Variational principle (Ch. 7) Only Sec. 7.1 Theorem: For an - PowerPoint PPT Presentation
EE201/MSE207 Lecture 16 Variational principle (Ch. 7) Only Sec. 7.1 Theorem: For an arbitrary | , the ground state energy satisfies inequality = Proof is simple. Let us expand =
EE201/MSE207 Lecture 16 Variational principle (Ch. 7) Only Sec. 7.1 Theorem: For an arbitrary |πβͺ , the ground state energy πΉ π satisfies inequality πΉ π β€ π πΌ π = β© πΌβͺ Proof is simple. Let us expand π = π π π |π π βͺ . Then since πΉ π β₯ πΉ π , we get πΌ = π π π 2 πΉ π β₯ πΉ π π π π 2 = πΉ π This theorem can be useful to estimate πΉ π (or at least to find an upper bound) Idea: Use trial wavefunctions |πβͺ with many adjustable parameters and minimize β© πΌβͺ . Hopefully min β© πΌβͺ is close to πΉ π . Extensions of this method can also be used to find |π π βͺ , first-excited state energy and wavefunction (using subspace orthogonal to |π π βͺ ), second-excited state, etc.
EE201/MSE207 Lecture 16 Band structure (back to Ch. 5) Band structure for electrons is a consequence of a periodic potential in a lattice (due to periodic arrangement of atoms). For simplicity let us consider 1D case π π¦ + π = π(π¦) (periodic with lattice constant π ) Blochβs theorem: If π π¦ + π = π(π) , then for an eigenstate of energy π(π¦ + π) = π ππΏπ π π¦ (almost periodic, β quasimomentum β βπΏ ) Therefore π π¦ = π ππΏπ¦ π(π¦) , π π¦ + π = π(π¦) with periodic π(π¦) Proof Introduce displacement operator πΈ , so that πΈπ(π¦) = π(π¦ + π) . It commutes with Hamiltonian, πΈ, πΌ = 0 , therefore common eigenfunctions. π π¦ + π = π π π¦ If π β 1 , then π would increase or decrease exponentially Therefore π = 1 , can denote π = π ππΏπ .
Periodic boundary condition for Blochβs theorem π(π¦ + π) = π ππΏπ π π¦ β π π¦ + π = π π Usually people use periodic boundary condition in using Blochβs theorem π π¦ + ππ = π(π¦) for π β 1 atoms in a (1D) sample Why? Because it does not matter, but makes calculations simpler πΏ = 2ππ ππ , π = 0, Β±1, Β±2, β¦ Then This gives π different values of πΏ (the same π ππΏπ if Ξπ = π ): π states in a band for π atoms Since π is very large, πΏ is almost continuous.
Simple example: βDirac combβ π βatomsβ Dirac comb: π π π¦ = π½ π=1 π(π¦ β ππ) (wrapped around) β β 2 π 2 π π¦ + π π¦ π π¦ = πΉ π π¦ ππ¦ 2 2π 0 < π¦ < π β π π¦ = π΅ sin(ππ¦) + πΆ cos ππ¦ , π = 2ππΉ/β From Blochβs theorem we know that at βπ < π¦ < 0 , π π¦ = π βππΏπ [π΅ sin π(π¦ + π) + πΆ cos π(π¦ + π) ] πΆ = π βππΏπ [π΅ sin ππ + πΆ cos(ππ)] π 0 + 0 = π 0 β 0 β π β² 0 + 0 β π β² 0 β 0 = ( 2ππ½ β 2 ) π(0) β ππ΅ β π βππΏπ [ππ΅ cos ππ β ππΆ sin(ππ) ] = ( 2ππ½ β 2 ) πΆ From these two equations we find (eliminating π΅ and πΆ ) cos πΏπ = cos ππ + ππ½π sin(ππ) β 2 ππ
Dirac comb (cont.) π π π¦ = π½ π=1 π(π¦ β ππ) cos πΏπ = cos ππ + ππ½π sin(ππ) β 2 ππ π states πΏπ = 2ππ π gap π states gap gap gap gap 2 nd band 1 st band 3 rd band π states ππ½π = 10 β 2 π states ππ (β πΉ) π states per band ( Γ 2 spin) Gaps become smaller, eventually continuum
Bands Gaps become smaller, eventually continuum If one electron per atom ( π = 1 ), then half a band is filled (good conductor) If π = 2 , then one band is filled completely (insulator or semiconductor; cannot slightly excite electrons) If π = 3 , then 1.5 bands are filled (good conductor) If π = 4 , then again insulator or semiconductor Etc. Metals usually have π = 1 π states per band ( Γ 2 spin)
Bands (cont.) β πΉ cos πΏπ = . . . βπΏ is quasimomentum (behaves as momentum) βπ/π π/π 2π/π (Brillouin zone) Periodic: 2π/π πΉ = β 2 π 2 For a free particle 2π Define effective mass π eff via ΞπΉ = β 2 πΏ 2 ΞπΉ = β 2 (ΞπΏ) 2 or even 2π eff 2π eff ππΏ 2 = β 2 π 2 πΉ (similar to bands in semiconductors) or even π eff
Quasimomentum βπΏ behaves as momentum Let us add small force πΊ (e.g., due to electric field acting on electron, πΊ = βπβ° ). Then Ξπ = βπΊπ¦ and therefore πΉ β πΉ β πΊπ¦ (for the same πΏ ). From Blochβs theorem we know π π¦ = π ππΏπ¦ π πΏ (π¦) β π ππΏπ¦ on the large scale Adding time dependence, we get (on the large scale) Ξ¨ π¦, π’ β π ππΏπ¦ π βπ πΉβπΊπ¦ π’ = π π πΏ+ πΊπ’ β π¦ π βππΉπ’/β β πΏ β πΏ + πΊ π(βπΏ) β π’ β = πΊ It means ππ’ We see that βπΏ behaves as momentum (so named quasimomentum) (for validity of this approach we need very small πΊ ) Actually, significant oversimplification in this approach; this rather a hint. π’ β² β) ππ’ β² π’ πΉ(πΏ+πΊ πΊπ’ β)π¦ π β π β π π(πΏ+ π πΏ+πΊ π’ β (π¦) More rigorously, 0 is an approximate solution of SE (straightforward to check). Also, makes sense for energy change: πΊπ€ ππ = β ππΏ β ππΏ = ππΉ ππΉ ππ’ . ππ’
End of material included into the final exam Following lectures are important, but not needed for the exam
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