I'm now interested in a Schrödinger's equation(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

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

[/tex]

where [itex]V[/itex] does not contain infinities, and satisfies [itex]V(x+R)=V(x)[/itex] with some [itex]R[/itex]. 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 [itex]\psi(x+R)=A\psi(x)[/itex] with some [itex]A[/itex], when it follows that [itex]\psi(x)=e^{Cx}u(x)[/itex] with some [itex]C[/itex] and [itex]u(x)[/itex], so that [itex]u(x+R)=u(x)[/itex]. This can be proven by setting

[tex]

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

[/tex]

and checking that this [itex]u(x)[/itex] is periodic.

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

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

[tex]

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

[/tex]

Consider then the following linear combinations.

[tex]

\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)

[/tex]

Direct calculations give

[tex]

\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)

[/tex]

and

[tex]

\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}}

[/tex]

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

[tex]

\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}

[/tex]

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

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

[tex]

\left(\begin{array}{cc}

0 & 1 \\

-\frac{A^2}{4} & A \\

\end{array}\right)

[/tex]

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

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Bloch's theorem and diagonalization of translation operator

Loading...

Similar Threads - Bloch's theorem diagonalization | Date |
---|---|

I Blochs theorem | Aug 21, 2016 |

I Blochs theorem | Apr 29, 2016 |

Derivation of Bloch's theorem | Dec 19, 2015 |

Proving Bloch's Theorem | Feb 26, 2012 |

Simultaneous diagonalisation of non-hermitian operator (Bloch theorem) | Jul 21, 2010 |

**Physics Forums - The Fusion of Science and Community**