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Translational Symmetry. Rotational Symmetry

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Translational Symmetry. Rotational Symmetry Physics and Material Science of Semiconductor Nanostructures PHYS 570P Prof. Oana Malis Email: [email protected] Course website: http://www.physics.purdue.edu/academic_programs/courses/phys570P/ Lecture Review of quantum mechanics, statistical physics, and solid state Band structure of materials Semiconductor band structure Semiconductor nanostructures Ref. Ihn Ch. 3, Yu&Cardona Ch. 2 The Bandstructure Problem • Begin with the one‐electron Hamiltonian: 2 H1e = (p) /(2mo) + V(r) p ‐iħ, V(r) Periodic Effective Potential • Solve the one‐electron Schrödinger Equation: H1eψn(r) = Enψn(r) In general, this is still a very complex, highly computational problem! The Bandstructure Problem (Many people’s careers over many years!) • Start with the 1 e‐ Hamiltonian: 2 H1e = ‐(iħ) /(2mo) + V(r) • Step 1: Determine the effective periodic potential V(r) • Step 2: Solve the one‐electron Schrödinger Equation: H1eψn(r) = Enψn(r) A complex, sophisticated, highly computational problem! • There are many schemes, methods, & theories to do this! • We’ll give a brief overview of only a few. • The 1 e‐ Hamiltonian is: 2 H1e = ‐(iħ) /(2mo) + V(r) NOTE Knowing the exact form of the effective periodic potential V(r) is itself a difficult problem! • However, we can go a long way towards understanding the physics behind the nature of bandstructures without specifying V(r). We can USE SYMMETRY. Group Theory A math tool which does this in detail. V(r) Periodic Crystal Potential. It has all of the symmetries of the crystal lattice. Translational symmetry. Rotational symmetry. Reflection symmetry • The most important symmetry is Translational Symmetry. • Using this considerably reduces the complexities of bandstructure calculations. We will illustrate bandstructure calculations with some model calculations first. Then, we will discuss real bandstructures. A One‐Dimensional Bandstructure Model: translational symmetry A One‐Dimensional Bandstructure Model ‐ 2 • 1 e Hamiltonian: H = (p) /(2mo) + V(x) p ‐iħ(∂/∂x), V(x) V(x + a) V(x) Effective Potential. Repeat Distance = a. • GOAL: Solve the Schrödinger Equation: Hψk(x) = Ekψk(x) k eigenvalue label • First it’s convenient to define The Translation Operator T. • For any function, f(x), T is defined by: T f(x) f(x + a) Translation Operator T. • For any f(x), T is defined by: T f(x) f(x + a) • For example, if we take f(x) = ψk(x) An eigenfunction solution to the Schrödinger Equation: T ψk(x) = ψk(x + a) (1) • Now, from translational symmetry: T ψk(x) λkψk(x) (2) λk An eigenvalue of T. • Using (1) & (2) together, it is fairly easy to show that: ika ikx λk e and ψk(x) e uk(x) with uk(x) uk(x + a) • For proof, see the texts by BW or YC, as well as Kittel’s Solid State Physics text & any number of other SS books • This shows that the translation operator applied to an eigenfunction of the Schrödinger Equation (or of the Hamiltonian H which includes a periodic potential) gives the result: ika Tψk(x) = e ψk(x) ψk(x) is also an eigenfunction of the translation operator T! • It also shows that the general form of ψk(x) must be: ikx ψk(x) = e uk(x) where uk(x) is a periodic function with the same period as the potential! That is uk(x) = uk(x+a) • In other words: For a periodic potential V(x), with period a, ψk(x) is a simultaneous eigenfunction of the translation operator T and the Hamiltonian H From the Commutator Theorem of QM, this is equivalent to [T,H] = 0. The commutator of T & H vanishes; they commute! They share a set of eigenfunctions. • The eigenfunction (the electron wavefunction!) always has the form: ikx ψk(x) = e uk(x) with uk(x) = uk(x+a) BLOCH’S THEOREM Bloch’s Theorem From translational symmetry • For a periodic potential V(x), the eigenfunctions of H (wavefunctions of e‐) always have the form: ikx ψk(x) = e uk(x) with uk(x) = uk(x+a) BLOCH FUNCTIONS • Recall, for a free e‐, the wavefunctions have the form: f ikx ψ k(x) = e (a plane wave) A Bloch function is the generalization of a plane wave for an e‐ in periodic potential. It is a plane wave modulated by a periodic function uk(x) (with the same period as V(x) ). Bandstructure A one dimensional model • The wavefunctions of the e‐ have in a periodic crystal must always have the Bloch function form ikx ψk(x) = e uk(x), uk(x) = uk(x+a) (1) It is conventional to label the eigenfunctions & eigenvalues (Ek) of H by the wavenumber k: p = ħk e‐ “quasi‐momentum” (rigorously, this is the e‐ momentum for a free e‐, not for a Bloch e‐) • So, the Schrödinger Equation is Hψk(x) = Ekψk(x), where ψk(x) is given by (1) and Ek The electronic “Bandstructure”. A One‐Dimensional Bandstructure Model: effect of periodicity Bandstructure: E versus k • The “Extended Zone scheme” A plot of Ek with no restriction on k ikx • But note! ψk(x) = e uk(x) & uk(x) = uk(x+a) Consider (for integer n): exp[i{k + (2πn/a)}a] exp[ika] The label k & the label [k + (2πn/a)] give the same ψk(x) (& the same energy)! In other words, the translational symmetry in the lattice Translational symmetry in “k space”! So, we can plot Ek vs. k & restrict k to the range ‐(π/a) < k < (π/a) “First Brillouin Zone” (BZ) (k outside this range gives redundant information!) Free e- “bandstructure” in the 1d extended zone scheme: 2 2 Ek = (ħ k )/(2mo) Bandstructure: E versus k • “Reduced Zone scheme” A plot of Ek with k restricted to the first BZ. • For this 1d model, this is ‐(π/a) < k < (π/a) k outside this range is redundant & gives no new information! • Illustration of the Extended & Reduced Zone schemes in 1d with the free electron energy: 2 2 Ek = (ħ k )/(2mo) • Note: These are not really bands! We superimpose the 1d lattice symmetry (period a) onto the free e‐ parabola. • The free e‐ “bandstructure” in the 1d reduced zone scheme: 2 2 Ek = (ħ k )/(2mo) For k outside the 1st BZ, take Ek & translate it into the 1st BZ by adding (2πn/a) to k Use the translational symmetry in k‐space as just discussed. (πn/a) “Reciprocal Lattice Vector” Simple bandstructure model: The Krönig‐Penney Model 19 Bandstructure To illustrate these concepts, we now discuss an EXACT solution to a 1d model calculation The Krönig‐Penney Model • This was developed in the 1930’s & is in MANY SS & QM books Why do this simple model? • The process of solving it shares contain MANY features of real, 3d bandstructure calculations. • Results of it contain MANY features of real, 3d bandstructure results! • The results are “easily” understood & the math can be done exactly A 21st Century Reason to do this simple model! It can be used as a prototype for the understanding of artificial semiconductor structures called SUPERLATTICES! Reminder: The 1d (finite) Rectangular Potential Well Reminder • We want to solve the Schrödinger Equation for: We want bound states: ε < Vo 2 2 2 [‐{ħ /(2mo)}(d /dx ) + V]ψ = εψ (ε E) V = 0, ‐(b/2) < x < (b/2); V = Vo otherwise Solve the Schrödinger Equation: 2 2 2 (½)b [‐{ħ /(2mo)}(d /dx ) + V]ψ = εψ -(½)b (ε E) V = 0, ‐(b/2) < x < (b/2) V = Vo otherwise V The bound states are in o Region II V= 0 In Region II: ψ(x) is oscillatory In Regions I & III: ψ(x) is exponentially decaying The 1d (finite) rectangular potential well A brief math summary! 2 2 2 2 Define: α (2moε)/(ħ ); β [2mo(ε‐Vo)]/(ħ ) The Schrödinger Equation becomes: (d2/dx2) ψ + α2ψ = 0, ‐(½)b < x < (½)b (d2/dx2) ψ ‐ β2ψ = 0, otherwise. Solutions: ψ = C exp(iαx) + D exp(‐iαx), ‐(½)b < x < (½)b ψ = A exp(βx), x < ‐(½)b ψ = A exp(‐βx), x > (½)b Boundary Conditions: ψ & dψ/dx are continuous SO: • Algebra (2 pages!) leads to: 2 2 (ε/Vo) = (ħ α )/(2moVo) ε, α, β are related to each other by transcendental equations. For example: tan(αb) = (2αβ)/(α 2‐β2) • Solve graphically or numerically. • Get: Discrete energy levels in the well (a finite number of finite well levels!) The Krönig‐Penney Model Repeat distance a = b + c. Periodic potential V(x) = V(x + na), n = integer Periodically repeated wells & barriers. Schrödinger Equation: 2 2 2 [‐{ħ /(2mo)}(d /dx ) + V(x)]ψ = εψ Periodic Potential Wells V(x) = Periodic potential (Kronig-Penney Model) The Wavefunction has the Bloch form: ikx ψk(x) = e uk(x); uk(x) = uk(x+a) Boundary conditions at x = 0, b: ψ, (dψ/dx) are continuous • Algebra: A MESS! But doable EXACTLY! Instead of an explicit form for the bandstructure εk or ε(k), we get: ‐1 k = k(ε) = (1/a) cos [L(ε/Vo)] OR L = L(ε/Vo) = cos(ka) WHERE L = L(ε/Vo) = L = L(ε/Vo) = cos(ka) ‐1< L< 1 ε in this range are the allowed energies (BANDS!) • But also, L(ε/Vo) = messy function with no limit on L ε in the range where |L| >1 These are regions of forbidden energies (GAPS!) (no solutions exist there for real k; math solutions exist, but k is imaginary) • The wavefunctions have the Bloch form for all k (& all L): ikx ψk(x) = e uk(x) For imaginary k, ψk(x) decays instead of propagating! Krönig‐Penney Results For particular a, b, c, Vo Each band has a finite well level “parent”. L(ε/V ) = cos(ka) Finite o Well ‐1< L< 1 Levels But also L(ε/Vo) = messy function with no limits For ε in the range ‐1 < L < 1 allowed energies (bands!) For ε in the range |L| > 1 forbidden energies (gaps!) (no solutions exist for real k) • Every band in the Krönig‐Penney model has a finite well discrete level as its “parent”! Evolution from the finite well to the periodic potential In its implementation, the Krönig‐Penney model is similar to the “almost free” e‐ approach, but the results are similar to the tightbinding approach! (As we’ll see).
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