Suppose there is a particle in space. When we measure the position of that particle, we get a particular value with a probability that can be calculated from the wave function. But, according to the Copenhagen interpretation, where was the particle before the measurement? Was it in a superposition of all possible positions?

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    $\begingroup$ Suppose you throw a dice 1000 times. You will get a probability distribution for the values of 1 to 6. Then suppose you throw the dice one time, the probability of getting, say, 4, is 1/6 as also read off the probability distribution.What number did the dice show before throwing this last time? Was the dice presenting up in all six faces before throwing this last time? $\endgroup$ – anna v Mar 29 '13 at 19:50
  • $\begingroup$ What anna v said. (+1!) $\endgroup$ – alexarvanitakis Mar 29 '13 at 20:49
  • $\begingroup$ @annav Can't a quantum die (a quantum system with six states) exists as a supperpostion of more than one state? $\endgroup$ – user774025 Mar 29 '13 at 21:19
  • $\begingroup$ The concept of probability has nothing to do with quantum or classical. It is a count of how many times something happens. It is the end result of calculations checked against measurements, not the input for them. the famous two slit experiment has been reproduced experimentally with a clever set up determining the slit the particle went through without disturbing it too much, and the interference pattern is retained:en.wikipedia.org/wiki/… . $\endgroup$ – anna v Mar 30 '13 at 4:15
  • $\begingroup$ The previous crude methods interfered destructively with the QM solutions when trying to detect the path. So the experiment tells us that the end measurable result of the quantum mechanical wave function is a probability distribution, not a particle spread out in space, as it is always a particle, though we do not know where it is. $\endgroup$ – anna v Mar 30 '13 at 4:16

Before the measurement the particle's quantum state is given by the wave function, normalised to some volume $V$ of space

$\psi({\bf r},t)=\frac{1}{\sqrt{V}}\exp[({\bf p.r}-Et)/\hbar]$,

This means that, according to the Copenhagen Interpretation of QM, before the measurement, the particle is everywhere, occupying every point inside the volume $V$.

This equation shows that the particle has well defined momentum (sharply defined,) but the position of the particle is completely undetermined, from the point of view that the particle can be found, with probability $P=1/V$, in any one position within the volume $V$. Given that the wave function is the only source of information we can have about the position of the particle, the particle is everywhere inside the volume V. In other words, no point ($x,y,z$) at any time $t$ has extra probability to be occupied by the particle.

If a free particle occupies some region, in the $x$-axis say, having width $\Delta x_0$ at $t=0$, then the particle is described by a wave-packet. The width $\Delta x_0$ of the wave-packet determines the width $\Delta p_0$ of all possible values of momentum the particle can have. This is an expression of Heisenberg's uncertainty principle. Since the particle is free, as time goes on the width expands $\Delta x(t)\rightarrow\infty$ while the momentum of the particle acquires a well defined value. I.e. in momentum space the wave packet reduces to a delta function. Conversely, as $\Delta x(t)\rightarrow 0$ the momentum of the particle is totally undetermined. This interplay is at the heart of quantum mechanics, and we "witness" it in the double slit (or the diffraction grating) experiment.

  • $\begingroup$ @Gugg Thanks for the comment. That is a valid point, but the discussion is about the Copenhagen interpretation indeed. A detailed discussion to cover all vital points, would take a whole book. I am Sorry if I have misinterpreted the point you are trying to make. -:) $\endgroup$ – JKL Mar 30 '13 at 17:08
  • $\begingroup$ @Gugg I hope the explicit mention of the Copenhagen interpretation in the edited answer (see bf text) will suffice? $\endgroup$ – JKL Mar 30 '13 at 20:15
  • $\begingroup$ Much better, but technically just off. (My opinion.) However, who (but I) cares? +1 $\endgroup$ – Keep these mind Mar 30 '13 at 20:24

Was it in a superpostion of all possible positions?

I cut out some (cat) bits from Wikipedia and got this:

In the Copenhagen interpretation, a system stops being a superposition of states and becomes either one or [an]other when an observation takes place. [...] [T]he nature of measurement, or observation, is not well-defined in this interpretation. [An] experiment can be interpreted to mean that [until an observation takes place] the system simultaneously exists in a superposition of the states [...] and that only when [...] an observation [is] performed does the wave function collapse into one of the [...] states.

Source (note the "However"-part that follows)


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