Engineering Physics 3(2+1)

LESSON 15. Concept of Bloch Function & Bloch Electron

Charged Particles

Electric charges, positive or negative, occur in multiples of the electronic charge. The electron is one of the fundamental particles constituting the atom. The charge of an electron is negative and is denoted by e.

The magnitude of  e is 1.6 Χ 10^-19 coulomb . The mass of an electron changes with its velocity in accordance with the theory of relativity. An electron moving with a velocity ν  has the mass

where c   is the velocity of light in free space (c = 3 \[\times {10^8}{{met} \over {sec}}\]).

If  ν<< c, \[m \approx {m_o}\], called the rest mass of the electron.

The rest mass of the electron has the value  m_{o = 9.11 Χ 10^-31 kg} .

Equation (1) shows that the mass m increases with the velocity ν and approaches infinity as ν → c . An electron, starting from rest and accelerated through a potential difference of 6kν , acquires a velocity of 0.15c . At this high speed, the mass of the electron increases only by 1 per cent. Therefore, the change of the mass of an electron with velocity can be neglected for accelerating potentials less than 6kν .

The radius of an electron is about 10^-15m  and that of an atom is 10^-10m . The radii are so small that electrons and atoms are ordinarily taken as point masses.

Electron Volt- A Unit of Energy

For energies involved in electron devices, ‘joule’ is too large a unit. Such small energies are conveniently measured in electron volt, abbreviated eV. The electron volt is the kinetic energy gained by an electron, initially at rest, in moving through a potential difference of 1 volt. Since

e = 1.6 Χ 10^-19 coulomb

1eV = 1.6 Χ 10^-19 coulomb Χ 1 volt = 1.6  Χ 10^-19 joule

Atomic Energy Levels

An atom of an element is generally made up of electrons, protons, and neutrons. The only exception is the hydrogen atom which possesses one electron and one proton, but no neutron. While an electron is negatively charged, the proton is a positively charged particle. The charge of a proton is numerically equal to that of an electron, but the mass of a proton is 1837 times that of an electron. A neutron is a neutral particle having a mass nearly equal to the proton mass. Because the protons and neutrons carry practically the entire mass of the atom, they remain almost immobile in a region, called the atomic nucleus. The electrons revolve round the nucleus in definite orbits which are circular or elliptical. The motion is analogous to that of planets around the sun. The atom is electrically neutral because the number of orbital electrons is equal to the number of protons in the nucleus. The atom of one element differs from that of another due to the differing numbers of protons, neutrons, and electrons in the atom. In the Bohr atomic model, the electrons are assumed to move about the nucleus in certain discrete circular orbits without radiating any energy. In any orbit, the angular momentum of the electron is equal to an integral multiple of  \[{h \over {2\pi }}\], where h is Planck’s constant (h = 6.62 Χ 10^-34 J.s ). The integral number n  has values 1, 2, 3 etc. for different orbits. The higher the value of    has the larger the radius of the orbit.

Velocity of Bloch Electron

• A Bloch wave or Bloch state, named after Swiss physicist Felix Bloch, is the wave-function of a particle (usually, an electron) placed in a periodic potential. Bloch's theorem states that the energy Eigen function for such a system may be written as the product of a plane wave envelope function and a periodic function (periodic Bloch function) un Κ(r)  that has the same periodicity as the potential, giving:

\[{\psi _{nk}}\left( r \right)={e^{ikr}}{u_{nk}}\left( r \right)\]

• The corresponding energy Eigen values are , periodic with periodicity Κ  of a reciprocal lattice vector.
• The energies associated with the index  n  vary continuously with wave vector  and form an energy band identified by band index n .

The Eigen values for given n  are periodic in Κ; all distinct values of   occur for Κ -values within the first Brillouin zone of the reciprocal lattice.

Statement of Bloch’s Function

Suppose that V(x) denotes the potential energy of an electron in a in a linear lattice of lattice constant a  and that V(x) = V(x + a) i.e the period of the potential is also a .

The wave functions of the electron in this potemtial are then obtained the Schrodinger equation

\[{{{\partial ^2}\Psi } \over {\partial {x^2}}}={{2m} \over {{\hbar ^2}}}\left[ {E - V\left( x \right)} \right]\Psi=0\]………….(1)

With reference to the solution of this equation there is an important theorem which predict that it has solution of the form

\[\Psi \left(x\right)={e^{ \pm iKx}}{u_K}\left(x \right)\]

\[{u_K}\left( x \right)={u_K}\left( {x + a} \right)\]…………..(2)

The solutions are plane waves  \[\Psi \left( x \right)={e^{ \pm iKx}}\] modulated by function  which depends in general on the wave vector Κ and have the periodically of the potential.

This theorem is known as the Bloch Theorem and the functions (2) as the Bloch solution.

Suppose g(x) and  ƒ(x) are two teal and independent solution of the Schrodinger equation (1)

Then most general solution is

\[\Psi \left( x \right)=Af\left( x \right) + Bg\left( x \right)\] ………… (3)

Where A and B are arbitrary constant

Since V(x) = V(x + a)  and only g(x) and  ƒ(x), but also g(x + a) and ƒ(x + a)  are the solution.

Now differential equation of the 2^nd order has only two independent solutions, these must be expressible as a linear combination of the independent ones.

We must have relation

\[f\left( {x + a} \right)={\alpha _1}f\left( x \right) + {\alpha _2}g\left( x \right)\]

\[g\left({x + a}\right)={\beta_1}f\left(x\right)+{\beta_2}g\left(x\right)\].........(4)

a1, a2 ,  and  \[{\beta_2}\]  are real functions of energy E

According to these solution, the general solution,

\[\Psi \left( {x + a} \right)=Af\left( {x + a} \right) + Bg\left( {x + a} \right)\]

Or

\[\Psi \left( {x + a} \right)=\left( {A{\alpha _1} + B{\beta _1}} \right)f\left( x \right) + \left( {A{\alpha _2} + B{\beta _2}} \right)g\left( x \right)\]………..(5)

Now if we choose A  and B such that

\[A{\alpha _1} + B{\beta _1}=\lambda A\]

\[{\alpha _2} + B{\beta _2}=\lambda B\]…….(6)

\[\lambda\] is  a constant then the functions \[\Psi \left( x \right)\]  will have the property

\[\Psi \left( {x + a} \right) = \lambda \Psi \left( x \right)\]………(7)

The equation (6) has non-vanishing solution for  A and B only if the determinant of their coefficient

\[\left|{\matrix{{{\alpha _1}-\lambda }&{{\beta _1}}\cr{{\alpha _2}}&{{\beta _2}-\lambda }\cr}}\right|\]= 0

\[{\lambda^2}-\left({{\alpha_1}+{\beta_2}}\right)\lambda+{\alpha _1}{\beta _2}-{\alpha _2}{\beta _1}\] = 0 ………..(8)

We can show that

\[{\alpha_1}{\beta_2}-{\alpha_2}{\beta_1}=1\] in the following manner.

Since ƒ(x) and g(x) are two real independent solution of equation (1)

\[{{{\partial^2}g\left(x\right)}\over{\partial{x^2}}}={{2m}\over{{\hbar^2}}}\left[{E-V\left(x\right)}\right]g\left(x\right)=0\]

\[{{{\partial^2}f\left(x\right)}\over{\partial{x^2}}}={{2m}\over{{\hbar^2}}}\left[{E-V\left(x\right)}\right]f\left(x\right)=0\]

Multiplying the former by  ƒ(x) and latter by g(x) and subtracting

\[f\left(x\right){{dg\left(x\right)}\over{dx}}-g\left(x\right){{df\left(x\right)}\over {dx}}=constant\]

LHS of this equation is called Wronskian  W(x), of the solutions and constant in this case.

Further, from equation (4) one can derive

\[W\left({x+a}\right)=\left({{\alpha_1}{\beta_2}-{\alpha_2}{\beta_1}}\right)W\left(x\right)\]………..(9)

The equation (8) becomes

\[{\lambda^2}-\left({{\alpha_1}+{\beta_2}}\right)\lambda+1=0\]………….(10)

\[{\alpha_1}+{\beta_2}\] is real function of energy  

In general there will be two functions

\[{\Psi _1}\left( x \right)\] and \[{\Psi _2}\left( x \right)\] which have the property (7)

Noting the product \[{\lambda _{1}}{\lambda _{2}}=1\]we may taken them as complex conjugates

\[{\lambda _{1}}={e^{iKa}}\] and \[{\lambda _{2}}={e^{iKa}}\]  ………….(11)

K is real

The corresponding functions     then have the property

\[{\Psi_1}\left({x + a}\right)={e^{iKa}}{\Psi_1}\left(x\right)\]

\[{\Psi_2}\left({x + a}\right)={e^{iKa}}{\Psi_2}\left(x\right)\]………….(12)

This is the property of Bloch Function is seen .

Put x + a = a in equation (2)

 Hence

\[\Psi \left({x + a}\right)=\lambda\Psi\left(x\right)\]  where, \[\lambda={e^{\pm iKa}}\]………..(13)

We see that (12) and (13) are same. Thus, we can write the solution of (1) and (2) and the Bloch Theorem is proved. The wave functions in this case are labeled by κ and are called the Bloch Wave functions.