# Coupled masses on springs – a solution using matrix diagonalization

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Exercise 1.8.11.

Here’s a practical example of how changing the basis by diagonalizing a hermitian matrix can make a problem easier to solve. Suppose we have two identical masses ${m}$ free to slide in one dimension on a frictionless horizontal surface. The two masses are connected to 3 springs, with the spring on the left attached to a solid support at one end and to mass #1 at the other, the middle spring connected between the two masses, and the spring on the right connected to mass #2 at one end and to a solid support at the other. The springs all have spring constant ${k}$. Define two coordinates ${x_{1}}$ and ${x_{2}}$ to be the positions of the two masses, with ${x_{i}=0}$ corresponding to the location at which mass ${i}$ is at rest in equilibrium.

Now suppose that the two masses are displaced from their respective equilibrium points, so that ${x_{1}}$ and ${x_{2}}$ are non-zero. The length of the spring to the left of mass 1 is changed (stretched or compressed, depending on the sign of ${x_{1}}$) by ${x_{1}}$, so exerts a force ${F_{1}=-kx_{1}}$ on mass 1. The length of the spring in the middle is changed by ${x_{2}-x_{1}}$, so it exerts a force ${F_{12}=k\left(x_{2}-x_{1}\right)}$ on mass 1, and an equal and opposite force ${F_{21}=-k\left(x_{2}-x_{1}\right)}$ on mass 2. Finally, the length of the spring on the right is changed by ${x_{2}}$ and exerts a force ${F_{2}=-kx_{2}}$ on mass 2. By applying Newton’s law ${F=ma}$, we get the set of equations of motion:

 $\displaystyle \ddot{x}_{1}$ $\displaystyle =$ $\displaystyle -2\frac{k}{m}x_{1}+\frac{k}{m}x_{2}\ \ \ \ \ (1)$ $\displaystyle \ddot{x}_{2}$ $\displaystyle =$ $\displaystyle \frac{k}{m}x_{1}-2\frac{k}{m}x_{2} \ \ \ \ \ (2)$

While it’s possible to solve such a coupled system directly, we can see how an easier method can be found by using matrix algebra. The 2 equations above can be written as a matrix equation

$\displaystyle \left|\ddot{x}\left(t\right)\right\rangle =\Omega\left|x\left(t\right)\right\rangle \ \ \ \ \ (3)$

If we use the basis in which the displacement of each mass is taken to be independent of the other, we have the two basis vectors

 $\displaystyle \left|1\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0 \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \left|2\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1 \end{array}\right] \ \ \ \ \ (5)$

In this basis

$\displaystyle \left|x\left(t\right)\right\rangle =x_{1}\left(t\right)\left|1\right\rangle +x_{2}\left(t\right)\left|2\right\rangle \ \ \ \ \ (6)$

Here, the ${x_{i}}$s are just numbers; the vector nature of the equation is delegated to the basis vectors.

In this basis, ${\Omega}$ is the operator whose matrix form is

$\displaystyle \Omega=\left[\begin{array}{cc} -2\frac{k}{m} & \frac{k}{m}\\ \frac{k}{m} & -2\frac{k}{m} \end{array}\right] \ \ \ \ \ (7)$

Since ${\Omega}$ is hermitian, it can be diagonalized by finding its eigenvalues and normalized eigenvectors, and forming a unitary operator ${U}$ whose columns are these eigenvectors. The basis vectors are now these eigenvectors ${\left|I\right\rangle }$ and ${\left|II\right\rangle }$ (I’m sticking to Shankar’s notation, even though it’s a bit clumsy), and they are found from ${\left|1\right\rangle }$ and ${\left|2\right\rangle }$ by applying the unitary transformation, that is

 $\displaystyle \left|I\right\rangle$ $\displaystyle =$ $\displaystyle U\left|1\right\rangle \ \ \ \ \ (8)$ $\displaystyle \left|II\right\rangle$ $\displaystyle =$ $\displaystyle U\left|2\right\rangle \ \ \ \ \ (9)$

These transformations can be inverted:

 $\displaystyle \left|1\right\rangle$ $\displaystyle =$ $\displaystyle U^{\dagger}\left|I\right\rangle \ \ \ \ \ (10)$ $\displaystyle \left|2\right\rangle$ $\displaystyle =$ $\displaystyle U^{\dagger}\left|II\right\rangle \ \ \ \ \ (11)$

Thus we can insert this into 3 and use ${UU^{\dagger}=I}$ to get

$\displaystyle U^{\dagger}\left|\ddot{x}\left(t\right)\right\rangle =U^{\dagger}\Omega UU^{\dagger}\left|x\left(t\right)\right\rangle \ \ \ \ \ (12)$

and ${U^{\dagger}\Omega U}$ is the diagonalized version of ${\Omega}$.

Shankar goes through the details of the calculation, with the results

 $\displaystyle \left|I\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 1 \end{array}\right]\ \ \ \ \ (13)$ $\displaystyle \left|II\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ -1 \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\\ 1 & -1 \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle U^{\dagger}\Omega U$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} -\omega_{1}^{2} & 0\\ 0 & -\omega_{2}^{2} \end{array}\right] \ \ \ \ \ (16)$

where

 $\displaystyle \omega_{1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{k}{m}}\ \ \ \ \ (17)$ $\displaystyle \omega_{2}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3k}{m}} \ \ \ \ \ (18)$

Using ${\left|I\right\rangle }$ and ${\left|II\right\rangle }$ as the basis, the differential equations become decoupled, and we have

$\displaystyle \ddot{x}_{i}+\omega_{i}^{2}x_{i}=0 \ \ \ \ \ (19)$

for ${i=I,II}$.

Second order ODEs require two initial conditions to be fully solved, and here we’re assuming that both masses start off at rest, so that ${\dot{x}_{i}\left(t\right)=0}$ for ${i=I,II}$. In this case, the solutions are

$\displaystyle x_{i}\left(t\right)=x_{i}\left(0\right)\cos\omega_{i}t \ \ \ \ \ (20)$

for ${i=I,II}$.

(A full, general solution would also have a ${\sin\omega_{i}t}$ term, but this disappears because we require ${\dot{x}_{i}\left(t\right)=0}$.)

The vector solution in the diagonal basis is therefore

$\displaystyle \left[\begin{array}{c} x_{I}\left(t\right)\\ x_{II}\left(t\right) \end{array}\right]=\left|I\right\rangle x_{I}\left(0\right)\cos\omega_{I}t+\left|II\right\rangle x_{II}\left(0\right)\cos\omega_{II}t \ \ \ \ \ (21)$

We now need to figure out what the coefficients ${x_{I}\left(0\right)}$ and ${x_{II}\left(0\right)}$ are. Assuming we know the initial position of each mass in the original basis as ${x_{1}\left(0\right)}$ and ${x_{2}\left(0\right)}$, we can find ${x_{I}\left(0\right)}$ and ${x_{II}\left(0\right)}$ by projecting ${x_{1}\left(0\right)}$ and ${x_{2}\left(0\right)}$ onto the basis ${\left|I\right\rangle }$ and ${\left|II\right\rangle }$. That is, we have

 $\displaystyle \left[\begin{array}{c} x_{I}\left(0\right)\\ x_{II}\left(0\right) \end{array}\right]$ $\displaystyle =$ $\displaystyle U\left[\begin{array}{c} x_{1}\left(0\right)\\ x_{2}\left(0\right) \end{array}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\\ 1 & -1 \end{array}\right]\left[\begin{array}{c} x_{1}\left(0\right)\\ x_{2}\left(0\right) \end{array}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} x_{1}\left(0\right)+x_{2}\left(0\right)\\ x_{1}\left(0\right)-x_{2}\left(0\right) \end{array}\right] \ \ \ \ \ (24)$

We get

 $\displaystyle \left[\begin{array}{c} x_{I}\left(t\right)\\ x_{II}\left(t\right) \end{array}\right]$ $\displaystyle =$ $\displaystyle \frac{x_{1}\left(0\right)+x_{2}\left(0\right)}{\sqrt{2}}\left|I\right\rangle \cos\omega_{I}t+\frac{x_{1}\left(0\right)-x_{2}\left(0\right)}{\sqrt{2}}\left|II\right\rangle \cos\omega_{II}t\ \ \ \ \ (25)$ $\displaystyle \left[\begin{array}{c} x_{1}\left(t\right)\\ x_{2}\left(t\right) \end{array}\right]$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} \left[x_{1}\left(0\right)+x_{2}\left(0\right)\right]\cos\sqrt{\frac{k}{m}}t+\left[x_{1}\left(0\right)-x_{2}\left(0\right)\right]\cos\sqrt{\frac{3k}{m}}t\\ \left[x_{1}\left(0\right)+x_{2}\left(0\right)\right]\cos\sqrt{\frac{k}{m}}t-\left[x_{1}\left(0\right)-x_{2}\left(0\right)\right]\cos\sqrt{\frac{3k}{m}}t \end{array}\right] \ \ \ \ \ (26)$

where in the last line we substituted using 13 to write everything in terms of the original basis ${\left|1\right\rangle }$ and ${\left|2\right\rangle }$.

For the special case where the initial positions are given by ${\left|1\right\rangle =\left[\begin{array}{c} 1\\ 0 \end{array}\right]}$, we have ${x_{1}\left(0\right)=1}$ and ${x_{2}\left(0\right)=0}$, so that

$\displaystyle \left[\begin{array}{c} x_{1}\left(t\right)\\ x_{2}\left(t\right) \end{array}\right]=\frac{1}{2}\left[\begin{array}{c} \cos\sqrt{\frac{k}{m}}t+\cos\sqrt{\frac{3k}{m}}t\\ \cos\sqrt{\frac{k}{m}}t-\cos\sqrt{\frac{3k}{m}}t \end{array}\right] \ \ \ \ \ (27)$

Going back to 26, we can write the solution as a matrix equation

$\displaystyle \left[\begin{array}{c} x_{1}\left(t\right)\\ x_{2}\left(t\right) \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc} \cos\sqrt{\frac{k}{m}}t+\cos\sqrt{\frac{3k}{m}}t & \cos\sqrt{\frac{k}{m}}t-\cos\sqrt{\frac{3k}{m}}t\\ \cos\sqrt{\frac{k}{m}}t-\cos\sqrt{\frac{3k}{m}}t & \cos\sqrt{\frac{k}{m}}t+\cos\sqrt{\frac{3k}{m}}t \end{array}\right]\left[\begin{array}{c} x_{1}\left(0\right)\\ x_{2}\left(0\right) \end{array}\right] \ \ \ \ \ (28)$

The matrix with the cosines is independent of the initial state, so that once we know this matrix, we can work out the general solution as a function of time for any initial state. The matrix is known as the propagator. [Although Shankar uses the symbol ${U\left(t\right)}$ to refer to the propagator, it’s not a unitary matrix. For example, its determinant is ${\cos\left(\sqrt{\frac{k}{m}}t\right)\cos\left(\sqrt{\frac{3k}{m}}t\right)\ne1}$ for ${t\ne0}$.]

## 6 thoughts on “Coupled masses on springs – a solution using matrix diagonalization”

1. Nilay Shrivastava

I didn’t get equation(12). Is this passive transformation? if yes, R Shankar says that we don’t change vectors in passive transformation.

1. gwrowe Post author

I’ve just multiplied eqn 3 by ${U^{\dagger}}$ on both sides and inserted a ${UU^{\dagger}=I}$ to get eqn 12. Since Shankar is actually transforming the vectors, I’d imagine this is an active transformation.