

8α



Theory


(iii) minimize 〈H〉 wrt parameter α

d

dα〈H〉 = 0 =⇒ α =

mω

2~



Theory



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~ x2



Theory



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~ 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

(mω~

)3/4



Theory




√2√π

(mω~

)3/4(ii) calculate 〈H〉 =

3~3

2mβ +

3mω2

8

1

β



Theory




√2√π

(mω~


3~3

2mβ +

3mω2

8

1

β

(iii) minimize 〈H〉 =⇒ β =mω

2~



Theory




√2√π

(mω~


3~3

2mβ +

3mω2

8

1

β


2~(iv) get minimal values

〈H〉min =3

2~ω

ψmin(x) =

√2√π

(mω~

)3/4xe−

mω2~ x2



Theory




√2√π

(mω~


3~3

2mβ +

3mω2

8

1

β


2~(iv) get minimal values

〈H〉min =3

2~ω

ψmin(x) =

√2√π

(mω~

)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

}=⇒

⟨ψ

b]trial (x)|ψgs

exact(x)⟩

= 0



Theory


In conclusion...




}=⇒

⟨ψ

b]trial (x)|ψgs

exact(x)⟩

= 0

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



Theory


In conclusion...




}=⇒

⟨ψ

b]trial (x)|ψgs

exact(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.



The ground state of helium

Contents

1 Theory



4 Literature




H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

(2

r1+

2

r2− 1

|~r1 −~r2|

)

A question

What does each of these terms mean?




H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

(2

r1+

2

r2− 1

|~r1 −~r2|

)

A question

What does each of these terms mean?




H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

(2

r1+

2

r2− 1

|~r1 −~r2|

)

kinetic energy of electrons 1 and 2




H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

(2

r1+

2

r2− 1

|~r1 −~r2|

)

electrostatic attraction between the nucleusand electrons 1 and 2




H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

(2

r1+

2

r2− 1

|r1 − r2|

)

electrostatic repulsion between the electrons1 and 2




H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

(2

r1+

2

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

(2

r1+

2

r2− 1

|~r1 −~r2|

)

A question

Can you identify the troublesome term in H?




H = − ~2

2m(∆1 + ∆2)

− e2

4πε0

(2

r1+

2

r2− 1

|~r1 −~r2|

)

A question

Can you identify the troublesome term in H?

Vee =e2

4πε0

1

|~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〉 =

(e2

4πε0

)(23

a3π

)2 ∫e−4

r1+r2a

|~r1 −~r2|d~r1d~r2






〈H〉 = 8E1 + 〈Vee〉

〈Vee〉 =

(e2

4πε0

)(23

a3π

)2 ∫e−4

r1+r2a

|~r1 −~r2|d~r1d~r2

A question

What do you expect for 〈Vee〉 and why?






〈H〉 = 8E1 + 〈Vee〉

〈Vee〉 =

(e2

4πε0

)(23

a3π

)2 ∫e−4

r1+r2a

|~r1 −~r2|d~r1d~r2 = 34 eV

HW

Calculate〈Vee〉 usingRefs. [2] and[5].






〈H〉 = 8E1 + 〈Vee〉

〈Vee〉 =

(e2

4πε0

)(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%




Zeff effective nuclearcharge

Trial w.f.

ψ1(~r1,~r2) =Z 3

eff

a3πe−Zeff

r1+r2a




Zeff effective nuclearcharge

Trial w.f.

ψ1(~r1,~r2) =Z3

eff

a3πe−Zeff

r1+r2a

Zeff - variationalparameter




Let us rewrite the Hamiltonian:

H = − ~2

2m(∆1 + ∆2)− e2

4πε0

(Zeff

r1+

Zeff

r2

)+

e2

4πε0

[Zeff − 2

r1+

Zeff − 2

r2− 1

|~r1 −~r2|

]





H = − ~2

2m(∆1 + ∆2)− e2

4πε0

(Zeff

r1+

Zeff

r2

)+

e2

4πε0

[Zeff − 2

r1+

Zeff − 2

r2− 1

|~r1 −~r2|

]

=⇒ 〈H〉 =

[−2Z 2

eff +27

4Zeff

]E1

For calculation details, see Ref. [2].





H = − ~2

2m(∆1 + ∆2)− e2

4πε0

(Zeff

r1+

Zeff

r2

)+

e2

4πε0

[Zeff − 2

r1+

Zeff − 2

r2− 1

|~r1 −~r2|

]

=⇒ 〈H〉 =

[−2Z 2

eff +27

4Zeff

]E1

Now minimizing 〈H〉 we getZmin

eff = 1.69





H = − ~2

2m(∆1 + ∆2)− e2

4πε0

(Zeff

r1+

Zeff

r2

)+

e2

4πε0

[Zeff − 2

r1+

Zeff − 2

r2− 1

|~r1 −~r2|

]

=⇒ 〈H〉 =

[−2Z 2

eff +27

4Zeff

]E1

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∑

i=1

ci |ψi 〉 , 〈ψi |ψj〉 = δij






Suppose

|ψ〉 =N∑

i=1


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






Suppose

|ψ〉 =N∑

i=1



H hermitian{|ψi 〉} real

}⇒ H symmetric






Suppose

|ψ〉 =N∑

i=1




}⇒ H symmetric

ψ normalized ⇒∑

i

c2i = 1






Suppose

|ψ〉 =N∑

i=1




}⇒ H symmetric


i

c2i = 1

the expectation value depends on cij :

=⇒ 〈ψ|H|ψ〉 =∑

ij

cijHij





i

c2i = 1


=⇒ 〈ψ|H|ψ〉 =∑

ij

cijHij

Unfortunately,∂

∂ck〈ψ|H|ψ〉 = 0 , k = 1, 2, . . . ,N

is unsolvable for ck are mutually dependent.





i

c2i = 1


=⇒ 〈ψ|H|ψ〉 =∑

ij

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

)=∑

ij

cicjHij − E

(∑i

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

)=∑

ij

cicjHij − E

(∑i

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∑

i

c2i = 1






i

c2i = 1

Then we have∂L∂ck

= 0 , k = 1, 2, . . . ,N − 1

but not necessarily∂L∂cN

= 0






i

c2i = 1


= 0 , k = 1, 2, . . . ,N − 1


= 0

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

∂L∂ck

= 0 , k = 1, 2, . . . ,N − 1,N






i

c2i = 1


= 0 , k = 1, 2, . . . ,N − 1


= 0


∂L∂ck

= 0 , k = 1, 2, . . . ,N − 1,N

On the other hand

∂L∂ck

=∑

j

cjHkj +∑

i

ciHik − 2Eck





= 0 , k = 1, 2, . . . ,N − 1


= 0


∂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




Or in matrix formHc = Ec

A question

What represents this equation?





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

i

cαi cβi = δαβ





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

i

cαi cβi = δαβ

Eαβ = Eαδαβ , Ciα = cαi =⇒ HC = EC





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

i

cαi cβi = δαβ


Solving gives N orthonormal solutions

|ψα〉 =N∑

i=1

cαi |ψi 〉 , α = 0, 1, . . . ,N − 1





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

i

cαi cβi = δαβ



|ψα〉 =N∑

i=1

cαi |ψi 〉 , α = 0, 1, . . . ,N − 1

What about E ’s:〈ψβ |H|ψα〉 = Eαδαβ




HC = EC


|ψα〉 =N∑

i=1

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∑

i=1

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

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.


Documents

The variational principle - UNIOSfizika.unios.hr/~ilukacevic/dokumenti/materijali_za_studente/qm2/... · The variational principle The variational principle Quantum mechanics 2 -