# Double Slit Experiment

I have a problem with the outcome of the Double Slit experiment , in that it was concluded that when under observation the electron particles behaved differently to when not directly under observation - conclusivly stating that it was the ACT of observing that changed the electron behaviour in the end . Why has it not been questioned or put forward that it was the detection/observing apparatus/equipment/device itself that interferred in the behaviour? Its not like we can see them with the naked eye! I just havnt seen anyone ask this question before....

You're 100% right. The observing equipment observes the electron by bouncing light off it. This light must be of a high enough frequency to resolve a "tiny" electron. Such a frequency of light has the effect of imparting some momentum onto the poor electron, ruining any interference it may have been having with it's self in private in it's own bedroom with the lights off and a box of kleenex handy. How rude.

THANK YOU Gendou2 Finally some clarification, knew someone must have wondered that before....

PS. Thanks for the laughs - made my afternoon

ZapperZ
Staff Emeritus
You're 100% right. The observing equipment observes the electron by bouncing light off it. This light must be of a high enough frequency to resolve a "tiny" electron. Such a frequency of light has the effect of imparting some momentum onto the poor electron, ruining any interference it may have been having with it's self in private in it's own bedroom with the lights off and a box of kleenex handy. How rude.

Er.. this is silly.

I don't need to bounce photons off electrons to detect them,. The moment electrons pass through a loop of wire, I can detect its signal, per Faraday Law.

Now it doesn't mean that, in some way, this doesn't effect the electrons. But the fallacy that I have to detect electrons ONLY via destroying its path via photon collision is faulty. It relegates the superposition principle to nothing more than an instrumentation issue, rather than something inherent to the system as indicated by QM.

Even with photon detection, one can show that even with http://physicsworld.com/cws/article/news/30913" [Broken], the system STILL collapses via QM's description. This clearly shows that it has nothing to do with the instrumentation of what we use, but rather it is an inherent property of the system.

Zz.

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Even with photon detection, one can show that even with http://physicsworld.com/cws/article/news/30913", the system STILL collapses via QM's description. This clearly shows that it has nothing to do with the instrumentation of what we use, but rather it is an inherent property of the system.
Zz.

Hey, maybe you can answer this question I've long had: in any of these multiple contingent path scenarios, what counts as detection? What are the characteristics of a detector or of the detection interaction that distinguish it from, say, a photon going through a half-silvered mirror and hitting a rock (which is stimulated in some way, though not in a way constructive to waveform breakdown, no?).

Thanks for helping.

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Even with photon detection, one can show that even with http://physicsworld.com/cws/article/news/30913" [Broken], the system STILL collapses via QM's description. This clearly shows that it has nothing to do with the instrumentation of what we use, but rather it is an inherent property of the system.

That article says... "For the first few measurements the answers are evenly distributed from zero to seven. This means that not enough information has been gathered to ascertain the number of photons, since the system is still in a superposition. But after many measurements the cumulative distribution of answers begins to centre on a particular number, revealing that the system is collapsing into a well-defined state."

This is averaging. Isn't this a property of mathematics calculating averages rather than a property of the system?

I am still wondering how one has determined that the act of measuring something changes it's state and why the device measuring it is not responsible for it.

As far as I can tell the interference of the measuring device is a predictable interference based on the methods of measurement.

How many different methods are there for measuring the quanta directly? Can it even be done or is it entirely inferred interpretation?

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ZapperZ
Staff Emeritus
That article says... "For the first few measurements the answers are evenly distributed from zero to seven. This means that not enough information has been gathered to ascertain the number of photons, since the system is still in a superposition. But after many measurements the cumulative distribution of answers begins to centre on a particular number, revealing that the system is collapsing into a well-defined state."

This is averaging. Isn't this a property of mathematics calculating averages rather than a property of the system?

I have no idea what you are saying here. Even if it is an 'average', why isn't this a physical property of the system since it is the system itself that is giving you the average value. What is <x> if it isn't the average value of the a system that you actually measured? Is <psi|x|psi> a property of mathematics, or the physical system?

You should read the actual paper (the exact references is in the article) rather than rely on a report of the paper.

Zz.

Er.. this is silly. I don't need to bounce photons off electrons to detect them,. The moment electrons pass through a loop of wire, I can detect its signal, per Faraday Law.

Faraday's Law is true because of the photon - the carrier of the electromagnetic force. A Feynman diagram of your loop of wire will show photons carrying a signal between the loop of wire and the electron. Right?

ZapperZ
Staff Emeritus
Faraday's Law is true because of the photon - the carrier of the electromagnetic force. A Feynman diagram of your loop of wire will show photons carrying a signal between the loop of wire and the electron. Right?

yeah, but you were talking about REAL photons, not virtual photons that are responsible for EM interaction. Do you think these are the SAME thing as in what you described earlier? I mean, did you had to send in a photon of a certain wavelength to detect the electron that had just passed through a loop of wire? No you didn't. You have no control over that.

It is still besides the main point, which is that the HUP has nothing to do with the instrumentation. I can make as accurate of a measurement as I want the single measurement of position and momentum based on the technology available. The uncertainty of each position and momentum measurement that I got has nothing to do with the HUP, no matter how I measured them.

This is the most common misconception of the HUP (try googling it).

Zz.

ZapperZ, you probably know better than I, but it has been my experience that comprehending the uncertainty principal involves explaining why all experiments confirm it's existence. So, in this case, the reality of the virtual photons wasn't "besides the main point" but it was the main point. Why do people call virtual photons different from regular photons? I'm for photon equality. Is there some way to tell wether a given photon is virtual or not? Do they have to wear an upside-down triangle pin?

ZapperZ
Staff Emeritus
ZapperZ, you probably know better than I, but it has been my experience that comprehending the uncertainty principal involves explaining why all experiments confirm it's existence. So, in this case, the reality of the virtual photons wasn't "besides the main point" but it was the main point. Why do people call virtual photons different from regular photons? I'm for photon equality. Is there some way to tell wether a given photon is virtual or not? Do they have to wear an upside-down triangle pin?

Yes. I hear they have some on sale at Macy's this week.

Then show me how the virtual photon actually affect the the trajectory of the electrons in the SAME way as photons that you use to collide with those electrons to detect its location or momentum. Go on, do it. You insisted that these are the SAME effect. So now show it before a misinformation infraction falls onto that upside-down triangle pin.

Zz.

You insisted that these are the SAME effect.
I don't insist. I just thought there ought to be some sameness, even if by analogy.
Any way you measure the electron in the double slit experiment, you got to interact with it.
I don't know how to show that the virtual photon interactions of a loop of wire. I'm not smart.

On the other hand, it occurs to me that when you pass an electron through a loop of wire, there's an induced current and thus a magnetic field, effecting the momentum.

I have a question on this experiment as well, if the detector was placed at some point beyond the double-slit then would the outcome be the same? Would there still be no interference pattern? The reason I ask is because if it is placed beyond the double-slit then wouldn't it have already interfered with itself by the time it reached the detector?

ZapperZ
Staff Emeritus
I don't insist. I just thought there ought to be some sameness, even if by analogy.
Any way you measure the electron in the double slit experiment, you got to interact with it.
I don't know how to show that the virtual photon interactions of a loop of wire. I'm not smart.

On the other hand, it occurs to me that when you pass an electron through a loop of wire, there's an induced current and thus a magnetic field, effecting the momentum.

Yeah, so? How does this destroys the original state of the electron the same way that you scatter off a photon? It is NOT the same, and in fact, this measurement using electrons can be made with very little disturbance to its trajectory. Is there some disturbance? Sure! I mentioned this already in the very beginning. But it is NOTHING like a photon collision as in Compton scattering! It is NOT the same!

The HUP is something INTRINSIC to the system. It isn't about the instrumentation. If this is something you cannot comprehend already by now, then you are welcome to continue in believing what you believe. I have no more patience to continue.

Zz.

ZapperZ
Staff Emeritus
I have a question on this experiment as well, if the detector was placed at some point beyond the double-slit then would the outcome be the same? Would there still be no interference pattern? The reason I ask is because if it is placed beyond the double-slit then wouldn't it have already interfered with itself by the time it reached the detector?

A "screen" is a detector. After the slit, it no longer matters anymore if you detect it or not. The superposition of path through the slit is preserved.

Zz.

Im sorry I didn't quite get your answer. Would there be an interference pattern or would there be two lines?

ZapperZ
Staff Emeritus
That would be an interference pattern. It is AFTER the slit, which is the "normal" way to detect the interference pattern, isn't it?

Zz.

DrChinese
Gold Member
Im sorry I didn't quite get your answer. Would there be an interference pattern or would there be two lines?

a. Interference pattern on the screen (which is a position detection instrument) results when we do NOT know which slit the particle passed through.

b. Two bar pattern on the screen results when we DO know which slit the particle passed through.

There are a lot of ways to convert from a. to b. For example, you could put a barrier down the middle of the apparatus so that a particle going through one slit cannot pass over to the other side. You can place crossed polarizers so that you can determine from the spin which slit the particle passed through. Or you can do it the old fashioned way, and simply cover up one slit or the other. Any way you choose to "look" at the path taken, that will result in the two bar pattern and not the interference pattern.

ZapperZ

You wrote..

yeah, but you were talking about REAL photons, not virtual photons that are responsible for EM interaction. Do you think these are the SAME thing as in what you described earlier? I mean, did you had to send in a photon of a certain wavelength to detect the electron that had just passed through a loop of wire? No you didn't. You have no control over that.

It is still besides the main point, which is that the HUP has nothing to do with the instrumentation. I can make as accurate of a measurement as I want the single measurement of position and momentum based on the technology available. The uncertainty of each position and momentum measurement that I got has nothing to do with the HUP, no matter how I measured them.

This is the most common misconception of the HUP (try googling it).

Both photons and material particles such as electrons create analogous interference patterns when passing through a double-slit experiment. For photons, this corresponds to the interference of a Maxwell light wave whereas, for material particles, this corresponds to the interference of the Schrödinger wave equation. Although this similarity might suggest that Maxwell's equations are simply Schrödinger's equation for photons, most physicists do not agree.[43][44] For one thing, they are mathematically different; most obviously, Schrödinger's one equation solves for a complex field, whereas Maxwell's four equations solve for real fields. More generally, the normal concept of a Schrödinger probability wave function cannot be applied to photons.[45] Being massless, they cannot be localized without being destroyed; technically, photons cannot have a position eigenstate |\mathbf{r} \rangle, and, thus, the normal Heisenberg uncertainty principle ΔxΔp > h / 2 does not pertain to photons. A few substitute wave functions have been suggested for the photon,[46][47][48][49] but they have not come into general use. Instead, physicists generally accept the second-quantized theory of photons described below, quantum electrodynamics, in which photons are quantized excitations of electromagnetic modes.

I'm not trying to split hairs here, but it seems clear that the HUP applies to particles with mass. Maybe this requires a higher level of understanding than that of which I dispose!

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...you are welcome to continue in believing what you believe. I have no more patience to continue.
I don't mean to irritate you, and I respect your perspective. You're probably making perfect sense and I'm missing the important point.

Carid said:
...the HUP applies to particles with mass
All particles have mass (not to be confused with the confusingly named "rest mass").

ZapperZ
Staff Emeritus
ZapperZ

You wrote..

I'm not trying to split hairs here, but it seems clear that the HUP applies to particles with mass. Maybe this requires a higher level of understanding than that of which I dispose!

I don't understand what's the issue here. The HUP applies to ANY quantum "particle", even photons, phonons, magnons, spinons, bipolarons, etc.. etc. Do those have "masses"?

Photons have no mass, as in invariant mass (see our FAQ). Are you claiming that the HUP does not apply to photons? Why is "mass" suddenly involved in this? I never brought it up as being relevant or irrelevant to the HUP.

Zz.

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ZapperZ

I don't understand what's the issue here. The HUP applies to ANY quantum "particle", even photons, phonons, magnons, spinons, bipolarons, etc.. etc. Do those have "masses"?

Photons have no mass, as in invariant mass (see our FAQ). Are you claiming that the HUP does not apply to photons? Why is "mass" suddenly involved in this? I never brought it up as being relevant or irrelevant to the HUP.

I'm not claiming anything. I'm trying to understand if what is written in the wikipedia article is correct, incorrect, or in need of clarification. It came as news to me that "the normal Heisenberg uncertainty principle ΔxΔp > h / 2 does not pertain to photons", to quote what is written in the article. This wikipedia article may be part of some erudite point-scoring between ecclesiastics of the church of QM or it may be a pretty good summary of the current state of knowledge. I'd like to know if what I read is of value.

ZapperZ
Staff Emeritus
A single-slit diffraction is the most obvious example of the HUP at work. Both photons and other particles such as electrons, protons, etc. exhibit this property.

As you make the slit smaller (i.e. your knowledge of the position $\Delta x$ at the slit becomes better), the particles (including the massless photons) that made it through the slit will acquire a larger spread of momentum $p_x$. This is manifested in the larger spread of the diffraction pattern. Your ability to predict what the next momentum would be from the next photon that passes through the slit becomes more uncertain as the slit is made smaller.

This is exactly the HUP! I have described this in more detail elsewhere. It is one of the most common misconception of the HUP.

I really have very little patience in trying to correct Wikipedia. It is an endless (and thankless) job. People who use it as their primary source deserve what they get.

Zz.

DrChinese
Gold Member
I'm not trying to split hairs here, but it seems clear that the HUP applies to particles with mass. Maybe this requires a higher level of understanding than that of which I dispose!

Wikipedia is great if you want to understand something in 10 seconds. It is designed for a different purpose than what you will find at PhysicsForums. Accordingly, we don't actually give any weight to a quote from Wikipedia. A standard textbook would be a much better place to start than Wikipedia.

And to be fair, you are splitting hairs! Photons absolutely obey the HUP. They have position and momentum operators, which do not commute. They also have spin operators that do not commute. Yes, it is true that a photon does not have exactly the same operators as electrons but that is because they are different particles.

For a photon, its energy (which implies mass) is proportional to its frequency (and inversely proportional to its wavelength). The speed is c (in a vacuum), and it moves in a direction so it has a velocity. Combine these elements and you have momentum. Problem solved!

Oh ok thx for the help.

What if you had two detectors beyond the double-slit. One detector along each path that the electron would travel if it had passed through eithier slits,and then fired electrons one at a time. Then you could tell which slit that the electron had passed through. Would you detect an electron on each path since it would be traveling both since we hadn't observed it yet? Would there be an interference pattern?

a. Interference pattern on the screen (which is a position detection instrument) results when we do NOT know which slit the particle passed through.

b. Two bar pattern on the screen results when we DO know which slit the particle passed through.
What "TWO BAR" pattern?
The interferance pattern is spread out over a dispersion pattern. If the interference pattern within the dispersion pattern (Bell Curve) goes away do to "looking" at at least one slits acctivity, the wide single dispersion pattern is what remains as if there were only one slit even though both are open.
NO "Two bar pattern" unless your in a "Near field" set up in which case a interfance pattern is not even possible.

DrChinese
Gold Member
What "TWO BAR" pattern?
The interferance pattern is spread out over a dispersion pattern. NO "Two bar pattern" unless your in a "Near field" set up in which case a interfance pattern is not even possible.

Of course it depends on the setup, but: In any normal Young type experiment, in which you know which slit the photon goes through, you will get the two bar pattern and no interference fringes. (This is not the same pattern you get from entangled photons, which produce the single dispersion pattern you describe.)

An example of such a setup would be: laser light source oriented at a 45 degree angle and aimed at 2 slits; place polarizers in front of each slit, one oriented at 0 degrees and the other at 90 degrees. You will see 2 bars, because any photon that can pass through one slit cannot pass through the other; and there can be no interference.

Of course it depends on the setup, but: In any normal Young type experiment, in which you know which slit the photon goes through, you will get the two bar pattern and no interference fringes. (This is not the same pattern you get from entangled photons, which produce the single dispersion pattern you describe.)

An example of such a setup would be: laser light source oriented at a 45 degree angle and aimed at 2 slits; place polarizers in front of each slit, one oriented at 0 degrees and the other at 90 degrees. You will see 2 bars, because any photon that can pass through one slit cannot pass through the other; and there can be no interference.
??????
Set the slits one inch apart and the source 1 foot away and yes you will see "Two Bars"
But remove the filters and you will still see Two Bars.
That is called a near field set up slits too wide and source too close.
Each slit must be so thin that a wide dispersion pattern is produced.
That has nothing to do with "entanglement"!
PLUS, the two slits must be so close to each other the two wide dispersion patterns completely overlap each other; to show only one twice as bright dispersion pattern.
And IF the source is far enough away to be in a "Far Field" set up - THEN you get a interferance pattern WITHIN the range of the one dispersion pattern.
At least until you put some filters in the way that mess up the SUPERPOSTION of the single photon going through.
Again nothing to do with "entanglement"!

DrChinese