Odd Form Of Eigenvalue - Coupled Masses

In summary, this homework problem is an example of applying eigenvalue methods to solve (classical) mechanical systems. The exact text glosses over this quickly and the problem becomes more difficult when trying to solve it without looking at the result.
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Odd Form Of Eigenvalue -- Coupled Masses

This isn't strictly homework, since it's something I'm trying to self-teach, but it seems to fit best here.

Homework Statement


It's an example of applying eigenvalue methods to solve (classical) mechanical systems in an introductory text to QM; specifically, 1.8.6 in Shankar. I can follow everything fine until it actually comes to calculating the eigenvalues; when I tried to do it without looking at the result, mine was out by a sign and square root.

The exact text is "...the eigenvalue of Ω is written as -ω^2 rather than ω in anticipation of the fact that Ω has eigenvalues of the form -ω^2, with ω real". How quickly it's glossed over makes me think I'm missing something obvious, but I can't for the life of me see it. I've reproduced my work up to that point below.

Homework Equations

&

The Attempt at a Solution


KPnzbHJ.png

"Determine x_1(t) and x_2(t) given the initial displacements x_1(0) and x_2(0) and that both masses are initially at rest."
Applying Hooke's law and F=ma:
[itex]
\\
{x}_1'' = -\frac{2k}{m}x + \frac{k}{m}x \\
{x}_2'' = \frac{k}{m}x_1 - 2\frac{k}{m}x_2 \\ \\
\varkappa \equiv \frac{k}{m} \\
\begin{bmatrix}
{x}_1''\\
{x}_2''
\end{bmatrix} = \begin{bmatrix}
-2\varkappa & \varkappa \\
\varkappa & -2\varkappa
\end{bmatrix}\begin{bmatrix}
x_1\\
x_2
\end{bmatrix} \\
\left|x''(t)\right\rangle= \Omega \left|x(t)\right\rangle
[/itex]

This is in the basis with one vector equal to a unit displacement of the first mass, and the other the same for the second; the coupled differential equation is a pain to solve, so I want to swap to a basis with no contribution from x_1 to x_2 and vice versa, so diagonalising the Hermitian Ω as usual (det(Ω - ωI) = 0 and making a unitary diagonalising matrix from the eigenvector components) I end up with [tex]\omega_1 = -\varkappa[/tex] and [tex]\omega_2 = -3\varkappa[/tex]. The book itself gets [tex]\omega_1 = \sqrt{\varkappa}[/tex] and [tex]\omega_2 = \sqrt{3\varkappa}[/tex], which fits with the assumed form of ω, but I don't know where that comes from.

Sorry if this is too much information, since a lot of it seems ancillary to the question of "Why is the eigenvalue taken to be that form?", but I wanted to err on the side of caution with my first post.
 
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  • #3
You're confusing yourself. What they've done is simply instead of calling the eigenvalues of the matrix eg. omega, they call them -omega^2. This should be trivial. You've called the eigenvalues of the matrix omega - so of course the two results (yours and the one in the book) for the omega are different!

Now, you just need to think about *why* they call it -omega^2..
 
  • #4
Yeah, it's the why that's confusing me; I was meaning to illustrate with that last part of the post that my approach was right in everything but not having the minus omega^2. It was unclear: sorry.

I'll give it some thought. It feels like a tip-of-the-tongue thing: it seems vaguely familiar, but there's some link that's not clicking. I continued on with the problem anyway and got the same basis vectors and end result, though; it's just why Shankar knows to use the (admittedly neater) -omega^2 notation from the outset that is throwing me.
 
  • #5
It's simple, really.
The usual equation is something like x'' = -k/m x. Solving this, we identify a characterstic frequency of the solution, namely sqrt(k/m). This is usually called omega. Only now, the equation is with matrices and all that, but the idea remains the same. The eigenvalues you find are really the k/m-eigenvalues, and then you 'translate' these to some frequency.
 
  • #6
Oh, God; thanks. I was getting too caught up in the linear algebra of it all just to realize the meaning of kappa. Shouldn't have gotten lazy and combined k/m into kappa; might have seen it then.
 

What is an odd form of eigenvalue?

An odd form of eigenvalue is a type of eigenvalue that results from solving systems of coupled masses. It is characterized by having a complex conjugate pair of eigenvalues with equal real parts and opposite imaginary parts.

How are odd form eigenvalues different from regular eigenvalues?

Unlike regular eigenvalues, which are usually real numbers, odd form eigenvalues always come in complex conjugate pairs. This is because odd form eigenvalues arise from solving systems of coupled masses, which involve complex numbers in their equations.

What are coupled masses?

Coupled masses refer to a physical system where multiple masses are connected to each other in a way that affects their motion. This can include systems such as pendulums or springs connected in series or parallel.

How are odd form eigenvalues used in physics?

Odd form eigenvalues are used in physics to analyze systems of coupled masses and determine their natural frequencies of oscillation. This is important in understanding the behavior of physical systems and predicting how they will respond to external forces.

Can odd form eigenvalues have real parts?

Yes, odd form eigenvalues can have real parts, but they always come in complex conjugate pairs with equal real parts and opposite imaginary parts. This is because the equations used to solve systems of coupled masses involve complex numbers, resulting in complex eigenvalues.

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