- #1
jostpuur
- 2,112
- 19
I'm now interested in a Schrödinger's equation
<br /> \Big(-\frac{\hbar^2}{2m}\partial_x^2 + V(x)\Big)\psi(x) = E\psi(x)<br />
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
<br /> u(x) = e^{-\frac{\log(A)}{R}x} \psi(x)<br />
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
<br /> \psi(x+2R) = A\psi(x+R) + B\psi(x).<br />
Consider then the following linear combinations.
<br /> \left(\begin{array}{c}<br /> \phi_1(x) \\ \phi_2(x) \\<br /> \end{array}\right)<br /> = \left(\begin{array}{cc}<br /> D_{11} & D_{12} \\<br /> D_{21} & D_{22} \\<br /> \end{array}\right)<br /> \left(\begin{array}{c}<br /> \psi(x) \\ \psi(x+R) \\<br /> \end{array}\right)<br />
Direct calculations give
<br /> \left(\begin{array}{c}<br /> \phi_1(x + R) \\ \phi_2(x + R) \\<br /> \end{array}\right)<br /> = \left(\begin{array}{cc}<br /> D_{11} & D_{12} \\<br /> D_{21} & D_{22} \\<br /> \end{array}\right)<br /> \left(\begin{array}{cc}<br /> 0 & 1 \\<br /> B & A \\<br /> \end{array}\right)<br /> \left(\begin{array}{c}<br /> \psi(x) \\ \psi(x+R) \\<br /> \end{array}\right)<br />
and
<br /> \left|\begin{array}{cc}<br /> -\lambda & 1 \\<br /> B & A - \lambda \\<br /> \end{array}\right| = 0<br /> \quad\quad\implies\quad\quad<br /> \lambda = \frac{A}{2}\pm \sqrt{B + \frac{A^2}{4}}<br />
This means, that if B + \frac{A^2}{4}\neq 0, then we can choose \boldsymbol{D} so that
<br /> \boldsymbol{D} \left(\begin{array}{cc}<br /> 0 & 1 \\<br /> B & A \\<br /> \end{array}\right)<br /> = \left(\begin{array}{cc}<br /> \lambda_1 & 0 \\<br /> 0 & \lambda_2 \\<br /> \end{array}\right) \boldsymbol{D}<br />
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
<br /> \left(\begin{array}{cc}<br /> 0 & 1 \\<br /> -\frac{A^2}{4} & A \\<br /> \end{array}\right)<br />
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?
<br /> \Big(-\frac{\hbar^2}{2m}\partial_x^2 + V(x)\Big)\psi(x) = E\psi(x)<br />
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
<br /> u(x) = e^{-\frac{\log(A)}{R}x} \psi(x)<br />
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
<br /> \psi(x+2R) = A\psi(x+R) + B\psi(x).<br />
Consider then the following linear combinations.
<br /> \left(\begin{array}{c}<br /> \phi_1(x) \\ \phi_2(x) \\<br /> \end{array}\right)<br /> = \left(\begin{array}{cc}<br /> D_{11} & D_{12} \\<br /> D_{21} & D_{22} \\<br /> \end{array}\right)<br /> \left(\begin{array}{c}<br /> \psi(x) \\ \psi(x+R) \\<br /> \end{array}\right)<br />
Direct calculations give
<br /> \left(\begin{array}{c}<br /> \phi_1(x + R) \\ \phi_2(x + R) \\<br /> \end{array}\right)<br /> = \left(\begin{array}{cc}<br /> D_{11} & D_{12} \\<br /> D_{21} & D_{22} \\<br /> \end{array}\right)<br /> \left(\begin{array}{cc}<br /> 0 & 1 \\<br /> B & A \\<br /> \end{array}\right)<br /> \left(\begin{array}{c}<br /> \psi(x) \\ \psi(x+R) \\<br /> \end{array}\right)<br />
and
<br /> \left|\begin{array}{cc}<br /> -\lambda & 1 \\<br /> B & A - \lambda \\<br /> \end{array}\right| = 0<br /> \quad\quad\implies\quad\quad<br /> \lambda = \frac{A}{2}\pm \sqrt{B + \frac{A^2}{4}}<br />
This means, that if B + \frac{A^2}{4}\neq 0, then we can choose \boldsymbol{D} so that
<br /> \boldsymbol{D} \left(\begin{array}{cc}<br /> 0 & 1 \\<br /> B & A \\<br /> \end{array}\right)<br /> = \left(\begin{array}{cc}<br /> \lambda_1 & 0 \\<br /> 0 & \lambda_2 \\<br /> \end{array}\right) \boldsymbol{D}<br />
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
<br /> \left(\begin{array}{cc}<br /> 0 & 1 \\<br /> -\frac{A^2}{4} & A \\<br /> \end{array}\right)<br />
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?