# Lecture 02

March 22, 2019 | Author: bgiangre8372 | Category: Band Gap, Electronvolt, Electrical Resistivity And Conductivity, Neutron, Electron

#### Short Description

Identical particles - 2...

#### Description

Lecture Lecture 2: Identica Identicall particles, applications (9/22/2005) Today, oday, we will ﬁrst ﬁnish the material material from the previous previous lecture lecture notes; it starts by exchange forces and continues with atomic physics. Today we look at two very primitive toy models of solid state physics, textbook applications of the Pauli exclusion principle:

•  electron gas theory of Sommerfeld •  Bloch’s theory. Recomme Recommended nded reading: reading: Griﬃths Griﬃths ch. 5.3.

Electron gas theory First, consider one electron in a box whose sides are lx , ly , lz . In term termss of a potential energy, the potential is V  = 0 for 0

≤x≤l , x

0

0

≤ y ≤ l

y

≤ z  ≤  ≤ l

z

and V  =  elsew  elsewher here. e. This This special special form of the potentia potentiall may may be writt written en as V  as  V x (x) + V y (y ) + V z (z ) where the terms only depend on x on  x,, y,z  respectively.   respectively. Consequently, the whole Hamiltonian splits in a similar way to three terms

∞

+  H y  + H   +  H z H  =  H x + H  that commute with each other: [H  [ H x , H y ] = 0 and so forth. forth. That’s That’s a typic typical al situation in which it is useful to split the original problem to three simpler problems. We will split the problem if we factorize the wavefunction  (y )Z (z ). ψ(x,y,z ) =  X (  X (x)Y  (y Unless we allow the energies to become inﬁnite, ψ (x , y , z)  must vanish outside the box. It’s because the average average potential potential energy  V  (  (r) ψ 2 d3 r  would otherw otherwise ise be inﬁnit inﬁnite. e. Once Once we assume assume that ψ(x,y,z ) vanishes outside the box, we may simply set V  set  V  (x  (x) = 0 (which is true inside, where  ψ  is nonzero). The time-independent Schr¨odinger odinger equation is then



||

¯ 2 d2X  h  d 2 Y   d 2Z   +  + 2 X Y   = E  =  E XY Z   Y Z  +  X Z  + 2m dx2 dy 2 dz

1

The partial derivatives could have been replaced by ordinary derivatives. Let’s divide the equation by XY Z  – a trick that we always can do (in our mind, once we understand it) whenever we factorize the wavefunction. We obtain ¯ 2  1 d2X  1 d2Y  1 d2 Z  h + +  = E  2m X  dx2 Y  dy 2 Z  dz 2

Because the right hand side does not depend on  x, y , z ,  the left hand side cannot either. It follows that each of three terms on the left hand side must be constant. Let’s call them  E x , E y , E z , for example E x  =

¯ 2 1 d2X  h 2m X  dx2

which is also one term adding to the total energy eigenvalue: E  = E x  + E y  + E z If you multiply the long equation for E x   above by X (x) again, you get a standard eigenstate equation:

¯ 2 d2X (x) h = E x X (x) 2m dx2

It can be solved easily in terms of sines: X (x) =

 2

lx

sin(kx x)

The square root normalization constant is chosen to have X  2dx = 1. We could not add the cosine because it does not vanish at x   = 0 which would either cause an inﬁnite potential energy from the box, if  X (x) were continuous, or inﬁnite kinetic energy, if  X (x) jumped to zero for x < 0 discontinuously. We must also require that X (lx ) = 0 which gives us the following quantization rule:

  |

kx  =

nx π , lx

|

nx  = 1, 2, 3, . . .

The negative values of  nx  give us the same functions, up to an overall normalization (minus sign) while nx  = 0 gives us X   = 0 and does not add any 2

new normalizable states to the Hilbert space. You can look at the deﬁnition of the operator E x  again and substitute our X (x) to see that the energy is ¯ 2 n2x π2 h E x  = . 2mlx2 Similar results hold for E y and E z  and the total energy may be written as ¯ 2k2 h E  = , 2m

  k = (kx , ky , kz ) =

 n π  n π  n π  x

lx

,

y

ly

,

z

lz

.

The energy eigenstates are parameterized by (nx , ny , nz ): imagine points in a lattice in an octant of a 3D space (draw a picture). Also, you can associate a small box of size π/l x π/l y π/lz

×

×

with each vertex in a one-to-one map so that the vertex will be the lower left front vertex of the box, for example. There is the same number of small boxes and vertices also because each box has 8 vertices, but each vertex is shared by 8 boxes. You can see that the volume of each state in the  k-space is simply π3 π3 = . lx ly lz V  Let’s now admit that these are real electrons with spin; there can be 0,1 or 2 electrons in each state with an allowed value of    k. You can’t squeeze them more than that. The electrons want to minimize the energy which depends on the distance from the origin in the    k-space. Therefore, they will ﬁll the octant of a ball whose radius is k F  where the equality of the volumes requires 1 8

×

4π 3 Nq   π 3 k = . 3 F  2 V

×

The left hand side is the volume of 1/8 of a ball. The right hand side is the total number of the   k-states: N   is the total number of atoms,  q  is the number of electrons per atom, and the denominator 2 means that we can squeeze 2 electrons to one   k-state because of the spin. In terms of the “electron density” ρ = Nq/V  we can write the maximum momentum as kF  = (3ρπ2 )1/3 3

and the corresponding energy is called the Fermi energy: ¯h2k 2 ¯2 h = (3ρπ 2)2/3 . E  = 2m 2m A funny fact is that this maximal occupied energy may be calculated for various materials:

Copper Its normal density is ρm  = 8.96 g/cm3; A = 63.5 g/mole. What’s q ? Copper’s atomic structure is (1s)2 (2s)2(2 p)6 (3s)2(3 p)6 (3d)10 (4s) One electron in the last shell (4s) means q  = 1. The electron density ρ = Nq/V  = N Avogadroρm /A 1 (recall N Avogadro = 6.023 atoms per mole) is then equal to 8.5 1022 electrons per cubed centimeter (check the numbers with a calculator and the formulae by dimensional analysis). The relation between E F  and ρ  allows us to calculate E F  and the result is

×

×

E F  = 1.13

× 10

−18

Joules = 7.04eV.

For comparison, a red photon around 650 nm of wavelength, has about E  = 0.3 eV. Recall that 1 eV is about 1.602 10−19 Joules. Electronvolts may be too obscure. It’s more interesting to reinterpret the result in terms of temperatures. Deﬁne the  Fermi temperature  to be simply T F  = E F /kB

×

Note that it is natural; the energy per degree of freedom is kB T /2, for example. For copper, you obtain T F  = 81, 700 kelvins, well above the melting point at 1356 kelvins.

Nuclear Fermi motion In the previous example we considered the box but because the volume cancelled, it’s clear that it works for any volume. For example a simple model of nuclei involves a ball with the inﬁnite potential wall. Inside, we have a proton gas and an independent neutron gas; a pretty good approximation for various questions. 4

Let’s use the following formula for the volume of the nuclei, obtained from electron scattering experiments:

≈ 4π3 r A, 2 0

r0  = 1.2

× 10

−15

meters.

In the gas model, one half of the nucleons are protons, one half are neutrons. The proton density is therefore 3 A/2 = . 4π/3 r03 A 8πr03

ρ =

×

For the Fermi momentum pF , for example, it means that  pF

≡

3 2 ¯hkF  = h ¯ 3 π 8πr03

1/3

 9  = π 8

1/3

¯ h . r0

With numbers, we obtain 1/3

pF

  9 = π 8

6.58 10−22 MeV s 3 1.2 10−15 meters

× ×

8

× 10 m/s = 251 MeV/c. c

We added 1 = (c/c) to get the resulting momentum in MeV/c   which is a pretty good unit in particle physics (where we often use units where one sets c = 1 anyway).

Back to a more general story of a fermion gas Note that most of the fermions are near the Fermi momentum because their number density increases with the momentum: V k2 N (k)dk = 2 dk π because k 2 comes from the surface of a spherical shell. Draw a parabola. We may also calculate the total energy of the nucleons: E tot

2

2

2

2

 h¯ k  V k h¯ V      = dk = kF

2m

0

But 2

kF  =

1/3

 3π Nq  V

2π2 m

π2

kF

0

5 ¯ 2V kF  h k dk = . 10π2m 4

¯ 2(3π2 Nq )5/3  − 2/3 h E tot  = V  . 10π 2m 5

Note the power law E  V  − 2/3 lowers the energy for increasing volume. The system thus tries to increase the volume which means that the fermionic nature acts as a pressure called “degeneracy pressure” (because the particles don’t want to be in the same low-energy state, i.e. they don’t want to be degenerate):

∼

p =

¯ 2(3π2 Nq )5/3 −5/3 (3π 2)2/3h ¯ 2 ρ5/3 dE  2 h = = V  . 3 10π2m 5m dV

For copper, it is better to calculate the bulk modules: B =

5 dp −V  dV  =  p = . . . = 6.4 × 10 3

10

N/m2 .

That’s not bad for such a rough model since the experimental value is only twice as large, 13.4 1010 N/m2 .

×

Applications to astrophysics A star such as the Sun maintains its volume by thermal pressure. Once it burns the fuel, gravity typically makes it collapse into a white dwarf whose size is stabilized by the electron degeneracy pressure. Above some mass, the Chandrasekhar bound, the pressure is not enough. At ever higher energies, the electrons carry too high an energy, and becomes energetically favored for the protons to absorb the electrons  p + e−

→ n + ν

and the star collapses into a neutron star whose volume is mostly stabilized by neutron degeneracy pressure. Above another bound, this pressure is not suﬃcient either, and gravity wins: a black hole is created and condensed matter physics breaks down; general relativity takes over.

Bloch’s theory In this case, the potential inside the “box” is not zero, but it is periodic. Let’s consider the one-dimensional case ﬁrst. We have V  (x + a) = V  (x)

6

Deﬁne the displacement operator ˆ = ψ(x + a) Dψ(x) ˆ In fact, one can write D as exp(iˆ  pa/¯h). It commutes with the kinetic energy ˆ also commutes with the potential energy. and because of the periodicity D  We have  ˆ H ] = 0 [D, ˆ and H   may be simultaneously diagonalized; ψ is an eigenvalue of  Again, D ˆ D: ˆ Dψ = λψ. ˆ 1000 has eigenvalue λ1000 and if it’s truly The displacement by 1000 atoms D ﬁnite, it’s clear that the absolute value of  λ must be one; it’s called  Bloch’s theorem  and it also trivially follows from “unitarity” of the displacement operator: λ = exp(iΛa). Because Λ is real, it follows that 2

2

|ψ(x)| = |ψ(x + a)| . We may also compactify the space into a large circle with N   periods by imposing ψ(x + N a) = ψ(x). This requires Λ to be quantized (and this discretization of the spectrum may sometimes be a useful psychological tool to imagine the states): Λ=

2πn Na

The integer n  only takes N   diﬀerent values; for other values we obtain an equivalent periodic condition for  ψ(x + a).

Dirac’s comb An example is Dirac’s comb potential – a string of negative delta-functions: N −1

V  (x) =

−α  δ (x − ja).  j =0

7

In the continuum limit, the sum would go from to + . The energy eigenstate is a function ψ(x). For 0 < x < a it’s simply a harmonic function

−∞

ψ(x) = A sin(kx) + B cos(kx),

k =

√ 2mE  ¯ h

For a single negative delta-function potential one could also ﬁnd exponentially decreasing bound states (negative energy solutions; choose k ik above) but they can’t be extended to the whole comb. The wavefunction for a < x  1 in some quasiperiodic “forbidden gaps” (that become narrower as z  increases). The allowed spectrum only lies in the “allowed bands”.

|

|

Conductors and insulators If a band is not full, we obtain conductor and physics qualitatively reduces to the fermion gas model. If it is full, we obtain an insulator. Even in this case, electrons can be “kicked” to a higher band by thermal excitations. Band gap of diamond is 6 eV. That’s a big enough energy so that the room temperature k B T  = 0.025 eV is not enough to cause conduction because the probability for an electron to jump over the gap is exp( ∆E/k B T ). Band gap of germanium is smaller, 0.72 eV, in which case we obtain an intrinsic (= “even without impurities”) semiconductor; a few electrons among 1013 may jump.

∼

9