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When we observe a quantum object does it collapse into a point? Or does it collapse into a smaller wave of area that is out of our range of accuracy? My gut tells me the latter.

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4 Answers 4

The concept of "collapse" is a misleading one. Objects which are in the dimensional range of quantum mechanics do not "collapse" like pricked deflated balloons.

Observation means any interaction a particle described by a wavefunction might have. The square of the wavefunction will tell us what the probability is to observe the particle at a specific (x,y,z,t). Elementary particles are considered as point particles and in that sense when they appear at a specific point they are a point.

One needs a distribution of several interactions to measure the probability distribution for specific reactions. This is what is being done at the LHC in CERN, measuring millions of interactions in order to study the probability distributions and compare them with predictions of theory.

But quantum mechanics describes also complex objects like molecules and crystals and even superconductors. Measurements on these objects follow the probability distributions of the complex interactions but the collective measurements will also reveal the size of atoms or molecules. A single scatter on an atom of a crystal will give a specific (x,y,z,t) point for it but a large statistical accumulation of scatterings will give a size for it which is within our measurement accuracies.

For example these nanotechnology atom arrangements:

nanotec

There are a large number of photons scattering off the atoms, each sampling the probability distribution that describes them, and they are within our measurement capacity.

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Well here is a follow up (to explain my idea a bit more clearly) lets say you did an experiment and you know the 10 m radius circle within which there is atom... your "observation" confirms that the atom is definitely not outside of that circle when you observed it... therefore at that moment in time, when you observed it the likelihood of finding the atom within the space you observed was 1... the atom was obviously where it was... but to the accuracy of a 1 cm radius circle you don't know where it is... so in that accuracy it is still a wave –  frogeyedpeas Jun 11 '13 at 20:04
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Your misconception lies in thinking of the wave as a physical wave where the energy/mass is spread in space time. The wave nature appears in the probability distribution , the square of the wave function. The wave function is not a space time distribution of the mass/energy of the particle. The particle is whole, a point if elementary, a specific spatial distribution if complex. The misunderstanding comes partly from the old double slit experiments (one hole no interference). There are newer experiments which detect the particle going through a single slit and still the pattern appears, –  anna v Jun 12 '13 at 3:29
    
See en.wikipedia.org/wiki/… .It is the change in the boundary conditions from going from one slit to two slits that changes the wavefunction and therefore the probability distributions. A one cm circle defines a classical boundary. as quantum mechanical behavior appears for particles/atoms at dimensions of order 10^-8cm or so your particle is very well defined within the circle, classically. –  anna v Jun 12 '13 at 3:31
    
I am having trouble reconciling point like particles... That would imply infinite density... For any non zero energy and therefore infinite gravitational pull for any non zero energy –  frogeyedpeas Jun 12 '13 at 4:10
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You should stop thinking classically when studying quantum mechanical entities. They are entities, sometimes appearing as particles, sometimes as probability waves. The mass spread out is a science fiction type visualizations and leads to mystical thinking. Density is not a quantum mechanical concept. –  anna v Jun 12 '13 at 4:18
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Indeed, the latter. What you measure is actually not a pure projection operator $|x\rangle \langle x|$ but something more smudged like $M_x = \int dy \, p_y |y\rangle \langle y|$. Further general measurements need not be matrices, they need to be linear operators, the most general physical form of which are completely positive maps.

This means there is some probability that your measurement assigned value $x$ to the position of an object which was actually at a different location. Operators of this form will not in general collapse the wavefunction into a point. But if you are not careful the information you did not collect will be erased by interaction with the environment, which can make your system nearly classical.

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As said by anna v, the "wave function" is not at all a real space-time wave function, it is only a probability amplitude, that is complex numbers which allow us to calculate probabilities.

The problem of "collapse thinking" comes because we often think in the Schrodinger representation, because in this representation, states depend on time.

To avoid this problem of "collapse thinking", a good way is to use the Heisenberg representation. In this representation, states do not depend on time (but operators depend on time).

In fact, in the Heisenberg representation, You can choose a matrix representation for the operators such as the states do no depend on "space", for instance, in the harmonic oscillator, the position operator is an infinite matrix which could be written : $$X(t) \sim a e^{+ i \omega t} + a^+ e^{- i \omega t}$$, with $a_{i,i+1} = \sqrt{i + 1}$, $a^+_{i + 1,i} = \sqrt{i + 1}$, with $i=0,1.....\infty$

That is, your state $|\Psi>$ is a set of constant complex numbers $\Psi_i$, $i=0,1.....\infty$, relative to some basis (the energy basis). Your state $|\Psi>$ is a complex (infinite) vector upon which the (infinite) matrix $X(t)$ acts.

From my point of view, a way to understand these $\Psi_i$ is to see them as initial conditions.

Suppose you make a measurement on this state $|\Psi>$, a measurement is an interaction. There is no reason why the initial conditions have to be conserved in this interaction.

To see a classical analogy, when you throw a ball against a wall, the initial conditions before and after the bouncing are not the same.

It is the same here, staying in the Heisenberg representation, after the measurement, there are new "initial conditions" $\Psi'_i$.

It is possible, for instance, that only one of the $\Psi'_i$ will be different of zero, for instance $\Psi'_0 \neq 0$ and $\Psi'_i = 0$ if $i \neq 0$, so you might think of some "collapse", but it is just a special expression that "initial conditions", after the measurement, can take.

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Keep in mind that, depending on which interpretation of quantum mechanics you prefer, wavefunction collapse might not even exist. Some interpretations, such as the Copenhagen interpretation (CI), have it, and some, such as the many-worlds interpretation (MWI) don't. Since all of these interpretations make the same predictions about the outcome of experiments, we can't say empirically which is true, and therefore we can't say empirically how a certain feature of one of them (wavefunction collapse) truly works.

However, within the framework of CI, there are some constraints required for consistency. It's not possible for any measurement of position to be exact. If it were, then the collapsed wavefunction would be a delta function in position space, which corresponds to an infinite sine wave in momentum space. But an infinite sine wave isn't normalizable, so it isn't a physically possible wavefunction.

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