What is the symmetry that corresponds to conservation of position?

We know that conserved quantities are associated with certain symmetries. For example conservation of momentum is associated with translational invariance, and conservation of angular momentum is associated with rotational invariance.

Now, if the particle position does not change, then the position of the particle is a conserved quantity. What is the symmetry that corresponds to conservation of position in this case?

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Conservation of impulse? Nothing can change speed without external impulse, so if the initial speed was zero, it will remain zero forever. –  Anixx Jan 2 '12 at 18:04
I don't think that trying frame inertia in the form of a conservation law is helpful or particularly meaningful. For one thing it would only apply in one frame. –  dmckee Jan 2 '12 at 18:29

Nature doesn't have this symmetry because your conservation law doesn't hold, either. According to the law of inertia, object keeps on moving with a constant velocity – which is however generically nonzero. In its own rest frame, it's zero, but in other frames, the velocity is nonzero.

If one studies the motion of the center-of-mass, it is indeed moving with a constant velocity. So the conserved quantity that is closest to your "conserved position" is the conserved velocity of the center-of-mass. This conservation law is directly linked, via Noether's theorem, to the Lorentz symmetry of the laws of physics – or, in the non-relativistic limit, to the Galilean symmetry. In the non-relativistic case, the generator of the Galilean symmetry is $\vec x_{\rm cm}$, the center-of-mass position, indeed: the generator of the symmetry is the conserved quantity itself.

If you designed boring laws in which the position has to be conserved, the symmetry would be generated by the conserved quantity $\vec x$. This symmetry generator generates translations in the momentum space. So the laws of physics (the Hamiltonian) would have to be effectively independent of the momentum. That would be pretty bad: you couldn't include the kinetic energy term to the total energy, among other things. That's related to the fact that the particles would have "infinite inertial mass", which would force them to sit at a single point. The whole term "dynamics" would be a kind of oxymoron because things wouldn't be changing with time.

Appendix

Consider the generator equal to the center-of-mass position $$\vec x_{\rm cm} = \frac{m_1 \vec x_1 + m_2 \vec x_2 +\dots +m_N \vec x_N}{m_1+m_2+\dots +m_N}$$ How do physical observables transform under the symmetry generated by it? Compute the commutators. The commutators of the position above with positions $x_i$ vanish, so positions (at $t=0$) don't transform. However, the commutator with $p_i$ is equal to $m_i \delta_{mn} / M_{\rm total}$, and if this is added to $p_i$ with an infinitesimal coefficient $\vec \epsilon \cdot M_{\rm total}$, you see that all velocities are changed by $$\vec v_i \to \vec v_i + \vec \epsilon$$ But if all velocities are just shifted by a constant, that's the Galilean transformation. let me emphasize that this simple transformation rule only holds at $t=0$. For $t\neq 0$, one would have to add extra terms proportional to $t$ to the generator (they would be similar for a Lorentz symmetry, too), namely $t\cdot \vec P_{\rm total}$. At any rate, the center-of-mass position is the generator of the Galilean transformations, the transformations switching from one inertial system to a nearby inertial system (which moves by a speed differing by $\delta \vec v$).

Note that the commutator of $\vec x_{\rm cm}$ with the Hamiltonian isn't quite zero, so according to some definitions, it isn't a symmetry. Instead, the commutator is proportional to the total momentum $\vec p$ which is a symmetry itself. So the commutators of various generators yield other generators – the standard form of a Lie algebra (Galilean/Lorentz in this case) in which the Hamiltonian isn't necessarily commuting with everyone else but is one of the generators of a non-Abelian group.

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Dear Lubos, thank you so much for such a beautifully crystal clear answer. –  Revo Jan 2 '12 at 19:59
Dear Lubos. I was wondering if I can generalize your answer by saying that not every conserved quantity can correspond to a useful symmetry. I mean, with the same token if a system is such that its temperature or number of particles is constant (such as in the canonical ensemble), the corresponding symmetries to T and N, if we can define them (I do not know what would that be in this case), would be useless as well? –  Revo Jan 3 '12 at 1:33
"I can generalize your answer by saying that not every conserved quantity can correspond to a useful symmetry" He just said you that the position is not conserved, otherwise the things could not move. –  Anixx Jan 3 '12 at 6:41
Dear @Revo, "I was wondering if I can generalize your answer by saying that not every conserved quantity can correspond to a useful symmetry." - My point was exactly the opposite, it was Noether's theorem. Whenever there is a [useful] conserved quantity, there is a [useful] symmetry, and vice versa. This rule is known as Noether's theorem. In this particular case, neither the symmetry nor the conservation law was valid for a sensible physical system. –  Luboš Motl Jan 3 '12 at 9:28
Constants such as $c$ or $\hbar$ or $\pi$, for that matter, are not considered conserved quantities because they're not really "quantities". By a quantity, we a priori mean a nonconstant function of the observables describing the system, such as $x_i$ and $p_i$, i.e. things that become operators (not $c$-numbers) in quantum mechanics. So the constants you mentioned are not conserved quantities. To say the least, they aren't independent conserved quantities. –  Luboš Motl Jan 3 '12 at 9:29

1) OP wrote:

What is the symmetry that corresponds to conservation of position?

For a Hamiltonian system, one can formally exchange the role of positions $q^i$ and momenta $p_i$ via a canonical transformation

$$Q^i ~=~ p_i , \qquad P_j~=~ - q^j .$$

Thus, in Hamiltonian systems, what you call positions and what you call momenta is up to you.

Now, I assume that OP is familiar with the statement, that translation symmetry in position space $q^i\to q^i+a^i$ leads to momentum conservation (at the classical level).

By $q\leftrightarrow p$ symmetry, one can then argue that translation symmetry in momentum space $p_i\to p_i+a_i$ leads to position conservation (at the classical level).

2) OP wrote further:

Now, if the particle position does not change, then the position of the particle is a conserved quantity. What is the symmetry that corresponds to conservation of position in this case?

One should differ between a symmetry of the action and a spontaneous symmetry, i.e. a symmetry of the state the system is in. Noether's theorem only applies to the former situation. For instance, a free particle $H=\frac{{\bf p}^2}{2m}$ could accidentally be at rest (wrt. some reference frame), although there is no translation symmetry in momentum space associated with it.

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Translation in momentum is galilean invariance, so this answer is really the same as Lubos's. –  Ron Maimon Jan 3 '12 at 7:59
I updated the answer. –  Qmechanic Jan 3 '12 at 12:44