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

EE201/MSE207 Lecture 12 Two-particle systems (Ch. 5) 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) 𝐼 = − ℏ 2 2 − ℏ 2 𝑗ℏ 𝜖Ψ 2 + 𝑊( 𝑠 𝜖𝑢 = 1 , 𝑠 2 , 𝑢 ) 𝛼 𝛼 2 𝐼Ψ SE: 1 2𝑛 1 2𝑛 2 𝜖 2 + 𝜖 2 + 𝜖 2 Ψ 2 : probability density (Laplacian) 𝜖𝑦 1 𝜖𝑧 1 𝜖𝑨 1 Ψ 2 𝑒 𝑠 1 𝑒 𝑠 2 = 1 Normalization If 𝑊( 𝑠 1 , 𝑠 2 ) (no time dependence for potential energy), then TISE: 𝐼𝜔 = 𝐹𝜔 Then general solution of SE: 𝑠 2 𝑓 −𝑗 𝐹 𝑜 𝑜 𝑑 𝑜 2 = 1 ℏ 𝑢 Ψ 𝑠 1 , 𝑠 2 , 𝑢 = 𝑑 𝑜 𝜔 𝑜 𝑠 1 , 𝑜

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): 𝑆 = 𝑛 1 𝑠 1 + 𝑛 2 𝑠 2 (moves freely by inertia) center of mass 𝑛 1 + 𝑛 2 𝑠 = 𝑠 1 − 𝑠 2 difference Force Evolution of 𝑠 : 𝐺 = −𝛼 𝑊( 𝑠) acts on both particles 𝑛 1 𝑛 2 𝐺 𝐺 𝐺 𝑛 1 +𝑛 2 𝐺 = 𝑠 = + = 𝜈 = 𝑛 1 𝑛 2 𝑛 1 𝑛 2 𝜈 𝑛 1 + 𝑛 2 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 ) ℏ 2 2 − ℏ 2 2 + 𝑊( 𝐼 = − 2(𝑛 1 +𝑛 2 ) 𝛼 𝑆 2𝜈 𝛼 𝑠 ) 𝑠 Derivation 𝜖 = 𝜖 𝜖𝑆 + 𝜖 𝜖𝑠 𝑛 1 𝜖𝑆 + 𝜖 𝜖 Kinetic energy for mass = 𝜖𝑠 𝜖𝑆 𝜖𝑠 𝜖𝑠 𝜖𝑠 𝑛 1 + 𝑛 2 𝜖𝑠 𝑛 1 + 𝑛 2 at center of mass, 1 1 1 𝜖 𝑛 2 𝜖𝑆 − 𝜖 𝜖 kinetic energy of mass 𝜈 at = position difference, and 𝜖𝑠 𝑛 1 + 𝑛 2 𝜖𝑠 2 2 𝜖 2 potential energy 𝜖 2 𝜖𝑆 2 + 𝜖 2 𝜖 2 𝑛 1 2𝑛 1 2 = 𝜖𝑠 2 + 𝑛 1 + 𝑛 2 𝑛 1 + 𝑛 2 𝜖𝑠 𝜖𝑆 𝜖𝑠 1 2 𝜖 2 𝜖 2 𝜖𝑆 2 + 𝜖 2 𝜖 2 𝑛 2 2𝑛 2 2 = 𝜖𝑠 2 − 𝑛 1 + 𝑛 2 𝑛 1 + 𝑛 2 𝜖𝑠 𝜖𝑆 𝜖𝑠 2 Separation of variables 𝜔 = 𝜔 𝑆 𝑆 𝜔 𝑠 𝑠 𝐹 = 𝐹 𝑆 + 𝐹 𝑠 𝐹 𝑆 is kinetic energy of free particle with mass 𝑛 1 + 𝑛 2 − ℏ 2 2𝜈 𝛼 2 𝜔 𝑠 For 𝐹 𝑠 we need to solve TISE 𝑠 + 𝑊 𝑠 𝜔 𝑠 𝑠 = 𝐹 𝑠 𝜔 𝑠 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 = 𝜔 𝑏 𝑠 1 𝜔 𝑐 ( 𝑠 2 ) First particle in state 𝑏 , second one in state 𝑐 (distinguishable particles) However, for indistinguishable particles 𝜔 𝑐 𝑠 1 𝜔 𝑏 ( 𝑠 2 ) corresponds to the same state Actually, such state is described by wavefunction 2 = 1 𝜔 ± 𝑠 1 , 𝑠 𝜔 𝑏 𝑠 1 𝜔 𝑐 𝑠 2 ± 𝜔 𝑐 𝑠 1 𝜔 𝑏 𝑠 2 2 + for bosons (integer spin), − for fermions (half-integer spin) Generally 𝜔 𝑠 1 , 𝑠 2 = ±𝜔 𝑠 2 , 𝑠 ( + for bosons, − for fermions) 1

Symmetry for identical particles Generally (no spin, i.e. the same spin) 𝜔 𝑠 1 , 𝑠 2 = ±𝜔 𝑠 2 , 𝑠 ( + for bosons, − for fermions) 1 (new development: “ anyons ”) With spin 𝜔 𝑠 1 , 𝑇 1 , 𝑠 2 , 𝑇 2 = ±𝜔 𝑠 2 , 𝑇 2 , 𝑠 1 , 𝑇 1 (exchange of two particles) Proof 𝑄 2 = Introduce exchange operator 𝑄 . It satisfies 1 and commutes with 𝐼 . Therefore common eigenstates, 𝜇 2 = 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 = 𝜔 𝑏 𝑦 1 𝜔 𝑐 (𝑦 2 )  distinguishable particles (2) 𝜔 = 1 2 [𝜔 𝑏 𝑦 1 𝜔 𝑐 𝑦 2 + 𝜔 𝑐 𝑦 1 𝜔 𝑏 (𝑦 2 )]  bosons (3) 𝜔 = 1 2 [𝜔 𝑏 𝑦 1 𝜔 𝑐 𝑦 2 − 𝜔 𝑐 𝑦 1 𝜔 𝑏 (𝑦 2 )]  fermions Let us show that bosons are closer to each other, fermions are farther away. Calculate 𝑦 1 − 𝑦 2 2 = 𝑦 1 2 + 𝑦 2 2 − 2〈𝑦 1 𝑦 2 〉 Case (1): distinguishable 2 = 𝑦 1 2 𝜔 𝑦 1 , 𝑦 2 2 𝑒𝑦 1 𝑒𝑦 2 = 𝑦 1 2 | 𝜔 𝑏 𝑦 1 | 2 𝑒𝑦 1 𝜔 𝑐 𝑦 2 2 𝑒𝑦 2 = 𝑦 1 = 𝑦 2 𝑏 × 1 = 𝑦 2 𝑏 2 = 𝑦 2 𝑐 𝑦 2 Similarly 2 𝑒𝑦 1 𝑒𝑦 2 = 𝑦 1 𝑦 2 = 𝑦 1 𝑦 2 𝜔 𝑦 1 , 𝑦 2 = 𝑦 1 | 𝜔 𝑏 𝑦 1 | 2 𝑒𝑦 1 𝑦 2 𝜔 𝑐 𝑦 2 2 𝑒𝑦 2 = 𝑦 𝑏 𝑦 𝑐 (uncorrelated)

Simple example of exchange correlation (cont.) 1 𝜔 = 2 [𝜔 𝑏 𝑦 1 𝜔 𝑐 𝑦 2 + 𝜔 𝑐 𝑦 1 𝜔 𝑏 (𝑦 2 )] Case (2): bosons 2 = 𝑦 1 2 𝜔 𝑦 1 , 𝑦 2 2 𝑒𝑦 1 𝑒𝑦 2 = 𝑦 1 2 | 𝜔 𝑏 𝑦 1 | 2 𝑒𝑦 1 𝜔 𝑐 𝑦 2 = 1 2 | 𝜔 𝑐 𝑦 1 | 2 𝑒𝑦 1 𝜔 𝑏 𝑦 2 1 2 𝑒𝑦 2 + 2 𝑒𝑦 2 2 𝑦 1 2 𝑦 1 1 1 ∗ 𝑦 2 𝜔 𝑏 𝑦 2 𝑒𝑦 2 + 2 𝜔 𝑏 ∗ 𝑦 1 𝜔 𝑐 𝑦 1 𝑒𝑦 1 𝜔 𝑐 + 1 2 𝑦 1 0 2 𝜔 𝑐 ∗ 𝑦 1 𝜔 𝑏 𝑦 1 𝑒𝑦 1 𝜔 𝑏 ∗ 𝑦 2 𝜔 𝑐 𝑦 2 𝑒𝑦 2 = + 1 2 𝑦 1 0 = 1 2 ( 𝑦 2 𝑏 + 𝑦 2 𝑐 ) 2 〉 = 1 2 ( 𝑦 2 𝑏 + 𝑦 2 𝑐 ) 〈𝑦 2 Similarly (quite natural, since each particle in both states)

Simple example of exchange correlation (cont.) 1 𝜔 = 2 [𝜔 𝑏 𝑦 1 𝜔 𝑐 𝑦 2 + 𝜔 𝑐 𝑦 1 𝜔 𝑏 (𝑦 2 )] Case (2): bosons 2 = 〈𝑦 2 2 〉 = 1 2 ( 𝑦 2 𝑏 + 𝑦 2 𝑐 ) 𝑦 1 2 𝑒𝑦 1 𝑒𝑦 2 = 𝑦 1 𝑦 2 = 𝑦 1 𝑦 2 𝜔 𝑦 1 , 𝑦 2 = 1 2 𝑦 1 | 𝜔 𝑏 𝑦 1 | 2 𝑒𝑦 1 𝑦 2 𝜔 𝑐 𝑦 2 2 𝑒𝑦 2 + 𝑦 𝑏 𝑦 𝑐 + 1 2 𝑦 1 | 𝜔 𝑐 𝑦 1 | 2 𝑒𝑦 1 𝑦 2 𝜔 𝑏 𝑦 2 2 𝑒𝑦 2 + same as line 1 ∗ 𝑦 2 𝜔 𝑏 𝑦 2 𝑒𝑦 2 + ∗ 𝑦 1 𝜔 𝑐 𝑦 1 𝑒𝑦 1 𝑦 2 𝜔 𝑐 + 1 2 𝑦 1 𝜔 𝑏 conjugate ∗ 𝑦 1 𝜔 𝑏 𝑦 1 𝑒𝑦 1 𝑦 2 𝜔 𝑏 ∗ 𝑦 2 𝜔 𝑐 𝑦 2 𝑒𝑦 2 = + 1 2 𝑦 1 𝜔 𝑐 same as line 3 ∗ 𝑦 𝜔 𝑐 𝑦 𝑒𝑦 2 = 𝑦 𝑏 𝑦 𝑐 + 𝑦 𝜔 𝑏 exchange term

Simple example of exchange correlation: summary 𝜔 = 𝜔 𝑏 𝑦 1 𝜔 𝑐 𝑦 2 Case (1): distinguishable particles 2 = 𝑦 2 𝑏 2 = 𝑦 2 𝑐 𝑦 1 𝑦 2 = 𝑦 𝑏 𝑦 𝑐 𝑦 1 𝑦 2 𝜔 = [𝜔 𝑏 𝑦 1 𝜔 𝑐 𝑦 2 + 𝜔 𝑐 𝑦 1 𝜔 𝑏 (𝑦 2 )]/ 2 Case (2): bosons 2 = 〈𝑦 2 2 〉 = 1 2 ( 𝑦 2 𝑏 + 𝑦 2 𝑐 ) 𝑦 1 2 ∗ 𝑦 𝜔 𝑐 𝑦 𝑒𝑦 𝑦 1 𝑦 2 = 𝑦 𝑏 𝑦 𝑐 + 𝑦 𝜔 𝑏 𝜔 = [𝜔 𝑏 𝑦 1 𝜔 𝑐 𝑦 2 − 𝜔 𝑐 𝑦 1 𝜔 𝑏 (𝑦 2 )]/ 2 Case (3): fermions 2 = 〈𝑦 2 2 〉 = 1 2 ( 𝑦 2 𝑏 + 𝑦 2 𝑐 ) 𝑦 1 ∗ 𝑦 𝜔 𝑐 𝑦 𝑒𝑦 2 𝑦 1 𝑦 2 = 𝑦 𝑏 𝑦 𝑐 − 𝑦 𝜔 𝑏 If 𝜔 𝑏 (𝑦) and 𝜔 𝑐 𝑦 do not overlap, then no difference. If 𝜔 𝑏 (𝑦) and 𝜔 𝑐 𝑦 overlap, then exchange term is non-zero. 𝑦 1 − 𝑦 2 2 is smaller (closer to each other) For case 2 (bosons), 〈𝑦 1 𝑦 2 〉 is larger  For case 3 (fermions), 𝑦 1 − 𝑦 2 2 is larger (like to be farther away from each other) Exchange correlation (“exchange interaction”, “exchange force”)

Molecule of hydrogen (H 2 ) 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 (antisymmetric to exchange), 2 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 (all symmetric), 2 therefore spatial part is antisymmetric (case 3)  electrons are farther away from each other  antibonding state (not stable) Ground state wavefunction 𝜔 = 1 ↑↓−↓↑ 𝜔 𝑏 𝑠 1 𝜔 𝑐 𝑠 2 + 𝜔 𝑐 𝑠 1 𝜔 𝑏 𝑠 2 2 2

Atoms (many electrons) 𝑎 𝑎 − ℏ 2 𝑎𝑓 2 𝑓 2 1 + 1 1 2 − 𝐼 = 2𝑛 𝛼 𝑘 4𝜌𝜁 0 2 4𝜌𝜁 0 𝑠 𝑠 𝑘 − 𝑠 𝑘 𝑙 𝑘=1 𝑘≠𝑙 and wavefunction must be antisymmetric for exchange of any two electrons (exchange of both positions and spins)