Handout 3 Free Electron Gas in 2D and 1D Electron Gases in 2D

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Handout 3

Free Electron Gas in 2D and 1D

In this lecture you will learn:

• Free electron gas in two dimensions and in one dimension • Density of States in k-space and in energy in lower dimensions

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electron Gases in 2D

• In several physical systems electron are confined to move in just 2 dimensions STM micrograph • Examples, discussed in detail later in the course, are shown below:

Semiconductor Quantum Wells: Graphene:

InGaAs GaAs quantum well (1-10 nm)

GaAs

Semiconductor quantum wells can be composed of Graphene is a single atomic layer pretty much any of carbon atoms arranged in a semiconductor from the TEM honeycomb lattice groups II, III, IV, V, and VI of micrograph the periodic table

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

1 Electron Gases in 1D

• In several physical systems electron are confined to move in just 1 dimension

• Examples, discussed in detail later in the course, are shown below:

Semiconductor Quantum Semiconductor Quantum Carbon Nanotubes Wires (or Nanowires): Point Contacts (Rolled Graphene (Electrostatic Gating): Sheets):

GaAs metal metal

InGaAs InGaAs Nanowire Quantum well

GaAs GaAs

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electrons in 2D Metals: The Free Electron Model

The quantum state of an electron is described by the time-independent Schrodinger equation:

2   2 r  V r  r  E  r 2m

Consider a large metal sheet of area A= Lx Ly :

A  LxLy Use the Sommerfeld model: Ly

• The electrons inside the sheet are confined in a Lx two-dimensional infinite potential well with zero potential inside the sheet and infinite potential outside the sheet V r  0 for r inside the sheet V r   for r outside the sheet free electrons • The electron states inside the sheet are given (experience no by the Schrodinger equation potential when inside the sheet)

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

2 Born Von Karman Periodic Boundary Conditions in 2D 2 Solve:   2 r  E  r 2m

Use periodic boundary conditions: y A  LxLy These imply that each  x  L ,y,z   x, y,z x L x edge of the sheet is y L  x, y  Ly ,z   x, y,z folded and joined to x the opposite edge

   1 i k . r 1 i k x x ky y  Solution is:  r  e  e A A

The boundary conditions dictate that the allowed values of kx , and ky are such that:

i kxLx 2 e  1  kx  n n = 0, ±1, ±2, ±3,……. Lx

i ky Ly 2 e  1  ky  m m = 0, ±1, ±2, ±3,……. Ly

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Born Von Karman Periodic Boundary Conditions in 2D Labeling Scheme: All electron states and energies can be labeled by the corresponding k-vector  2 2  1 i k . r  k   r  e Ek   k A 2 m 2 2    Normalization: The wavefunction is properly normalized:  d r  k r  1

Orthogonality: Wavefunctions of two different states are orthogonal:   i k k'. r 2  *   2  e  d r   r   r   d r     k' k A k' , k Momentum Eigenstates: Another advantage of using the plane-wave energy eigenstates (as opposed to the “sine” energy eigenstates) is that the plane-wave states are also momentum eigenstates    Momentum operator: pˆ    pˆ   r     r  k   r i k i k  k Velocity:   k 1  Velocity of eigenstates is:    vk   k Ek m 

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

3 States in 2D k-Space ky 2 k-space Visualization: Lx The allowed quantum states states can be visualized as a 2D grid of points in the entire “k-space” 2 2 kx  n ky  m Lx Ly kx 2 n, m = 0, ±1, ±2, ±3, ……. Ly

Density of Grid Points in k-space:

Looking at the figure, in k-space there is only one grid point in every small area of size:  2  2  2 2         Lx  Ly  A

A  There are grid points per unit area of k-space Very important 2 2 result

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 2D at Zero Temperature - I • Suppose we have N electrons in the sheet.

• Then how do we start filling the allowed quantum states? N y A  LxLy • Suppose T~0K and we are interested in a filling scheme x Ly that gives the lowest total energy. Lx

The energy of a quantum state is: ky 2 2 2 2 2   kx  ky  k Ek    2 m 2 m Strategy: • Each grid-point can be occupied by two electrons (spin up and spin down)

• Start filling up the grid-points (with two electrons kx each) in circular regions of increasing radii until k you have a total of N electrons F

• When we are done, all filled (i.e. occupied) quantum states correspond to grid-points that are

inside a circular region of radius kF

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

4 The Electron Gas in 2D at Zero Temperature - II ky • Each grid-point can be occupied by two electrons (spin up and spin down)

kF • All filled quantum states correspond to grid-points that are inside a circular region of radius kF kx 2 Area of the circular region =  kF Fermi circle A 2 Number of grid-points in the circular region =  kF 2 2

Number of quantum states (including A 2 A 2 spin) in the circular region = 2   kF  kF 2 2 2 But the above must equal the total number N of electrons inside the box: A N  k 2 2 F 2 N kF Units of the electron  n  electron density   2 A 2 density n are #/cm 1 2  kF  2 n

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 2D at Zero Temperature - III k • All quantum states inside the Fermi circle are filled (i.e. y occupied by electrons) • All quantum states outside the Fermi circle are empty kF Fermi Momentum: kx The largest momentum of the electrons is: kF This is called the Fermi momentum Fermi momentum can be found if one knows the electron density: 1 Fermi circle 2 kF  2 n Fermi Energy: 2 2 The largest energy of the electrons is:  kF 2m 2k2 This is called the Fermi energy E : E   F F F 2m

2   n m Also: E  or n  E F m 2 F  

Fermi Velocity: k The largest velocity of the electrons is called the Fermi velocity v : v   F F F m

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

5 The Electron Gas in 2D at Non-Zero Temperature - I ky dkx A Recall that there are grid points per unit area of k- dk 2 y space 2

kx  So in area d k x d k y of k-space the number of grid points is: A A 2  dkx dky  d k 2 2 2 2

 The summation over all grid points in k-space can be replaced by an area integral  d 2k   A   2 all k 2 Therefore:   d 2k  N  2  f k  2  A f k    2  all k 2

 f k is the occupation probability of a quantum state

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 2D at Non-Zero Temperature - II   The probability f  k  that the quantum state of wavevector k is occupied by an electron is given by the Fermi-Dirac distribution function: 1 2 2 2 2 2    kx  ky  k f k   Where: Ek    1 eEk Ef KT 2 m 2 m Therefore:   d 2k  d 2k 1 N 2 A f k 2 A    2    2  2 2 1 eEk Ef KT Density of States: The k-space integral is cumbersome. We need to convert into a simpler form – an energy space integral – using the following steps: 2  2k2 2k d k  2 k dk and E    dE   dk 2 m m Therefore:  d 2k  k dk  m 2  A  2  A   A  2 dE 2 0  0  

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

6 The Electron Gas in 2D at Non-Zero Temperature - III  d 2k 1  1 N  2  A   A dE g E  2 E k E KT  2D E E KT 2 1 e f 0 1 e f

m Where: g2D E  2 Density of states function is constant   (independent of energy) in 2D 2 g2D(E) has units: # / Joule-cm ky The product g(E) dE represents the number of quantum states available in the energy interval between E and (E+dE) per cm2 of the metal

Suppose E corresponds to the inner circle from the relation: k 2k 2 x E   2 m And suppose (E+dE) corresponds to the outer

circle, then g2D(E) dE corresponds to twice the number of the grid points between the two circles

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 2D at Non-Zero Temperature - IV  1  N  A dE g E  A dE g E f E  E  2D  E E KT  2D  f 0 1 e f 0 g2D E  m Where: g2D E  2   f E  E The expression for N can be visualized as the  f  integration over the product of the two functions: Ef E Check: Suppose T=0K:  Ef N  A  dE g2D E f E  Ef  A  dE g2D E f E  Ef  0 0 1 m T = 0K  A 2 Ef   0 Ef E m  n  2 Ef   Compare with the previous result at T=0K: m n  E At T=0K (and only at T=0K) the Fermi level 2 F  E is the same as the Fermi energy E   f F

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

7 The Electron Gas in 2D at Non-Zero Temperature - V

For T ≠ 0K:

Since the carrier density is known, and does not change with temperature, the Fermi level at temperature T is found from the expression  Ef   1 m n  dE g E  K T log1 eK T   2D  E E KT 2 0 1 e f      

In general, the Fermi level Ef is a function of temperature and decreases from EF as the temperature increases. The exact relationship can be found by inverting the above equation and recalling that: m n  2 EF   to get:

 EF  KT Ef T  KT loge  1    

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Total Energy of the 2D Electron Gas

The total energy U of the electron gas can be written as:    d 2k   U  2   f k E k  2  A f k E k     2   all k 2  Convert the k-space integral to energy integral:U  A  dE g2D E f E  Ef E 0 U  The energy density u is:u    dE g2D E f E  Ef E A 0 Suppose T=0K: E F m 2 u   dE g2D E E  2 EF 0 2 

m Since: n  2 EF  

1 We have: u  n E 2 F

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

8 2D Electron Gas in an Applied Electric Field - I ky e   ky  E    E  Ex xˆ f k    e   kt    k  E 

kx kx

Electron distribution in k-space Electron distribution is shifted in when E-field is zero k-space when E-field is not zero    e    Distribution function: f k  Distribution function: f k  E    

Since the wavevector of each electron is shifted by the same amount in the presence of the E-field, the net effect in k-space is that the entire electron distribution is shifted as shown  E

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

2D Electron Gas in an Applied Electric Field - II e   ky Current density (units: A/cm)  E  2    d k   e     E J  2 e   2 f k  E  vk 2    Do a shift in the integration variable: kx 2   d k    e   J  2 e   2 f k v k  E  2    Electron distribution is shifted in   e   k-space when E-field is not zero 2   k  E   d k       e    J  2 e   2 f k Distribution function: f k  E  2 m      e2  d 2k    J  2   f k  E m 2  2  electron density = n (units: #/cm2)  n e2   J  E   E m Same as the Drude result - but n e2 Where:   units are different. Units of  are m Siemens in 2D

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

9 Electrons in 1D Metals: The Free Electron Model

The quantum state of an electron is described by the time-independent Schrodinger equation:

2 2    x  V x  x  E  x 2m x2 L Consider a large metal wire of length L :

Use the Sommerfeld model:

• The electrons inside the wire are confined in a one-dimensional infinite potential well with zero potential inside the wire and infinite potential outside the wire V x  0 for x inside the wire V x   for x outside the wire free electrons • The electron states inside the wire are given by (experience no the Schrodinger equation potential when inside the wire)

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Born Von Karman Periodic Boundary Conditions in 1D 2 2 Solve:    x  E  x 2m x2 Use periodic boundary conditions: L These imply that each  x  L, y,z   x, y,z facet of the sheet is folded and joined to the opposite facet

1 i k x Solution is:  x  e x L

The boundary conditions dictate that the allowed values of kx are such that:

i k L 2 e x  1  k  n n = 0, ±1, ±2, ±3,……. x L

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

10 States in 1D k-Space k-space Visualization: The allowed quantum states states can be visualized as a 1D grid of points in the entire 2 “k-space” L 2 kx  n L 0 kx

n = 0, ±1, ±2, ±3, …….

Density of Grid Points in k-space:

Looking at the figure, in k-space there is only one grid point in every small length of size:  2     L 

 There are L grid points per unit length of k-space Very important 2 result

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 1D at Zero Temperature - I • Each grid-point can be occupied by two electrons (spin up and spin down) k k • All filled quantum states correspond to grid-points that F F are within a distance kF from the origin k 0 x Length of the region = 2kF L Fermi points Number of grid-points in the region =  2k 2 F

Number of quantum states (including L 2  2k spin) in the region = 2 F But the above must equal the total number N of electrons in the wire: 2k N  L F  N 2k  n  electron density   F Units of the electron L  density n are #/cm  n  k  F 2

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

11 The Electron Gas in 1D at Zero Temperature - II • All quantum states between the Fermi points are filled (i.e. occupied by electrons) • All quantum states outside the Fermi points are empty k 0 x Fermi Momentum: The largest momentum of the electrons is: kF This is called the Fermi momentum Fermi points Fermi momentum can be found if one knows the electron density:  n k  F 2 Fermi Energy: 2 2 The largest energy of the electrons is:  kF 2m 2k2 This is called the Fermi energy E : E   F F F 2m

2 2 n2 Also: E   8m F 8 m or n  EF  

Fermi Velocity: k The largest velocity of the electrons is called the Fermi velocity v : v   F F F m

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 1D at Non-Zero Temperature - I

Recall that there are L grid points per unit length of k- space 2 dkx

kx  So in length dk x of k-space the number of 0 grid points is: L dk 2 x

 The summation over all grid points in k-space can be replaced by an integral  dk   L x   all k  2 Therefore:  dk N  2  f k  2 L x f k  x  x all k  2

f kx  is the occupation probability of a quantum state

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

12 The Electron Gas in 1D at Non-Zero Temperature - II

The probability f  k x  that the quantum state of wavevector k x is occupied by an electron is given by the Fermi-Dirac distribution function: 1 2 2 f k    kx x Ek E K T Where: Ek  1 e x f 2 m Therefore:  dk  dk 1 N  2 L x f k  2 L x  x  Ek E KT  2  2 1 e x f Density of States: The k-space integral is cumbersome. We need to convert into a simpler form – an energy space integral – using the following steps:

  2 2 2 dkx dk and  k  k 2 L   2 L  2  E   dE  dk  2 0 2 2 m m Therefore:  dk  2m 1 2 L  x  L  dE  2 0   E

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 1D at Non-Zero Temperature - III

 dk 1  1 N  2 L x  L dE g E  Ek E KT  1D  E E KT  2 1 e x f 0 1 e f 2m 1 Where: g1D E  Density of states function in 1D   E

g1D(E) has units: # / Joule-cm

The product g(E) dE represents the number of quantum states available in the energy interval between E and (E+dE) per cm of the metal

Suppose E corresponds to the inner points from the relation: 2k 2 E   2 m 0 kx And suppose (E+dE) corresponds to the outer

points, then g1D(E) dE corresponds to twice the number of the grid points between the points (adding contributions from both sides)

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

13 The Electron Gas in 1D at Non-Zero Temperature - IV  1  N  L dE g E  L dE g E f E  E  1D  E E KT  1D  f 0 1 e f 0 2m 1 g1D E  Where: g1D E    E f E  Ef  The expression for N can be visualized as the integration over the product of the two functions: Ef E

Check: Suppose T=0K:  Ef N  L  dE g1D E f E  Ef  L  dE g1D E f E  Ef  0 0 1 8m T = 0K  L Ef   0 Ef E 8m  n  Ef   Compare with the previous result at T=0K:

8m At T=0K (and only at T=0K) the Fermi level n  E F  E is the same as the Fermi energy E   f F

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Electron Gas in 1D at Non-Zero Temperature - V

For T ≠ 0K:

Since the carrier density is known, and does not change with temperature, the Fermi level at temperature T is found from the expression  1 n  dE g E  1D  E E KT 0 1 e f

In general, the Fermi level Ef is a function of temperature and decreases from EF as the temperature increases.

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

14 Total Energy of the 1D Electron Gas

The total energy U of the electron gas can be written as:  dk U  2  f k E k  2 L x f k E k  x x  x x all k  2  Convert the k-space integral to energy integral: U  L  dE g1D E f E  Ef E 0 U  The energy density u is:u    dE g1D E f E  Ef E L 0 Suppose T=0K:

EF 3 2 8m EF u   dE g1D E E  0   3

8m Since: n  EF  

1 We have: u  n E 3 F

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

1D Electron Gas in an Applied Electric Field - I e    E  E  E xˆ f kx  e   x kx t    kx  Ex 

kx kx

Electron distribution in k-space Electron distribution is shifted in when E-field is zero k-space when E-field is not zero Distribution function: f k   e   x Distribution function: f kx  Ex    

Since the wavevector of each electron is shifted by the same amount in the presence of the E-field, the net effect in k-space is that the entire electron distribution is shifted as shown  E

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

15 1D Electron Gas in an Applied Electric Field - II

Current (units: A) e   E  x  dkx  e   E  Ex xˆ I  2 e   f  kx  Ex  vkx  2    Do a shift in the integration variable: k  x dkx  e  I  2 e   f kx v kx  Ex  Electron distribution is shifted in 2     k-space when E-field is not zero  e   e    kx  Ex  Distribution function: f kx  Ex  dkx      I  2 e   f kx   2 m e2   dk  I  2   x f k E m  2 x  x    electron density = n (units: #/cm) n e2   I  E   E m Same as the Drude result - but n e2 Where:   units are different. Units of  are m Siemens-cm in 1D

ECE 407 – Spring 2009 – Farhan Rana – Cornell University