Bloch's theorem and diagonalization of translation operator

In summary, the conversation discusses the Schrödinger's equation and Bloch's theorem. The equation involves a wave function and a potential, and the conversation mentions the need for the potential to not contain infinities and satisfy a specific periodicity condition. The conversation also mentions the existence of two linearly independent solutions to the equation, and the task of showing that they can be written in a specific form. The conversation ends with a question about the possibility of a certain matrix not being diagonalizable and its implications for proving Bloch's theorem.
  • #1
jostpuur
2,116
19
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?
 
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  • #2
jostpuur said:
By basic theory of DEs, there exists two linearly independent solutions [itex]\psi_1,\psi_2[/itex] 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...
 

Related to Bloch's theorem and diagonalization of translation operator

1. What is Bloch's theorem?

Bloch's theorem is a fundamental principle in solid state physics that states that the wavefunction of a particle in a periodic potential can be written as a product of a periodic function and a plane wave.

2. How is Bloch's theorem related to the diagonalization of the translation operator?

The translation operator is used to describe the periodicity of the potential in Bloch's theorem. By diagonalizing this operator, we can find the eigenstates of the system, which are known as Bloch states.

3. What is the significance of diagonalizing the translation operator?

Diagonalizing the translation operator allows us to find the eigenstates of the system, which are the solutions to the Schrödinger equation. These eigenstates are known as Bloch states and are important in understanding the behavior of particles in periodic potentials.

4. How does Bloch's theorem apply to real-world systems?

Bloch's theorem is widely used in solid state physics to study the electronic properties of materials, such as metals and semiconductors. It also has applications in other fields, such as quantum optics and condensed matter physics.

5. Are there any limitations to Bloch's theorem?

Bloch's theorem is based on several assumptions, such as the periodicity of the potential and the absence of interactions between particles. These assumptions may not hold in certain real-world systems, making Bloch's theorem less applicable in those cases.

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