# Quantum Coherence - Decoherence Question

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1. Mar 11, 2015

### burke142

So I understand that as the number of entangled particles increases, observable quantum mechanical properties decrease to the extent that the mass of particles collectively loses its wave-particle character and behaves classically.

In other words, the particles' collective position-space wavefunction is effectively collapsed.

My question is this: if a classical object is functionally a collapsed position-space wavefunction, doesn't the uncertainty principle hold that the object would have a correspondingly expanded momentum-space wavefunction?

If so, would a classical object, let's say a baseball, have a single measurable momentum space wavefunction?

2. Mar 11, 2015

### Demystifier

A wave function of a macroscopic object such as baseball will be extended in both position and momentum. The point is that both extensions are negligibly small when compared with the size of the object and average momentum of the object. So for all practical purposes you may pretend that there is no extension at all.

3. Mar 11, 2015

### Staff: Mentor

Only if the wave functions of all the individual particles are coherent - but of course they aren't, which is why we call it "decoherence".

[warning! - what follows is a very handwavy and semi-classical way of thinking about a multi-particle quantum system - do not build too much on top of it! Read the wikipedia article on quantum decoherence if you want the real thing!]

There is some uncertainty as to the position of each individual particle but on average the uncertainties cancel out. Some particles will be a bit to the left of where they "belong" and others a bit to the right, and at any given moment the numbers of both will be approximately equal so the baseball as a whole stays very close to its expected position with a very high probability.

How high? Well, the baseball contains something on the order of $10^{25}$ atoms and the probability of them all randomly "deciding" to be a noticeable distance to the left at the same time is on the order of $2^{(-10^{25})}$, a number that is unimaginably (quite literally - you can't imagine it) small.