# Bloch's theorem and diagonalization of translation operator

#### jostpuur

I'm now interested in a Schrödinger's equation

$$\Big(-\frac{\hbar^2}{2m}\partial_x^2 + V(x)\Big)\psi(x) = E\psi(x)$$

where $V$ does not contain infinities, and satisfies $V(x+R)=V(x)$ with some $R$. I have almost already understood the Bloch's theorem! But I still have some little problems left. I shall first describe what I already know, and then what's the problem.

If a wave function satisfies a relation $\psi(x+R)=A\psi(x)$ with some $A$, when it follows that $\psi(x)=e^{Cx}u(x)$ with some $C$ and $u(x)$, so that $u(x+R)=u(x)$. This can be proven by setting

$$u(x) = e^{-\frac{\log(A)}{R}x} \psi(x)$$

and checking that this $u(x)$ is periodic.

By basic theory of DEs, there exists two linearly independent solutions $\psi_1,\psi_2$ to the Schrödinger's equation, and all other solutions can be written as a linear combination of these. (This is done with fixed energy $E$.) Now the real task is to show, that $\psi_1,\psi_2$ can be chosen to be of form $e^{C_1x}u_1(x)$ and $e^{C_2x}u_2(x)$.

Suppose that at least other one of $\psi_1,\psi_2$ is not of this form, and denote it simply with $\psi$. Now $\psi(x)$ and $\psi(x+R)$ are linearly independent solutions to the Schrödinger's equation, so there exists constants $A,B$ so that

$$\psi(x+2R) = A\psi(x+R) + B\psi(x).$$

Consider then the following linear combinations.

$$\left(\begin{array}{c} \phi_1(x) \\ \phi_2(x) \\ \end{array}\right) = \left(\begin{array}{cc} D_{11} & D_{12} \\ D_{21} & D_{22} \\ \end{array}\right) \left(\begin{array}{c} \psi(x) \\ \psi(x+R) \\ \end{array}\right)$$

Direct calculations give

$$\left(\begin{array}{c} \phi_1(x + R) \\ \phi_2(x + R) \\ \end{array}\right) = \left(\begin{array}{cc} D_{11} & D_{12} \\ D_{21} & D_{22} \\ \end{array}\right) \left(\begin{array}{cc} 0 & 1 \\ B & A \\ \end{array}\right) \left(\begin{array}{c} \psi(x) \\ \psi(x+R) \\ \end{array}\right)$$

and

$$\left|\begin{array}{cc} -\lambda & 1 \\ B & A - \lambda \\ \end{array}\right| = 0 \quad\quad\implies\quad\quad \lambda = \frac{A}{2}\pm \sqrt{B + \frac{A^2}{4}}$$

This means, that if $B + \frac{A^2}{4}\neq 0$, then we can choose $\boldsymbol{D}$ so that

$$\boldsymbol{D} \left(\begin{array}{cc} 0 & 1 \\ B & A \\ \end{array}\right) = \left(\begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{array}\right) \boldsymbol{D}$$

and then we obtain two linearly independent solutions $\phi_1,\phi_2$ which satisfy $\phi_k(x+R)=\lambda_k\phi_k(x)$, $k=1,2$.

Only thing that still bothers me, is that I see no reason why $B + \frac{A^2}{4} = 0$ could not happen. The matrix

$$\left(\begin{array}{cc} 0 & 1 \\ -\frac{A^2}{4} & A \\ \end{array}\right)$$

is not diagonalizable. It could be, that for some reason $B$ will never be like this, but I cannot know this for sure. If $B$ can be like this, how does one prove the Bloch's theorem then?

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#### Klockan3

By basic theory of DEs, there exists two linearly independent solutions $\psi_1,\psi_2$ to the Schrödinger's equation
This is wrong, it depends on the potential. The Schrödinger equation is not a normal DE and there are for most values of E no solution at all to a specific potential which is the whole deal with quantum mechanics and often there is just one solution for a specific E.

Edit: Wrong topic, was answering on another topic...