- #1

jostpuur

- 2,116

- 19

[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?