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I was going trough the paper "Classical Time Crystals" by A. Shapere and F. Wilczek, Physical Review Letters 109 (2012).

At the beginning they state that it is easy to construct Hamiltonians or Lagrangians whose lowest-energy state is a spatial crystal. Quoting, "With $\phi (x)$ and angular variable, the potential energy functions $$ V_1(\phi) = - \kappa_1 \frac{\mathrm{d} \phi}{\mathrm{d}x} + \frac{\lambda_1}{2}\Big(\frac{\mathrm{d} \phi}{\mathrm{d}x}\Big)^2 $$ $$V_2(\phi) = - \frac{\kappa_2}{2} \Big(\frac{\mathrm{d} \phi}{\mathrm{d}x}\Big)^2 +\frac{\lambda_2}{4} \Big(\frac{\mathrm{d} \phi}{\mathrm{d}x}\Big)^4$$

with all the Greek coefficients positive, are minimised for $\frac{\mathrm{d} \phi_1}{\mathrm{d}x} = \frac{\kappa_1}{\lambda_1} $, $\frac{\mathrm{d} \phi_2}{\mathrm{d}x} = \pm \sqrt{\frac{\kappa_2}{\lambda_2}}$ respectively".

Now, I shamelessly fail to understand why these solutions would represent a crystal. I have seen some Hamiltonians with periodic potential energies used to represent crystals, but nothing like such simple ones. Let us take the first one, $\frac{\mathrm{d} \phi_1}{\mathrm{d}x} = \frac{\kappa_1}{\lambda_1} $, it represents a continuous change, at constant rate, of an angular variable. I fail to see where is the broken symmetry.

As a basic example for laymen like myself, Kepler elliptic solution is quoted as a symmetry breaking example, with respect to the original, circularly-symmetric equation. In this sense, a circular orbit is not symmetry-breaking.

The solution reported in the paper, the constant change of an angular variable, seems to me analogous to a circular orbit.

Would be grateful if anybody could point out what is that I am missing, thanks.

My best attempt to understand is as follows. The Hamiltonian in question is translationally-invariant, continuosly. As the angular variable grows linearly, the solution is periodic, which breaks the continuous symmetry. Is this correct?

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In the first example with potential $V_1(\phi)$, the symmetry being broken is the spatial translation symmetry, $x \rightarrow x + a$. The potential doesn't change under this transformation, but since $d\phi/dx$ is non-zero, $\phi(x)$ does change when we change $x$. In other words, $\phi(x + a) \neq \phi(x)$. This symmetry is also broken in the second example in addition to the inversion symmetry as mentioned by the authors.

I wonder if the description of $\phi$ as an angular variable is causing some confusion. I think it would help to just think about $\phi(x)$ as some function defined along the $x$-axis, without worrying necessarily about what it represents. It should be clear then that this function changes as we change $x$, since the rate of change of $\phi$ with $x$ is non-zero.

Note that in the Kepler problem, the hamiltonian is symmetric under rotational symmetry in three dimensions. For the case of a circular orbit, rotational symmetry in the plane of the circle is preserved, but the full rotational symmetry is broken because a preferred direction is selected by the angular momentum vector.

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  • $\begingroup$ thanks makes sense. Following your explanation, one must ask then, is there any non-trivial, constant function that preserves translational symmetry? If breaking $\phi(x+a) \neq \phi(x)$ is sufficient, then one wonders if any "iinteresting" function at all exists, that does not break it. Thanks $\endgroup$
    – Smerdjakov
    Aug 5, 2021 at 8:15

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