Kinetic Energy in terms of Conjugate Momenta

[MisterX]

Thanks if you take the time to read this.

Homework Statement

Homework Equations

The Attempt at a Solution

The problem I'm getting is that I'm not getting the kinetic energy diagonal when I convert to the coordinates that diagonalize the potential energy. If you scroll all the way down you should see there's a cross derivative term involving the partial derivatives w.r.t.$q_1$and $q_1$. I'm not able to spot any significant mistakes. So is there some other way to diagonalize both the potential and the conjugate momenta in the new coordinates? Is there something I'm missing? How does the $\frac{1}{2(2M + m_2)}$ factor come into play?

First we make a transformation in the x coordinates so that the $a$'s go away. Namely
$$\begin{bmatrix} x_1^\prime \\x_2^\prime \\ x_3^\prime\end{bmatrix} = \begin{bmatrix} x_1 \\x_2 \\ x_3 \end{bmatrix} - \begin{bmatrix} a/2 \\-a/2 \\ -3a/2\end{bmatrix}$$
And I will refer to the $x_i^\prime$ coordinates without the prime hereafter.
$$(x_2 - x_1)^2 + (x_3 - x_2)^2 = x_1^2 +x_2^2 - 2x_1x_2 + x_2^2+ x_3^2 - 2x_2x_3$$
There are multiple choices for the matrix which below which would produce the above algebraic expression. We want to choose the one which Hermitian, so that the corresponding transformation is unitary.
$$= \begin{bmatrix} x_1 &x_2 & x_3\end{bmatrix}\begin{bmatrix} 1 &-1 &0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\x_2 \\ x_3\end{bmatrix}$$
$$det\begin{bmatrix} 1 - \lambda &-1 &0 \\ -1 &2- \lambda &-1 \\ 0 & -1 & 1- \lambda \end{bmatrix} = \left(1 - \lambda \right)^2\left(2 - \lambda \right) - \left(1 - \lambda \right) - \left(1 - \lambda \right)$$
\begin{align*} = \left(1 - \lambda \right)\left[\left(1 - \lambda \right)\left(2 - \lambda \right) -2 \right] &= \left(1 - \lambda \right)\left[2 - 3\lambda + \lambda^2 -2 \right] \\
& =\lambda\left(1 - \lambda \right)\left(\lambda - 3 \right) \end{align*}
So the set of eigenvalues is $\left\{0, 1, 3 \right\}$.
$$U = \begin{bmatrix} 1/\sqrt{3} &1/\sqrt{3} &1/\sqrt{3} \\ 1/\sqrt{2} &0 &-1/\sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \end{bmatrix}$$
$$U^{T}\begin{bmatrix} 0 &0 &0 \\ 0 &1 &0 \\ 0 & 0 & 3 \end{bmatrix}U = \begin{bmatrix} 1 &-1 &0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}$$
$$\begin{bmatrix} q_1 \\q_2 \\ q_3\end{bmatrix} = U \begin{bmatrix} x_1 \\x_2 \\ x_3\end{bmatrix}$$
Next begins the calculation of the partial derivatives of $x_i$ in the new coordinate frame, as the first step to finding the kinetic energy in terms of conjugate momenta.
$$\begin{bmatrix}\frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3}\end{bmatrix} = \begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix}U$$$$KE = -\hbar^2\begin{bmatrix}\frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3}\end{bmatrix} \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix} \begin{bmatrix}\frac{\partial }{\partial x_1} \\ \frac{\partial }{\partial x_2} \\ \frac{\partial }{\partial x_3}\end{bmatrix}$$
$$= -\hbar^2\begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix}U \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix}U^{T}\begin{bmatrix}\frac{\partial }{\partial q_1} \\ \frac{\partial }{\partial q_2} \\ \frac{\partial }{\partial q_3}\end{bmatrix}$$
$$U \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix}U^{T}$$
$$= \begin{bmatrix} 1/\sqrt{3} &1/\sqrt{3} &1/\sqrt{3} \\ 1/\sqrt{2} &0 &-1/\sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \end{bmatrix}\begin{bmatrix} \frac{1}{2M\sqrt{3}} & \frac{1}{2M\sqrt{2}} &\frac{1}{2M\sqrt{6}} \\ \frac{1}{2m_2\sqrt{3}} & 0 &-\frac{2}{2m_2\sqrt{6}} \\ \frac{1}{2M\sqrt{3}} & -\frac{1}{2M\sqrt{2}} & \frac{1}{2M\sqrt{6}} \end{bmatrix}$$
$$= \begin{bmatrix} \frac{1}{3M} + \frac{1}{6m_2} & 0 & \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right) \\ 0 & \frac{1}{2M} & 0 \\ \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right) & 0 & \frac{1}{6M} + \frac{1}{3m_2} \end{bmatrix}$$
$$\frac{KE}{-\hbar^2} = \begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix} \begin{bmatrix}\left(\frac{1}{3M} + \frac{1}{6m_2} \right)\frac{\partial }{\partial q_1} + \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial }{\partial q_3}\\ \frac{1}{2M}\frac{\partial }{\partial q_2} \\ \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial }{\partial q_1}+ \left(\frac{1}{6M} + \frac{1}{3m_2} \right)\frac{\partial }{\partial q_3}\end{bmatrix}$$$$= \left(\frac{1}{3M} + \frac{1}{6m_2} \right)\frac{\partial^2 }{\partial q_1^2} + \frac{1}{2M}\frac{\partial^2 }{\partial q_2^2} + \left(\frac{1}{6M} + \frac{1}{3m_2} \right)\frac{\partial^2 }{\partial q_3^2}$$$$+ \frac{\sqrt{2}}{3}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial^2 }{\partial q_1 \partial q_3}$$

 

[MisterX]

I have gathered from reading Goldstein that my choice of $U$ was not unique.

We shall reserve a discussion of this exceptional case of degeneracy for a later time. For the present, suffice it to state that a set of $a_{jk}$ coefficients can always be found that satisfies both the eigenvalue conditions Eqs. (6.10), and Eq. (6.22a), so that Eq. (6.23) always holds.

see pages 241-245

So I shall have to have another go at finding U.