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

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  4. 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.

    Reply
    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.

      Reply
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