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Yeah, I haven't quite understood, or been told, what happens to, for example an electron and it's wavefunction, when you stop to measure it?

I mean, an electron has a wave function describing it's position and so on in the region of space it is confined. When I measure it, I get an output, and the wavefunction collapses - as far as I understand. But then, when I stop measuring, the electron has not been "destroyed" has it? So does it resume it's original wavefunction, and continue as nothing has happened, or have I totally misunderstood something? :)

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The wavefunction exists just fine before, during and after measurement. At the moment of measurement, it does "collapse" in the Copenhagen interpretation of QM, meaning it discontinuously changes from a given probability amplitude distribution to a Dirac delta function-like distribution (sharply peaked). Immediately after the measurement, the wave function (continuously) starts to broaden out again, generally into a different probability amplitude distribution. –  Wouter Dec 10 '13 at 1:26
    
Measurement $|\psi (t_0) \rangle \rightarrow | n (t_0) \rangle$; Time-evolution after measurement $ \hat{U}(t,t_0)| n (t_0) \rangle $ –  user26143 Dec 10 '13 at 1:34

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Assuming wave-function collapse is correct (which can be a relatively hefty philosophical claim in some circles), then think of measurement this way:

When you measure an observable on a system, you collapse the wave-function of the system into a Dirac delta function in the eigenbasis for that observable.

If you measure position, you get a delta function in position-space. If you measure momentum, you get a delta function in momentum-space (or a sine wave in position space). If you measure energy, you get an energy eigenfunction.

Then - after the collapse - the system begins evolving according to Schroedinger's Equation once again, but this time your initial conditions for the system are whatever shape you collapsed the wave-function into with your measurement.

Remember, particles obey Schroedinger's Equation. It tells you what they do in Quantum Mechanics - just like Newton's 2nd Law tells you what they do in Classical Mechanics. Give me the Hamiltonian and the initial conditions of a system, and I will tell you how it evolves in time. That is the name of the game for much of Quantum Mechanics.

(As an interesting side-note: if you make another measurement of the same observable very quickly after making the first measurement (and I mean VERY quickly), you will get back the same result because the wave-function has not had time to evolve away from that state yet.)

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So basically, before the measurement, I have $\Psi(t_{0})$, then when I measure, the wavefunction "collapses" into a Dirac delta function, and then when I stop measuring, the new wavefunction, $\Psi(t)$, has been altered to look like the Dirac delta function, and then evolves accordingly, or...? –  Denver Dang Dec 10 '13 at 1:40
    
Let's say we've got a particle in a one-dimensional, infinite square well of width $L$. At time $t=0$, we give the particle some initial condition (say, $\Psi (t=0)=\frac{1}{\sqrt{L}}$ - equally likely to be anywhere in the well). We drop that initial $\Psi$ into Schroedinger's Equation ($\hat H\Psi=i\partial_t{\Psi}$) and let it evolve for some time $t=t_0$ at which point we make a measurement of its position. It collapses into a delta function (i.e. we know exactly where it is). (cont.; 1 of 2) –  Geoffrey Dec 10 '13 at 1:58
    
If we stop measuring its position and allow the particle to evolve in time once more, this process begins again except this time $\Psi (t=0)=\delta(x-a)$, where $a$ is the location we found the particle when we measured its position. Maybe after waiting a while we measure its energy. Now it's in an energy eigenstate (we know its exact energy) - say, the ground state (a sine wave with roots at the walls of the well). If we let it evolve again, the process starts again. This time $\Psi (t=0)=\sin(\frac{\pi}{L}x)$. This can go on forever. Do you understand what's happening? Is this making sense? –  Geoffrey Dec 10 '13 at 2:04
    
Yup, I think I got it now :) Thanks for clearing it up ! Cheers. –  Denver Dang Dec 10 '13 at 2:05
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It won't quite be a dirac delta in the real world, because that would mean an infinitely precise measurement. It would still be a localized packet that quickly flies away, however. –  Kevin Kostlan Dec 10 '13 at 3:16

Two cents from an experimentalist.

It is always good to keep in mind that a wavefunction for a real particle in the lab is a solution of Schrodinger's equation with specific boundary conditions given by the experimental setup that makes the measurement. Every measurement changes the boundary conditions for the solution that describes the particle's (x,y,z,t) and (p_x,p_y,p_z,E) which are the vectors we can normally measure.

It is also good to keep in mind that the solution of S's equation that describes the specific particle in the lab is a function whose square gives the probability of finding the specific measurement one finds with the experiment. That is the reason one does not try to devise experiments chasing after the "same" electron, because a single measurement in space and time ( or momentum and energy) cannot give any information on probability distributions and whether one has the correct potentials in the S equation ( or the more advanced formalisms of quantum mechanics). We do scattering experiments with beams with an enormous number of particles for that reason. Same boundary conditions and a plethora of particles will give us the probability function, and thus help us discriminate between theories, which is the reason for experiments.

After the measurement each particle is described by a new probability function given by the new boundary conditions , because each measurement changes the boundary conditions.

And finally it should also be stressed that the probability distribution describing a particle is just that , a distribution in space (or energy momentum space) of the probability of finding the particle whole when you measure it at that specific coordinate. It is not a solution with the particle's mass spread out like a splash in coordinate space. Thus the concept of "collapse" is a misleading concept. The "collapse" happens in probability space not in real space, in the same way that when one throws the dice, each of the 6 numbers is spread in probability space equally and the throw "collapses it " to a specific number. Nothing material gets collapsed. It is not a balloon being punctured.

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The electron doesn't get destroyed when you measure it (though photons usually do), but its wavefunction doesn't go back to how it was before. Instead it gets a new wavefunction, different from the old one. If you measured the position of the electron, this new wavefunction will be a delta function (a single infinitely sharp spike) centred at the position you measured. This change in the wavefunction is what's meant by "collapse".

If this didn't happen then you would be able to measure both the position and the momentum simultaneously, by making multiple measurements: first measure the position, and then measure the momentum. But in reality you can't do this, because the first measurement changes the wavefunction. The delta function it turns into doesn't have a well-defined momentum (i.e. its momentum could be anything), and this is essentially how the Heisenberg uncertainty principle works.

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