Identical particles - Quantum mechanics 2 - Lecture 1 · 2016. 9. 30. · 2m (1 + 2) + V = E A question What happens with S.E. if we interchange the coordinates of particles? S.E

Identical particles

Identical particlesQuantum mechanics 2 - Lecture 1

Igor Lukačević

UJJS, Dept. of Physics, Osijek

14. prosinca 2011.

Igor Lukačević Identical particles

Identical particles

Contents

1 Two-particle wave equation

2 Symmetric and antisymmetric solutions

3 Example: He spectrum

4 Antisymmetry principle

5 Example: H molecule

6 Literature


Identical particles

Two-particle wave equation

Contents






6 Literature


Identical particles


Which interactions exist here and what istheir nature?


Identical particles


H =1

2m1p21 +

1

2m2p22 + V (r1, r2)


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H =1

2m1p21 +

1

2m2p22 + V (r1, r2)

pk →~i∇k


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H =1

2m1p21 +

1

2m2p22 + V (r1, r2)

pk →~i∇k

− ~2m1

∆1Ψ−~

2m2∆2Ψ + V (r1, r2)Ψ = EΨ


Identical particles


H =1

2m1p21 +

1

2m2p22 + V (r1, r2)

pk →~i∇k

− ~2m1

∆1Ψ−~

2m2∆2Ψ + V (r1, r2)Ψ = EΨ

Ψ = Ψ(x1, y1, z1, x2, y2, z2) visualization is lost


Identical particles

Symmetric and antisymmetric solutions

Contents






6 Literature


Identical particles


Assumption: no interaction!

electron 1 electron 2


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− ~2m1

∆1Ψ +V1(r1)Ψ−~

2m2∆2Ψ +V2(r2)Ψ = EΨ Ψ(r1, r2) = u(r1)v(r2)

@@@R

��)

Separationofvariables

+r1 → 1r2 → 2

?

− ~2m1

∆1u + V1(1)u = E1u

− ~2m2

∆2v + V2(2)v = E2v

}99K E1 + E2 = E


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No interaction:

probability density Ψ∗Ψ = u(1)∗v(2)∗u(1)v(2) = u(1)∗u(1)v(2)∗v(2)

particles do not correlate


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Assumption: allow interaction!

nucleus electrons


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1 e-e: V12 =e2

r12, r12 =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

2 n-e: V1 = −2e2

r1, V2 = −

2e2

r2

3 total: V = −2e2

r1−

2e2

r2+

e2

r12

S.E. ⇒ − ~2

2m(∆1 + ∆2)Ψ + VΨ = EΨ


Identical particles



1 e-e: V12 =e2

r12, r12 =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

2 n-e: V1 = −2e2

r1, V2 = −

2e2

r2

3 total: V = −2e2

r1−

2e2

r2+

e2

r12

S.E. ⇒ − ~2

2m(∆1 + ∆2)Ψ + VΨ = EΨ

A question

What happens with S.E. if we interchange the coordinates of particles?


Identical particles



1 e-e: V12 =e2

r12, r12 =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

2 n-e: V1 = −2e2

r1, V2 = −

2e2

r2

3 total: V = −2e2

r1−

2e2

r2+

e2

r12

S.E. ⇒ − ~2

2m(∆1 + ∆2)Ψ + VΨ = EΨ

A question

What happens with S.E. if we interchange the coordinates of particles?S.E. is symmetrical wrt that interchange!


Identical particles



1 e-e: V12 =e2

r12, r12 =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

2 n-e: V1 = −2e2

r1, V2 = −

2e2

r2

3 total: V = −2e2

r1−

2e2

r2+

e2

r12

S.E. ⇒ − ~2

2m(∆1 + ∆2)Ψ + VΨ = EΨ

Second question

What about its solutions then?


Identical particles



1 e-e: V12 =e2

r12, r12 =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

2 n-e: V1 = −2e2

r1, V2 = −

2e2

r2

3 total: V = −2e2

r1−

2e2

r2+

e2

r12

S.E. ⇒ − ~2

2m(∆1 + ∆2)Ψ + VΨ = EΨ

Second question

What about its solutions then?

Ψ(1, 2)

Ψ(2, 1)

Ψ = c1Ψ(1, 2) + c2Ψ(2, 1)


Identical particles



1 e-e: V12 =e2

r12, r12 =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

2 n-e: V1 = −2e2

r1, V2 = −

2e2

r2

3 total: V = −2e2

r1−

2e2

r2+

e2

r12

S.E. ⇒ − ~2

2m(∆1 + ∆2)Ψ + VΨ = EΨ

Third question

How many linear combinations arethere and which of them can wechoose?

Ψ(1, 2)

Ψ(2, 1)

Ψ = c1Ψ(1, 2) + c2Ψ(2, 1)


Identical particles


A principle

Electrons are identical particles...they cannot be distinguished betweeneachother.


Identical particles


A principle




Identical particles


A principle

Electrons are identical particles...they cannot be distinguished betweeneachother.=⇒ probability density must be unchanged if we interchage their coordinates


Identical particles


A principle


Fourth question

Which linear combinations satisfy this condition?


Identical particles


A principle


Fourth question

Which linearcombinations satisfythis condition?

There are two possibilities (Hund & Wigner)

Ψ+ = Ψ(1, 2) + Ψ(2, 1) - symmetric solutionΨ− = Ψ(1, 2)−Ψ(2, 1) - antisymmetric solution


Identical particles

Example: He spectrum

Contents






6 Literature


Identical particles


Energy values are relative to the ground state of

He+, i.e. one has to subtract 54.4 eV.


Identical particles




Bohr’s theory cannot explain this, although it was known that toparahelium and ortohelium belonged the singlet (antiparallel spin) andtriplet (parallel spin) states


Identical particles




Bohr’s theory cannot explain this, although it was known that toparahelium and ortohelium belonged the singlet (antiparallel spin) andtriplet (parallel spin) states

Solution Heisenberg in 1926.


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Assumptions:

un = electron wave functions in state n! (n, l ,m)

again, no interaction ⇒{

Ψ = u0(1)un(2)E = E0 + En

e-e interaction is a small perturbation


Identical particles


Assumptions:



Ψ = u0(1)un(2)E = E0 + En


What do we need, if we want to estimate the e-e interaction energy?


Identical particles


Assumptions:



Ψ = u0(1)un(2)E = E0 + En



Ψ+ =

1√2

[u0(1)un(2) + u0(2)un(1)]

Ψ− =1√2

[u0(1)un(2)− u0(2)un(1)]


Identical particles


Assumptions:



Ψ = u0(1)un(2)E = E0 + En



Ψ+ =

1√2

[u0(1)un(2) + u0(2)un(1)]

Ψ− =1√2

[u0(1)un(2)− u0(2)un(1)]

1√2

are normalization

factors


Identical particles


Expectation of e-e interaction energy

E ′ =

∫ ∫Ψ∗

e2

r12Ψdτ1dτ2


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E ′ =

∫ ∫Ψ∗

e2

r12Ψdτ1dτ2

E ′ = Eex ± Ecorr , Eex > 0, Ecorr > 0

For calculation details seeRefs. [1] and [3].


Identical particles



E ′ =

∫ ∫Ψ∗

e2

r12Ψdτ1dτ2

E ′ = Eex ± Ecorr , Eex > 0, Ecorr > 0

For calculation details seeRefs. [1] and [3].

Terms

+ 99K paraterms− 99K ortoterms


Identical particles

Antisymmetry principle

Contents






6 Literature


Identical particles


Spin s =

{+ 1

2, spin up, parallel to the outer mag. field

− 12, spin down, antiparallel to the outer mag. field


Identical particles


Spin s =

{+ 1



wave function:

Ψ(x , y , z , s) =

{Ψ(x , y , z ,+ 1

2) = Ψ+(x , y , z)

Ψ(x , y , z ,− 12) = Ψ−(x , y , z)


Identical particles


Spin s =

{+ 1



wave function:

Ψ(x , y , z , s) =

{Ψ(x , y , z ,+ 1

2) = Ψ+(x , y , z) = Ψ(x , y , z)α

Ψ(x , y , z ,− 12) = Ψ−(x , y , z) = Ψ(x , y , z)β

α� spin wave function for parallel spinβ � spin wave function for antiparallel spin


Identical particles


Assumptions

z we have 2 electrons: α(1) and β(2)

z their spins are independent


Identical particles


Assumptions



A question

What will total spin wave functions look like?


Identical particles


Assumptions



Total spin wave functions

Spin orientationelectron 1 electron 2

α(1)α(2) ↑ ↑β(1)β(2) ↓ ↓α(1)β(2) ↑ ↓β(1)α(2) ↓ ↑

Second question

Can you guess any other?


Identical particles



Spin orientationelectron 1 electron 2

α(1)α(2) ↑ ↑β(1)β(2) ↓ ↓α(1)β(2) ↑ ↓β(1)α(2) ↓ ↑


S MSα(1)α(2) 1 1

α(1)β(2) + α(2)β(1) 1 0 symmetric ⇒ tripletβ(1)β(2) 1 -1

α(1)β(2)− α(2)β(1) 0 0 antisymmetric ⇒ singlet


Identical particles


Ok, let us now construct the total wave function:

♠ spatial wave function♠ spin wave function


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Remember

Ψr =

{u(1)v(2) + u(2)v(1), symmetricu(1)v(2)− u(2)v(1), antisymmetric


Identical particles




Total wave function

Ψr,s =

u(1)v(2) + u(2)v(1)×

α(1)α(2) symmα(1)β(2) + α(2)β(1) symmβ(1)β(2) symmα(1)β(2)− α(2)β(1) antisymm

u(1)v(2)− u(2)v(1)×

α(1)α(2) antisymmα(1)β(2) + α(2)β(1) antisymmβ(1)β(2) antisymmα(1)β(2)− α(2)β(1) symm


Identical particles


Total wave function

Ψr,s =

u(1)v(2) + u(2)v(1)×

α(1)α(2) symmα(1)β(2) + α(2)β(1) symmβ(1)β(2) symmα(1)β(2)− α(2)β(1) antisymm

u(1)v(2)− u(2)v(1)×

α(1)α(2) antisymmα(1)β(2) + α(2)β(1) antisymmβ(1)β(2) antisymmα(1)β(2)− α(2)β(1) symm

A question

Which of these w.f. come in nature?

A hint...

Use Pauli’s exclusion principle.


Identical particles


Possible total wave functions

Ψr,s =

u(1)v(2) + u(2)v(1)× α(1)β(2)− α(2)β(1) antisymm

u(1)v(2)− u(2)v(1)×

α(1)α(2) antisymmα(1)β(2) + α(2)β(1) antisymmβ(1)β(2) antisymm


Identical particles


Possible total wave functions

Ψr,s =

u(1)v(2) + u(2)v(1)× α(1)β(2)− α(2)β(1) antisymm

u(1)v(2)− u(2)v(1)×

α(1)α(2) antisymmα(1)β(2) + α(2)β(1) antisymmβ(1)β(2) antisymm

Antisymmetry principle (generalized Pauli’s principle)

Total wave function of electrons has to be antisymmetric, wrt the interchangeof their (spatial and/or spin) coordinates.


Identical particles


If we turn back to helium...

He electron wave functions

ΨHer,s =

u(1)v(2) + u(2)v(1)× α(1)β(2)− α(2)β(1) parahelium

u(1)v(2)− u(2)v(1)×

α(1)α(2)α(1)β(2) + α(2)β(1) ortoheliumβ(1)β(2)

Antisymmetry principle (generalized Pauli’s principle)

Total wave function of electrons has to be antisymmetric, wrt the interchangeof their (spatial and/or spin) coordinates.


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Generalizations

♣ to N-particle systems Ψ(1, 2, 3, . . .) = −Ψ(2, 1, 3, . . .)♣ to all fermions


Identical particles


Generalizations


A question

What about symmetric wave functions Ψ(1, 2, 3, . . .) = Ψ(2, 1, 3, . . .)?They describe the particles with spin 0, 1, 2, . . .


Identical particles


Generalizations


In conclusion

Spin Symmetry Statistics

0,1,2,. . . symmetric Einstein-Bose

1

2,

3

2, . . . antisymmetric Fermi-Dirac


Identical particles

Example: H molecule

Contents






6 Literature


Identical particles

Example: H molecule

Let us first look at rotational spectrum of H molecule.(Mecke - first experimental observation; Heisenberg & Hund - theoreticalexplanation in 1928.)


Identical particles

Example: H molecule


The movement of nuclei determines the following properties:

vibration

rotation

spin


Identical particles

Example: H molecule


The movement of nuclei determinesthe following properties:

vibration

rotation

spin

These properties are independent

Ψ = ψvibψrotψspin


Identical particles

Example: H molecule



A question

What happens with this w.f. if weinterchange the coordinates?


Identical particles

Example: H molecule



A question


vibration

rotation

spin

ψvib = ψvib(r12), r12 = |r2 − r1|

~r → −~r =⇒ ψvib = ψvib


Identical particles

Example: H molecule



A question


vibration

rotation

spinψrot 99K Y

ml =

{Y ml , l even−Y ml , l odd

~r → −~r =⇒ ψrot = (−1)lψrot


Identical particles

Example: H molecule



A question


vibration

rotation

spin


Identical particles

Example: H molecule

Quantum # l Spin orientation Total spin Statistical weigth NameEven ↑↓ 0 1 para H2Odd ↑↑ 1 3 orto H2


Identical particles

Example: H molecule


Note

There can be no transitions between the states with different symmetrycharacter.


Identical particles

Example: H molecule


Note

There can be no transitions between the states with different symmetrycharacter. para H2 : orto H2 = 1 : 3


Identical particles

Example: H molecule


Note


A question

What happens at T = 0 K?


Identical particles

Example: H molecule


Note


A question

What happens at T = 0 K? All hydrogen molecules go to para-state.


Identical particles

Example: H molecule

Another confirmation (Dennison 1928.)


Identical particles

Literature

Contents






6 Literature


Identical particles

Literature

Literature

1 I. Supek, Teorijska fizika i struktura materije, II. dio, Školska knjiga,Zagreb, 1989.

2 D. J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., PearsonEducation, Inc., Upper Saddle River, NJ, 2005.

3 A. Szabo, N. Ostlund, Modern Quantum Chemistry, Introduction toAdvanced Electronic Structure theory, Dover Publications, New York,1996.


ContentsTwo-particle wave equationSymmetric and antisymmetric solutionsExample: He spectrumAntisymmetry principleExample: H moleculeLiterature

0.0: 0.1: 0.2: 0.3: 0.4: 0.5: 0.6: 0.7: 0.8: 0.9: 0.10: 0.11: 0.12: anm0: 1.0: 1.1: 1.2: 1.3: 1.4: 1.5: 1.6: 1.7: 1.8: 1.9: 1.10: 1.11: 1.12: 1.13: 1.14: 1.15: 1.16: 1.17: 1.18: 1.19: 1.20: 1.21: 1.22: 1.23: anm1:

Documents

Identical particles - Quantum mechanics 2 - Lecture 1 · 2016. 9. 30. · 2m (1 + 2) + V = E A question What happens with S.E. if we interchange the coordinates of particles? S.E