The Bandstructure ProblemA one-dimensional model (“easily generalized” to 3D!)

A One-Dimensional Bandstructure Model

• 1 e- Hamiltonian: H = (p)2/(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 defineThe 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)

Eigenfunction solution to the Schrödinger Equation: T ψk(x) = ψk(x + a) (1)

• Now, find the eigenvalues of T:T ψk(x) λkψk(x) (2) λk An eigenvalue of T.

• Using (1) & (2) together, it is fairly easy to show that: λk eika and ψk(x) eikx 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 (of the Hamiltonian H which includes a periodic potential) gives:

Tψk(x) = eika 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:

ψk(x) = eikx 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.• In other words: The eigenfunction (the electron

wavefunction!) always has the form:

ψk(x) = eikx uk(x) with uk(x) = uk(x+a)

“Bloch’s Theorem”

Bloch’s TheoremFrom translational symmetry

• For a periodic potential V(x), the eigenfunctions of H (wavefunctions of e-) always have the form:

ψk(x) = eikx uk(x) with uk(x) = uk(x+a)

“Bloch Functions”• Recall, for a free e-, the wavefunctions have the form:

ψfk(x) = eikx (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)).

BandstructureA one dimensional model

• The wavefunctions of the e- have in a periodic crystal

Must always have the Bloch function formψk(x) = eikxuk(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 isHψk(x) = Ekψk(x)

where ψk(x) is given by (1) and

Ek The Electronic “Bandstructure”.

Bandstructure: E versus k• The “Extended Zone scheme”

A plot of Ek with no restriction on k

• But note!ψk(x) = eikx 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!)

• “Reduced Zone sScheme” 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:

Ek = (ħ2k2)/(2mo)• Note: These are not really bands! We superimpose the

1d lattice symmetry (period a) onto the free e- parabola.

Bandstructure: E versus k

Free e- “bandstructure” in the 1d extended zone scheme: Ek = (ħ2k2)/(2mo)

• Free e- “bandstructure” in the 1d reduced zone scheme:

Ek = (ħ2k2)/(2mo)

•For k outside the 1st BZ,take Ek & translate it intothe 1st BZ by adding

(πn/a) to k•Use the translationalsymmetry in k-space asjust discussed.

(πn/a) “Reciprocal Lattice

Vector”

Bandstructure To illustrate these concepts, we now discuss an EXACTsolution to a 1d model calculation (BW Ch. 2, S, Ch. 2)

The Krönig-Penney Model• 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 Krönig-Penney Model• Developed in the 1930’s & is in MANY SS & QM books

Why do this simple model?• The results are “easily” understood & the math can be

done exactly• We won’t do this in class. It is in the books!

A 21st Century Reasonto do this simple model!

It can be used as a prototype for the understanding of artificial semiconductorstructures called SUPERLATTICES!

(Later in the course?)

QM Review: The 1d (finite) Rectangular Potential Well In most QM texts!!

[-{ħ2/(2mo)}(d2/dx2) + V]ψ = εψ (ε E)

V = 0, -(b/2) < x < (b/2); V = Vo otherwise

We want bound states: ε < Vo

• We want to solve the Schrödinger Equation for:

• Solve the Schrödinger Equation:

[-{ħ2/(2mo)}(d2/dx2) + V]ψ = εψ(ε E), V = 0, -(b/2) < x < (b/2)

V = Vo otherwise

The bound states arein Region II

• In Region II:ψ(x) is oscillatory

• In Regions I & III:ψ(x) is exponentially

decaying

-(½)b (½)b

Vo

V = 0

The 1d (finite) rectangular potential wellA brief math summary!

• Define: α2 (2moε)/(ħ2); β2 [2mo(ε - Vo)]/(ħ2)

• 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:(ε/Vo) = (ħ2α2)/(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!)

• Even eigenfunction solutions (a finite number):

Circle, ξ2 + η2 = ρ2, crosses η = ξ tan(ξ)

Vo

o

o

b

• Odd eigenfunction solutions:

Circle, ξ2 + η2 = ρ2, crosses η = -ξ cot(ξ)

|E2| < |E1|

b

b

o

oVo

The Krönig-Penney ModelRepeat distance a = b + c. Periodic potential V(x) = V(x + na), n = integer

Periodically repeated

wells & barriers.

Schrödinger Equation:

[-{ħ2/(2mo)}(d2/dx2)

+ V(x)]ψ = εψ

V(x) = Periodic potential

The Wavefunction has the Bloch form:

ψk(x) = eikx uk(x); uk(x) = uk(x+a)

Boundary conditions at x = 0, b:

ψ, (dψ/dx) are continuous

Periodic Potential Wells(Kronig-Penney Model)

• Algebra: A MESS! But doable EXACTLY!Instead of an explicit form for the bandstructure εk or ε(k), we get:

k = k(ε) = (1/a) cos-1[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):

ψk(x) = eikx 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/Vo) = cos(ka)

-1< L< 1

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)

Finite Well Levels

• Every band in the Krönig-Penney model has a finite well discrete level as its “parent”!

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.

Evolution from the finite well to the periodic potential

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, as we’ll see, to some simple tightbinding results.

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).

Lowest Krönig-Penney BandsIn the linear approximation

for L(ε/Vo) ε = (ħ2k2)/(2m0)