1 free electron fermi gas

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Solid State Physics

for Illumination Engineering II

Lecture I

Prof. Shavkat Yuldashev

Dongguk University, September 2010


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Free Electron Fermi Gas I

Electron Waves

In the classical picture, electrons are particles that follow Newton's laws of mechanics. They arecharacterized by their mass m0, their position = ( x , y , z ) , and their velocity However, this

intuitive picture is not sufficient for describing the behavior of electrons within solid crystals, where

it is more appropriate to consider electrons as waves. The wave-particle duality is one of the

fundamental features of quantum mechanics. Using complex numbers, the wave function for a free

electron can be written as

(1)

with the wave vector The wave vector is parallel to the electronmomentum

(2)

and it relates to the electron energy E as

with


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In general, electron wave functions need to satisfy the Schrodinger equation

(3)

where the potential represents the periodic semiconductor crystal. This equation is often writtenas

(4)

with H called the Hamiltonian. The Schrodinger equation is for just one electron; all other electrons

and atomic nuclei are included in the potential . For the free electron, = 0 and the solution

to the Schrodinger equation is of the simple form given by Eq. (1.1).

The boundary conditions are n(0) = 0; n (L) = 0, as imposed by the infinite potential energy

barriers. They are satisfied if the wavefunction is sinelike with an integral numbern of half-

wavelengths between 0 and L:

(5)

where A is a constant. We see that (1.5) is the solution of (1.4), because


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(6)

11.

According to the Pauli exclusion principle no two

electrons can have all their quantum numbers

identical. That is each orbital can be occupied by

at most one electron. This applies to electrons in

atoms, molecules, or solids.


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The Fermi energy F is defined as the energy of the topmost filled level in the ground

state of the N electron system. By (1.6) with n = nF we have in one dimension:

(7)

The Fermi Dirac distribution gives the probability that the orbital at energy will be

occupied in an ideal electron gas in thermal equilibrium:

(8)

The quantity is the chemical potential, and we see that at absolute zero temperature the

chemical potential is equal to the Fermi energy, defined as the energy of the topmost filled

orbital at absolute zero.


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The kinetic energy of the electron gas increases as the temperature is increased: some energy

levels are occupied which were vacant at absolute zero, and some levels are vacant which were

occupied at absolute zero temperature.


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(9)

(10)

We now require the wavefunctions to be periodic in x, y, z with the period L. Thus

(11)

With the wavevector component k satisfy

(12)


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Any component of kis of the form 2n/L , where n is a positive or negative integer.

The components of k are the quantum numbers of the problem, along with quantum number

ms for the spin direction. We confirm that these values of kx satisfy (1.11), for

(13)==

The energy

(1.14)


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We see that there is one allowed wavevector that is, one distinct triplet of quantum

numbers kx , ky , kz - for the volume element (2/L)3 of kspace.

Thus in the sphere of volume 4kF3 /3 the total number of orbitals is

(15)

where the factor 2 on the left comes from the two allowed values ofms , the spin quantum number,

for each allowed value of k. Then

(16)

which depends only on the particle concentration.

Using (1.14)(17)


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(19)

(20)


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(21)


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If N is the total number of electrons, only a fraction of the order of T / TF can be excited

thermally at temperature T, because only these lie within an energy range of the order of kBT

of the top of the energy distribution (Fig.5).

(22)

(23)


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It is good approximation to evaluate the density of statesD () at F and take it outside of the integral:


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Solution:


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Solution:


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Solution:


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Solution:


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The number of moles per cm3 is so that the concentration is

atoms cm3 . The mass of an atom of He3 is

Thus

Solution:


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Solution:


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Solution:


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Solution:


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Solution:


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Rsq (ohms) = 10-9 c2 Rsq(gaussian) (30)(137)ohms 4.1k.

Solution:

1 free electron fermi gas

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