In a double-slit experiment, interference implies that the particle has traveled different distances from the 2 slits. Since the speed of light is constant, it seems to me that the particle would arrive at the detector at a different time if it went through one slit as opposed to the other slit. What am I missing here?
15
-
1$\begingroup$ There are no particles in these experiments, there are only quanta. Quanta don't travel, they simply describe the state of the system at the time of the measurement. Between measurements quanta are not defined, so there are no paths to begin with. $\endgroup$– CuriousOneCommented Jan 14, 2016 at 17:29
-
3$\begingroup$ @CuriousOne - in the context of quantum mechanics, the words "particles" and "quanta" are basically synonymous and be sure that both are used - one of the reasons why the part of "quantum physics" studying events at the LHC etc.is known as "particle physics". It makes no sense to say that one exists and the other doesn't so I am afraid that your comment only conveys confusion, not any useful information that would have anything to do with the fact that we can't get the "which slit" information from the timing. $\endgroup$– Luboš MotlCommented Jan 14, 2016 at 17:50
-
$\begingroup$ @LubošMotl: Only in the context of high energy particles when weak measurements are performed. How "particles" emerge from a field theory has been successfully explained in 1929 by Mott, the paper simply hasn't garnered much attention and some still think that "particle" has some fundamental physical meaning. It doesn't have any more meaning in quantum mechanics than it does in classical mechanics, where "particle" simply means that we can simplify the motion of an extended piece of matter to the motion of its center of mass. For orbital mechanics even Jupiter can be treated as a "particle". $\endgroup$– CuriousOneCommented Jan 14, 2016 at 17:55
-
2$\begingroup$ @CuriousOne - the very claim that there exists something like "weak measurements" is a speculative and highly controversial subfield of "foundations of quantum physics" and virtually no particle physicist would actively say that they're doing "weak measurements". The LHC detectors are making standard measurements of the energy of the final particles. ... Jupiter is an extended object which may be approximated by a point mass. But particles in quantum field theory are exactly point-like, this is how quantum field theory differs e.g. from string theory, so they're nothing like Jupiter. $\endgroup$– Luboš MotlCommented Jan 14, 2016 at 17:58
-
1$\begingroup$ There is absolutely nothing "weak" about the measurements of the energy of particles at the LHC. - BTW the term "particle" isn't just some modern deviation of particle physicists. Check Wikipedia - all elementary particles like photon, electron, even proton (despite compositeness) etc.- the particles that the OP is probably talking about - are defined as "particles" of some sort in the first sentence. Fermions were always called particles; bosons (photons in particular) were called particles as soon as the confusion about the existence of light quanta evaporated. $\endgroup$– Luboš MotlCommented Jan 14, 2016 at 18:05
|
Show 10 more comments
2 Answers
$\begingroup$
$\endgroup$
1
We can't determine the "which slit" information from the timing because whenever the interference pattern is being built at all, the time that the particle needs to get to the screen is the same, within the uncertainty, for both interfering paths.
Indeed, if the duration of the journey depended on the chosen slit, the interference pattern would disappear. The interference pattern is there because the nonzero wave function contributed by slit A (or its support) is overlapping in a region of the spacetime with the wave function contributed by slit B (or its support). And this overlap means that there is a large probability that the particle - whose motion is described by the wave function - has a high enough chance to get to the final point by both slits at the same moment.
In practice, the interference pattern often appears from a combination of waves that exist for a very long time. Whenever the energy of the particle is sufficiently well-defined, comparable to $\Delta E$ which is small, the wave packet must exist at a given place for some time $$\Delta t \gt \frac{\hbar}{2 \cdot \Delta E} $$ which is somewhat analogous to the uncertainty principle for $x$ and $p$. So if the energy is accurate, the timing will not be known accurately and your discrimination can't be done. It doesn't help to make $\Delta E$ large, either. If $\Delta E$ is too large, the velocity is uncertain and the predicted time needed for the journey is uncertain, so the discrimination can't be done, either.
As long as the uncertainty inequality above is obeyed, you could indeed get a situation in which the discrimination could be done by the timing. But if you could discriminate - because the time would be significantly longer for slit A - it would mean that there would be no interference. (The amount of interference is decreasing continuously, just like the reliability of your method to discriminate would increase continuously.)
Geometrically: the interference pattern appears on the photographic plate close enough to the points which as "about equally far" from both slits - more precisely, where you need the same time for the journeys through both slits. If you look at places where the lengths (more precisely durations) would be too different for both slits, the interference pattern would largely fade away there.
The textbook examples calculating the interference pattern assume that the particle is described by waves that are basically independent of time, up to a phase. When it's so, it means that the timing is completely unknown, $\Delta t\to\infty$, and your method can't be used at all. In this textbook scenario, the detection of a particle on the screen is the first event which gives one some information about the timing.
In some more realistic situations, we know the timing when the particle entered the interference experiment with some accuracy and the wave functions are "wave packets". But they usually contain a sufficiently high number of periods of the wave so that the approximation with the "stationary" waves is good enough and $\Delta t$ is large enough.
answered Jan 14, 2016 at 17:39
-
$\begingroup$ Thanks. Very interesting conversation, it'll take me a bit to digest this. $\endgroup$ Commented Jan 14, 2016 at 18:33
$\begingroup$
$\endgroup$
It’s always should be remembered that in the quantum world there’s not just particle but a duality called wave-particle (a quanta). There’s no objective preference to either one of the 2 phases (wave and particle) of this duality. Which phase we shall reveal at any time is dependent upon the manner we choose to trace or measure each quanta. Furthermore, recent theories like the Transactional Interpretation (TI) of Quantum Mechanics assert that the waveform is the only real authentic nature of any quanta as long as it does not interact with any other quanta. It entails that the single particle (quanta) you are asking about, approaches the slits as a real wave and behaves precisely like the waves we know in the Classical Physics, like water waves, etc.
It’s already well known how does a single water wave behave when it gets split and passes through two paths. At the exit of each path, a new and independent water wave gets created and it interferes with the new water wave which got created at the exit of the other path. If the TI theory is correct, the same thing occurs with a quanta in the 2 slits experiments. As long as it does not interact with anything in the slits or with any particles of measuring-instruments, the wave (quanta) gets split into 2 new waves (in the 2 slits) which then interfere with each other.
It means your particle (quanta) always passes through both slits as long as it has not interacted with anything (remaining as wave) and so there should not be any time difference.
