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Once measuring momentum, the wavefunction "collapses" into something that looks like this

enter image description here

If you were to then measure the position, couldn't it be literally anywhere? What am I missing? Is it even possible to measure momentum perfectly?

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    $\begingroup$ You're not missing anything at all! You have hit upon exactly the right conclusion. $\endgroup$ – probably_someone Dec 27 '16 at 5:40
  • $\begingroup$ So... you're saying measuring momentum or whatever instantly teleports the particle anywhere in the universe? That doesn't sound right... Shouldn't it somehow be constrained by the speed of light or something $\endgroup$ – Farzher Dec 27 '16 at 5:50
  • $\begingroup$ Why do you assume that the particle has a definite location in the first place? $\endgroup$ – probably_someone Dec 27 '16 at 5:51
  • $\begingroup$ Not a definite location, but I know it's in my lab for example, and there's only an infinitesimal chance it isn't. The wavefunction I showed gives it an equal chance to be anywhere, which like I said, doesn't sound right $\endgroup$ – Farzher Dec 27 '16 at 6:00
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    $\begingroup$ The very fact that you can confine the particle to your lab means you can no longer measure momentum infinitely precisely. This is a consequence of the Heisenberg Uncertainty Principle. $\endgroup$ – probably_someone Dec 27 '16 at 6:17
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We call a wavefunction that has a precisely defined momentum a momentum eigenstate. For a free particle the momentum eigenstates are infinite plane waves like the one you show in your graph:

$$ \psi = e^{i(\mathbf p\cdot\mathbf x - \omega t)} $$

And as you say in your question for this eigenstate the position of the particle is completely undefined, or put another way $\Delta x = \infty$.

But how are you ever going to make a measurement that results in an infinite plane wave? What possible physical process could achieve this? Any measurement necessarily takes place within some finite region so the best you can achieve is to end up with a wavepacket that is about the size of your system:

Wave packet

where $x$ is some length scale determined by how you did the measurement. The resulting wavefunction will be:

$$ \psi = \mathcal F(\mathbf x,t) e^{i(\mathbf p\cdot\mathbf x - \omega t)} $$

where $\mathcal F(\mathbf x,t)$ is the envelope function. However this wavepacket no longer has a precisely defined momentum because it is not an infinite plane wave so it isn't a momentum eigenstate. In fact the momentum spread will be roughly given by:

$$ \Delta p = \frac{\hbar}{2x} $$

i.e. just the uncertainty principle. So as a consequence of your measuring apparatus having a limited extent in space you can only measure the momentum to a limited precision. Your measurement does not cause the wavefunction to collapse to a momentum eigenfunction and the resulting particle cannot be anywhere in space.

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  • $\begingroup$ I've always heard it explained as "when you measure some observable, your measurement operator acts on your wavefunction, turning it into an eigenstate. The eigenvector is what you observe" It seems like that's not realistic, and it doesn't actually turn into an eiganstate. I'll go look up more details about uncertainty principle $\endgroup$ – Farzher Dec 27 '16 at 7:00
  • $\begingroup$ @StephenBugsKamenar: yes, that is generally what is taught. However this is an idealised concept of a measurement i.e. we consider the measurement to be represented by an operator without considering whether this is actually possible in practice. And, well, it is in principle possible to collapse the wavefunction to a momentum eigenstate as the operator suggests, but you would need an infinitely big apparatus :-) $\endgroup$ – John Rennie Dec 27 '16 at 7:02
  • $\begingroup$ It seems like my main confusion comes from measurement. uncertainty principle makes sense to me. Can you list some things for me to research to understand how measurements actually works in a non-ideal situation. Something else I've wondered about is, if I try to detect a particle, and notice it's not there. How does that affect the wavefunction? (there's now a 0% chance it's where I just looked (well maybe there's a small chance my detector didn't catch it, but the wavefunction changed in some way, even though I "didn't detect anything")) $\endgroup$ – Farzher Dec 27 '16 at 7:15
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Each observable corresponds to a mathematical operator in Hilbert Space. There are pairs of observables which are called conjugate variables, these cannot both be known accurately at the same time. The measurement of one immediately makes the measurment of the other impossible. Position and momentum are such a pair. It is possible to measure one to a lesser precision, and then measure the corresponding conjugate variable in a similar manner. The relationship between the accuracy of the two measurments is given by the Heisenburg uncertainty.

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Real measuring devices all have a granularity. Your experiment will never tell you that a particle has momentum exactly $4.03752\,\mathrm{MeV}/c$; it will tell you (assuming it is very precises indeed) that it has momentum $(4.037 \pm .014)\,\mathrm{MeV}/c$ which is completely compatible with the particle still being found withing the apparatus at the end of the measurement.

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"We have measured sample's momentum with absolute accuracy, so nobody knows where it is now!"

You are right and not missing anything.

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