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The variational principle The variational principle Quantum mechanics 2 - Lecture 5 Igor Lukaˇ cevi´ c UJJS, Dept. of Physics, Osijek November 8, 2012 Igor Lukaˇ cevi´ c The variational principle

The variational principle - unios.hr...The variational principle The variational principle Quantum mechanics 2 - Lecture 5 Igor Luka cevi c UJJS, Dept. of Physics, Osijek November

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  • 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

    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

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

    〈H〉 = ~2α

    2m+

    mω2

    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√

    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√

    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