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I understand that the double slit experiment (i.e. the creation of interference pattern) holds also when the "particle" is not just a single particle but any item, experimentally proven even for a C60 molecule.

I am wondering what is the correct interpretation of the double slit experiment when the particle under test is a macroscopic object, like a space capsule with people inside.

I think that in order to obtain the interference pattern the capsule shouldn't be hit by any photon coming from the external environment (or any other particle). So the people inside the capsule are not able to understand which slit they are going through since they can't receive any information from the outside world.

People outside the capsule are not able either to see the capsule during his flight otherwise the wave function of the capsule will collapse to a specific position.

I guess the experiment setup would require extremely low temperature of the objects involved (at least on their external surface) and a very controlled environment.

Is this interpretation correct?

Are the people inside the capsule able to make some internal experiment to cause the capsule wave function to collapse (and to avoid any interference pattern when they hit the final wall?) even without interacting with the external environment?

Thanks

[Edit] Thanks for your answers. Actually I want to clarify my question. I am not asking why we don't usually see interference pattern on macroscopic objects. I am wondering what's the interpretation of the interference pattern we see (do we see it?) when we setup an experiment with such macroscopic object. Obviously it is a thought experiment and it doesn't really matter if it's technically unfeasible.

In example, considering the numbers that tom proposed, we have that _lambda is 4e-38m. Now, if the slits are 10m wide and they are 1000m apart, the interference pattern should emerge for angles of about 4e-41 radiants. So if the wall is like 1e45m far (let's forget about cosmology) then we should see the crests of the interference pattern kilometers apart.

So what really happens in such scenario? I can think of the following options: 1) Interference pattern emerge, passengers and external observer can't tell which slit the ship went through since the experiment itself requires no interaction between the ship and the environment 2) Interference pattern doesn't emerge even if the ship doesn't interact with the other objects as it's too complex and its wavefunction collapses anyway. 3) ?

We can make the experiment cosmologically more realistic. We can use a nanorobot (able to record some observations and perform little experiments, so conceptually equivalent to a human being) with a mass of 1e-18kg and make it travel at 1mm/s. If the slits are .1m apart, we should expect interference crests .1m apart after a Sun-Earth distance.

Qmechanic
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Juggernaut
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    See on the validity of the deBroglie wavelength for macroscopic objects for a more abstract discussion. Also, have you calculated the actual deBroglie wavelength of such a capsule, and examined whether size and position of the slits that would be required for interference are such that the object actually can pass through them? (Spoiler: I seem to recall you get a ludicrous setup that is impossible to realize) – ACuriousMind Mar 22 '15 at 14:48
  • You have to realize that the grand majority of quantum mechanical effects appear at the nanometer level. This is mainly due to the smallness of h, the planck's constant, which enters in all quantum mechanical solutions. Macroscopic manifestations happen with very special initial conditions, as in superconductivity and superfluidity. – anna v Mar 23 '15 at 12:03
  • This is an excellent question and not at all a trivial one. It is not at all a duplicate of the one linked above, and nor is it answered by that question's answers. (some people have cast close votes.) – N. Virgo Mar 23 '15 at 12:18
  • I have edited the answer to try to address your question about the physical interpretation. – tom Mar 24 '15 at 18:25

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The de Broglie wavelength, $\lambda$, is given by $$\lambda = {h \over p}$$ where $h$ is planck's constant and $p$ is momentum.

If we take the mass to be $160$ kg and speed to be $100$ ms$^{-1}$ then we get $\lambda = 4\times10^{-38}$m given a slow light spaceship and only one person (more people and speed would increase the momentum and decrease the value of $\lambda$).

Now to see intereference typically the slit width should be similar in magnitude to the size of the wavelength. Clearly it will be difficult to pass a spaceship with a person inside through a slit so small - and if it is pass through a larger slit then diffraction would not be visible until the spaceship had travelled a hugely long distance.

A relevant equation here is single slit diffraction where $n\lambda = d sin(\theta)$ where $\theta$ is the angle of the minimum for $n = \pm1, \pm2, \pm3....$, $d$ is the slit width and $\lambda$ is the waveldght. We get the pattern below, but for the spaceship passing through a $10$ m slit the angle would for $n=\pm1$ would be $\sim4\times10^{-39}$ radians and to have the minimum $1$ m from the maximum would require travelling $2.5\times10^{38}$ m. enter image description here

For double slit diffraction the equation is the same except $d$ is the separation between the two slits - but we run into the issue the person on board might observe which slit the spaceship travelled through, but event in that case we should still get single slit diffraction.

The interpretation is that the diffraction pattern above is a probability distribution. The spaceship would not be smeared out, but lots of spaceships observed would have this probability pattern.

This interpretation can be seen in the intereference pattern generated by a Young's Slit experiment for molecules which generated the image below enter image description here

[This makes me think a modern update of the 'camel through they eye of a needle' might be 'observation of diffraction following passing camel through a small slit' ]

This answer was hinted at in the comment of ACuriousMind, but I though it might be helpful to put some numbers in.

(Just spotted the possible duplicate question has a useful answer by JohnRennie that addresses this issue of diffraction)

tom
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  • "Now to see intereference typically the slit width should be similar in magnitude to the size of the wavelength. Clearly it will be difficult to pass a spaceship with a person inside through a slit so small - and if it is pass through a larger slit then diffraction would not be visible until the spaceship had travelled a hugely long distance." I think your answer would be improved by giving the formulas that you get these statements from, and estimating the value of the long distance that would be needed. – Hypnosifl Mar 24 '15 at 02:05
  • (cont) probably the relevant formula is the one saying that for a double slit, constructive interference is seen on the screen at an angle $\theta$ from the slits (measured relative to a line that's normal to the plane of the slits and screen) such that $\sin \theta = m\lambda / d$ and destructive at $\sin \theta = (m + 1/2)\lambda / d$ where $m$ is any integer, and $d$ is the distance from slits to screen. At large distances $\theta$ is close to 0 so you can approximate $\sin \theta$ as $\theta$. – Hypnosifl Mar 24 '15 at 02:08
  • @Hypnosifl - thanks for the useful comments I will edit – tom Mar 24 '15 at 18:04