Double slit -- simple question

[lukephysics]

TL;DR Summary

whys the gun low precision?

With the double slit experiment the first thing I have to ask that is never explained, is why would an electron not always go through one slit? One does not design a gun to fire randomly over an area, but to be fairly precise and accurate in it's firing.

Is there an assumption of the firing pattern of an electron gun, and does each gun have imperfections in its firing accuracy or precision?

I would think that would make a huge difference to the statistics on the measurement. Maybe one gun always does hit one slit due to high precision and low accuracy. Do they do a pre-experiment check of this when they set up the experiments?

Also, another quick q, when they are firing at the slots, do most of the electrons get blocked by the slots? You would think if its low precision, most will pepper the wall around the slots, and only a few get though that happen to line up with the slots. is that correct?

 

[lukephysics]

should i understand the experiment as firing a wavefunction at the slots? and the wavefunction is wide enough to envelope the double slit? if so how can they can make a double slit small enough to encompass the wave function, wouldn't it be the size of an atom? when its in an atom its small and and contained in areas around the shells. what does the wavefunction of a free electron coming out of a gun look like in size and shape?

 

[PeterDonis]

why would an electron not always go through one slit?

It will, if you use the right kind of electron source that can aim precisely enough. But then you will see a different kind of pattern on the detector screen. (Look up "electron diffraction". The first experiment I'm aware of along those lines was the Airy experiment.) Which kind of source you use will depend on what kind of quantum phenomenon you are trying to investigate.

when they are firing at the slots, do most of the electrons get blocked by the slots?

There's no way of knowing in most experiments because individual electrons are not measured; only the pattern formed by a beam of a huge number of electrons is. One could, I suppose, compare the calibrated intensity of the source with the cumulative intensity of what is detected, but I don't know if any existing experiments have looked at that.

If you could run the experiment with a source that could emit single electrons, one at a time, spaced far enough apart that you could measure them separately, then yes, you would expect that some electrons (I don't know if it would be "most") would hit the screen that has the slits in it, instead of going through the slits. You would know this happened if an electron got fired by the source but nothing showed up on the detector screen.

 

[PeterDonis]

One does not design a gun to fire randomly over an area, but to be fairly precise and accurate in it's firing.

You may be being misled by the term "electron gun". Electrons are not like bullets; they are quantum objects, not classical objects.

The kind of source normally used for a double slit experiment is one in which the momentum of the electrons coming out is as precise as possible, in both magnitude and direction. But because of the quantum uncertainty principle, the more precisely the momentum is determined at the source, the less precisely the position is determined. So a beam of electrons with very precisely determined momentum will have a large uncertainty in position. That places a fundamental limit on how finely you can "aim" an electron if you want it to have a precisely determined momentum.

One could also push the tradeoff in the other direction, trying to make the position more precisely determined, and sacrificing precision in momentum. But that won't help you with "aiming" more precisely, because "aim" is really a matter of direction of travel, not position, and direction of travel is part of momentum, not position. Or, to put it another way, you could try to determine the position of the electron more precisely at the source, but the resulting larger uncertainty in momentum would mean you would have less control over where it went once it left the source.

 

[PeterDonis]

It will, if you use the right kind of electron source that can aim precisely enough.

That places a fundamental limit on how finely you can "aim" an electron if you want it to have a precisely determined momentum.

Btw, these two statements of mine are not actually contradictory. The spacing of the slits in most double slit experiments, AFAIK, allows quite a bit of leeway before the fundamental limit imposed by the uncertainty principle is reached. So one could imagine changing the source to a beam that could be aimed precisely enough to show diffraction from one slit, instead of interference from two slits, without changing the rest of the experiment. But practically speaking this might be prohibitively difficult, since the limitations imposed our current technology are much stricter in many ways than any fundamental limits imposed by the laws of physics.

 

[PeterDonis]

should i understand the experiment as firing a wavefunction at the slots? and the wavefunction is wide enough to envelope the double slit?

In the normal case, yes, this is one way of looking at it. More precisely, the position uncertainty in the wave function is large enough that both slits are within the beam.

if so how can they can make a double slit small enough to encompass the wave function, wouldn't it be the size of an atom?

No. Free electrons in beams are very different from bound electrons in atoms.

what does the wavefunction of a free electron coming out of a gun look like in size and shape?

It looks approximately like a plane wave, i.e., a series of planar wave fronts that are all perpendicular to the direction of motion, and wide enough to include both slits.

 

[sophiecentaur]

In the normal case, yes, this is one way of looking at it. More precisely, the position uncertainty in the wave function is large enough that both slits are within the beam.

I was waiting to read something like this as I strolled down the thread. The de Broglie wavelength is h/p so wouldn't diffraction at a narrow gun 'barrel' be relevant. Wouldn't that affect the fineness of any beam it could produce with a given electron momentum? A wide gun would give 'ideal' illumination and a long collimation tube would produce the best coherence. (I'm thinking in terms of non laser monochromatic sources of old as we don't have 'electron lasers'?)

 

[PeterDonis]

wouldn't diffraction at a narrow gun 'barrel' be relevant.

If the gun barrel were narrow enough, yes, you would expect diffraction effects to reduce collimation instead of improving it.

 

[PeterDonis]

A wide gun would give 'ideal' illumination and a long collimation tube would produce the best coherence.

You can't make the "gun" too wide, though, or it won't collimate. There is some optimal "width" at which collimation is as good as it can get. (As I understand it collimation is usually done by magnetic fields so the term "gun" might give the wrong image, but hopefully you see what I mean.)

 

[lukephysics]

why does the electron wavefunction not get entangled with the blocker between the slots as it goes through both slots? i thought its very hard to keep unentangled quantum states. and flying a wavefunction through a big block of classical material (the material between the two slots) wouldn't that completely entangle the particle causing no interference? if that wouldn't entangle it what would?!

 

[PeterDonis]

why does the electron wavefunction not get entangled with the blocker between the slots as it goes through both slots?

Getting entangled with the blocker between the slots would mean it never goes through the slots, but gets stopped by the blocker instead. Which, as I noted in an earlier post, will probably happen with plenty of the electrons from the source. But any electron that does get through the slits was not entangled with the blocker.

 

[lukephysics]

Getting entangled with the blocker between the slots would mean it never goes through the slots, but gets stopped by the blocker instead. Which, as I noted in an earlier post, will probably happen with plenty of the electrons from the source. But any electron that does get through the slits was not entangled with the blocker.

You can make detectors that don’t stop the electron yet entangle with it. why did you suggest ‘stopping’ it is a key criteria for entanglement for the wall?

I understand to measure something you have to copy it, and you can't copy a quantum system, so that's why you get a 'classical' result on measurement.

i guess i am asking why the block between the slit doesn't 'measure' the wavefunction since the wavefunction goes through it, but the wall at the back does?

 

[PeterDonis]

people can make detectors that don’t stop the electron but that entangle with it.

Can you give an example?

 

[PeterDonis]

I’m not sure why ‘stopping’ it is a key criteria for entanglement?

What the "key criteria for entanglement" are will depend on the specific scenario. There is no single general rule that you can always apply.

 

[lukephysics]

Can you give an example?

I am probably mistaken about non-stopping detectors.

What the "key criteria for entanglement" are will depend on the specific scenario. There is no single general rule that you can always apply.

ok thanks.

so the idea with these electron guns is to increase momentum precision which makes the position-wavefunction wider, big enough to cover two slits, which answers the original question.

 

[PeterDonis]

the idea with these electron guns is to increase momentum precision which makes the position-wavefunction wider, big enough to cover two slits

Basically, yes.

 

[sophiecentaur]

But any electron that does get through the slits was not entangled with the blocker.

I am probably mistaken about non-stopping detectors.

Is it not a trivial task to detect the signal from a probe as a fast moving electron goes past it? That would involve some change in energy though.

 

[Grinkle]

Basically, yes.

Engineering-wise, do you know how this is done? In other words, what kind of emitter has a tight spread of momentum and what type of emitter has a tight spread of position?

 

[lukephysics]

fyi Peter you can entangle without stopping the electron

For example you can flip the spin on one slit, and it will destroy the interference. you just get the classical pattern.

See figure 6 here:

http://adamilab.blogspot.com/2015/07/on-quantum-measurement-part-6-quantum.html

 

[PeterDonis]

For example you can flip the spin on one slit, and it will destroy the interference.

Not just flipping the spin, no. Read the article you linked to carefully. It says if you flip the spin on one slit, and then measure the spin after the electron goes through the slits, you descroy the interference. Just flipping the spin alone is not enough. You have to measure it. The spin measurement, because of the flip at one of the slits, is a "which-way" measurement, i.e., it tells you which slit the electron went through. And doing that destroys the interference, because the interference is only present if you don't measure which slit the electron went through.

you can entangle without stopping the electron

Using a "which-way" measurement to destroy interference has nothing to do with "entangling without stopping the electron". The flipping of the electron's spin doesn't entangle it with anything. And the key part about the which-way measurement is the measurement. Entanglement is part of measurement, but not the crucial part: the crucial part is that an irreversible record is made of which slit the electron went through (by measuring its spin).

 

[jartsa]

Small uncertainty of position of a particle means that the uncertainty of the momentum of the particle is large, which means that the uncertainty of the direction of motion of the particle is large, which means that after said particle has traveled a quite large distance into said uncertain direction, then the position of the particle has become highly uncertain.

So, when a small electron gun shoots at a distant double slit screen target, both slits get illuminated by the electron probability wave.

 

[vanhees71]

Not just flipping the spin, no. Read the article you linked to carefully. It says if you flip the spin on one slit, and then measure the spin after the electron goes through the slits, you descroy the interference. Just flipping the spin alone is not enough. You have to measure it. The spin measurement, because of the flip at one of the slits, is a "which-way" measurement, i.e., it tells you which slit the electron went through. And doing that destroys the interference, because the interference is only present if you don't measure which slit the electron went through.Using a "which-way" measurement to destroy interference has nothing to do with "entangling without stopping the electron". The flipping of the electron's spin doesn't entangle it with anything. And the key part about the which-way measurement is the measurement. Entanglement is part of measurement, but not the crucial part: the crucial part is that an irreversible record is made of which slit the electron went through (by measuring its spin).

I guess you refer to a setup where you work with polarized single photons and put quarter-wave plates with orientations differing by $90^{\circ}$. Then you have 100% which-way information encoded in the polarization state of the photon behind the slit. You don't need to measure it, but it's clear that behind the screen the partial waves coming from the one or the other slit do not interfere anymore, because the polarization states are orthogonal. So you don't need to measure the which-way information to destroy the interference pattern.

This is important for the delayed-choice quantum eraser experiments, where you can decide whether you observe an ensemble of photons which have which-way information (and no double-slit interference pattern) or not (and a double-slit interference pattern) by selecting partial ensembles after all the photons's positions on the photoplate are irreversibly stored. Particularly illustrative is the realization by Walborn et al. Here's a simplified talk about this, I've given some time ago:

https://itp.uni-frankfurt.de/~hees/publ/habil-coll-talk-en.pdf

 

[PeterDonis]

I guess you refer to a setup where you work with polarized single photons and put quarter-wave plates

Yes, and by continuously varying the relative angle between the plates at the two slits, you can continuously vary how much interference is observed. With the plates parallel, you get full interference; with them perpendicular, you get zero interference. With an intermediate angle, you get partial interference, with the amount depending on the angle.

you don't need to measure the which-way information to destroy the interference pattern.

Yes, agreed. The which-slit information gets generated and decohered whether or not a human-readable record of it is made.

 

[lukephysics]

Yes, agreed. The which-slit information gets generated and decohered whether or not a human-readable record of it is made.

That was my point that you can entangle without terminating the electron/photon.

the which way device is a form of entanglement with the wider world

and this is why I ask why isn’t the block in between the slits going to entangleme the WF going through it but a flipper does

 

[DrChinese]

That was my point that you can entangle without terminating the electron/photon.

the which way device is a form of entanglement with the wider world

and this is why I ask why isn’t the block in between the slits going to entangleme the WF going through it but a flipper does

The double slit experiment does not traditionally involve any entanglement in the normal sense of that word. The double slit experiment is a demonstration of such concepts as wave/particle duality (as mentioned already: position uncertainty vs momentum uncertainty) and particle self-interference. It is not a demonstration of a particle entangling with a slit. Any particle going through the slit(s) is not entangled with the slit apparatus.

 

[PeterDonis]

the which way device is a form of entanglement with the wider world

In the sense that decoherence does this, yes. But you can't measure this kind of entanglement; the whole point of decoherence is that the entanglement with the degrees of freedom in the environment can't be tracked. The which-way devices don't provide any kind of measurable entanglement, as for example we see in a Bell-type experiment on a pair of entangled particles whose joint state is the singlet state.

this is why I ask why isn’t the block in between the slits going to entangleme the WF going through it but a flipper does

As has already been noted, the block between the slits, if it interacts with the electron, just stops it. This is actually also an example of decoherence: the electron gets absorbed by the block and you can no longer track it individually, it's just one of zillions of electrons inside the block. But the only decoherent outcome in this case is "electron got absorbed by the block, nothing useful to be gained from this run of the experiment".

So you can think of an electron that gets stopped by the block as being "entangled" with the block, but only in the sense of decoherence as described above. And, as noted above, the same is true of the which-way device. Neither one provides the kind of entanglement you see in a Bell-type experiment.

 

[vanhees71]

That was my point that you can entangle without terminating the electron/photon.

the which way device is a form of entanglement with the wider world

and this is why I ask why isn’t the block in between the slits going to entangleme the WF going through it but a flipper does

In the example of the Wallenborn quantum eraser with entangled photon pairs the which-way information is stored in the photons's polarization state, i.e., the setup with the quarter-wave plates in the slits is such that there's entanglement between "which-way the photon went" and "polarization is helicity +1 or -1".