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Consider an ion lattice of N atoms, modelled as a linear chain of length L

$$ H = \sum_n \big( \frac{p_n^2}{2m} +\frac{m \omega^2}{2}(x_n-x_{n-1})^2 \big).$$

The Hamiltonian diagonalizes by going to fourier coordinates $$ x_n = \frac{1}{\sqrt N}\sum_q x_q e^{iqna}, \, p_n = \frac{1}{\sqrt N}\sum_q p_q e^{iqna}.$$ where $a$ is the equilibrium distance between the atoms. Introducing the creation and annihilation operators $$ b_q= \sqrt{\frac{m \omega_q}{2}}x_q+i\frac{1}{\sqrt{2m\omega_q}}p_q, \, b^\dagger_q=\sqrt{\frac{m \omega_q}{2}}x_{-q}-i\frac{1}{\sqrt{2m\omega_q}}p_{-q} \\ H=\sum_q\omega_q(b^\dagger_qb_q+\frac{1}{2}),$$ with $$\omega_q=2\omega |Sin(\frac{qa}{2})|,$$ and the well known occupation number eigenstates.

By writing $$x_n = \sum_q (\frac{1}{2mN\omega_q})^{1/2}(b_q \, e^{iqna}+b^\dagger_q \, e^{-iqna})$$

we can calculate the fluctuations $\langle x_n^2\rangle $ in the ground state

$$ \langle x_n^2\rangle=\langle 0|\sum_{qp}\frac{1}{2mN\sqrt{\omega_q \omega_p}}(b_q \, e^{iqna}+b^\dagger_q \, e^{-iqna})(b_p \, e^{ipna}+b^\dagger_p \, e^{-ipna}) |0 \rangle \\ =\sum_q \frac{1}{2mN \omega_q}. $$

In the thermodynamic limit, $L\rightarrow \infty, N/L \rightarrow$ const, the sum can be written as an integral $$\langle x_n^2\rangle = \frac{L}{N} \frac{1}{8\pi m\omega}\int_{-\pi/a}^{\pi/a} \frac{1}{|Sin(\frac{qa}{2})|} = \infty $$

Thus, the variance of $x_n$ shows an infrared divergence due to the quantum groundstate fluctuations of the infinite number of modes. We have a vacuum catastrophy, analogous to the free scalar field in QFT. However here the operator in question is the completely regular position operator, as compared to the singular $\phi$ in QFT.

How do we make sense of this?

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    $\begingroup$ Related post by OP: physics.stackexchange.com/q/599889/2451 $\endgroup$
    – Qmechanic
    Commented Dec 11, 2020 at 16:42
  • $\begingroup$ In the classical system you can readily devise states that evolve to have unbounded positions for all particles. From my point of view the quantum system is similar enough. The interparticle potential does nothing to prevent a single-particle position to become unbounded,it just constrains the relative positions. You could have similar phenomenon for a single particle in 1D; the states there would be in $L^2(\mathcal{R})$ but of course some of them are not square integrable when you multiply them by $x^2$ to get the variance of the position. commenting bc idk if it's enough for an answer $\endgroup$
    – tbt
    Commented Dec 11, 2020 at 17:02
  • $\begingroup$ The $x_n$ are the relative positions. And the conclusion is the same for any eigenstate of the Hamiltonian, the fluctuations will always diverge... it seems to me like a complete failure of the theory. What are we supposed to make of a model that predicts that all our position measurements are infinite? $\endgroup$
    – curio
    Commented Dec 11, 2020 at 18:15
  • $\begingroup$ Are you sure $x_n$ is a relative position between two particles and not a displacement from a fixed point? $x_n$ being a relative position is inconsistent with your definition of $a$ as equilibrium distance between atoms because from the hamiltonian the equilibrium distance is zero. The fluctuation of the actual relative positions/displacements $\langle (x_n -x_{n+1})^2\rangle $ does not look divergent to me $\endgroup$
    – tbt
    Commented Dec 14, 2020 at 13:54
  • $\begingroup$ Sorry, I mean $x_n$ is the positon of the nth atom relative to its equilibrium positon. And since all those have large fluctuations, also the relative positions must have large fluctuations. $\endgroup$
    – curio
    Commented Dec 14, 2020 at 17:43

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