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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 $$\psi(x+a)=C\psi(x)$$ where C is a constant . Then on going once around the ring $$\psi(x+Na)=\psi(x)=C^N\psi(x)$$ because $\psi(x)$ must be single valued. It follows that C is one of the N roots of unity or

$C=exp(i2\pi s/N)$ ; s=$0,1,2,...N-1 .$ $(1)$

We use $(1)$ to see that

$\psi(x)=u_k(x)exp(i2\pi sx/Na)$ $(2)$

$(2)$ is the bloch result

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

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To clarify, the Bloch result that you wish to show is that $\psi(x) = e^{i k x}u_k(x)$ where $k = 2\pi s / Na$ 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 $u_k(x+a) = u_k(x)$:

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

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  • $\begingroup$ I see that k can be written as $2\pi s/Na$ but how is C required in the proof. $\endgroup$ Commented Apr 11, 2021 at 17:54
  • $\begingroup$ Sorry I didn't make that clearer. $C = e^{ika}$ is required in the proof: $u_k(x+a) = e^{-ik(x+a)}\psi(x+a) = e^{-ik(x+a)}C\psi(x) = e^{-ikx}e^{-ika}C\psi(x) = e^{-ikx}e^{-ika}e^{ika}\psi(x) = e^{-ikx}\psi(x) = u_k(x)$. We needed the relationship $\psi(x+a) = e^{ika}\psi(x)$ in the proof, given by how $C$ is defined in (1). $\endgroup$ Commented Apr 11, 2021 at 18:16
  • $\begingroup$ Hi So are you already not assuming that $\psi(x)=u_k (x) e^{-ikx} $. We want to derive that right. I understand that using it we are able to satisfy the properties. $\endgroup$ Commented Apr 12, 2021 at 5:32
  • $\begingroup$ Any $\psi(x)$ can be written as $\psi(x)=u_k (x) e^{-ikx}$ for some function $u_k (x)$. What we have shown here is that for any $\psi(x)$ satisfying $\psi(x+a) = C\psi(x)$, the corresponding $u_k(x)$ has the same periodicity as the potential. $\endgroup$ Commented Apr 12, 2021 at 19:34
  • $\begingroup$ yes, got it now $\endgroup$ Commented Apr 13, 2021 at 7:18

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