# Are subatomic movements random

1. Jun 3, 2013

### susskind99

In the Cosmic Landscape, Susskind writes:

My question is if subatomic movements are random then does the conservation of momentum law break down at the quantum level? My hunch is yes. The conservation laws only apply at the classical level. Some people say that Noether's Theorem proves the conservation laws but that was devised in 1918 before the HUP was devised.

2. Jun 3, 2013

Whenever a particle is confined it doesn't have a well defined momentum. Momentum in a certain understanding doesn't make sense for a confined particle. The Schrödinger equation conserves momentum therefore it is conserved, but if you start with a particle which is confined the equation will conserve the probability distribution of the momentum. The particle doesn't have one momentum which changes but it never had a well defined momentum to begin with.

3. Jun 3, 2013

### VantagePoint72

Classical conservation laws still apply at the quantum level.

If you prepare a state in a momentum eigenstate and the momentum operator commutes with the Hamiltonian (the requirement for momentum conservation), then it will stay in that eigenstate as it time-evolves. Since, in practice, we don't work with particles in momentum eigenstates, there is uncertainty about its value before a measurement is made. Thus, we usually discuss conservation laws in terms of the expectation values of the corresponding operators.

In the Heisenberg picture of quantum mechanics (where operators depend on time), the momentum operator associated with the system is exactly conserved when the corresponding classical momentum would be. The proof comes via Noether's theorem, just as in the classical case. As a consequence, the expectation value of momentum measurements also obey conservation of momentum, and it evolves in time due to external forces exactly as described by Newton's second law law. Similarly for quantum fields, with the appropriate generalization to relativistic momentum. Every particle is associated with a quantum field, and that field has a conjugate momentum that is conserved in the same circumstances as a classical field would be. Noether's theorem is the proof.

4. Jun 3, 2013

### daveyrocket

Noether's theorem derives conservation laws from symmetries of the Hamiltonian (or Lagrangian). If you take the traditional example of dropping a ball with a constant gravitational field in the y direction, $H = p_x^2/2m + p_y^2/2m + mgy$, the you will find with Noether's theorem that momentum in the y direction isn't conserved because of the external potential. The key idea here is if you neglect the change in momentum of the earth that occurs when the ball's gravity pulls on the earth, you no longer get conservation of momentum (in that direction).

If you rewrite the example to be contain the earth and use a central potential between the earth and the ball, $H = p_b^2/2m_b + p_E^2/2m_E + Gm_bm_E/(r_b - r_E)$ then you find that total momentum is conserved via Noether's theorem. In classical mechanics you check the Poisson bracket {P,H} where P = p_b + p_E.

The same thing is true in quantum mechanics. If you have an external potential, you won't get any momentum eigenstate so you won't have momentum be conserved. This is because your particle is constantly interacting with its environment through the potential or the boundary conditions and you've neglected any effect the particle has on the environment. Thus, momentum is not conserved.

If you write a quantum Hamiltonian with two particles that interact with each other through a central potential, you will find that the total momentum is conserved. This is because $[P,H] = [P,V(r_1-r_2)] = [p_1,V(r_1-r_2)] + [p_2,V(r_1-r_2)] = 0$. That last quantity is zero because the potential depends only on the difference $r_1-r_2$ which, if you assume any sort of series expansion of V you will find that $[p_1,V(r_1-r_2)] = -[p_2,V(r_1-r_2)]$ since $[p_1,r_2] = 0$.

5. Jun 3, 2013

### Chronos

The problem is nearly all of physics involves large numbers of particles. Statistical mechanics is the only known comprehensible way to model such systems. The motion of individual particles may be deterministic, but, rapidly exceeds human ability to comprehend when other particles are added to the mix. It is still frightfully difficult to analytically solve even 'simple' 3 body problems.

6. Jun 4, 2013

### susskind99

I have a real hard time believing that. If the momentum cannot even be defined at the quantum level then how do you know it's deterministic? Plus in the text Susskind clearly says that atomic movements are random.

Also, to other members, I didn't get an answer regarding whether subatomic movements are random.

7. Jun 4, 2013

### daveyrocket

The thing is, when you say "subatomic movements are random" it implies that things like position and momentum have well defined values all the time and you can track them well enough to say "see, the motion is random!" But that isn't true. You can't identify these quantities without taking a measurement, and taking a measurement perturbs the system (causes a wavefunction collapse). The results of measurements are random, although the statistics of those measurements is determined by the wavefunction.

8. Jun 4, 2013

### Staff: Mentor

Yea - interesting questions.

First what subatomic objects are doing, or even what properties they have, when not being observed QM is silent about - but various interpretations have different takes.

Also when you observe it in general you cant predict the outcome - only probabilities. These are determined by the system state and a mathematical object called an operator that depends on what you are measuring. The operator for the total momentum (or you can consider the operator fixed and the state changing) is in general conserved - ie remains the same. But the actual measurements are still only given by probabilities.

Thanks
Bill

Last edited: Jun 4, 2013
9. Jun 4, 2013

### susskind99

See, that's what I thought. If you can't know the properties of an object then how do they know that their movements are random?

At the same time, the structure that gives rise to the galaxies supposedly is determined by the UP. And the distribution of the galaxies throughout the universe is random so it stands to reason that that which galaxies are composed of is also random.

10. Jun 4, 2013

### Crazymechanic

Maybe it is not random and there is a great chance it isn't , but to us and our atleast current understanding it seems random , just like when you were a child many things seemed different to you than they seem now when you grow up.
Some things never make sense , that's life.

11. Jun 4, 2013

### VantagePoint72

You didn't get an answer to that because it's not what you asked. What you asked was:

That is a very different question, and is completely independent of whether or not quantum motion is truly random. The answer is that even if they are truly random (in sense permitted by QM's formalism) then conservation of momentum still doesn't break down at the quantum level.

12. Jun 5, 2013

### daveyrocket

Saying that atomic movements are random is illustrative, but imprecise. In the paragraphs that you quote in the original post, the author is making an analogy with classical physics.

In classical physics, you can construct a model of atomic motions by taking a bunch of billiard balls and giving them some sort of attractive potential (pairwise is easiest). If you do this, you will see that at temperature their motions are more-or-less random. They're not actually random, they're governed by Newton's laws, but with many particles, the correlations in their movements have a very short time scale. So their motions might as well be random, and as far as doing statistical mechanics is concerned, that's a reasonably good approximation for the right temperatures and pressures.

If one considers a harmonic approximation for the interatomic potentials, then in the classical system you can compute the normal modes of the system of atoms. These are essentially the standing waves of the atoms. These are also the quantum vibrational modes of the system of atoms. So the analogy with classical physics is not so farfetched. The only thing is, the quantum oscillator does not run out of kinetic energy at zero temperature.

13. Jun 7, 2013

### Khashishi

Conservation of momentum is not broken by uncertainty. That's a misunderstanding of how uncertainty works. If we have a closed box with two particles inside, and we know for certain that the total momentum in the box is 2, if we have uncertainty in the momentum of one particle, we have a corresponding uncertainty in the other particle. If we know one particle has a 20% amplitude of having a momentum of 1.5, then the other particle has a 20% amplitude of having a momentum of 0.5, but the total is still 2. If we have uncertainty in the total momentum of the box, that uncertainty is still conserved if you consider a bigger box where the momentum is known. The momenta of the contents of the box will still add up in a mathematically consistent way. (Think of the wavefunction as a set of possible scenarios. Each possible scenario conserves momentum, but you don't know which scenario you are in, or you are in some combination of multiple scenarios.)