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?

  • $\begingroup$ 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?" $\endgroup$ – Brian Moths Sep 9 '13 at 21:16
  • $\begingroup$ Point taken, I changed the title and added the question on the bottom. $\endgroup$ – 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.

Our hope is that if we change to a different basis the matrix in the equation giving the time derivatives of the coordinates with respect to the new coordinates will be simpler. In fact, we hope it will be diagonal.

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.

  • $\begingroup$ 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. $\endgroup$ – Astrum Sep 9 '13 at 21:52
  • 1
    $\begingroup$ The solving for the eigenvalues part is what I actually did when I said I made a lucky guess. To find the new basis, you have to find the eigenvalues, then you find the eigenvectors. The two eigenvectors will be the new basis vectors. The new coefficient matrix is just the diagonal matrix formed by the eigenvectors. $\endgroup$ – Brian Moths Sep 9 '13 at 22:02

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