PH575 Spring Lecture #13 Free electron theory: Sutton Ch. 7 pp 132 -> 144; Kittel Ch. 6. 3/2 " # % & D( E) = V E 1/2. 2π 2.

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1 PH575 Spring 2014 Lecture #13 Free electron theory: Sutton Ch. 7 pp 132 -> 144; Kittel Ch. 6. E( k) = 2 k 2 D( E) = V 2π 2 " # $ 2 3/2 % & ' E 1/2

2 Assumption: electrons metal do not interact with each other, not with the potential due to the ions EXCEPT that the ion potential confines them to the region of space occupied by the material. We will see that this assumption leads to a spherical Fermi surface and explains the properties of simple metals. It is remarkable that the theory works as well as it does, and it improves the classical Drude model that is still used with success. Paul Drude ( )

3 Free electron bands & LCAO (tight binding) bands are qualitatively similar, with important exception of gaps. (Sutton, Fig. 7.6, Ge e.g.) So free electron model works well only for metals. FE bands (Na) LCAO band (Na)

4 Energy "Jellium" - free electrons a uniform positive BG 0 PE +PBC -V 0 0 vacuum Metal, L vacuum L Ĥ ψ k 2 # % $ = E k ψ k d 2 dx 2 + d 2 dy 2 + d 2 dz 2 & ( ' ψ k = E k ψ k ψ k 1 ( r ) = Similar to Bloch form V ei k r

5 2 " $ # d 2 dx 2 + d 2 dy 2 + d 2 dz 2 % ' & ψ ( k r ) = E k ψ k ( r ) ψ k r r = xˆx + yŷ + zẑ; r = r = x 2 + y 2 + z 2 k = k x ˆx + k y ŷ + k z ẑ; k = k = k x 2 + k y 2 + k z 2 ( ) = 1 V ei k r E( k ) =?; k x =? k y =? k z =? E( k) = 2 k 2 k x = n x 2π L ; k y = n y 2π L ; k z = n z 2π L E 0 k x k y

6 Constant energy surfaces are spheres in k space Each state occupies 8π 3 /L 3 in k-space When all N valence electrons have filled up k-states, we reach the Fermi Energy N 8π 3 V 1 2 4πk 3 F 3 E F = 2 k F 2 Is the factor of 2 from spin (spin => 2 electrons/state) in the right place? E( k) = 2 k 2 k x = n x 2π L ; k y = n y 2π L ; k z = n z 2π L E 0 k x k y

7 Fermi parameters (see Kittel Ch.6, Table I): " k F = 3π 2 n # $ # e/vol E F = 2 2 k F % & ' 1/3 k F = cm -1 : E F = 3.2 ev: kinetic energy of Fermi level e T F = E F k B T F = 37,500 K: temp. above which e obey classical statistics (can ignore quantum effects) v F = k F m v F = cm s -1 : 1% c Sodium: n = e - /cm 3

8 Density of states (number of states per unit energy at a particular energy): We discussed this in a previous lecture (but then without spin). S D ( E) = 4πk3 3 ( E) = ds de = V 2π 2 2V 8π = V 3 3π 2! # " 2 $ & %! # " 3/2 2 $ & % E 1/2 3/2 E 3/2 Important result: density of states for free electrons varies as E 1/2 ; also electrons near VBM and CBM in semiconductors

9 3-D Density of states for free electron: S counts the states. E = 2 k 2 ; de dk = 2 k m E 0 E 0 + de k y D( E) = ds de k x ds = D(E)dE ds = D (k) annular vol of k-space = 2V ( 2π ) 4πk 2 dk 3 D D D ( E)dE = 2V 2π ( E) = 8πk 2 V 2π ( E) = V 2π 2 ( ) 3 4πk 2 dk de ( ) 3 dk " # $ 2 3/2 % & ' E 1/2

10 Example: Find total energy at T = 0 (homework) Will use this later for electronic specific heat. E Tot = E F 0 E D( E)dE E 0 E 0 + de Total energy at T 0 k x E Tot = 0 0 = E E f ( E)D( E)dE 1 ( )/k T B +1 D( E)dE e E E F k y

11 Free electron bands: square lattice E E = 2 k + G 2 k x (k y = 0) Real space a 1 = a 1,0 ( ) a 2 = a( 0,1) Reciprocal g 1 = 2π a 0,1 ( ) g 2 = 2π ( a 1,0 ) G = 2π ( a 0,0) G = 2π a ±1,0 ( ) G = 2π a 0,1 ( ) G = 2π a ±1,±1 G = 2π a 0,2 ( ) ( ) G = 2π ( a ±2,0)

12 Free electron bands: square lattice, reduced zone scheme E E = 2 k + G 2 k x (k y =0)

13 Free electron bands: bcc lattice E = 2 k + G 2

14 Free electron bands: fcc lattice E = 2 k + G 2 From Hummel, Electronic Properties of Materials

15 Free electron bands: fcc lattice Example Copper E = 2 k + G 2 <- Calculated with pseudopotential methods. Note d-bands! <- Free electron bands From Tsymbal (UNL)

16 Free electron bands & LCAO (tight binding) bands are qualitatively similar, with important exception of gaps. (Sutton, Fig. 7.6, Ge e.g.) So free electron model works well only for metals. FE bands (Na - bcc) LCAO band (Na)

17 Free electron bands & LCAO (tight binding) bands are qualitatively similar, with important exception of gaps. (Sutton, Fig. 7.6, Ge e.g.) So free electron model works well only for metals. FE bands (Na - bcc) LCAO band (Na)

18 There is a gap in the energy spectrum! This same result can be obtained by treating the periodic ion potential as a perturbation of the smooth potential (we won't do this, but it's discussed in Kittel, for example)

19 Free electrons (plane waves) don't interact with the lattice much until the wave vector becomes comparable with 1/a, then they are Bragg reflected as we discussed before (nearly free) and we have interference between a plane wave and its oppositely directed counterpart. These superpositions are standing waves with the same kinetic energy, but total energy is different e ikx e ikx sin( kx) e ikx + e ikx cos( kx)

20 Lower energy state piles up electron density on the atoms (stronger Coulomb attraction to cores), while the higher energy state piles up electron density between the atoms. True plane waves have constant electron density at all positions (this energy is intermediate). Similar to the LCAO approach in this limit. Thus there is a gap in the energy spectrum! e ikx e ikx sin( kx) e ikx + e ikx cos( kx)

21 Fermi Function Probability that a state of energy E is occupied at temperature T. f(µ) = 0.5 Varies sharply over width of k B T about E F. Step function T = 0. f ( E) = 1 E µ k e T B + 1 µ = 1eV T = ev =?K T = 0.3 ev =?K f E k B T k B T k B T E ev 0.0 E

22 Electronic specific heat Electrons are a repository of energy in metals and contribute to the specific heat. Dominant contribution at low T where phonons are frozen out. If electrons are a "classical gas" (no QM), equipartition theorem => electronic energy is E elec = (3/2)k B T. Electronic specific heat is C elec = de elec /dt= (3/2)k B Actual values in metals at low temperatures are less than 1% of this, and are NOT temperature independent.

23 Electronic specific heat Nearly free electrons are a repository of energy in metals and contribute to the specific heat. At low temperatures, when phonons (lattice vibrations) are frozen out, this contribution dominates. E Tot = 0 0 = E D( E) = V 2π 2 E f ( E)D( E)dE 1 ( )/kbt +1 D( E)dE e E µ " # $ 2 3/2 % & ' E 1/2 many functions df ( E,T ) dt k B T E ev

24 C v = de Tot dt 0 ( ) df E,T = E dt D( E)dE Area under black curve is electronic specific heat. Very slight asymmetry is crucial! Easy enough to calculate numerically, but let's look for analytical insight. What is dominant contribution? Weakest? many functions E ev E F = 1 ev k B T = 0.1 ev

25 C v = de Tot dt 0 ( ) df E,T = E dt D( E)dE D(E) varies slowly - remove from integral (but what value? Why can lower limit of integration be extended to -? df ( E,T ) dt =? Change variables: x = E µ 2k B T Helpful: x 2 dx = π 2 - cosh2 x 6

26 C v = de Tot dt 0 ( ) df E,T = E dt D( E)dE C v = π 2 2 D ( E F )k 2 B T Interpretation: Depends linearly on temperature - NOT independent Size depends only on density of states at Fermi surface - small for semimetals; large for so-called heavy fermions Also interpreted as each electron near Fermi surface contributing roughly k B to heat capacity: C v = D( E F )k B T k B

27 C v = de Tot dt 0 ( ) df E,T = E dt D( E)dE C v = π 2 2 D ( E F )k 2 B T C v,fe = π 2 2 Nk B T T F Much smaller than expected from classical theory: C v,class = 3 2 Nk B

28 C v,tot = electronic phonon γ T + AT 3 C v,tot C v,tot T Temperature Slope A γ expt (mj/mol/k 2 ) γ fe (mj/mol/k 2 ) Na K Cu Zn Ag Intercept γ T 2