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I'm reading this paper Quantum Time Crystals, Frank Wilczek, Center for Theoretical Physics, about time crystals, and in the situation described there is a particle of unit mass and charge $q$, confined to a ring on which there is a current of $\frac{2\pi\alpha}{q}$.

Abstract Quantum Time Crystals Frank Wilczek Center for Theoretical Physics arXiv:1202.2539v2 [quant-ph] 11 Jul 2012 Some subtleties and apparent diculties associated with the notion of spontaneous breaking of time translation symmetry in quantum mechanics are identied and resolved. A model exhibiting that phenomenon is displayed.

The paper states the Lagrangian and angular momentum of the particle to be: $$L=\frac{1}{2}\dot{\phi}^2+\alpha\dot{\phi}, \qquad \pi_\phi=\dot{\phi}+\alpha \tag{2}$$


  1. My question is how these equations were developed? I understand basic analytical mechanics so I guess just electric part is new to me,
  2. What is the parameter $\alpha$?
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  • $\begingroup$ Hi proton I added an abstract, as people sometimes like to know more before going off site, and you may consider explicity stating that a PDF download is involved, otherwise users will think your link is faulty. I did :). Best of luck with it $\endgroup$ – user108787 Nov 5 '16 at 0:17
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The Lagrangian for a charged particle in a field is written this way:

$$L=\frac{1}{2}m\dot{\boldsymbol{r}}\cdot\dot{\boldsymbol{r}}+q(\boldsymbol{A}\cdot\dot{\boldsymbol{r}}-\phi)$$

where $\boldsymbol{A}$ is the magnetic vector potential, and $\phi$ is the electric scalar potential. Now, given a magnetic flux $\Phi_{B}$, the magnetic vector potential can be obtained with the formula:

$$\Phi_{B}=\oint \boldsymbol{A}\cdot d\boldsymbol{l}$$

  1. Notice that the paper you post, states that $2\pi\alpha/q$ is a flux passing through the ring, so in that case the vector potential would be like:

    $$A_{\varphi}=\Phi_{B}/2\pi$$

    where we are using the asumption that the ring has unit radius. Also, in the Lagrangian we would have $m=1$ and $\phi=0$ (there is no electric field). Because the particle is confined to the ring, it has only 1 degree of freedom, which can be described by the coordinate $\varphi$. With all of this, you recover the Lagrangian given in the paper:

    $$L=\frac{1}{2}\dot{\varphi}^2+q\bigg(\frac{\Phi_{B}}{2\pi}\dot{\varphi}\bigg)=\frac{1}{2}\dot{\varphi}^2+\alpha\dot{\varphi}$$

  2. I think the parameter $\alpha$ is exactly that... a parameter. Of course it must have the correct dimensions given by $\Phi_{B}=2\pi\alpha/q$.

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