# Can conservation of momentum be violated?

The law of the conservation of momentum has been established for hundred of years. Even in Quantum field theory every particle collision must be momentum-conserving if there is homogenity in space. Can this theorem still be violated?

If yes, what requirements must have a momentum-non-conserving theory? Is Heisenberg's uncertainity principle $\Delta x \Delta p \geq \frac{\hbar}{2}$ the possible answer? (when one considers physical Systems in which $\Delta x$ is very small)?

• momentum isn't conserved globally in GR – Jim Apr 18 '15 at 14:29

If the theory is invariant under translations in space, then linear momentum is conserved by Noether's theorem. If the theory is quantum, conservation holds only on the level of the expectation values (because that's the only meaningful level where you can talk about momentum as a number that's conserved in time), but it still holds.

There is no way out. You must break homogeneity/translation invariance to break momentum conservation. Heisenberg's uncertainty principle has nothing to do with it, as it is just a statement about standard deviations, not expectation values, and hence has no influence on the quantum version of conservation.

• And how one can add inhomogenity of space in a theory without violating elementary consistence conditions? – kryomaxim Feb 26 '15 at 12:58
• You say that it's only meaningful to talk about expectation values, but isn't the whole probability spectrum conserved? Conservation of the spectrum would be a much stronger condition. – kristjan Feb 26 '15 at 12:59
• @kryomaxim: I did not claim that that is possible. – ACuriousMind Feb 26 '15 at 13:03
• @kristjan: I didn't really say that it is the only meaningful level, it's the only meaningful level at which momentum is a number. The full quantum version of Noether's theorem are the Ward identities. I don't understand what you mean by "probability spectrum" or its conservation. – ACuriousMind Feb 26 '15 at 13:05
• @kryomaxim: Nothing is an absolutely true fact. That does not change the fact that there is no way within current scientific knowledge to write down a consistent theory that does not obey Noether's theorem and/or has a non-conserved momentum, nor is there any experimental indication that we should look for such a theory. – ACuriousMind Feb 26 '15 at 13:23

From my readings; the key to conservation of momentum appears to be based on defining a closed system to see if any mass crosses the boundaries of the system.

If the quantum state has no defined momentum, as in the considered case, the standard law of conservation of momentum cannot applied evidently. What is conserved during the time evolution of the state, provided the Hamiltonian operator is translationally invariant (in particular it is the free Hamiltonian) is the probabilty of measuring a certain momentum, for every choice of that value of the momentum.

In particular, the expectation value is conserved together with all moments of the distribution of possible outcomes of the measurement of the momentum observable as well as the quantum implementation of the Noether theorem (regarding translational invariance).

In fact, spatial homogeneity aka translational invariance of the quantum dynamics of a given quantum system means $$V_x H V_x^\dagger = H \tag{0}\quad \mbox{for all x\in \mathbb{R},}$$ where $$V_x$$ is the unitary implementation of translations and $$H$$ the Hamiltonian operator of the system. Eq.(0) implies (actually is equivalent to) $$V_x U_t = U_t V_x\tag{1}\quad \mbox{for all x,t \in \mathbb{R}}$$ where $$U_t := e^{-itH}$$ is the time evolutor of states. By definition the generator of $$V_x$$ is the momentum $$P$$ (along $$x$$) $$V_x = e^{-ixP}$$ (1) is equivalent to $$e^{-i xU_t^\dagger P U_t} = e^{-ixP}\mbox{for all x,t \in \mathbb{R}}$$ which, in turn, via Stone's theorem is equivalent to $$U_t P U_t^\dagger = P\:.\tag{2}$$ Since $$U_t$$ is unitary (thus bounded) and decomposing according to its spectral decomposition $$P = \int_{\mathbb{R}} p \:dQ^{(P)}(p)$$, (2) implies $$U_t Q_E^{(P)}U_t^\dagger = Q_E^{(P)}\tag{3}$$ for every (Borel) set $$E$$ for instance of the form $$(a,b)$$. If $$\psi_0$$ is the normalized state vector at time $$0$$, $$\langle \psi_0 |Q_E^{(P)} \psi_0 \rangle = \|Q_E^{(P)} \psi_0\|^2$$ is the probability that the outcome $$p$$ of a momentum measurement at $$t=0$$ belongs to $$E$$. From (3) we finally have, defining $$\psi_t := U_t \psi_0$$ the state at time $$t$$, $$\langle \psi_0 |Q_E^{(P)} \psi_0 \rangle =\langle \psi_t |Q_E^{(P)} \psi_t \rangle\:.\tag{4}$$ In other words: the probability to find the value of $$P$$ in $$E$$ does not depend on time. It is possible to prove that this sort of invariance implies (0) and (1), so that (4) is the deepest physical meaning of translational invariance in quantum physics.

(4) implies that, for instance, the expectation value of $$P$$ is conserved along the time evolution, since (provided the integral in the right-hand side is defined) $$\langle P\rangle_{\psi_t} = \int_{\mathbb R} p \: d\langle \psi_t |Q_E^{(P)} \psi_t \rangle = \langle P\rangle_{\psi_0}\:.$$ The same result is therefore valid for every moment of the said probability distribution, $$\langle P^n\rangle_{\psi_t} = \int_{\mathbb R} p^n \: d\langle \psi_t |Q_E^{(P)} \psi_t \rangle =\langle P^n\rangle_{\psi_0}\:,$$ so that, for instance, the standard deviation is also conserved in time.

Conservation of momentum & energy don't hold true in a non-uniform gravitational field. Lets think 2 masses A & B free falls. Mass A is situated far away from mass B on earth such the non uniform nature of the gravitational field becomes significant. According to mass A, when mass B free falls under gravity, mass A thinks mass B is accelerating, instead of finding mass B at rest with respect to mass A. This means mass A thinks mass B is gaining momentum out of nowhere. Really the momentum that appears-to-be-gained is due to its motion in curved spacetime. Motion in curved line even if its with uniform velocity has acceleration as there is change in the direction of velocity. So when there is acceleration, velocity changes. When there is change in velocity, momentum changes. It is as if the spacetime is giving the mass B a gain in momentum. The same goes for energy. Mass A thinks mass B is gaining energy from nowhere or as if the spacetime is giving it (though again its the same aforementioned reason). (energy=momentum*velocity). So its according to mass A that there is violation of conservation laws of momentum & energy concerning the motion of mass B. This is basically due to inhomogeneity of spacetime in a non uniform gravitational field.

Also both conservation laws hold true only in closed systems. If you believe or assume a system is a closed one but find violation of conservation of momentum & energy, most probably there might be an external force or external source/sink of energy that you must have missed to identify.

• Can I know if there is anything wrong with this answer? – CuriousMind9 Nov 5 '19 at 19:02