# Couple Masses - Change in Basis

I'm having trouble with the linear algebra used to solved a coupled mass problem.

$\ddot{x}_1 = -(2k/m)x_1 + (k/m)x_2$ and $\ddot{x}_2 = (k/m)x_1 - (2k/m)x_2$

Shankar then sets the equation up in matrix form.

$$\left \lbrack \matrix{ \ddot{x}_1 \cr \ddot{x}_2 }\right \rbrack = \left \lbrack \matrix{ \Omega _{1,1} & \Omega _{1,2} \cr \Omega _{2,1} & \Omega _{2,2}} \right \rbrack \left \lbrack \matrix{x_1 \cr x_2} \right \rbrack$$

Where $\Omega _{i,j}$ equals -$(2k/m)$ and $(k/m)$ as in the original equations.

This is the part I don't understand, he goes on to say that this equation (the matrix above) is set in a particular basis. The general equation here is $\left| \ddot{x}(t) \right \rangle = \Omega \left| x(t) \right \rangle$..

His explanation is "The equation is obtained by projecting (the ket equation above) on the basis vectors $\left| 1 \right \rangle$ and $\left| 2 \right \rangle$ which have the following significance: $\left| 1 \right \rangle$ first mass displace by unity, second mass undisplaced, $\left| 2 \right \rangle$ first mass undisplaced, second mass displaced by unity.

Hew goes on to talk about how we need to find a matrix tin a basis that is diagonalized.

How does the change of basis work?

• I think you will be more likely to get an answer if you state a question. I think your question must be something like, "How do you represent a point in multidimensional Cartesian space by a coordinate vector given with respect to some basis?" and/or "How does a change of basis work?" – Brian Moths Sep 9 '13 at 21:16
• Point taken, I changed the title and added the question on the bottom. – Astrum Sep 9 '13 at 21:19

I will write a vector given with respect to the initial basis $[x_1,x_2]_{B_1}$. $[x_1,x_2]_{B_1}$ represents the configuration where the first object is at position $x_1$ and the second object is at position $x_2$. We have found the equation giving the time derivatives of the coordinates in this basis has the form $$\left \lbrack \matrix{ \ddot{x}_1 \cr \ddot{x}_2 }\right \rbrack_{B_1} = \left \lbrack \matrix{ a & b \cr b & a} \right \rbrack \left \lbrack \matrix{x_1 \cr x_2} \right \rbrack_{B_1}.$$ Notice all four entries of the matrix are non-zero.
For the sake of this answer, we will make a lucky guess. We will say that $[\Sigma,\Delta]_{B_2}$ represents the state where $x_1 + x_2 = \Sigma$ and $x_1 - x_2 = \Delta$. We find that the equation in this new basis is $$\left \lbrack \matrix{ \ddot{\Sigma} \cr \ddot{\Delta} }\right \rbrack_{B_2} = \left \lbrack \matrix{ a+b & 0 \cr 0 & a-b} \right \rbrack \left \lbrack \matrix{\Sigma \cr \Delta} \right \rbrack_{B_2}.$$ This differential equation is easier and Shankar probably says how to solve it and gives the solution. I am not sure if this will clear up your confusion. Leave a comment if it doesn't.
• Hm, I'm still confused. We want a diagnol matrix, because it's easier to solve. Shankar does: $\Omega \left| I \right \rangle = - \omega ^2 _1 \left|I \right \rangle$ and he does the analogues for the second ket $\left| II \right \rangle$. So he solves this for it's eigen values, and expands $\left| x(t) \right \rangle$ in the new basis. I think that makes a little bit more sense. – Astrum Sep 9 '13 at 21:52