Note to self, the following three problem are equivalent (Thanks to the notes of Dr. Robert Hunt, Cambridge University http://www.damtp.cam.ac.uk/user/reh10/lectures/)
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1) Find the eigenvalues and eigenfunction of the following Sturm-Liouville problem
\begin{eqnarray}
-\frac{d}{dx}\left[p(x)y^\prime \right]+q(x)y &=& \lambda w(x) y
\end{eqnarray}
for $a<x<b$ where neither $p$ nor $w$ vanish.
2) Find the functions for which the following functional is stationary
\begin{eqnarray}
F[y] &=& \int_a^b (p(y^\prime)^2+qy^2)dx
\end{eqnarray}
subject to the constraint
\begin{eqnarray}
G[y] &=& \int_a^b wy^2 dx =1
\end{eqnarray}
If the constraint is satisfied, then the eigenvalues of the Sturm-Liouville problem are given by the values of $F[y]$. The eigenvalue of the system will have the smallest eigenvalue.
3) Find the functions $y(x)$ for which the functional
\begin{eqnarray}
\Lambda[y] &=& \frac{F[y]}{G[y]}
\end{eqnarray}
is stationary; the eigenvalues of the first problem are then values of $\Lambda[y]$ -- without normalization of $G[y]$.
It can be seen in the functional $F[y]$ and $G[y]$ that if $w,p$ and $q$ are all positive functions of $x$, then the eigenvalues will be greater than zero. Or, the Sturm-Liouville operator is positive definite.
-----------------
Using these facts, I was playing around with the one-dimensional Schrodinger equation, and reformed it as a Sturm Liouville problem,
\begin{eqnarray}
\frac{-\hbar^2}{2m}\psi^{\prime \prime} + V(x)\psi &=& i\hbar \frac{\partial \psi}{\partial t}\\
\end{eqnarray}
If one assumes separability, and that $\frac{\partial \psi}{\partial t}=E_n \psi$, then
\begin{eqnarray}
\frac{-\hbar^2}{2m}\psi^{\prime \prime} + V(x)\psi &=& \lambda \psi
\end{eqnarray}
so that we can write $w=1$, $q(x)=V(x)$, $p(x)=\frac{\hbar^2}{2m}$, and now write the functional to be minimized -- assuming $\psi$ is complex:
\begin{eqnarray}
F[\psi] &=& \int \left( \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x}\frac{\partial \psi^\star}{\partial x} + V(x) \psi^\star \psi \right)dx \\
F[\psi] &=& \lambda \\
\int \left( -T+V \right)dx&=& \lambda \\
\int \left( \mathcal{L} \right)dx&=& -\lambda
\end{eqnarray}
This is just the action! I don't know if I made some awful assumptions here, but if we take $F[\psi]=-S[\psi]=-\lambda$, we can now write the time dependence of our `separable' equation as
\begin{eqnarray}
\psi &=& C_1 e^{\frac{iS[\psi]}{\hbar}t}
\end{eqnarray}
Redundant, but interesting. Note also that the Sturm-Liouville construction takes into account normalization -- which I didn't mention before -- since
\begin{eqnarray}
G[\psi] &=& \int \psi \psi^\star dx = 1 \\
\Rightarrow F[\psi] &=& \lambda
\end{eqnarray}
-------
In problem construction (3), one can make guesses at the solution function -- or eigenfunction -- of the problem $y$, and find successive values of $\lambda$. For p,q,w all greater than zero, the lowest value of $\lambda$ wins, and so one reduces the Sturm-Liouville to a minimization problem. This is called the Rayleigh-Ritz method.
-------
1) Find the eigenvalues and eigenfunction of the following Sturm-Liouville problem
\begin{eqnarray}
-\frac{d}{dx}\left[p(x)y^\prime \right]+q(x)y &=& \lambda w(x) y
\end{eqnarray}
for $a<x<b$ where neither $p$ nor $w$ vanish.
2) Find the functions for which the following functional is stationary
\begin{eqnarray}
F[y] &=& \int_a^b (p(y^\prime)^2+qy^2)dx
\end{eqnarray}
subject to the constraint
\begin{eqnarray}
G[y] &=& \int_a^b wy^2 dx =1
\end{eqnarray}
If the constraint is satisfied, then the eigenvalues of the Sturm-Liouville problem are given by the values of $F[y]$. The eigenvalue of the system will have the smallest eigenvalue.
3) Find the functions $y(x)$ for which the functional
\begin{eqnarray}
\Lambda[y] &=& \frac{F[y]}{G[y]}
\end{eqnarray}
is stationary; the eigenvalues of the first problem are then values of $\Lambda[y]$ -- without normalization of $G[y]$.
It can be seen in the functional $F[y]$ and $G[y]$ that if $w,p$ and $q$ are all positive functions of $x$, then the eigenvalues will be greater than zero. Or, the Sturm-Liouville operator is positive definite.
-----------------
Using these facts, I was playing around with the one-dimensional Schrodinger equation, and reformed it as a Sturm Liouville problem,
\begin{eqnarray}
\frac{-\hbar^2}{2m}\psi^{\prime \prime} + V(x)\psi &=& i\hbar \frac{\partial \psi}{\partial t}\\
\end{eqnarray}
If one assumes separability, and that $\frac{\partial \psi}{\partial t}=E_n \psi$, then
\begin{eqnarray}
\frac{-\hbar^2}{2m}\psi^{\prime \prime} + V(x)\psi &=& \lambda \psi
\end{eqnarray}
so that we can write $w=1$, $q(x)=V(x)$, $p(x)=\frac{\hbar^2}{2m}$, and now write the functional to be minimized -- assuming $\psi$ is complex:
\begin{eqnarray}
F[\psi] &=& \int \left( \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x}\frac{\partial \psi^\star}{\partial x} + V(x) \psi^\star \psi \right)dx \\
F[\psi] &=& \lambda \\
\int \left( -T+V \right)dx&=& \lambda \\
\int \left( \mathcal{L} \right)dx&=& -\lambda
\end{eqnarray}
This is just the action! I don't know if I made some awful assumptions here, but if we take $F[\psi]=-S[\psi]=-\lambda$, we can now write the time dependence of our `separable' equation as
\begin{eqnarray}
\psi &=& C_1 e^{\frac{iS[\psi]}{\hbar}t}
\end{eqnarray}
Redundant, but interesting. Note also that the Sturm-Liouville construction takes into account normalization -- which I didn't mention before -- since
\begin{eqnarray}
G[\psi] &=& \int \psi \psi^\star dx = 1 \\
\Rightarrow F[\psi] &=& \lambda
\end{eqnarray}
-------
In problem construction (3), one can make guesses at the solution function -- or eigenfunction -- of the problem $y$, and find successive values of $\lambda$. For p,q,w all greater than zero, the lowest value of $\lambda$ wins, and so one reduces the Sturm-Liouville to a minimization problem. This is called the Rayleigh-Ritz method.
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