The Bandstructure Problem A One-dimensional model (

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The Bandstructure ProblemA One-dimensional model
(easily generalized to 3D!)
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Bandstructure Problem A One Dimensional Model

• One e- Hamiltonian H (p)2/(2mo) V(x)
• p ? -ih(?/?x), V(x) ? V(x a), V(x) ?
  Effective Potential.
• V has translational symmetry with repeat distance
  a.
• GOAL Solve the Schrödinger Equation
• H?k(x) Ek?k(x), k ? Eigenvalue Label
• Ek Electronic Energy of the e- in state k
• ?k(x) Wavefunction of the e- in state k
• Define a Translation operator ? T.
• T is defined, for any function f(x), as
• T f(x) ? f(x a)

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• Now consider the Translation Operator T for this
  one dimensional solid T f(x) ? f(x a)
• Take the special case for which f(x) ?k(x)
• (That is for f(x) an eigenfunction solution of
  the Schrödinger Equation)
• The definition of T is
• T ?k(x) ?k(x a) (1)
• Now, look for the eigenvalues of T
• T ?k(x) ? ?k?k(x) (2)
• ?k ? Eigenvalue of T. It can be shown using (1)
  (2) that
• ?k ? eika and ?k(x) ? eikx uk(x)
• with uk(x) ? uk(x a)
• (see Kittels book for proof)

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• This Shows The translation operator applied to
  an eigenfunction of the Schrödinger Equation (or
  of the Hamiltonian H, with a periodic potential)
  gives
• T?k(x) eika ?k(x)
• ? ?k(x) is also an eigenfunction of the
• translation operator T!
• This also shows that the general form of ?k(x) is
• ?k(x) eikx uk(x)
• where uk(x) uk(xa)
• uk(x) a periodic function with the same
• period as the potential!

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• In other words For a periodic potential V(x),
  with period a, ?k(x) is a simultaneous
  eigenfunction of the translation operator T and
  the Hamiltonian H.
• The Commutator Theorem of QM tells us that 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 (electron wavefunction!) has
  the form
• ?k(x) eikx uk(x) with uk(x) uk(xa)
• ? Blochs Theorem

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Blochs TheoremFrom translational symmetry

• For a periodic potential V(x), the eigenfunctions
  of H (wavefunctions of the e-) have the form
• ?k(x) eikx uk(x) with uk(x) uk(xa)
• ? Bloch Functions
• Recall, for a free e-, the wavefunctions have the
  form
• ?fk(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)).

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Bandstructure A one dimensional model

• So, the wavefunctions of the e- in a perfect,
  periodic crystal MUST have the Bloch Function
  form
• ?k(x) eikxuk(x), uk(x) uk(x a) (1)
• This is easily generalized to proven in 3 D!!
  Label
  the eigenfunctions eigenvalues (Ek) by the
  wavenumber k
• p hk ? the e- Quasi-Momentum or Crystal
  Momentum.
• Note! p hk is the electron momentum for FREE
  e-s ONLY!
• Free e- wavefunctions are plane waves ?fk(x)
  eikx, which are also eigenfunctions of the
  momentum operator p ? -ih(?/?x) with eigenvalue
  hk.
• However, the wavefunctions for e- s in bands are
  Bloch Functions ?k(x)
  eikxuk(x). Bloch Functions are NOT eigenfunctions
  of the momentum operator.
• The e- momentum for a Bloch Electron state ?k(x)
  is found by calculating the QM expectation value
  of the momentum operator in that state
• ltpgt lt?k(x)p?k(x)gt integral of (?k(x)) p
  ?k(x) over all x

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• The Schrödinger Equation for an electron in a
  periodic potential is
• H?k(x) Ek?k(x)
• where ?k(x) must have the Bloch Function form
• ?k(x) eikxuk(x), uk(x) uk(x a)
• Ek ? The Electronic Bandstructure.
• One way to plot Ek is in
• The Extended Zone Scheme
• ? A plot of Ekwith no restriction on k

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• Ek ? The Electronic Bandstructure
• The wavefunctions ?k(x) must be Bloch Functions
• ?k(x) eikxuk(x), uk(x) uk(x a) (1)
• Another way to plot Ek is to first consider the
  Bloch Function in (1) look at the identity
• expik (2pn/a)a ? expika (integer n)
• ? The label k the label k (2pn/a) give the
  same ?k(x) ( the same
  energy)!
• In other words, Translational symmetry in the
  lattice
• ? Translational symmetry in the reciprocal
  Lattice k space!
• So, we can plot Ek vs. k restrict k to the
  range
• -(p/a) lt k lt (p/a) ? First Brillouin Zone
  (BZ)
• (k outside this range gives redundant
  information!)
• ? The Reduced Zone Scheme

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Bandstructure E versus k

• Example Illustration
• The Extended Reduced Zone schemes in 1d with
  the free electron energy
• Ek (h2k2)/(2mo)
• Note Obviously, for free e-s there are no
  bands! In what follows, the 1d lattice symmetry
  (with period a) is imposed onto the free e-
  parabola.

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Free e- bandstructure in the 1d extended zone
scheme Ek (h2k2)/(2mo)
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• Free e- bandstructure in the 1d reduced zone
  scheme
• Ek (h2k2)/(2mo)
• For k outside of the 1st BZ,
• take Ek translate it into
• the 1st BZ by adding
• ?(pn/a) to k
• Use the translational symmetry
• in k-space just discussed.
• ?(pn/a) ?
• Reciprocal Lattice Vector

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BandstructureNow, illustrate these concepts with
an EXACT 1d model calculation (Kittel Ch. 7)

• The Krönig-Penney Model
• Developed in the 1930s.
• In MANY Solid State Physics Quantum Mechanics
  books.
• Why do this simple model?
• Its solution contains MANY features of real, 3d
  bandstructures!
• The results are easily understood.
• The math can be done exactly.
• We wont do this in class. It is in many books,
  including Kittel!
• 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!

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

• We want to solve the Schrödinger Equation for
• -h2/(2mo)(d2/dx2) V? e? (e ? E)
• V 0, -(b/2) lt x lt (b/2) V Vo otherwise

We want bound states e lt Vo
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• Solve the Schrödinger Equation
• -h2/(2mo)(d2/dx2) V? e?
• (e ? E) V 0, -(b/2) lt x lt (b/2)
• V Vo otherwise
• Bound States are
• in Region II
• Region II
• ?(x) is oscillatory
• Regions I III
• ?(x) is decaying

(½)b
-(½)b
Vo
V 0
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The 1d (finite) Rectangular Potential WellA
brief math summary!

• Define a2 ? (2moe)/(h2) ß2 ? 2mo(e -
  Vo)/(h2)
• The Schrödinger Equation becomes
• (d2/dx2) ? a2? 0, -(½)b lt x lt (½)b
• (d2/dx2) ? - ß2? 0, otherwise.
• Solutions
• ? C exp(iax) D exp(-iax),
  -(½)b lt x lt (½)b
• ? A exp(ßx), x lt -(½)b
• ? A exp(-ßx), x gt (½)b
• Boundary Conditions
• ? ? d?/dx are continuous. So

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• Algebra (2 pages!) leads to
• (e/Vo) (h2a2)/(2moVo)
• e, a, ß are related to each other by
  transcendental equations.
• For example
• tan(ab) (2aß)/(a 2- ß2)
• Solve graphically or numerically.
• Get Discrete energy levels in the well (a finite
  number of finite well levels!)

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• Even eigenfunction solutions (a finite number)
• Circle, ?2 ?2 ?2, crosses ? ? tan(?)

Vo
o
o
b
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• Odd eigenfunction solutions
• Circle, ?2 ?2 ?2, crosses ? -? cot(?)
• E2 lt E1

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

• Periodically repeated
• wells barriers.
• Schrödinger Equation
• -h2/(2mo)(d2/dx2)
• V(x)? e?
• V(x) Periodic potential
• ? The Wavefunctions must have the Bloch Form
• ?k(x) eikx uk(x) uk(x) uk(xa)
• Boundary conditions at x 0, b
• ?, (d?/dx) are continuous ?

Periodic Potential Wells (Krönig-Penney Model)
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• Algebra Calculus give A MESS!
• But doable EXACTLY!
• Instead of an explicit form for the bandstructure
  ek or e(k), we get
• k k(e) (1/a) cos-1L(e/Vo)
• OR
• L L(e/Vo) cos(ka)
• WHERE L L(e/Vo)

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• L L(e/Vo) cos(ka) ? -1lt Llt 1
• The e in this range are the allowed energies
• (The Allowed BANDS! )
• But also, L(e/Vo) a messy function with no
  limit on L
• The ks in the range where L gt1 are imaginary.
• ? These are regions of forbidden energy.
• (The Forbidden 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)
• ?k(x) eikx uk(x)
• ? For imaginary k, ?k(x) decays instead of
  propagating!

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Krönig-Penney Results For particular a, b, c, Vo

• Each band has a finite
• well level parent.
• L(e/Vo) cos(ka)
• ? -1lt Llt 1
• But also L(e/Vo)
• a messy function
• with no limits.
• For e in the range
• -1 lt L lt 1 ? Allowed Energies (Bands!)
• For e in the range
• L gt 1 ? Forbidden Energies (Gaps!)
• (no solutions exist for real k)

? Finite Well Levels
?
?
?
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• 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 well see). Each band is associated
  with an atomic level from the well.

Evolution from the finite well to the periodic
potential
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More on Krönig-Penney Solutions

• L(e/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
  e. Use this to simplify the results
• For (say) the lowest band, let e ? e1 (L -1)
  e ? e2 (L 1) use the linear approximation for
  L(e/Vo). Invert this get
• e-(k) (½) (e2 e1) - (½)(e2 - e1)cos(ka)
• For the next lowest band,
• e(k) (½) (e4 e3) (½)(e4 e3)cos(ka)
• In this approximation, all bands are cosine
  functions!!!
• This is identical, as well see, to some simple
  tightbinding results.

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The Lowest Krönig-Penney Bands
In the linear approximation for L(e/Vo)
e (h2k2)/(2m0)

• All bands are cos(ka)
• functions! Plotted in the
• extended zone scheme.
• Discontinuities at the BZ
• edges, at k ?(np/a)
• Because of the periodicity
• of e(k), the reduced zone scheme
• (red) gives the same information
• as the extended zone scheme (as is true in
  general).