<s>[INST] Coupled Harmonic Oscillator - Solve by diagonalization I have a problem where I have two massive Particles $M$ and one particle with mass $m<M$. The two particles on the outside are coupled with the one in the middle with two springs.
The Hamiltonian for the system is given by:
\begin{equation}
H=\frac{p_1^2}{2M}+\frac{p_2^2}{2M}+\frac{p_3^2}{2m} + \frac{1}{2}k(x_3-x_1-d)^2+\frac{1}{2}k(x_2-x_3-d)^2
\end{equation}
($d$ is the length of the spring at equilibrium.)
I have replaced
\begin{equation}
q_1 = x_1\\
q_2 = x_2-2d\\
q_3 = x_3 -d
\end{equation}
and rewrote the potential as a Matrix equation:
\begin{equation}
\frac{k}{2}\vec{q}^T A\vec{q}
\end{equation}
with
\begin{equation}
A = \begin{pmatrix}
1 & 0 &-1\\
0 & 1 & -1\\
-1 & -1& 1\\
\end{pmatrix}\qquad \vec{q} = \begin{pmatrix}q_1\\q_2\\q_3\end{pmatrix}
\end{equation}
I can find  diagonal representation:
\begin{equation}
A = \begin{pmatrix}
1 & 0 &0\\
0 & 1+\sqrt{2} & 0\\
0 & 0& 1-\sqrt(2)\\
\end{pmatrix}
\end{equation}
With the corresponding eigenvectors
\begin{equation}
\begin{pmatrix}
-1\\1\\0
\end{pmatrix}\qquad
\begin{pmatrix}
-\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\\1
\end{pmatrix}\qquad
\begin{pmatrix}
\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\1
\end{pmatrix}
\end{equation}
This probably means to switch to new coordinates
\begin{equation}
y_1 = -q_1+q_2\\
y_2 = -\frac{1}{\sqrt{2}}q_1-\frac{1}{\sqrt{2}}q_2+q_3\\
y_3 = \frac{1}{\sqrt{2}}q_1+\frac{1}{\sqrt{2}}q_2+q_3
\end{equation}
But what do I do with the momentum operators. They should change accordingly but I am confused as to how exactly
 [/INST] You solve the last 3 equations for the $x_i$ and express them as functions of the normal coordinates $y_i$. Then you differentiate these equations and multiply the velocities with the masses $m_i$ to obtain the momenta $$p_i=m_i\frac{dx_i}{dt}=m_i f_i(dy_1/dt,dy_2/dt,dy_3/dt)$$ These expression for the momenta in terms of the new coordinate velocities you insert into the original Hamiltonian.
</s>