Two-particle systems (Ch. 5) Generalization to two-particles is very - - PowerPoint PPT Presentation

EE201/MSE207 Lecture 12

Two-particle systems (Ch. 5)

𝑗ℏ 𝜖Ψ 𝜖𝑢 = 𝐼Ψ

Generalization to two-particles is very natural. A two-particle state is characterized by wavefunction Ψ

𝑠

1,

𝑠

2, 𝑢

(for a moment forget about spin) SE:

𝐼 = − ℏ2 2𝑛1 𝛼

1 2 − ℏ2

2𝑛2 𝛼2

2 + 𝑊(𝑠

1, 𝑠2, 𝑢)

𝜖2 𝜖𝑦1 + 𝜖2 𝜖𝑧1 + 𝜖2 𝜖𝑨1

(Laplacian)

Ψ 2: probability density

Normalization

Ψ 2 𝑒 𝑠

1𝑒

𝑠

2 = 1

If 𝑊(

𝑠

1,

𝑠2) (no time dependence for potential energy), then TISE:

𝐼𝜔 = 𝐹𝜔

Then general solution of SE:

Ψ 𝑠

1, 𝑠2, 𝑢

=

𝑜

𝑑𝑜𝜔𝑜 𝑠

1,

𝑠2 𝑓−𝑗 𝐹𝑜

ℏ 𝑢

𝑜 𝑑𝑜 2 = 1

Simplification if 𝑊 = 𝑊( 𝑠

1 −

𝑠

2)

Then the problem reduces to one-particle problem (may be important for excitons) Similar situation in astronomy: two-body problem is essentially one-body problem Classical mechanics Introduce two variables (new coordinates): center of mass

𝑆 = 𝑛1 𝑠

1 + 𝑛2

𝑠

2

𝑛1 + 𝑛2

(moves freely by inertia) difference

𝑠 = 𝑠

1 −

𝑠2

Evolution of 𝑠: Force 𝐺 = −𝛼 𝑊( 𝑠) acts on both particles

𝑠 = 𝐺 𝑛1 + 𝐺 𝑛2 = 𝐺 𝑛1 +𝑛2 𝑛1𝑛2 = 𝐺 𝜈 𝜈 = 𝑛1𝑛2 𝑛1 + 𝑛2

Therefore, evolution of

𝑠 is as for a mass 𝜈 in potential 𝑊

(this is how Earth-Moon problem is analyzed)

Quantum mechanics for 𝑊 = 𝑊( 𝑠

1 −

𝑠

2)

𝐼 = − ℏ2 2(𝑛1+𝑛2) 𝛼𝑆

2 − ℏ2

2𝜈 𝛼

𝑠 2 + 𝑊(

𝑠) Derivation 𝜖 𝜖𝑠

1

= 𝜖 𝜖𝑆 𝜖𝑆 𝜖𝑠

1

+ 𝜖 𝜖𝑠 𝜖𝑠 𝜖𝑠

1

= 𝑛1 𝑛1 + 𝑛2 𝜖 𝜖𝑆 + 𝜖 𝜖𝑠 𝜖 𝜖𝑠

2

= 𝑛2 𝑛1 + 𝑛2 𝜖 𝜖𝑆 − 𝜖 𝜖𝑠 𝜖2 𝜖𝑠

1 2 =

𝑛1 𝑛1 + 𝑛2

2 𝜖2

𝜖𝑆2 + 𝜖2 𝜖𝑠2 + 2𝑛1 𝑛1 + 𝑛2 𝜖2 𝜖𝑠 𝜖𝑆 𝜖2 𝜖𝑠

2 2 =

𝑛2 𝑛1 + 𝑛2

2 𝜖2

𝜖𝑆2 + 𝜖2 𝜖𝑠2 − 2𝑛2 𝑛1 + 𝑛2 𝜖2 𝜖𝑠 𝜖𝑆 Kinetic energy for mass 𝑛1 + 𝑛2 at center of mass, kinetic energy of mass 𝜈 at position difference, and potential energy Separation of variables

𝜔 = 𝜔𝑆 𝑆 𝜔𝑠 𝑠 𝐹 = 𝐹𝑆 + 𝐹𝑠

𝐹𝑆 is kinetic energy of free particle with mass 𝑛1 + 𝑛2 For 𝐹𝑠 we need to solve TISE

− ℏ2 2𝜈 𝛼2𝜔𝑠 𝑠 + 𝑊

𝑠 𝜔𝑠

𝑠 = 𝐹𝑠 𝜔𝑠 𝑠

If 𝑊 ≠ 𝑊( 𝑠), then need to solve much more complicated 2-particle TISE

Identical particles

In quantum mechanics two electrons are indistinguishable (postulate), similarly two protons, two holes, etc. Surprisingly, this leads to non-trivial consequences. Simple case (two particles in two states, no interaction, no spin)

𝜔 𝑠

1,

𝑠

2 = 𝜔𝑏

𝑠

1 𝜔𝑐(

𝑠

2)

First particle in state 𝑏, second one in state 𝑐 (distinguishable particles) However, for indistinguishable particles 𝜔𝑐

𝑠

1 𝜔𝑏(

𝑠2)

corresponds to the same state Actually, such state is described by wavefunction

𝜔± 𝑠

1,

𝑠

2 = 1

2 𝜔𝑏 𝑠

1 𝜔𝑐

𝑠

2 ± 𝜔𝑐

𝑠

1 𝜔𝑏

𝑠

2

+ for bosons (integer spin), − for fermions (half-integer spin) Generally

𝜔 𝑠

1,

𝑠2 = ±𝜔 𝑠2, 𝑠

1

(+ for bosons, − for fermions)

Symmetry for identical particles

Generally (no spin, i.e. the same spin)

𝜔 𝑠

1,

𝑠2 = ±𝜔 𝑠2, 𝑠

1

(+ for bosons, − for fermions)

(new development: “anyons”)

With spin

𝜔 𝑠

1, 𝑇1,

𝑠2, 𝑇2 = ±𝜔 𝑠2, 𝑇2, 𝑠

1, 𝑇1

(exchange of two particles) Proof Introduce exchange operator 𝑄. It satisfies 𝑄2 = 1 and commutes with 𝐼. Therefore common eigenstates, 𝜇2 = 1  𝜇 = ±1. Remark: no such symmetry for different particles (e.g., proton and electron) Pauli exclusion principle Two fermions cannot occupy the same state (“cannot sit on the same chair”) Proof: otherwise 𝜔 = 0 This symmetry changes average distance between particles (exchange correlation, “exchange interaction”)

Simple example of exchange correlation

Consider two particles in 1D, occupying states 𝑏 and 𝑐. 3 cases (1) 𝜔 = 𝜔𝑏 𝑦1 𝜔𝑐(𝑦2)  distinguishable particles (2) 𝜔 = 1

2 [𝜔𝑏 𝑦1 𝜔𝑐 𝑦2 + 𝜔𝑐 𝑦1 𝜔𝑏(𝑦2)]

 bosons (3) 𝜔 = 1

2 [𝜔𝑏 𝑦1 𝜔𝑐 𝑦2 − 𝜔𝑐 𝑦1 𝜔𝑏(𝑦2)]

 fermions Let us show that bosons are closer to each other, fermions are farther away. Calculate 𝑦1 − 𝑦2 2 = 𝑦1

2 + 𝑦2 2 − 2〈𝑦1𝑦2〉

Case (1): distinguishable

𝑦1

2 = 𝑦1 2 𝜔 𝑦1, 𝑦2 2 𝑒𝑦1𝑒𝑦2 = 𝑦1 2 | 𝜔𝑏 𝑦1 |2𝑒𝑦1 𝜔𝑐 𝑦2 2𝑒𝑦2 =

= 𝑦2 𝑏× 1 = 𝑦2 𝑏 𝑦2

2 = 𝑦2 𝑐

Similarly

𝑦1𝑦2 = 𝑦1𝑦2 𝜔 𝑦1, 𝑦2

2 𝑒𝑦1𝑒𝑦2 =

= 𝑦1| 𝜔𝑏 𝑦1 |2𝑒𝑦1 𝑦2 𝜔𝑐 𝑦2

2𝑒𝑦2 = 𝑦 𝑏 𝑦 𝑐

(uncorrelated)

Case (2): bosons Similarly

Simple example of exchange correlation (cont.)

𝜔 =

1 2 [𝜔𝑏 𝑦1 𝜔𝑐 𝑦2 + 𝜔𝑐 𝑦1 𝜔𝑏(𝑦2)]

𝑦1

2 = 𝑦1 2 𝜔 𝑦1, 𝑦2 2 𝑒𝑦1𝑒𝑦2 =

= 1

2 𝑦1 2 | 𝜔𝑏 𝑦1 |2𝑒𝑦1 𝜔𝑐 𝑦2 2𝑒𝑦2 +

1 2 𝑦1 2 | 𝜔𝑐 𝑦1 |2𝑒𝑦1 𝜔𝑏 𝑦2 2𝑒𝑦2

+ 1

2 𝑦1 2 𝜔𝑏 ∗ 𝑦1 𝜔𝑐 𝑦1 𝑒𝑦1 𝜔𝑐 ∗ 𝑦2 𝜔𝑏 𝑦2 𝑒𝑦2 +

+ 1

2 𝑦1 2 𝜔𝑐 ∗ 𝑦1 𝜔𝑏 𝑦1 𝑒𝑦1 𝜔𝑏 ∗ 𝑦2 𝜔𝑐 𝑦2 𝑒𝑦2 =

1 1

= 1 2 ( 𝑦2 𝑏 + 𝑦2 𝑐) 〈𝑦2

2〉 = 1

2 ( 𝑦2 𝑏 + 𝑦2 𝑐)

(quite natural, since each particle in both states)

Case (2): bosons

Simple example of exchange correlation (cont.)

𝜔 =

1 2 [𝜔𝑏 𝑦1 𝜔𝑐 𝑦2 + 𝜔𝑐 𝑦1 𝜔𝑏(𝑦2)]

𝑦1𝑦2 = 𝑦1𝑦2 𝜔 𝑦1, 𝑦2

2 𝑒𝑦1𝑒𝑦2 =

= 1

2 𝑦1 | 𝜔𝑏 𝑦1 |2𝑒𝑦1 𝑦2 𝜔𝑐 𝑦2 2𝑒𝑦2 +

+ 1

2 𝑦1 𝜔𝑏 ∗ 𝑦1 𝜔𝑐 𝑦1 𝑒𝑦1 𝑦2 𝜔𝑐 ∗ 𝑦2 𝜔𝑏 𝑦2 𝑒𝑦2 +

exchange term

𝑦1

2 = 〈𝑦2 2〉 = 1 2 ( 𝑦2 𝑏 + 𝑦2 𝑐)

+ 1

2 𝑦1| 𝜔𝑐 𝑦1 |2𝑒𝑦1 𝑦2 𝜔𝑏 𝑦2 2𝑒𝑦2 +

+ 1

2 𝑦1 𝜔𝑐 ∗ 𝑦1 𝜔𝑏 𝑦1 𝑒𝑦1 𝑦2 𝜔𝑏 ∗ 𝑦2 𝜔𝑐 𝑦2 𝑒𝑦2 =

𝑦 𝑏 𝑦 𝑐

same as line 3 conjugate same as line 1

= 𝑦 𝑏 𝑦 𝑐 + 𝑦 𝜔𝑏

∗ 𝑦 𝜔𝑐 𝑦 𝑒𝑦 2

Case (2): bosons

Simple example of exchange correlation: summary

𝜔 = [𝜔𝑏 𝑦1 𝜔𝑐 𝑦2 + 𝜔𝑐 𝑦1 𝜔𝑏(𝑦2)]/ 2

𝑦1

2 = 〈𝑦2 2〉 = 1 2 ( 𝑦2 𝑏 + 𝑦2 𝑐)

𝑦1𝑦2 = 𝑦 𝑏 𝑦 𝑐 + 𝑦 𝜔𝑏

∗ 𝑦 𝜔𝑐 𝑦 𝑒𝑦 2

Case (1): distinguishable particles 𝜔 = 𝜔𝑏 𝑦1 𝜔𝑐 𝑦2

𝑦1

2 = 𝑦2 𝑏

𝑦2

2 = 𝑦2 𝑐

𝑦1𝑦2 = 𝑦 𝑏 𝑦 𝑐

Case (3): fermions 𝜔 = [𝜔𝑏 𝑦1 𝜔𝑐 𝑦2 − 𝜔𝑐 𝑦1 𝜔𝑏(𝑦2)]/ 2

𝑦1

2 = 〈𝑦2 2〉 = 1 2 ( 𝑦2 𝑏 + 𝑦2 𝑐)

𝑦1𝑦2 = 𝑦 𝑏 𝑦 𝑐 − 𝑦 𝜔𝑏

∗ 𝑦 𝜔𝑐 𝑦 𝑒𝑦 2

If 𝜔𝑏(𝑦) and 𝜔𝑐 𝑦 do not overlap, then no difference. If 𝜔𝑏(𝑦) and 𝜔𝑐 𝑦 overlap, then exchange term is non-zero. For case 2 (bosons), 〈𝑦1𝑦2〉 is larger 

𝑦1 − 𝑦2 2 is smaller (closer to each other)

For case 3 (fermions), 𝑦1 − 𝑦2 2 is larger (like to be farther away from each other) Exchange correlation (“exchange interaction”, “exchange force”)

Molecule of hydrogen (H2 )

𝑏 𝑐

Now need to take spin into account What is spin state of the ground state? (Show that singlet) Ground state: both electrons have 𝑜 = 1, but also have spins If total spin is 0 (singlet), then spin state is

↑↓−↓↑ 2

(antisymmetric to exchange), therefore spatial part should be symmetric (so that total is antisymmetric), therefore case (2)  electrons closer to each other  covalent bond (actually, electrons repel each other, but attraction to protons is more important) If total spin is 1 (triplet), then spin state is ↑↑ or ↓↓ or

↑↓+↓↑ 2

(all symmetric), therefore spatial part is antisymmetric (case 3)  electrons are farther away from each other  antibonding state (not stable) Ground state wavefunction

𝜔 = 1 2 𝜔𝑏 𝑠

1 𝜔𝑐

𝑠

2 + 𝜔𝑐

𝑠

1 𝜔𝑏

𝑠2 ↑↓−↓↑ 2

Atoms (many electrons)

𝐼 =

𝑘=1 𝑎

− ℏ2 2𝑛 𝛼

𝑘 2−

1 4𝜌𝜁0 𝑎𝑓2 𝑠

𝑘

+ 1 2 1 4𝜌𝜁0

𝑘≠𝑙 𝑎

𝑓2 𝑠

𝑘 − 𝑠 𝑙

and wavefunction must be antisymmetric for exchange of any two electrons (exchange of both positions and spins)

Atom of helium (He)

Z=2

What are spins of lowest energy states? If we neglect e-e interaction (approximation), then just two electrons in hydrogen-like atom (𝑎 = 2 instead of 𝑎 = 1)

11 parahelium (𝑡 = 0) 12 orthohelium (𝑡 = 1) 12 parahelium (𝑡 = 0)

Ground state (𝑜 = 1 and 𝑜 = 1)  symmetric spatial part of wavefunction  antisymmetric spin part, i.e. 𝑡 = 0 (singlet, parahelium) (↑↓−↓↑)/ 2 First excited state (𝑜 = 1 and 𝑜 = 2) If 𝑡 = 0 (singlet, parahelium)  spin part of wavefunction is antisymmetric  spatial part is symmetric  electrons closer to each other  higher energy of e-e interaction  higher energy If 𝑡 = 1 (triplet, orthohelium)  spin part is symmetric  spatial part is antisymmetric  electrons farther away  lower energy