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Bloch's theorem and diagonalization of translation operator

  1. Jan 25, 2009 #1
    I'm now interested in a Schrödinger's equation

    [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?
     
  2. jcsd
  3. Apr 21, 2010 #2
    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...
     
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