TransWikia.com

Problem in Proof of Bloch theorem

Physics Asked by Shaurya Anand on April 22, 2021

In kittel’s book on solid state physics a proof of bloch theorem is given . It says:

We consider N identical lattice points on a ring of length Na. the potential energy is periodic in a width $U(x)=U(x+sa$), where s is an integer.
Let us be guided by the symmetry of the ring to look for solutions of the wave equation such that
$$psi(x+a)=Cpsi(x)$$
where C is a constant . Then on going once around the ring
$$psi(x+Na)=psi(x)=C^Npsi(x)$$
because $psi(x)$ must be single valued. It follows that C is one of the N roots of unity or

$C=exp(i2pi s/N)$ ;
s=$0,1,2,…N-1 .$ $(1)$

We use $(1)$ to see that

$psi(x)=u_k(x)exp(i2pi sx/Na)$ $(2)$

$(2)$ is the bloch result

I couldnt follow the part where $(1)$ is used to arrive at the bloch result

One Answer

To clarify, the Bloch result that you wish to show is that $psi(x) = e^{i k x}u_k(x)$ where $k = 2pi s / Na$ and $u_k(x)$ is a function which has the same periodicity as the potential $U(x)$, that is, $u_k(x) = u_k(x+a)$.

We can directly verify that (2) fulfills these conditions. From (2), we have $u_k(x) = e^{-ikx}psi(x)$. To verify that this has the same periodicity as $U(x)$, show that $u_k(x+a) = u_k(x)$:

$u_k(x+a) = e^{-ik(x+a)}psi(x+a) = e^{-ik(x+a)}e^{i k a}psi(x) = e^{-ikx}psi(x) = u_k(x)$

Answered by Anonymous physicist on April 22, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP