4:20 you can definitely freeze photons and you should do a video about it! Look into the experiment of Susanne Yelin and Mikhail Lukin(if you haven't already). This actually had a great impact on quantum information theory so it would fit right into the current topic. Thumbs up so he can see.
I had this idea for turn signals. Making the turn would collapse the wave function and make the appropriate signal trigger before the turn is made. That way people would never need to manually indicate their turns.
If you could make a car warn people of things that would kill the driver when there is still time to react and even after the reaction takes place, voila, no more traffic deaths.
I wanted to thank all of you at pbs space time. I've watched all of your videos many times through over the last 3 years and have loved everyone. You teach me the things I'm curious about and keep me grounded. while I wish I had payed more attention in math in high school to better understand the foreground and join in the discussions I couldn't be happier to be your student. Thank you all so much!
I think the problem here is that the interference patterns of C and D overlap on the screen. When we turn on the quantum eraser, we get both C and D patterns on the screen. Heres the key point: by overlapping, they produce a pattern that looks like a non-interference A or B pattern; this is why he explicitly mentions, that Cs peaks fall into Ds valleys - they cancel eachother out. Only by backtracking which Photons correlate to C and which to D, we can extract the two Interference patterns. However, past me doesnt have that information. All it sees is non-Interference (A or B), or the result of the two quantum eraser patterns overlapping(C or D), which also looks like non-interference. The lack of information whether detector C or D fired up is the reason why we cant win stuff...
Lord Of Blood No it literally does. It absolutely 110% works like that. I mean, someone might have thought I was joking, but you come along, realise I'm being serious, then embarrass yourself with how wrong you are about the Quantum Eraser effect.
It could be that RUclips and he now uses quantum computers made in such a way that the qbits values of the RUclips server with that comment is altered because of an entangled pair that he had was exposed to the observer effect. ;) P.S - This comment has no substance
Carcinogenic Radiotherapy Yes, it's a pity that one needs a postscript to negate a comment that was clearly a joke in case someone doesn't get it and goes all Professor Frink...
since the challenge is over... here's the answer You can't read the interference patterns until the entangled pairs hit their respective detectors. Without the information about which detectors have been triggered the detection screen just looks like static. It's only after the fact once you know which detectors triggered can you separate out the interference patterns from the rest of the signal on the detection screen.
Could not you solve this problem if you fire many photons for every bit of information? Lets say you want to send 1 so you let enough photons to enter eraser to produce interference patern, then you wait for some time and send another bit and another group of photons, than wait for some time and send another bit and so on. You should get many paterns or mess diveded by silence.
Let's say I just want to send past self a single true/false boolean. No interference pattern is a small cluster right? And both c and d interference patterns are spread out. If I shoot one photon, then in the future choose to detect with A, or scramble with CD, can't I probabilistically tell? If the single spot on the paper is within the small cluster area of "no interference pattern zone" future self probably chose to detect with A. If the spot is on a far edge of the paper, future self probably chose to send to CD. What am I missing here?
Because you can only control weather it hits A or C/D. In the past you don't know which photons triggered C or D and their interference patterns are line up peaks to troughs you can't see any interference pattern. After the photons hit the detectors and you now which photons on the screen they correspond to you can see it when you separate them. Basically the patterns from A and C mixed with D, because we can't separate them in the past, are indistinguishable. In the future we can separate them and can see the interference patterns again.
i figured it out! the reason is that there is no love in it when ou're trying to get money! as we learned in interstellar, only love can transcend both space and time! i guess that means to shirt for me
***** Oh come on! So not only has this timeline NOT invented diet bacon, but climate change is still a thing? Well at least Clinton won't be president, her opponent has to be better than Cruz this ti- OH GOD DAMMIT!
late to the question, but i solved it. entanglement is not a one way street. if you look at the results as soon as they appear then the other particle will have been decided. to simplify it take it down to a coin toss. if it displays a pattern then the outcome is heads, if it displays two piles then the outcome is tails. when you look at entangled particle 1 (seen as the results) it will tell entangled particle 2 what to do regardless of what has been done to particle 2. in short the entangled pair do not care which of the two are being observed. the only way to get tomorrows coin toss right is to check the answer after you have already put the correct answer in (with this set up atleast)
@@eriknelson2559 I may not know what I'm talking about here but I have a question anyways. do you think this could be circumvented with the use of polarized light?
the question is philosophical one, are you prepared to create a working model prove it works but purposefully don't claim the winnings , tear up the ticket and repeat the process over and over until you come up with a better bigger time machine that a person can get into, because that's the end game we all seek.
@@unneccry2222 Not sure if he ever posted it the answer, but the solution is that the past version of himself just sees a single blob of particles on the screen. That blob is made up of 2 different interlocking interference patterns from detector C and D combined. For that data to be useful, you need to know which detectors activate corresponding to specific particles on the screen. If you only look at the particles on the screen that correspond to what hit a single detector, THEN you can see an interference pattern by ignoring the particles that reached from the other detector. In simpler terms, you need the future information to unencrypt the past.
Build the machine. Start it, and retrieve your six numbers. You see, you all ready saw the winning numbers and placed them in the machine in your multiverse copy of yourself tomorrow; so that, you get it today. So just wait for them to come thru. Play the numbers and win. But you do need to build your time machine.
Yes we cannot read the numbers but we can still get a YES or NO signal. If there was a signal it would be a YES. We can do it this manner. 2 signals for a 1....and 1 signal for a 0. There you have it....a binary code.
You can "freez" the 100 us laser puls in Pr3+:Y2SiO5 crystal using EIT to store the twin photons up to 60 seconds. No sendin photons from Earth to Moon needed. Using two paralel time machines you can create feedback loop.
Solution: Delay space between making entangled particles, and the detectors, by a large enough amount to make it several minute delay or something. Set up 8 of these experiments. Thus, you can make a byte from 1s and 0s. If you want to send a 1, you detect *NONE* of the particles at all, thus giving you an interference pattern to see. If you want to send a 0, you detect *ALL* of the particles, giving you a random landing set that doesnt make a pattern. Common mistakes: "You need data from the detectors to know what particals to filter out/to filter out particles, as half interfere and half dont" - You mis unstand, Either one of two cases happens for each experiment: 0. Send *ALL* the particles to detectors A and B only, thus knowing exactly where each one went, and seeing a random landing spot of particles 1. Send *NONE* of the particles to any detector at all, thus not knowing at all what slit they went through, and thus seeing an interference pattern on that experiment without anything else there, there is no data later you need, as there was never any particles detected for that experiment besides in the interference
Because: if you receive the photons from the future you would just buy lots of tickets and forgot to send the message informing your past-self of the lottery numbers the next day, thus making your past self not receive information from the future, what would make you disappointed for not receiving information from your future-self.
joining the party late, but what everyone is missing here is the fact that you need the information on the detectors A,B,C,D to discern the random pattern that shows up on the which way detector. Without the "future" information from A,B,C,D all you can see is a randomly distributed points with no head or tail.
Similar to using entangled particles to communicate instantaneously, the key needed to decode the message is bound within the message itself. Without it, it's just random noise.
Why do you even need a key to decode it if you're sending a message to yourself in the past? Just plan out ahead of time what method you're going to use to communicate the information. (For example, if I'm playing roulette I can decide ahead of time that I'll measure the which-path information to say "Bet on red," and I won't measure it to say "Bet on black.")
I would just like to say that I am so freaking excited that you decided to go over the double slit experiment finally. My obsession over the weirdness of this experiment led me to search for a good explanation of it on RUclips. This some how led me to this awesome station even though a year ago quantum physics wasn't part of it. My patience has been rewarded with a great visual interpretation of the experiment.
It's funny I thought of the same general idea a long time ago, and when I found I found out about the weal or woe spell in D&D I immediately devised a way to use it for backwards information transfer.
This will create a causal loop, to prevent that from happening you will need to use automated computer to flip the switches, if you send it yourself your message from future will affect your past actions and you will never be able to do the same actions in exact same order, and we will have 2 scenarios: 1- an infinite time loop will occur, 2- Or time travelling signal will collapse. But the problem is that you need to feed the numbers to the computer yourself, and thus you will need to be involved. Any conscious involvement will break down the system.
I think the answer is that the interference pattern from the eraser would just be a clump too, until you could separate detectors C and D, which you wouldn't be able to know until after the entangled photon had landed there in the future. So it doesn't work with this kind of eraser. We need a better eraser design.
All the binary numbers would hit the interference screen detector at the same instant. The time delay between the binary numbers information tapped out, to make it decodable is not sent back.
The reason it doesn’t work is because the photon screen is gibberish until you read the detectors and coincidence counter. To fix this you need to replace the Quantum eraser with another photon screen. So now you will get two interference patterns (the photons making the patterns should be entangled due to the beam splitter). Now the original (closer) photon screen is no longer gibberish but shows an interference pattern. Now we can send the photon’s entangled twin on a long journey before we decide to let it either hit a more distant photon screen (creating an interference pattern on both) or we could detect “which way” the photon went and there by collapsing both interference patterns.
It's a trick question because you already are your own grandfather and the cat was never in the box in the first place. Another challenge victory for Scowlie.
In the experiment described, the numbers are encoded sequentially in time. The arrival of photons at the screen are not said to be time-stamped. The overall pattern on the screen will be like 2 blobs (for every '0' bit), blurrily mixed with interference pattern (for every '1' bit). But, this explanation still allows for communication of one bit of information back in time. All you need is a bunch of these machines all next to each other to communicate a multibit message. So I think there's another, more fundamental, aspect of the situation I don't see yet.
That good feeling when my friends peek at my screen and read the video title and think i'm some genius even though i kinda only understand 20% of what i watch
well the Right answer is that our current understanding of Quantum Theory is lacking. But well, the answer they want to hear is propperly just a simple time paradox like "if you receive the numbers you would have no reason to send them in the first place"
For those unfamiliar: The Bootstrap Paradox is a theoretical paradox of time travel that occurs when an object or piece of information sent back in time becomes trapped within an infinite cause-effect loop in which the item no longer has a discernible point of origin, and is said to be “uncaused” or “self-created” - www.astronomytrek.com/the-bootstrap-paradox-explained/
but if time is relative just like space then ...relative to the time traveler. so it or the person wouldn't necessarily appear in our timeline but continue in its/there own... im not a physicist this is just my take on a back to the future-ish view. and its just an idea.
According to Quantum Theory, it is the knowledge/observation which causes collapse of the wave function. In other words, startling but true, observation in the present builds up a back history of events for the past. So the past depends on how you choose to look in the present. In the context of this video, the guy is trying to know the past from the interference screen without making an observation on the screen A/B. Which means that the collapse has not occurred on the interference screen. So instead of finding the lottery numbers he will still find an interference pattern (or the lottery numbers in a blurred out fashion). Only once the observation is made in the future (which is applicable to all not just to the guy in the video), will the lottery numbers appear on the interference screen. Else it would be a violation of Quantum Theory as we know it. And the headlines in the newspapers the next day wouldn't be about a guy who guessed lottery numbers but that our whole work of the last 100 years in Quantum physics just got quashed!
The only way the eraser works is if you don't look at the interference patterns first. If you look at the interference patterns first, you've already collapsed the waveform and it can't be erased later.
He said that the past self can read the interference pattern. And i'm pretty sure it only stops working if you detect which slit it came through, not the interference pattern itself.
Florian Fahrenberger Actually I sent an email with what I think was the correct answer. I really don't want to spoil it because I'm absolutely sure that it was the right answer but it has to do with the interference patterns of receiver C and D
Correct me if I'm wrong but wouldn't the combined interference pattern of both c and d look the same as a or b since the nodes and antinodes overlap creating a even distribution?
But they could fire one photon at a time, and see which sensor it hits and where it landed. So combined they could still overlap and cancel each other out, but you would still be able to know what the pattern looks like from each sensor. I think.
But if the interference pattern for c and d canceled out to look the same as a and b, then they wouldn't look like an interference pattern at all. They would look like standard particle bands. But I thought we knew for a fact that that does not happen if you erase the which-path information?
Anthony Trupiano Firing one at a time wouldn't also be definitive since it is still semi random. The photon may land in a node area (dark area) giving no real proof that it landed in either C or D.
It is because future interference pattern ( or which path information) can only be made sense retroactively (assuming faster than light travel is not possible).
This video actually made the quantum entanglement video make much more sense, the whole thing is simpler(at least to me) when you don't include detector B.
The numbers for the lottery would be all scrambled as long as the lottery numbers themselves were random and not already pre-selected. The machine you invented would not only start a butterfly effect changing random occurrences and you would be altering a time line but which one. The message would be sent and received by all dimensions; then the signal could quiet possible be broadcast not from just Point A (initial message) to Point B (Past self). But, from Point A to the beginning of time & to the end of time.
The original lottery numbers are chosen at random and are like an interference pattern. Measuring them will change the results of the drawing just like measuring witch slit a photon passes through. I think.
They are macro objects so i think they are a different kind of pseudorandom. You could with enough information use classical physics to measure all the balls and the paddle spinning them and make an accurate prediction. If you had impossibly vast information down to the smallest detail and every parameter. In QM you cant do that even hypothetically.
This might be just me being stupid, but isn't the answer simply that when the interference pattern is created, where the individual photons land on the screen is fundamentally random and can only be described in terms of probablilities, and therefore can't be used to transmit information?
I've sent an answer, here's what I know (let's discuss how we could make this experiment work!) Original message: As far as I know, this experiment only works when analysing the data/measurements at a later time. The value that is measured is the time at which an photon hits the 'screen' and when it's entangled partner hits a detector, by firing one photon at a time. As for the interference pattern, you stated that the patterns from detectors C and D were the inverse of eachother. Combined, they will produce a regular pattern, which seems not to have any interference patterns. However, the quantum eraser will 'create' an interference pattern, only because it is possible to extract the which-path information from the analysed data -> I believe it is called 'joint detection'. Basically, you could only get your sent information back if you knew the whole dataset; the data for the interference pattern is only from a subset of all the photons. I could take this further and say that it doesn't seem like data is being sent back in time. The interference pattern from the quantum eraser is just hidden within that regular pattern.
Sorry for being a smart-arse, but even though he did not explicitly say that we shouldn't discuss answers online, I guess this rule still holds as for every challenge question... Maybe you delete your comment to not spoil the fun for the others ;)
We could know the whole dataset since we could count the photons and the time the photon struck the sensor, and if we were using 8 bit binary for the Pick 8, we would have a total of 64 photons to represent the numbers. I believe the problem lies in the fact that our past selves will be measuring the photons so no matter the placement of the mirrors we will end up with piles of photons instead of patterns. The waveform collapse is apparently coming from the conscious act of interacting and or intent.
Nothing here has to do with consciousness. The problem lies in the fact that the pattern will always be the exact same. As stated in the video @ 5:00 -> the signal (which is just a non-interfering pattern) does contain the interference patterns! However, the two patterns from detector C and D are shifted by π, or one stripe. This causes is what causes that non-interfering pattern. It is just that the two waves exactly add up to a single even distribution. Only when we use the analysed data, we could determine which of the subset of the photons are causing which interference pattern. Basically, the clue is in the video itself. It should also be noted that this experiment only 'show interference patterns' when we look at the data, as far as I know. (They don't actually show up on a screen like the standard experiment)
The lottery is a random even. The wave function is host of mathematical possibilities. Once observed the the wave function collapses and one of the possibilities is brought into our universe. The lottery rem as ones a random event and would reranfomize for every observation.
You haven't created a way to interpret interference patterns as numerical data. A way you COULD do it is by having each mirror trigger independently and having each assigned to either 1 or 0. Now you can send binary messages into the past by measuring where the peaks and troughs are and at what time they occurred.
Say I am going through space in a warp drive, I have a telephone connected to a quantum entangled transmitter/receiver (connected to another on Earth) and I am using it to talk to someone. How would the communication work?
I believe people on earth wouldn't be able to gain any information from the telephone. they might see it do something but they wouldn't be able to decode it. that's my best guess
I think that the inevitability of "time being just a co-product" comes to play here. I think that it doesn't really "go back in time" to change anything. The wave function just updates the screen whenever the wave function changes. Since the energy packets are entangled, it sorta "refreshes" the universe to reflect the changes and rearranges which molecules on the screen get existed instead.
I think the answer is before you actually release the frozen photon, the interference screen will only show the same result as there is no observation. The next day when you really insert the lottery information by switching the switch, the photon changed in the same moment but not yesterday. The past hasn't been changed by the machine, it only changed the result now, even if you have that result yesterday. (Sorry, my English might not be good enough to describe the whole situation since I'm not a scientist, but still, hope my guess is understandable and correct.)
If the position of the landed photon changes after landing ("back in time"), how does the change look like? Are we able to detect this change which occurs at a fraction of a second before the paired photon is (not) detected? If yes...does the first photon (i.e. the photon trail left at the detector) "teleport" itself to another position?
I know this is 4 years late but it goes where it would have gone, the light either forms an interference pattern or it doesn't based on if the wave function collapsed. The point of the quantum eraser is that this collapse can happen at any point in space/time and its like the wave function knew what was going to happen all along.
@@theslavegamer you didn’t understand the above question because you assume we only observe the pattern after we use the detector. Try to think about what happens with the pattern during the time it isn’t observed.
Solution is that the photons would either interact with other particles or would be absorbed or altered there for resulting in skewed results. Also I personally believe this to be entangled pair results issue of time relativity between the two photon pairs but that's just my opinion. We need to understand time before we try to understand quantum physics.
I propose an addition to the quantum eraser experiment. Create a way look at the interference screen before the entangled pair hits A through D. This means that all particles are observed at the interference screen as it hits and nothing is yet known at the entangled pair. This should create an interference pattern as nothing was measured yet and the interference screen has been hit. Would this force all the entangled pairs to wind up in C or D?
If you acknowledge that the signal does contain the interference pattern from your future choices with perfect fidelity, I don't see why you wouldn't be able to use it to convey information. Maybe the lottery number / Morse code example is needlessly complicated, but the thought experiment can be simplified to a single roulette spin. (e.g. You measure the which-path information to say "bet on red," and don't measure it to say "bet on black.") If the signal actually contains the interference pattern based on your future decision, then that's the same as it also containing the information about which color to bet on. I'm curious to see Matt's solution, because it's a question that I've pondered for a long time. I don't really see why it shouldn't work.
Because both sets of photons are hitting the first detector. Because each interference pattern is offset from the other, the combined patterns are indistinguishable from a clumping pattern and its only once you know which photons hit the second set of detectors can you go back and see if their entangled pair made the pattern or not.
What if the 50% silvered mirror creating the quantum eraser in between Detectors C and D was only 25% silvered....or 0.00000001%? Technically it still creates enough uncertainty to erase the information. I’d still expect to see an interference pattern. But then the question is, at what point does the photon consider the information “conserved” and produce two lines again?
Actually, if the photon lands anywhere outside of the two bands that are formed under a situation where "which path" information is obtained then you can pretty confidently say it is an interference pattern. So it should only take a few photons to determine interference and a few more than that to confidently determine wave function collapse.
But there is too much overlap. The interference pattern only occurs with A and B, right? But A's interference pattern and B's interference pattern together form the whole picture. And the "single pile" of A and B is also going to be contained in one of those two interference patterns-- so because of the overlap of area it can fall into, you won't be able to distinguish between no interference pattern and one of the two interference patterns. Moreover, the patterns only appear after you have fired thousands of protons at it. A single proton can land anywhere, it just has a higher chance of landing in one of the areas you see as a stripe in the pattern.
So are we implying that there's a time delay in quantum entanglement communication when traveling through dimensions? Which wouldnt be all that surprising as time is relative no?
The tapping out of the lottery numbers informs the switching between particle and wave modes at the screen. While, the particle modes (A,B) will do their part in encoding the lottery values at the screen, the wave modes (C,D) encodings at the screen are indeterminate. That is, while in wave mode the particle will land randomly on the screen thus will manifest no information about the lottery value that originated it. About half of the lottery information will be lost at the screen.
I think you won't be able to know which detector was fired (C or D) when you use the binary bit correlated with activating your switch, so you won't know which interference pattern to look for with each individual photon, basically ruining one of your two binary numbers. and then what's worse, the non-interfering pattern landing zone could even overlap one or both of the interference pattern landing zones, possibly ruining all the information.
Even though for some reason I cant read the actual numbers I would still get some signal? In other words I would be 100% positive that future me sending me some unreadable information, if so, that is amazing.
My intuition says that pattern for each photon in such a case will be an interference pattern and it wound never be a particle pattern. Once photon hits detector A, entangled photon which is between Earth and moon might disappear?
Does anyone here have a deeper understanding of the quantum eraser? I'm struggling to wrap my head around this. Is this an oversimplification? Is it really as simple as moving the two beam splitters out of the way and an interference pattern magically appears on the screen? Or is there some analysis of the data involved that isn't mentioned here?
The pattern seen on the screen is always the same (a non-interfering pattern). Only when you later analyse data about detector+time, you can find which part of the screen is the interference pattern. In the video, it was stated that the patterns are offset by one stripe (π). Together, they form a regular pattern. So, the information is indeed there, but the key is having 'which-path' information by examining the detector+time data.
if A and C/D are swapped is the result the same? do only interfering photon wavefronts pass through the splitter? are all photons entangled by the delayed choice crystal? if there are multiple As, do they interfere?
He sent the information as binary to his past, his past doesn't know that his future self is trying to communicate with him, so his past just sees this as random patterns. If his past is expecting the information then as soon as he realises what each binary code means the future entangled electrons will untangle themselves and form a random pattern again.
It wouldn't work because an interference pattern is a probability chart. If you took a photon, did all this stuff with it, then looked where it hit the graph, you wouldn't know if it was part of an interference pattern or not.
"Why am I going to be sadly disappointed after turning on my invention?" Because research has shown a correlation between depression and winning the lottery thanks to "hedonic adaptation"?
the reason C and D are mirror images of each other is obvious: the ones that get reflected get flipped horizontally, as all mirrors do. thus, their peaks and valleys are mirror images of that recorded by the other sensor.
Doesn't this mean that information CAN travel faster than the speed of light? so we could (in theory) use this to create long range transmitters that transmit across light years instantaneously.
Yes, except that the machine doesn't actually work. I won't spoil it for you though, so if you want to find out before next week, you'll need to do some research. :}
This can still be done through Quantum Entanglement outside of photons, just on a molecular level. Also, Quantum tunnelling allows for faster than light communication.
Would be great if you can also make an episode on the current state and different types of quantum computers like Xanadu, Dwave, IBM..., what challenges are ahead in developing commercially viable QC and debunking myths on the actual applications of Quantum
The only way to win is never to play :) The information you get back is just noise. Nature prevents a paradox to occur, "Sender" and "Receiver" must be distinguished from each other and cannot send or receive information simultaneously in this way (quantum eraser experiment) Your stuck in two versions of reality (parallel universes), the person who knows the outcoming of lottery can never collect the money because he is not a part of that reality. (because of complementarity)
Plot twist: He asks us a really hard physics problem and then publish the results in a scientific paper.
You figure it out and then he appears 60 years earlier and assassinates your grandparents. He never has to share that lottery money.
Malcolm in the Middle had an episode with that plot.
Lmfao!!
Or even worse, a patent!
Decoding quantum states indeed requires specific information, which might only be accessible within the constraints of causality.
4:20 you can definitely freeze photons and you should do a video about it! Look into the experiment of Susanne Yelin and Mikhail Lukin(if you haven't already). This actually had a great impact on quantum information theory so it would fit right into the current topic.
Thumbs up so he can see.
Freezing photon is an act of observing or interfering, and that would colapse the wave
Favorite number 😂
I had this idea for turn signals. Making the turn would collapse the wave function and make the appropriate signal trigger before the turn is made. That way people would never need to manually indicate their turns.
genius
If you could make a car warn people that they're about to die in a car crash, that would be cool. Enough for a "tell my wife I..."
If you could make a car warn people of things that would kill the driver when there is still time to react and even after the reaction takes place, voila, no more traffic deaths.
If you could just make self-driving cars universal so nobody ever dies in car wrecks life would be grand.
If your quantum eraser time travelling messenger device worked, you'd know that that was inevitable anyway.
I wanted to thank all of you at pbs space time. I've watched all of your videos many times through over the last 3 years and have loved everyone. You teach me the things I'm curious about and keep me grounded. while I wish I had payed more attention in math in high school to better understand the foreground and join in the discussions I couldn't be happier to be your student. Thank you all so much!
I think the problem here is that the interference patterns of C and D overlap on the screen. When we turn on the quantum eraser, we get both C and D patterns on the screen. Heres the key point: by overlapping, they produce a pattern that looks like a non-interference A or B pattern; this is why he explicitly mentions, that Cs peaks fall into Ds valleys - they cancel eachother out. Only by backtracking which Photons correlate to C and which to D, we can extract the two Interference patterns. However, past me doesnt have that information. All it sees is non-Interference (A or B), or the result of the two quantum eraser patterns overlapping(C or D), which also looks like non-interference. The lack of information whether detector C or D fired up is the reason why we cant win stuff...
Did you get the t-shirt? How about using only one of the C or D detectors?
@@Hhuimklf it again won't show any interference, as you'll be able to figure out the slit it passed through
Right!
Thanks to the Quantum Eraser, this comment is actually first.
Doesn't work like that.
Lord Of Blood No it literally does. It absolutely 110% works like that. I mean, someone might have thought I was joking, but you come along, realise I'm being serious, then embarrass yourself with how wrong you are about the Quantum Eraser effect.
It could be that RUclips and he now uses quantum computers made in such a way that the qbits values of the RUclips server with that comment is altered because of an entangled pair that he had was exposed to the observer effect. ;)
P.S - This comment has no substance
Carcinogenic Radiotherapy Yes, it's a pity that one needs a postscript to negate a comment that was clearly a joke in case someone doesn't get it and goes all Professor Frink...
+LisztyLiszt Jest and seriousness are entangled, as you just proved in your previous comment.
since the challenge is over... here's the answer
You can't read the interference patterns until the entangled pairs hit their respective detectors. Without the information about which detectors have been triggered the detection screen just looks like static. It's only after the fact once you know which detectors triggered can you separate out the interference patterns from the rest of the signal on the detection screen.
Could not you solve this problem if you fire many photons for every bit of information? Lets say you want to send 1 so you let enough photons to enter eraser to produce interference patern, then you wait for some time and send another bit and another group of photons, than wait for some time and send another bit and so on. You should get many paterns or mess diveded by silence.
@@martinpavlicek2299 no both of the interference pattern in c and d are complimentary
You can't distinguish between them
@@rajajagati5610 what if there is detector e?
3a3a3a At what location in the setup? Remember, in this experiment, it’s all about location. Everything is geometrically set up.
Let's say I just want to send past self a single true/false boolean. No interference pattern is a small cluster right? And both c and d interference patterns are spread out. If I shoot one photon, then in the future choose to detect with A, or scramble with CD, can't I probabilistically tell? If the single spot on the paper is within the small cluster area of "no interference pattern zone" future self probably chose to detect with A. If the spot is on a far edge of the paper, future self probably chose to send to CD. What am I missing here?
Because you can only control weather it hits A or C/D. In the past you don't know which photons triggered C or D and their interference patterns are line up peaks to troughs you can't see any interference pattern. After the photons hit the detectors and you now which photons on the screen they correspond to you can see it when you separate them.
Basically the patterns from A and C mixed with D, because we can't separate them in the past, are indistinguishable. In the future we can separate them and can see the interference patterns again.
i figured it out! the reason is that there is no love in it when ou're trying to get money! as we learned in interstellar, only love can transcend both space and time!
i guess that means to shirt for me
Dennis Haupt when you’re watching that movie with people who are eating it up 😱
love AND gravity, don't forget that important part - or else he'd never been able to communicate across time
You still have to figure out the equation for love tho
@@LuisSierra42 easy. 1+1=1
Lmao the next video is so true I agree wow good point. Oh right you guys haven't seen it yet.
I saw it in the future but since I'm in the now I completely forgot it already
Are you talking about the next video, or the one after that? I mean, it was ok, but just a filler really.
Haha, joke's on you! After your comment we decided not to make the next video that you already saw, thus creating a paradox.
Is it OK if I call you et. Al?
*****
Oh come on! So not only has this timeline NOT invented diet bacon, but climate change is still a thing? Well at least Clinton won't be president, her opponent has to be better than Cruz this ti-
OH GOD DAMMIT!
Ugh I'm really eagerly awaiting the solution of this challenge!!! It's gonna be such a long week :/
I didn't have to wait. ^^
+Tim de Vries lol
lol so am I
I'll be at burning man by this time next week, so I'll have to wait 3 weeks to find out. :(
Well...I built a quantum eraser info-time machine to go forward to the next video...but I couldn't read the answer. :/
late to the question, but i solved it. entanglement is not a one way street. if you look at the results as soon as they appear then the other particle will have been decided. to simplify it take it down to a coin toss. if it displays a pattern then the outcome is heads, if it displays two piles then the outcome is tails. when you look at entangled particle 1 (seen as the results) it will tell entangled particle 2 what to do regardless of what has been done to particle 2. in short the entangled pair do not care which of the two are being observed. the only way to get tomorrows coin toss right is to check the answer after you have already put the correct answer in (with this set up atleast)
Hands down, best science channel on YT
Half of any lottery winnings...with a straight face. Nice
We would see the sum of interference patterns C + D which results in no interference pattern at all.
Relative displacement of C+D implies a relative phase shift of the wave components emerging from the slits towards the detector screen
@@eriknelson2559 I may not know what I'm talking about here but I have a question anyways. do you think this could be circumvented with the use of polarized light?
the question is philosophical one, are you prepared to create a working model prove it works but purposefully don't claim the winnings , tear up the ticket and repeat the process over and over until you come up with a better bigger time machine that a person can get into, because that's the end game we all seek.
This is actually the most interesting challenge yet. Can't wait for the answer from Matt.
where is the answer?
@@unneccry2222 Not sure if he ever posted it the answer, but the solution is that the past version of himself just sees a single blob of particles on the screen. That blob is made up of 2 different interlocking interference patterns from detector C and D combined. For that data to be useful, you need to know which detectors activate corresponding to specific particles on the screen.
If you only look at the particles on the screen that correspond to what hit a single detector, THEN you can see an interference pattern by ignoring the particles that reached from the other detector. In simpler terms, you need the future information to unencrypt the past.
5:11 - You aren't actually sending information back into the past, and the patterns are just a result of retroactively filtering the data.
Excatly!!why arent there more comments on this !! I thought i had got the whole thing wrong!!
He just forgot to put on his reading glasses. Where's my T-shirt?
Build the machine. Start it, and retrieve your six numbers. You see, you all ready saw the winning numbers and placed them in the machine in your multiverse copy of yourself tomorrow; so that, you get it today. So just wait for them to come thru. Play the numbers and win.
But you do need to build your time machine.
Yes we cannot read the numbers but we can still get a YES or NO signal. If there was a signal it would be a YES. We can do it this manner. 2 signals for a 1....and 1 signal for a 0. There you have it....a binary code.
You can "freez" the 100 us laser puls in Pr3+:Y2SiO5 crystal using EIT to store the twin photons up to 60 seconds. No sendin photons from Earth to Moon needed. Using two paralel time machines you can create feedback loop.
Solution:
Delay space between making entangled particles, and the detectors, by a large enough amount to make it several minute delay or something. Set up 8 of these experiments. Thus, you can make a byte from 1s and 0s. If you want to send a 1, you detect *NONE* of the particles at all, thus giving you an interference pattern to see. If you want to send a 0, you detect *ALL* of the particles, giving you a random landing set that doesnt make a pattern.
Common mistakes:
"You need data from the detectors to know what particals to filter out/to filter out particles, as half interfere and half dont" -
You mis unstand, Either one of two cases happens for each experiment:
0. Send *ALL* the particles to detectors A and B only, thus knowing exactly where each one went, and seeing a random landing spot of particles
1. Send *NONE* of the particles to any detector at all, thus not knowing at all what slit they went through, and thus seeing an interference pattern on that experiment without anything else there, there is no data later you need, as there was never any particles detected for that experiment besides in the interference
Because: if you receive the photons from the future you would just buy lots of tickets and forgot to send the message informing your past-self of the lottery numbers the next day, thus making your past self not receive information from the future, what would make you disappointed for not receiving information from your future-self.
I need to watch the last video 20 more times before I can even attempt to use the quantum eraser.
Same ,but you 'll understand it better
joining the party late, but what everyone is missing here is the fact that you need the information on the detectors A,B,C,D to discern the random pattern that shows up on the which way detector. Without the "future" information from A,B,C,D all you can see is a randomly distributed points with no head or tail.
Similar to using entangled particles to communicate instantaneously, the key needed to decode the message is bound within the message itself. Without it, it's just random noise.
don't spoil
You Sir are correct
Can't you just use an agreed message as a guide that you send as the first part of the message. Something that allows you to "crack" the encryption?
Exactly! Couldn't I just say that I'll use ascii???
Why do you even need a key to decode it if you're sending a message to yourself in the past? Just plan out ahead of time what method you're going to use to communicate the information. (For example, if I'm playing roulette I can decide ahead of time that I'll measure the which-path information to say "Bet on red," and I won't measure it to say "Bet on black.")
I would just like to say that I am so freaking excited that you decided to go over the double slit experiment finally. My obsession over the weirdness of this experiment led me to search for a good explanation of it on RUclips. This some how led me to this awesome station even though a year ago quantum physics wasn't part of it. My patience has been rewarded with a great visual interpretation of the experiment.
It's funny I thought of the same general idea a long time ago, and when I found I found out about the weal or woe spell in D&D I immediately devised a way to use it for backwards information transfer.
This will create a causal loop, to prevent that from happening you will need to use automated computer to flip the switches, if you send it yourself your message from future will affect your past actions and you will never be able to do the same actions in exact same order, and we will have 2 scenarios:
1- an infinite time loop will occur,
2- Or time travelling signal will collapse.
But the problem is that you need to feed the numbers to the computer yourself, and thus you will need to be involved. Any conscious involvement will break down the system.
I think the answer is that the interference pattern from the eraser would just be a clump too, until you could separate detectors C and D, which you wouldn't be able to know until after the entangled photon had landed there in the future. So it doesn't work with this kind of eraser. We need a better eraser design.
All the binary numbers would hit the interference screen detector at the same instant. The time delay between the binary numbers information tapped out, to make it decodable is not sent back.
The reason it doesn’t work is because the photon screen is gibberish until you read the detectors and coincidence counter. To fix this you need to replace the Quantum eraser with another photon screen. So now you will get two interference patterns (the photons making the patterns should be entangled due to the beam splitter). Now the original (closer) photon screen is no longer gibberish but shows an interference pattern. Now we can send the photon’s entangled twin on a long journey before we decide to let it either hit a more distant photon screen (creating an interference pattern on both) or we could detect “which way” the photon went and there by collapsing both interference patterns.
It's a trick question because you already are your own grandfather and the cat was never in the box in the first place. Another challenge victory for Scowlie.
In the experiment described, the numbers are encoded sequentially in time. The arrival of photons at the screen are not said to be time-stamped. The overall pattern on the screen will be like 2 blobs (for every '0' bit), blurrily mixed with interference pattern (for every '1' bit).
But, this explanation still allows for communication of one bit of information back in time. All you need is a bunch of these machines all next to each other to communicate a multibit message. So I think there's another, more fundamental, aspect of the situation I don't see yet.
Project cyber-leap is a special project
That good feeling when my friends peek at my screen and read the video title
and think i'm some genius even though i kinda only understand 20% of what i watch
well the Right answer is that our current understanding of Quantum Theory is lacking. But well, the answer they want to hear is propperly just a simple time paradox like "if you receive the numbers you would have no reason to send them in the first place"
I was wondering about this. Very well explained. Thank you.
Hi, pi.
Hi :-)
I thought pi was 3.14159265358979...
+Shannon Wiggins it is. you can't use periods in usernames if I am not mistaken
+Daniel Blair The last digit differs, and I'm not sure which one is correct. For all I know, they both might be correct in some sense.
I like how this guys explains things.
It sound like if it worked it would create a paradox.
boot strap paradox
For those unfamiliar: The Bootstrap Paradox is a theoretical paradox of time travel that occurs when an object or piece of information sent back in time becomes trapped within an infinite cause-effect loop in which the item no longer has a discernible point of origin, and is said to be “uncaused” or “self-created”
- www.astronomytrek.com/the-bootstrap-paradox-explained/
but if time is relative just like space then ...relative to the time traveler. so it or the person wouldn't necessarily appear in our timeline but continue in its/there own... im not a physicist this is just my take on a back to the future-ish view. and its just an idea.
Not if there's such a thing as a 'delayed-many-worlds' interpretation.
hubes69 I thought of this, but even if it create another time-line / world. In this world the person would also be stuck in a paradox
THIS IS THE BEST CHALLENGE YET
And then God said; "Let there be Light." And there was Light.
Light then replied: "OI FUCK I GOTTA GO BACK!!"
Quantum entanglement in a nutshell.
Some day in the far future, we're going o be telling kids:
"Alright, so the 21st century is where shit starts getting real spooky."
"What went wrong"
Photons forgot lottery numbers.
According to Quantum Theory, it is the knowledge/observation which causes collapse of the wave function. In other words, startling but true, observation in the present builds up a back history of events for the past. So the past depends on how you choose to look in the present. In the context of this video, the guy is trying to know the past from the interference screen without making an observation on the screen A/B. Which means that the collapse has not occurred on the interference screen. So instead of finding the lottery numbers he will still find an interference pattern (or the lottery numbers in a blurred out fashion). Only once the observation is made in the future (which is applicable to all not just to the guy in the video), will the lottery numbers appear on the interference screen. Else it would be a violation of Quantum Theory as we know it. And the headlines in the newspapers the next day wouldn't be about a guy who guessed lottery numbers but that our whole work of the last 100 years in Quantum physics just got quashed!
is there a follow up video? I really want to know more about why this wouldnt work
Because to decode the signal you need to know the position of the switch. Which means you have to know the lotto numbers ahead of time
The only way the eraser works is if you don't look at the interference patterns first. If you look at the interference patterns first, you've already collapsed the waveform and it can't be erased later.
He said that the past self can read the interference pattern. And i'm pretty sure it only stops working if you detect which slit it came through, not the interference pattern itself.
This is the correct answer, but also a major spoiler. Please let people know up front. :-)
Florian Fahrenberger Actually I sent an email with what I think was the correct answer. I really don't want to spoil it because I'm absolutely sure that it was the right answer but it has to do with the interference patterns of receiver C and D
No it's not.
+Col Peps Yes it is, I solved it and it does.
I used this system today, sent information to myself yesterday and won the lottery tomorrow.
Correct me if I'm wrong but wouldn't the combined interference pattern of both c and d look the same as a or b since the nodes and antinodes overlap creating a even distribution?
But they could fire one photon at a time, and see which sensor it hits and where it landed. So combined they could still overlap and cancel each other out, but you would still be able to know what the pattern looks like from each sensor.
I think.
yes it would and thats also the key to the correct answer
you are getting closer to the right answer with each coment. Now there is a less chance that I'll get my t-shirt :(
But if the interference pattern for c and d canceled out to look the same as a and b, then they wouldn't look like an interference pattern at all. They would look like standard particle bands. But I thought we knew for a fact that that does not happen if you erase the which-path information?
Anthony Trupiano Firing one at a time wouldn't also be definitive since it is still semi random. The photon may land in a node area (dark area) giving no real proof that it landed in either C or D.
It is because future interference pattern ( or which path information) can only be made sense retroactively (assuming faster than light travel is not possible).
This video actually made the quantum entanglement video make much more sense, the whole thing is simpler(at least to me) when you don't include detector B.
Uh no
The numbers for the lottery would be all scrambled as long as the lottery numbers themselves were random and not already pre-selected. The machine you invented would not only start a butterfly effect changing random occurrences and you would be altering a time line but which one. The message would be sent and received by all dimensions; then the signal could quiet possible be broadcast not from just Point A (initial message) to Point B (Past self). But, from Point A to the beginning of time & to the end of time.
The original lottery numbers are chosen at random and are like an interference pattern. Measuring them will change the results of the drawing just like measuring witch slit a photon passes through. I think.
They are macro objects so i think they are a different kind of pseudorandom. You could with enough information use classical physics to measure all the balls and the paddle spinning them and make an accurate prediction. If you had impossibly vast information down to the smallest detail and every parameter. In QM you cant do that even hypothetically.
It’s fixed anyway, don’t know how they do it but computers are much more advanced so it’s possible.
Decoding quantum states indeed requires specific information, which might only be accessible within the constraints of causality.
This might be just me being stupid, but isn't the answer simply that when the interference pattern is created, where the individual photons land on the screen is fundamentally random and can only be described in terms of probablilities, and therefore can't be used to transmit information?
I've sent an answer, here's what I know (let's discuss how we could make this experiment work!) Original message:
As far as I know, this experiment only works when analysing the data/measurements at a later time. The value that is measured is the time at which an photon hits the 'screen' and when it's entangled partner hits a detector, by firing one photon at a time. As for the interference pattern, you stated that the patterns from detectors C and D were the inverse of eachother. Combined, they will produce a regular pattern, which seems not to have any interference patterns. However, the quantum eraser will 'create' an interference pattern, only because it is possible to extract the which-path information from the analysed data -> I believe it is called 'joint detection'.
Basically, you could only get your sent information back if you knew the whole dataset; the data for the interference pattern is only from a subset of all the photons.
I could take this further and say that it doesn't seem like data is being sent back in time. The interference pattern from the quantum eraser is just hidden within that regular pattern.
Yeah, looks like this is the answer :)
Sorry for being a smart-arse, but even though he did not explicitly say that we shouldn't discuss answers online, I guess this rule still holds as for every challenge question... Maybe you delete your comment to not spoil the fun for the others ;)
You Sir are correct :-)
We could know the whole dataset since we could count the photons and the time the photon struck the sensor, and if we were using 8 bit binary for the Pick 8, we would have a total of 64 photons to represent the numbers. I believe the problem lies in the fact that our past selves will be measuring the photons so no matter the placement of the mirrors we will end up with piles of photons instead of patterns. The waveform collapse is apparently coming from the conscious act of interacting and or intent.
Nothing here has to do with consciousness. The problem lies in the fact that the pattern will always be the exact same. As stated in the video @ 5:00 -> the signal (which is just a non-interfering pattern) does contain the interference patterns! However, the two patterns from detector C and D are shifted by π, or one stripe. This causes is what causes that non-interfering pattern. It is just that the two waves exactly add up to a single even distribution. Only when we use the analysed data, we could determine which of the subset of the photons are causing which interference pattern.
Basically, the clue is in the video itself. It should also be noted that this experiment only 'show interference patterns' when we look at the data, as far as I know. (They don't actually show up on a screen like the standard experiment)
The lottery is a random even. The wave function is host of mathematical possibilities. Once observed the the wave function collapses and one of the possibilities is brought into our universe. The lottery rem as ones a random event and would reranfomize for every observation.
BECAUSE QUANTUM ERASERS CANT MELT STEEL BEAMS!!!!
*****
'Positively charged'?
You haven't created a way to interpret interference patterns as numerical data. A way you COULD do it is by having each mirror trigger independently and having each assigned to either 1 or 0. Now you can send binary messages into the past by measuring where the peaks and troughs are and at what time they occurred.
Say I am going through space in a warp drive, I have a telephone connected to a quantum entangled transmitter/receiver (connected to another on Earth) and I am using it to talk to someone. How would the communication work?
I believe people on earth wouldn't be able to gain any information from the telephone. they might see it do something but they wouldn't be able to decode it. that's my best guess
I think that the inevitability of "time being just a co-product" comes to play here. I think that it doesn't really "go back in time" to change anything. The wave function just updates the screen whenever the wave function changes. Since the energy packets are entangled, it sorta "refreshes" the universe to reflect the changes and rearranges which molecules on the screen get existed instead.
Why can't we take a snap of the screen every second(assuming the binary numbers are entered every second) and interpret our lottery number from that??
I think the answer is before you actually release the frozen photon, the interference screen will only show the same result as there is no observation.
The next day when you really insert the lottery information by switching the switch, the photon changed in the same moment but not yesterday.
The past hasn't been changed by the machine, it only changed the result now, even if you have that result yesterday.
(Sorry, my English might not be good enough to describe the whole situation since I'm not a scientist, but still, hope my guess is understandable and correct.)
If the position of the landed photon changes after landing ("back in
time"), how does the change look like? Are we able to detect this change
which occurs at a fraction of a second before the paired photon is
(not) detected? If yes...does the first photon (i.e. the photon trail
left at the detector) "teleport" itself to another position?
I know this is 4 years late but it goes where it would have gone, the light either forms an interference pattern or it doesn't based on if the wave function collapsed. The point of the quantum eraser is that this collapse can happen at any point in space/time and its like the wave function knew what was going to happen all along.
@@theslavegamer you didn’t understand the above question because you assume we only observe the pattern after we use the detector. Try to think about what happens with the pattern during the time it isn’t observed.
Solution is that the photons would either interact with other particles or would be absorbed or altered there for resulting in skewed results. Also I personally believe this to be entangled pair results issue of time relativity between the two photon pairs but that's just my opinion. We need to understand time before we try to understand quantum physics.
What's the physical phenomenon that causes photons detected by sensors C and D be out of phase?
Maybe Quantum Entanglement through the usage of Quantum Tunneling.
I propose an addition to the quantum eraser experiment. Create a way look at the interference screen before the entangled pair hits A through D. This means that all particles are observed at the interference screen as it hits and nothing is yet known at the entangled pair. This should create an interference pattern as nothing was measured yet and the interference screen has been hit. Would this force all the entangled pairs to wind up in C or D?
If you acknowledge that the signal does contain the interference pattern from your future choices with perfect fidelity, I don't see why you wouldn't be able to use it to convey information. Maybe the lottery number / Morse code example is needlessly complicated, but the thought experiment can be simplified to a single roulette spin. (e.g. You measure the which-path information to say "bet on red," and don't measure it to say "bet on black.") If the signal actually contains the interference pattern based on your future decision, then that's the same as it also containing the information about which color to bet on.
I'm curious to see Matt's solution, because it's a question that I've pondered for a long time. I don't really see why it shouldn't work.
look up wheeler experiments ;)
didn't have to \O/
Because both sets of photons are hitting the first detector. Because each interference pattern is offset from the other, the combined patterns are indistinguishable from a clumping pattern and its only once you know which photons hit the second set of detectors can you go back and see if their entangled pair made the pattern or not.
What if the 50% silvered mirror creating the quantum eraser in between Detectors C and D was only 25% silvered....or 0.00000001%? Technically it still creates enough uncertainty to erase the information. I’d still expect to see an interference pattern. But then the question is, at what point does the photon consider the information “conserved” and produce two lines again?
Yes...I was wondering that....and what if 49% one way and 51% the others way......when does that become apparent? At p=0.05? Madness!!!
Only 5 t-shirts? come on PBS, I am dying for that t-shirt, give us more chance :)
is there a way to do an erasure test at home
This is the my favorite Challenge Question ever! (so far)
Because you can't differentiate the difference between an interference pattern and a non-interference pattern with single photons?
It can be done with lots of photons though.
Actually, if the photon lands anywhere outside of the two bands that are formed under a situation where "which path" information is obtained then you can pretty confidently say it is an interference pattern. So it should only take a few photons to determine interference and a few more than that to confidently determine wave function collapse.
first: but why is that.
second: dont place your answer here, report them on the correct page. or you blow the whole quiz
But there is too much overlap. The interference pattern only occurs with A and B, right? But A's interference pattern and B's interference pattern together form the whole picture. And the "single pile" of A and B is also going to be contained in one of those two interference patterns-- so because of the overlap of area it can fall into, you won't be able to distinguish between no interference pattern and one of the two interference patterns.
Moreover, the patterns only appear after you have fired thousands of protons at it. A single proton can land anywhere, it just has a higher chance of landing in one of the areas you see as a stripe in the pattern.
TheHobgobyn
Right, but you probably mean C&D, instead of A&B.
I won the Nuclear Physics question, but I have absolutely no idea for this one.
Hooray! I'm so glad I won a t-shirt!
Did you really win one?
More like a t-shirt, half of the lottery winnings, and a Nobel Prize
A working time communication device is the great filter that keeps intelligent life from colonizing the universe.
So are we implying that there's a time delay in quantum entanglement communication when traveling through dimensions? Which wouldnt be all that surprising as time is relative no?
now the movie primer is bouncing about in my head, its half the reason i can follow anything said in these videos
Great movie!
"parabolas are important"
That movie made by brain hurt when I watched it next week.
The tapping out of the lottery numbers informs the switching between particle and wave modes at the screen. While, the particle modes (A,B) will do their part in encoding the lottery values at the screen, the wave modes (C,D) encodings at the screen are indeterminate. That is, while in wave mode the particle will land randomly on the screen thus will manifest no information about the lottery value that originated it. About half of the lottery information will be lost at the screen.
if only half the information is lost then just repeat the lottery numbers enough times to get the full sequence
I was told if I say...
"Keep the comment at an even number of likes"
That I would get lots of likes, IS it true?
No
Yes
Most people give an even amount of shits about your likes.
Artem Borisovskiy Really? That's odd.
actually if the answer is no that is odd.if not even then odd.
I think you won't be able to know which detector was fired (C or D) when you use the binary bit correlated with activating your switch, so you won't know which interference pattern to look for with each individual photon, basically ruining one of your two binary numbers. and then what's worse, the non-interfering pattern landing zone could even overlap one or both of the interference pattern landing zones, possibly ruining all the information.
Even though for some reason I cant read the actual numbers I would still get some signal? In other words I would be 100% positive that future me sending me some unreadable information, if so, that is amazing.
My intuition says that pattern for each photon in such a case will be an interference pattern and it wound never be a particle pattern. Once photon hits detector A, entangled photon which is between Earth and moon might disappear?
Does anyone here have a deeper understanding of the quantum eraser? I'm struggling to wrap my head around this. Is this an oversimplification? Is it really as simple as moving the two beam splitters out of the way and an interference pattern magically appears on the screen? Or is there some analysis of the data involved that isn't mentioned here?
Its worth a nobel prize to find out. It really is as nonsensical as it sounds.
You should be able to crack the challenge just with the data mentioned and your logic. You need to play those videos few times tho.
The pattern seen on the screen is always the same (a non-interfering pattern). Only when you later analyse data about detector+time, you can find which part of the screen is the interference pattern. In the video, it was stated that the patterns are offset by one stripe (π). Together, they form a regular pattern. So, the information is indeed there, but the key is having 'which-path' information by examining the detector+time data.
you're spoiling the challenge
Well, I've just found this channel and this is my first time taking part in this. I'll wait until the answer videos to do my commenting next time.
if A and C/D are swapped is the result the same? do only interfering photon wavefronts pass through the splitter? are all photons entangled by the delayed choice crystal? if there are multiple As, do they interfere?
Those of us with a predilection for the Everett model have already won the lotteries we've played by simply buying the tickets. :)
......an infinite number of times.
I'm one of those people ^_~ Many-Worlds has the preponderance of evidence in comparison to other QM theories.
He sent the information as binary to his past, his past doesn't know that his future self is trying to communicate with him, so his past just sees this as random patterns.
If his past is expecting the information then as soon as he realises what each binary code means the future entangled electrons will untangle themselves and form a random pattern again.
"Go watch this episode first" *gets there* "you should watch the first episode first"... What?
Click bait and switch tactic, also segmented 2 part video here.. I think uploader is desperate
no. you won't understand jiggly shit if you don't watch those.
this is *quantum mechanics* after all.
It wouldn't work because an interference pattern is a probability chart. If you took a photon, did all this stuff with it, then looked where it hit the graph, you wouldn't know if it was part of an interference pattern or not.
"Why am I going to be sadly disappointed after turning on my invention?" Because research has shown a correlation between depression and winning the lottery thanks to "hedonic adaptation"?
This comment is pure bliss
thank you.
the reason C and D are mirror images of each other is obvious: the ones that get reflected get flipped horizontally, as all mirrors do. thus, their peaks and valleys are mirror images of that recorded by the other sensor.
Doesn't this mean that information CAN travel faster than the speed of light?
so we could (in theory) use this to create long range transmitters that transmit across light years instantaneously.
rip, he already said that...
Yes, except that the machine doesn't actually work. I won't spoil it for you though, so if you want to find out before next week, you'll need to do some research.
:}
kk thnx
This can still be done through Quantum Entanglement outside of photons, just on a molecular level. Also, Quantum tunnelling allows for faster than light communication.
Final Spartan: Source?
Would be great if you can also make an episode on the current state and different types of quantum computers like Xanadu, Dwave, IBM..., what challenges are ahead in developing commercially viable QC and debunking myths on the actual applications of Quantum
where the fuck is its answer !
Is it because the pattern on the 'interference screen' is only decipherable when combined with information from detectors A, B, C and D?
said simply, the one who gets the information from the future is now in a parallel universe
The only way to win is never to play :)
The information you get back is just noise.
Nature prevents a paradox to occur, "Sender" and "Receiver" must be distinguished from each other and cannot send or receive information simultaneously in this way (quantum eraser experiment)
Your stuck in two versions of reality (parallel universes), the person who knows the outcoming of lottery can never collect the money because he is not a part of that reality. (because of complementarity)