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). Each band is associated with an “atomic” level from the well. More on Krönig‐Penney Solutions

• L(ε/Vo) = cos(ka)  BANDS &GAPS! • The Gap Size: Depends on the c/b ratio • Within a band (see previous Figure) a good approximation is that L ~ a linear function of ε. Use this to simplify the results:

• For (say) the lowest band, let ε  ε1 (L = ‐1) & ε  ε2 (L = 1) use the linear approximation for L(ε/Vo). Invert this & get:

ε‐(k) = (½) (ε2+ ε1) ‐ (½)(ε2 ‐ε1)cos(ka) For next lowest band,

ε+(k) = (½) (ε4+ ε3) + (½)(ε4 – ε3)cos(ka) In this approximation, all bands are cosine functions!!! This is identical, to some simple tightbinding results. Lowest Krönig‐Penney Bands In the linear approximation ε = (ħ2k2)/(2m ) for L(ε/Vo) 0 All bands are cos(ka) functions! Plotted in the extended zone scheme. Discontinuities at the BZ edges, at k = (nπ/a) Because of the periodicity of ε(k), the reduced zone scheme (red) gives the same information as the extended zone scheme (as is true in general). Effective mass

32 Brief, General Bandstructure Discussion (1d, but easily generalized to 3d) Relate bandstructure to classical electronic transport

Given an energy band ε(k) (a Schrödinger Equation eigenvalue): The Electron is a Quantum Mechanical Wave • From Quantum Mechanics, the energy ε(k) & the frequency ω(k) are related by: ε(k)  ħω(k) (1) • Now, from Classical Wave Theory, the wave group velocity v(k) is defined as: v(k)  [dω(k)/dk] (2) • Combining (1) & (2) gives: ħv(k)  [dε(k)/dk] • The QM wave (quasi‐) momentum is: p  ħk • Now, a simple “Quasi‐Classical” Transport Treatment! – “Mixing up” classical & quantum concepts! • Assume that the QM electron responds to an EXTERNAL force, F CLASSICALLY (as a particle). That is, assume that Newton’s 2nd Law is valid: F = (dp/dt) (1) • Combine this with the QM momentum p = ħk & get: F = ħ(dk/dt) (2) Combine (1) with the classical momentum p = mv: F = m(dv/dt) (3) Equate (2) & (3) & also for v in (3) insert the QM group velocity: v(k) = ħ‐1[dε(k)/dk] (4) • So, this “Quasi‐classical” treatment gives F = ħ(dk/dt) = m(d/dt)[v(k)] = m(d/dt)[ħ‐1dε(k)/dk] (5) or, using the chain rule of differentiation: ħ(dk/dt) = mħ‐1(dk/dt)(d2ε(k)/dk2) (6) Note!! (6) can only be true if the e‐ mass m is given by 2 2 2 m  ħ /[d ε(k)/dk ] (& NOT mo!) (7) m  EFFECTIVE MASS of e‐ in the band ε(k) at wavevector k.

Notation: m = m* = me • The Bottom Line is: Under the influence of an external force F The e‐ responds Classically (According to Newton’s 2nd Law)

BUT with a Quantum Mechanical Mass m*, not mo! • m  The EFFECTIVE MASS of the e‐ in band ε(k) at wavevector k m  ħ2/[d2ε(k)/dk2] • Mathematically, m  [curvature of ε(k)]‐1 • This is for 1d. It is easily shown that:

m  [curvature of ε(k)]‐1 also holds in 3d!! In that case, the 2nd derivative is taken along specific directions in 3d k space & the effective mass is actually a 2nd rank tensor. m  [curvature of ε(k)]‐1  Obviously, we can have m > 0 (positive curvature) or m < 0 (negative curvature) • Consider the case of negative curvature:  m < 0 for electrons For transport & other properties, the charge to mass ratio (q/m) often enters.  For bands with negative curvature, we can either

1. Treat electrons (q = ‐e) with me < 0

Or 2. Treat holes (q = +e) with mh > 0 Consider again the Krönig‐Penney Model

In the Linear Approximation for L(ε/Vo). The lowest 2 bands are:

           Positive Negative m me e • The linear approximation for L(ε/Vo) does not give accurate effective masses at the BZ edge, k =  (π/a).

 For k near this value, we must use the exact L(ε/Vo) expression. • It can be shown that, in limit of small barriers

(|Vo| << ε), the exact expression for the Krönig‐Penney effective 2 2 2 ‐1 mass at the BZ edge is: m = moεG[2(ħ π )/(moa )  εG] with: mo = free electron mass, εG = band gap at the BZ edge. +  “conduction band” (positive curvature) like:

‐  “valence band” (negative curvature) like: For Real Materials, 3d Bands The Krönig‐Penney model results (near the BZ edge): 2 2 2 ‐1 m = moεG[2(ħ π )/(moa )  εG] This is obviously too simple for real bands!

• A careful study of this table, finds, for real materials, m  εG also! NOTE: In general (m/mo) << 1 More Bandstructure Discussion

“Almost free” electron Model Bandstructure Problem One‐dimensional, “almost free” electron model (easily generalized to 3D!) (BW, Ch. 2 & Kittel’s book, Ch. 7) • “Almost free” electron approach to bandstructure. ‐ 2 1 e Hamiltonian: H = (p) /(2mo) + V(x); p  ‐iħ(d/dx) V(x)  V(x + a) = Effective potential, period a (lattice repeat distance)

GOAL • Solve the Schrödinger Equation: Hψ(x) = εψ(x) Periodic potential V(x)  ψ(x) must have the Bloch form: ikx ψ k(x) = e uk(x), with uk(x) = uk(x + a) • Lets look at in more detail at k near (but not at!) the BZ boundary to get the k dependence of ε near the BZ boundary: Messy! Student exercise (see Kittel) to show that the Free Electron Parabola SPLITS into 2 bands, with a gap between: 2 2 2 ε(k) = (ħ π )/(2a mo)  V 2 2 2 2 2 + ħ [k‐ (π/a) ]/(2mo)[1  (ħ π )/(a moV)] 2 This also assumes that |V| >> ħ (π/a)[k‐ (π/a)]/mo. For the more general, complicated solution, see Kittel! Almost Free e- Bandstructure: (Results, from Kittel for the lowest two bands)

2 2 ε = (ħ k )/(2mo)

V V