# If wave packets spread, why don't objects disappear?

If you have an electron moving in empty space, it will be represented by a wave packet. But packets can spread over time, that is, their width increases, with it's uncertainty in position increasing. Now, if I throw a basketball, why doesn't the basketball's packet spread as well? Wouldn't that cause its uncertainty in position to increase so much to the point it disappears?

EDIT: I realize I wasn't clear what I meant by disappear. Basically, suppose the wave packet is spread over the entire Solar System. Your field of vision covers only an extremely tiny part of the Solar System. Therefore, the probability that you will find the basketball that you threw in your field of vision is very small.

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Are you asking why macroscopic objects don't show quantum behaviour? If so, have a quick search of this site as that question has been asked many times. Try a search for "decoherence". –  John Rennie Nov 10 '12 at 11:07
–  Qmechanic Nov 10 '12 at 20:20
because the wavepackets also collapse. –  Ron Maimon Nov 11 '12 at 4:17

It is true that the spreading depends on the mass as @twistor59 has already noticed, but the more important fact is that the basketball is an open system and interaction with its surrounds makes that (due to decoherence) the state of the basketball is not described by quantum wavefunction theory [*]. Using the Wigner-Moyal formulation of quantum mechanics it is possible to show that the basketball always have a well-define position $x(t)$ at each instant.

[*] Wavefunction theory only applies to isolated quantum systems.

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What if throw the basketball in a perfect vacuum, completely empty of anything or any field? –  Ignacio Nov 10 '12 at 17:50
@Ignacio: The basketball is not in a pure state. If you manage to put it in a perfect vacuum (which is impossible) it will remain in the same non-pure state because a perfect vacuum cannot change a state. –  juanrga Nov 10 '12 at 18:12
@juanrga: So when you cut all "interaction with its surrounds", it remains in a mixed state? –  Vladimir Kalitvianski Nov 10 '12 at 19:30
@VladimirKalitvianski: I cannot say. It depends on the system, the kind of mixed state, and the dynamics. –  juanrga Nov 11 '12 at 12:55
@juanrga So from what I understand it's wrong to say that macroscopic objects exhibit quantum behaviour, but with effects that are too small to be detected. Instead, they really aren't quantum at all. Is this correct? –  Ignacio Nov 11 '12 at 20:33
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AFAIK there is no answer to this question, that is why there are few "theories" that try to answer this question like GRW

GRW

EDIT:

Let me just elaborate because I gave you the most interesting answer without directly addressing your question. There is no need for you to throw the ball. The atoms of the ball are constrained by their mutual potential so the waves do not spread in the sense of free particles. @Twistor59 answer is heuristic usually given in textbooks, but obviously there are no 100 g particles as such. The main issue is why we don’t see the ball at 100 meters or whatever since wavefunctions have non zero probability throughout space. The measurement problem is a close relation to your “proper “question but a bit different. GRW is more concerned with that.

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Spreading the wave packet does not mean spreading and disappearing the electron.

If a basketball has initially uncertainty $\Delta V$ of its velocity, then with time the ball position uncertainty will grow as $\Delta V\cdot t$.. With time this uncertainty gets so large that the ball disappears from your sight (I mean you will not find it where you expect it to be without knowing the initial velocity spread).

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Slightly longer answer is that, if I start with a Gaussian wavepacket with width $a$, then after time $t$, the width will have spread to $$\sqrt{\frac{a^2+\hbar^2t^2/m^2}{a}}$$ The incredible smallness of $\hbar$ makes the spread negligble for something as massive as a basketball.
That's true, but if you put some numbers in ($\sqrt{a}=10^{-10}meter; m=0.1kg$, you can calculate how many times the age of the universe you have to wait for it to spread by a significant fraction. The $\hbar$ kills you. (However, @juanrga 's argument is probably stronger). –  twistor59 Nov 10 '12 at 18:21