Odd Form Of Eigenvalue - Coupled Masses

AI Thread Summary
The discussion centers on understanding the eigenvalue notation used in a problem involving coupled masses in classical mechanics. The original poster struggles with the sign and form of the eigenvalues, specifically why they are expressed as -ω^2 instead of ω. It is clarified that the negative sign arises from the mathematical formulation of the problem, where the eigenvalues correspond to the characteristic frequencies of the system. The conversation highlights the importance of recognizing the relationship between the eigenvalues and the physical parameters of the system, such as k/m. Ultimately, the confusion stems from a misinterpretation of the notation rather than a fundamental misunderstanding of the underlying mechanics.
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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


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"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:
<br /> \\<br /> {x}_1&#039;&#039; = -\frac{2k}{m}x + \frac{k}{m}x \\<br /> {x}_2&#039;&#039; = \frac{k}{m}x_1 - 2\frac{k}{m}x_2 \\ \\<br /> \varkappa \equiv \frac{k}{m} \\<br /> \begin{bmatrix}<br /> {x}_1&#039;&#039;\\ <br /> {x}_2&#039;&#039;<br /> \end{bmatrix} = \begin{bmatrix}<br /> -2\varkappa &amp; \varkappa \\ <br /> \varkappa &amp; -2\varkappa<br /> \end{bmatrix}\begin{bmatrix}<br /> x_1\\ <br /> x_2<br /> \end{bmatrix} \\<br /> \left|x&#039;&#039;(t)\right\rangle= \Omega \left|x(t)\right\rangle<br />

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 \omega_1 = -\varkappa and \omega_2 = -3\varkappa. The book itself gets \omega_1 = \sqrt{\varkappa} and \omega_2 = \sqrt{3\varkappa}, 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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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..
 
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.
 
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.
 
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.
 
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