Bloch assumed that electrons move in a perfect periodic potential. He considered one dimensional array of lattice. The potential of electron at positive ion site is zero and is maximum in between. So long any line passing through the centers of positive ions, the potential variation must be as shown in below figure.
So Bloch gave a condition which is
𝚿(x+Na)=𝚿(x) .............................................................................................................(1)
It is considered as boundary condition.
Consider Schrodinger wave equation for one dimensional lattice.
(d²𝚿(x)/dx²) + (2m/ħ²)*[E-V(x)]*𝚿(x) = 0 .................................................................(2)
The Schrodinger equation for an electron in the potential at x+a is
[d²𝚿(x+a)/d(x+a)²] + (2m/ħ²)*[E-V(x+a)]*𝚿(x+a) = 0 ..............................................(3)
Because of periodicity,
[d/d(x+a)] = d/dx ; V(x+a) = V(x)
With this, eqn (3) reduces to
[d²𝚿(x+a)/dx²] + (2m/ħ²)*[E-V(x+a)]*𝚿(x+a) = 0 ...................................................(4)
This is Schrodinger equation at x+a.
as 𝚿 at x+a is also obeying Schrodinger wave equation as 𝚿 at x there should exist a relation between 𝚿(x+a) & 𝚿(x).
Let 𝚿(x+a) = A𝚿(x)..................................................................................................(5)
𝚿(x+2a) = A²𝚿(x) [i.e. A𝚿(x+a) = A.A𝚿(x) = A²𝚿(x)]
𝚿(x+na) = Aⁿ𝚿(x)
from eqn(1), Aⁿ =1 [i.e by using bloch condition]
Aⁿ =exp(2πij) [i.e. exp(2πij) =1 for j=01,2............]
or
A=exp(2πij/n)
Therefore, 𝚿(x+a) = exp(2πij/n)*𝚿(x) --------------------------------------------------(6) [ from eqn 5]
𝚿(x) can be written in terms of other function Uk(x )
𝚿(x) = exp(ikx)*Uk(x) where k=(2πj/n) ..................................................................(7)
From eqns (6) & (7),
exp[ik(x+a)]*Uk(x+a) = exp(2πij/n)*exp(ikx)*Uk(x)
exp[ika]*Uk(x+a) = exp(2πij/n)*Uk(x)
noting that Ka = 2πj/n,
we can write that Uk(x+a) = Uk(x) ..........................................................................(8)
So Bloch gave a condition which is
𝚿(x+Na)=𝚿(x) .............................................................................................................(1)
It is considered as boundary condition.
Consider Schrodinger wave equation for one dimensional lattice.
(d²𝚿(x)/dx²) + (2m/ħ²)*[E-V(x)]*𝚿(x) = 0 .................................................................(2)
The Schrodinger equation for an electron in the potential at x+a is
[d²𝚿(x+a)/d(x+a)²] + (2m/ħ²)*[E-V(x+a)]*𝚿(x+a) = 0 ..............................................(3)
Because of periodicity,
[d/d(x+a)] = d/dx ; V(x+a) = V(x)
With this, eqn (3) reduces to
[d²𝚿(x+a)/dx²] + (2m/ħ²)*[E-V(x+a)]*𝚿(x+a) = 0 ...................................................(4)
This is Schrodinger equation at x+a.
as 𝚿 at x+a is also obeying Schrodinger wave equation as 𝚿 at x there should exist a relation between 𝚿(x+a) & 𝚿(x).
Let 𝚿(x+a) = A𝚿(x)..................................................................................................(5)
𝚿(x+2a) = A²𝚿(x) [i.e. A𝚿(x+a) = A.A𝚿(x) = A²𝚿(x)]
𝚿(x+na) = Aⁿ𝚿(x)
from eqn(1), Aⁿ =1 [i.e by using bloch condition]
Aⁿ =exp(2πij) [i.e. exp(2πij) =1 for j=01,2............]
or
A=exp(2πij/n)
Therefore, 𝚿(x+a) = exp(2πij/n)*𝚿(x) --------------------------------------------------(6) [ from eqn 5]
𝚿(x) can be written in terms of other function Uk(x )
𝚿(x) = exp(ikx)*Uk(x) where k=(2πj/n) ..................................................................(7)
From eqns (6) & (7),
exp[ik(x+a)]*Uk(x+a) = exp(2πij/n)*exp(ikx)*Uk(x)
exp[ika]*Uk(x+a) = exp(2πij/n)*Uk(x)
noting that Ka = 2πj/n,
we can write that Uk(x+a) = Uk(x) ..........................................................................(8)
Conclusion
Bloch Theorem is a mathematical theorem and it gives us the form of electron wave function in a periodic potential.
𝚿(x) = exp(ikx)*Uk(x) represents Plane Wave
Thus, electron in a one dimensional lattice behaves a a plane wave.It only gives Wave nature of electron.
