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I have trying to show that the continuum limit of N quantum harmonic oscillators gives rise the the klein-gordon field. However, instead of a usual finite string, I want to do it on a ring. Hence, my Lagrangian is

$$L=\frac{m}{2}(\dot{q_1}^2+\dot{q_2}^2+.... \dot{q_n}^2)-\frac{m \omega^2}{2}[(q_1-q_2)^2+(q_2^2-q_3^2)+....(q_n-q_1)^2]$$

So that the matrix for V is

$V=\begin{pmatrix} 2 & -1 &0 &. &. &. &-1\\-1&2&-1\\0 &-1 &2 &-1 \\.\\.\\.\\-1 &&.&.&.&-1 & 2 \end{pmatrix}$

So that $L=\frac{m}{2}[\dot{x}^2-\omega^2 x^{T}V{x}]$. All the quantities here are matrices.

How do I find the eigenvalues of this matrix?

I tried to find a recursion relation between the characteristic polynomial of $N$ and $N-1$ dimensional matrix, but I failed. Is this the correct method? What other method is there?

After finding the eigenvalues, the lagrangian can be written separated into its normal modes, and the propagator or kernel can be found out easily using that of the free particle. The limit of this as $N\to \infty$ should be the klein gordon field.

But I am stuck on this. Any help will be appreciated.

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In some cases (and I think that yours can also work out), it is sufficient to use the ansatz $q_l=\Re\exp(ikl-i\omega t),\,1\leq l\leq n$. Try to solve for $k$ and show that you have just the right number of solutions ($n$ solutions). Note, however, that in your case you have to consider the zero mode $q_l=vt$ -- the rigit rotation -- separately, so you have to look for $n-1$ solutions only. –  Peter Kravchuk May 25 '13 at 11:21
    
First, remove the "2" diagonal. Then you have a matrix $V′$ with $V′=(a+a^{−1})$ where $a^n=Id$. So, you have easily the eigenvalues of $a$. And a eigenvector of a is also an eigeinvector of $V'$. So you have the eigenvalues of $V'$, and adding 2, of $V$. –  Trimok May 25 '13 at 11:26
    
Sorry, I forgot a $(-1)^n$ factor : $a^n = (-1)^n Id$ –  Trimok May 25 '13 at 11:34
    
@Trimok, this is nice, but you have to invent some argument in order to show that there is no degeneracy and you have every root of unity in the spectrum of $a$. –  Peter Kravchuk May 25 '13 at 11:47
    
I wonder whether there is any deep reason the matrix $V$ is the extended Cartan matrix for $\mathfrak{sl}(N,\mathbb C) = \mathfrak{su}(N)^{\mathbb C}$, or more accurately the affine Lie algebra $\hat{\mathfrak{sl}}(N,\mathbb C)$. –  Heidar May 26 '13 at 22:35
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1 Answer 1

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You want to find the eigenfrequences of this system. First, note the existence of zero mode: $$ q_l=vt+\phi_l, $$ it is the 'equilibrium' rotating around with arbitrary velocity.

Next, we have the equations $$ \ddot{q}_l=-\omega^2(2q_l-q_{l-1}-q_{l+1}) $$ and the periodicity condition $q_{l+n}=q_l$. Let us use the ansatz for the eigenvectors $$ q_l=\Re A\exp(ikl-i\Omega t). $$ Substituition reads $$ \Omega^2=\omega^2(2-e^{-ik}-e^{ik})=2\omega^2(1-\cos k), $$ while the periodicity condition is $nk=2\pi m,\,m\in\mathbb{Z}$. Let us restrict $k$ to $(0,2\pi)$, in order to exclude double-counting the eigenvectors and the zero mode. Then $m=1,\dots,n-1$, and the eigenfrequency reads: $$ \Omega^2=2\omega^2(1-\cos \frac{2\pi m}{n}),\,m\in\{1\dots n-1\}. $$ Thus we have $n-1$ vectors of the form $q_l=\Re A\exp(ikl-i\Omega t)$ with the above eigenfrequency, and one zero mode given by $q_l=vt+l\delta$, $\delta$ being the equilibrium distance, with zero frequency. So, the full spectrum reads as $$ \Omega^2=2\omega^2(1-\cos \frac{2\pi m}{n}),\,m\in\{0\dots n-1\}. $$ (Note that this is the full spectrum since we have found all $n$ eigenvectors).

Edit: note that while $n-m$ corresponds to the same eigenvalue as $m$, we have two different eigenvectors for each eigenvalue, because $A$ can be complex. E.g. we can take $q_l=\cos(kl-\Omega t)$ and $q_l=\sin(kl-\Omega t)$, $k=\frac{2\pi m}{n}$.

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