Does spatial entanglement reduce total classical entropy?

[LarryS]

In classical mechanics, if a system consisting of one particle suddenly became two particles, the entropy of the system would increase because the number of spatial degrees of freedom would double. But, in QM, I believe, when one particle decays into two particles, the two new particles would be entangled spatially. Thus the spatial entanglement could be viewed as a phenomena that reduces overall spatial entropy. Does that make sense?

Thanks in advance.

 

[PeterDonis]

if a system consisting of one particle suddenly became two particles

Can you give an example?

the entropy of the system would increase because the number of spatial degrees of freedom would double

I think you're overlooking conservation laws.

in QM, I believe, when one particle decays into two particles, the two new particles would be entangled spatially

Their states are related by conservation laws, but as above, that's also the case in classical physics.

 

[vanhees71]

I don't know what "spatial entropy" might be.

What of course happens in a two-particle decay is that the two daughter particles are entangled concerning the conserved observables. E.g., if you are in the center-momentum frame, where the mother particle is at rest, the momenta of the daughter particles are entangled, i.e., if you measure $\vec{p}_1$ for the momentum of one of the particles you necessarily must measure $\vec{p}_2=-\vec{p}_1$ for the other, no matter how far they have gone from their origin (supposed there are no interactions with other particles in the environment).

 

[LarryS]

I don't know what "spatial entropy" might be.

What of course happens in a two-particle decay is that the two daughter particles are entangled concerning the conserved observables. E.g., if you are in the center-momentum frame, where the mother particle is at rest, the momenta of the daughter particles are entangled, i.e., if you measure $\vec{p}_1$ for the momentum of one of the particles you necessarily must measure $\vec{p}_2=-\vec{p}_1$ for the other, no matter how far they have gone from their origin (supposed there are no interactions with other particles in the environment).

I understand that, because of the conservation of momentum, if you know the momentum of one of the daughter particles then you will always know the momentum of the other but is that the same thing as true quantum entanglement? To be entangled in the momentum property, the quantum states of the two daughter particles (the combined 2-particle state), when expressed in their respective momentum eigenbases, must not be separable in those momentum eigenbases, i.e. there must be at lease one complex coefficient in the tensor product of the two states that is not the product of any two coefficients in the two individual states. Does conservation of momentum imply that?

 

[jfizzix]

In classical mechanics, if a system consisting of one particle suddenly became two particles, the entropy of the system would increase because the number of spatial degrees of freedom would double. But, in QM, I believe, when one particle decays into two particles, the two new particles would be entangled spatially. Thus the spatial entanglement could be viewed as a phenomena that reduces overall spatial entropy. Does that make sense?

Thanks in advance.

In a closed system, if one particle decays into two particles, the quantum entropy of the total system will be the same before and after the decay. The single wavefunction of one particle would become a single wavefunction of a pair of particles. In general, this wavefunction is entangled due to correlations from momentum conservation and a common place of origin.

Indeed, if the position and momentum correlations are strong enough, you can prove entanglement by demonstrating the EPR paradox. This would be accomplished if the conditional position and conditional momentum uncertainty are below the Heisenberg limit.

All particles obey they Heisenberg uncertainty relation for position and momentum
\sigma(x)\sigma(p)\geq\frac{\hbar}{2}
..but the corresponding conditional relation between particles A and B can be violated:
\sigma(x_{A}|x_{B})\sigma(p_{A}|p_{B})\geq\frac{\hbar}{2}

To be clear, strong momentum correlations alone is not sufficient to prove entanglement. When a particle decays into a pair of other particles, it has high momentum correlations because of conservation of momentum, but it has high position correlations as well because they must come from the same point of origin, even though the point itself has its own position uncertainty.

 

[LarryS]

The the Spontaneous Parametric Down-Conversion experiment, the two output photons are correlated (vs anti-correlated) in position - they are spatially entangled.

I probably should have posted the following question first: In normal particle decay, are the new output particles also correlated in position?

 

[king vitamin]

In both classical and quantum statistical thermodynamics, entropy is microscopically conserved. If you track your total knowledge of the system, either through evolution in phase space in classical physics or unitary time-evolution in quantum physics, the entropy is constant.

So your example of a classical particle becoming two particles shouldn't involve increasing the entropy unless you actually "lost track" of how the decay proceeded. As other have mentioned, even if you had incomplete knowledge of the original particle, you should have information about the final two-particle state due to knowing the physics by which the decay occurred, as opposed to not knowing anything about the two particles.

Of course, if you do not know how the decay occurs microscopically, you really have lost information and entropy does increase. But the same can be said about the quantum system if you didn't know the interactions and had no way of tracking the entanglement in the final state.

 

[vanhees71]

The the Spontaneous Parametric Down-Conversion experiment, the two output photons are correlated (vs anti-correlated) in position - they are spatially entangled.

I probably should have posted the following question first: In normal particle decay, are the new output particles also correlated in position?

Photons have no position observable. So I still don't get what you mean by "position entanglemen". The spontaneous parametric-down converted photons are entangled in momentum and polarization.

 

[jfizzix]

The the Spontaneous Parametric Down-Conversion experiment, the two output photons are correlated (vs anti-correlated) in position - they are spatially entangled.

I probably should have posted the following question first: In normal particle decay, are the new output particles also correlated in position?

If you look at the two-particle wavefunction just after decay, the two particles will be correlated in position and anti-correlated in momentum.

If you take a two-particle wavefunction highly correlated in position, its width of the x1-x2 direction will be much smaller than in the x1+x2 direction. If you take the Fourier transform to express this wavefunction in momentum space, the small width in x1-x2 becomes a large width in p1-p2, and the large width in x1+x2 becomes a small width in p1+p2.

A small width in x1-x2 indicates the initial positions of the two particles, though uncertain individually, are not far from one another since they came from the same point of origin.
A small width in p1+p2 indicates the uncertainty in the total momentum of the two particles after decay is not much larger than the momentum uncertainty of the initial particle before decay.

The actual values of these widths will vary from process to process, and it is possible for the particles to be uncorrelated afterwords depending on the nature of the interaction.

 

[LarryS]

Photons have no position observable. So I still don't get what you mean by "position entanglemen". The spontaneous parametric-down converted photons are entangled in momentum and polarization.

I was referring to that which makes Quantum Imaging possible. See this article in Nature.

 

[LarryS]

Can you give an example?

I was being extremely hypothetical regarding this situation in classical mechanics. But i guess one example might be a structurally flawed bullet that broke into two pieces during flight.