I think about weak measurement like this: we've got a hockey puck on the ice, of unknown dimension; if we poke it we gain some information about its diameter at a point, but it moves/slides by an unknown amount, confounding our efforts to learn the diameter by poking it again on the opposite side. BUT: if we poke it SOFTLY ENOUGH, it won't slide at all. Now it may be difficult to gain the same amount of information from our gentlest of pokes as from our hard poke, but we can do this over and over, and take averages. That's the idea, but by the time you get don't propagating the error my money says you're right back at Heisenberg's limit
I should first point out that I'm a biochemistry student and my knowledge of physics is fairly limited so I apologize in advance if I get some details wrong. But seeing how all of these particles have some mass, and gravity propagates in a wave-like manner, is it possible to measure the gravitational field around an isolated particle so carefully that you can establish the exact center or origin? Or, if that's not possible right now, is it at least plausible? From what I understand, gravitons haven't been confirmed but if they were, could the detection of gravitons be used to pinpoint the position of a particle with mass? The graviton would be emitted by the particle you're trying to measure, as opposed to a photon that hits the particle and is then detected.
I really don't think that's possible at all, for starters gravity is incredibly weak, so weak that the gravitational force created by a particle is so immeasurably small it can easily be cancelled out by its surroundings at the quantum scale all the way to the planetary scale, if gravitons are confirmed and depending on how they are detected the issue would be trying to figure out if they originate from the particle or from the background noise, I think it would be like trying to measure the temperature of a one millimetre squared patch of the suns photosphere to the nearest micro-kelvin.
I think i got it.. particle interaction is like making subtle waves of known wavelength to create a picture of defined position.. but i can also see looping itself.. it's the first time that i really have the grasp on what "uncertainty" really means.. thanks
The trick to not disturbing the particle would be to detecting where the particle isn't. Sort of a template field. Greater resolution would be derived by narrowing how close you get to the particle without interacting with it.
You can't precisely bound the location of where the particle can be like that. See Quantum Tunneling. You could in fact check all of the locations inside the box and not find the particle because it can be slightly outside the box.
Hi Brady, could you ask this question for Prof Michael? If it was possible, if the particle was to be measured by 2 photons hitting the particle at the same time on exactly symmetrical sides instead of only 1 photon wouldn't this neglect the disturbance on the particle and there for would be able to do the measurements correctly?
This, unfortunately, hits the relativity laws, because you're dealing with very high speeds. There's no simultaneity when something travels at a near-C speed, so the electron would happen to.... wwell, it's goose-damn'd complicated.
There's actually another uncertainty principle which says you can't measure a photons energy and the time it arrives at a location perfectly. So you'd run into the same problem.
***** That only applies if you're trying to measure a free photon's energy/location. If you create a photon with an exact amount of energy at an exact moment of time, you can use that photon to "see" another particle's position or momentum.
I don't think it is a good idea to talk about the observer effect and Heisenberg uncertainty principle in the same video without making it clear those are two different things that have for a very long time been talked about interchangeably. the former is debateable enough to be attracting research into novel ways of measurement while the latter is pretty much mathematical fact from fourier analysis and the deBroglie relation.
Maybe the Heisenberg uncertainty principle is "uncertain" for us, 'cause we are too "big" to see small particles without perturbate their states. If we are small like particles we can see how they really "want" to move and where, and the world would be predictable... anyway right or no, I think that uncertainty and unpredictability is better than being an oracle.
Could you try this weak measurement with particles that have been chilled to very very low temperatures. As we know that the momentum is going to be very low, we could assume that it is zero but this is physics so I guess we will still have the momentum before to consider. But the momentum could just become negligible. Hence we can measure the disturbance and hence you know the momentum from the disturbance and the position from the measurement. I'm no quantum physicist, just a guy doing a Maths Degree, who also did physics at A-level and watches videos online about it.
I think you need to sum all of the disturbances to say how much you disturb the system. Each individual disturbance isn't much, but if you are using a lot of disturbances, to create an average, then you also need to add all the disturbances. Perhaps Heisenberg's principal can be modified slightly to account for averaging of values?
I still have some trouble understanding the problem of this meta measurement he is talking about at 1:27, which is trying to determine how much the particle has been disturbed by measuring. Here are my thoughts on this. I'm making some assumptions on things of which I'm not sure if/how they work, please correct me wherever I'm wrong: In the example, a particle is measured by a single photon. When the photon is fired at the particle, we know the momentum/wavelength of this photon, obviously. After hitting the particle, getting deflected and hitting the lens/detector, the photon is measured. When the photon hit the particle, it disturbed it by transferring some of its momentum to the particle, which means, the photon lost momentum and therefore changed its wavelength. Why can't we infer from measuring the photon after the collision how much we disturbed the particle? After all, we know how much of the momentum was given to the particle (change in wavelength) and we even know at which angle the photon was deflected, because we know the angle at which the photon hit the detector. TL;DR: We know the initial momentum and the initial angle of the photon in the beginning. We can measure the momentum and the angle of the photon after the collision. Where is the problem with the measurement of the particle?
I am not a physicist, but what would happen if you put a particle in a strong magnetic field which would guarantee that a particle stays in a region (or maybe at a fixed point) and you would try to measure it. Would there still be a problem (maybe because of the applied forces or would the particle oscillate?) or does it no longer makes sense to measure it?
Couldnt you messure the gravitational pull of the electron? It wouldnt interfere with the electron but still give information about the position. You simply have to make two messurements in a short time.
+Iasonas Zotos The method of measurement can't involve nearby mass, as then you are changing the position of the electron. Maybe send a photon and measure the change of direction from the electron ?
8 лет назад+2
Also, we would need a theory of quantum gravity for that to work. We don't have that yet.
Can anyone answer me this....you have a lazer you can precisely aim at a target, shooting individual photons of a known wavelength. Wouldn't you know each photon's speed, energy, and when it arrives at a given point? So you'd know both its speed and position along its path?
Nope. The more accurately you know the photon's frequency, the less accurately you know when it left the laser (the less accurately you know its position).
Ive just learnt about this today from a fellow youtuber and ive got to say im excited and im thinking on ways around Heisenberg's Principle. Like does light make a diffraction pattern traveling through a fluid, and will single photons still make that pattern. How can one record a photons path without a disturbance to that photon. I dont buy photons being in multiple places at the same time. I wonder if that 3rd video was ever made? This is so cool
If you had a negatively charged spherical shell and put one electron inside it, would that electron not end up right at the centre after a while? Then we would know its exact position (centre of the sphere) and speed (which would be zero) ?
what if you hit a particle you are measuring with more than one photon at the same time in a manner that they both cancel out the momentum imparted onto the measured object 8) E.g., if you hit a billiard ball with two other balls with the same energy on the same axis but on opposite sides, the middle ball won't move. Sounds impossibly precise to do, but if you got stupid lucky, could you? That's assuming the particle is stationary though, which it probably isn't...
In order to create two photons like that, you'd have to be able to precisely specify the position and the energy of both of the photons. You can't do that for the very same reason.
Brady, did you make two videos from one interview just to inflate your publishing numbers compared to your podcast co-host? (Jk love the RUclips channels and Hello Internet)
If you had some insanely accurate detectors, couldn't you detect the gravitational field of the electron as it passes by and use that to determine both it's position and vector (and from that you can determine it's momentum)? I know that this isn't currently possible to do, but it's the only thing I can think of that doesn't involve bouncing a photon off the electron first. So would that work?
My question is if we took that electron and did something like throw it in a particle accelerator, couldn't we control both its speed and location? I know its not the same thing is measuring both the speed and location of a random particle but doesn't that count somehow?
The thing with quantum cryptography I don't understand is why doesn't the interceptor just create a new message with the contents they have seen and pass that on? The receiver will then get a copy of the original, which as far as they know is the original, that looks completely fine. Is there something I'm missing here?
I'm by no means an expert on the matter but I believe that because of the No-cloning theorem you can never create an identical copy of a unknown quantum state. "In physics, the no-cloning theorem is a no-go theorem of quantum mechanics that forbids the creation of identical copies of an arbitrary unknown quantum state." en.wikipedia.org/wiki/No_cloning
There is this theorem, the "No-cloning theorem" which says that you cannot copy quantum states. So, you can't copy the message, read one copy and send the other copy to the receiver. And if you read the message, some bytes will be randomly wrong, so you'll send the wrong bytes too, and the receiver then reads out the message again and adds more errors. Therefore, the receiver can with some statistics notice more errors than there should be.
The idea is good, but in practice, it cannot be done. There are two ways quantum cryptography is accomplished (that i am aware of). One way it so send a message via conventional means but encoded with a quantum algorithm. You would not be able to replicate it as it is inherently impregnated with a unique quantum signature that only exists as the message is initially sent or decoded. The other is by quantum entanglement, where I have a quantum bit in my lab and my correspondence has the matching twin in his lab. When I 'flip' my bit his will spontaneously 'flip' as well. You would not be able to intercept the message without access to either bit.
How about this: what if you surround the area that the electron is in in some sort of electron detection grid. Then when your photon disturbed the location of the electron you'll be able to detect where that electron tries to escape the detection grid. Then using the image you've gotten and the time it took to hit the grid wouldn't that tell you its position?
If Heisenbergs uncertainty principle isn't correct, would that mean a particles groundstate energy could be 0? I was always taught that the reason a particle's minimum energy is greater than 0 is due to uncertainty.
Couldn't you send an electron across and have stuff that detects electric charges to know the momentum and position at the same time? does merely detecting electric charge influence its momentum? You're not shooting anything at it that way... It sounds WAY too obvious, but still curious
What if it was possible to get around Heisenberg's uncertainty principle? I mean, what would be possible to do knowing precisely the position and the momentum of an electron?
Watched a few videos on Heisenberg’s Microscope. Why can’t you measure the electron with a single photon to see the position, why can’t you have multiple lenses 90* to measure the momentum. Example. 1 photon bumps the electron. Then bombard the electron with 100 photons from 2/3 angled lenses to measure the momentum?
You mention at the end the fact that you could make 5000 measurements to get subtle information, this makes more sense to me. You also mentioned that it does not violate the Heisenberg uncertainty principle. Is this because it’s using the principle but “outsmarting” it by using smaller energy to measure the system. I think about it like the electron is a bowling ball and the measuring particle is a bb. If you hit a bowling bowl with a bb it’s unlikely to effect its position or momentum. If I hit a bowling ball through 5000 walls of bb’s I’m likely to gain some level of information from the bowling ball. Do I understand this correctly?
Perhaps in the future, when we get to use (if ever) the particles we can't measure, produce or handle, like gravitons, neutrinos or dark matter, there could be a way to measure them.
Any sort of measurement has an intrinsic error due to the device itself. For example, when you put a thermometer into a hot sample, the thermometer will read a value lower than what the sample is because the room-temperature thermometer will have lowered the temperature of the sample. Even something as simple as placing a ruler against an object to measure its length is flawed, because the mass of the ruler causes it to gravitationally attract the object and distort its shape ever so slightly.
Has nothing to do with what is being discussed in the video. They are talking about fundamental limits in our ability to measure due to the Heisenberg uncertainty principle. And a ruler is not going to affect the length of any object due to gravity, that is just wrong.
+Meta Tron I never stated that I was referring to the Uncertainty principle. I was mentioning an additional source of error from any measuring device, regardless of whether or not it emits particles to do the detecting. And all objects with mass slightly distort the shapes of masses around it, because no object is perfectly rigid. There is a differential of force from all gravitating bodies. The ruler attracts the near side of an object more stronger than the far side, ever so slightly, and this creates a shear which can distort the shape of an object. The effect is incredible small but it does occur.
+Anthony Khodanian the gravitational force is so weak that it is negligible when doing everyday QM. It only becomes apparent when dealing with weird QM mechanical/massive gravitational objects aka a black hole. The measuring device itself is part of the entire QM system when measuring whatever particle you please, thus is also subject to the uncertainty principle. This is why you can not eliminate all the noise out of a QM system. So your statement about gravity affecting the QM system is meaningless unless it is in the extreme circumstances where gravity is actually strong enough to influence the systems. Once you do your QM corrections for Gravity you see that the force of gravity is negligible in most cases.
Is the zero sum game which he talked about at the end solved with the average ? Think of an pascals triangle, as you get lower the chances of you being in the center increase.
To me it always sounds like Uncertainty merely exists because we humans can't measure a particle properly. But I always hear that in a system there is even an uncertainty in every system without a human interacting with it. What's up with what? Do all particles interact with each other on their own so that they cause their own uncertainty?
Could you send a particle through a nano-tube and measure the path the particle took by how it interacted with the atoms of the tube along the way? You should be able to deduct both position and speed.
How about if you catch the particle? You'd know where the particle is, obviously, and you'd know it's momentum by the force of the impact as you caught it?
Same force, photon interaction. It's electromagnetism that makes matter hard, when you press a finger against the table its the electrons repelling eachother via exchangeing photons that keeps your finger from going through the table..
In my ignorance I had a thought. If a particle is a wave doesn't that wave cause an interference similar to the wave(s) that are transmitted from a speaker. If the wave(s) are in a range that can be "heard" or if you have a device to interpret the wave and convert it to a wave range that can be heard (or seen as in an oscilloscope)...can you also do the same for a particle? Basically listening to the wave as it interferes with what is around and use this to measure? Instead of bouncing a photon off it. Admittedly, I am only a little read on this subject anf have come to grasp a small glimpse, thanks in no small part to "Sixty Symbols!"
Nothing can be detected without being interacted with. The sound waves you speak of is the air being pushed and pulled by the speaker membrane, only to interact with your ear after a while. The membrane pushing the air is the same as the photon scattering off the electron in the video. Any interaction or "interference" as you mention will affect the particle we are trying to measure, putting us right back at the issue the video is all about.
You both are missing my point. Ok 1st, imagine the pond ripple, you can infer the mass of whatever is dropped into the water by the ripples. Now imagine if you can't see the whole picture of those same ripples but just, let's say 20% +/- one can still infer an approximation with fairly high accuracy! 2nd imagine yourself listening to music, you don't need to see or interact with the sound coming out from the speaker to infer what is playing nor how loud it is AND to top it off without interacting/disturbing the soundwaves you could almost "see" the walls of a room because the sound is bouncing off the walls...finally 3rd, since a particle is an excitation of a field i.e. an electron is an excitation of the field or could also be considered a very dense or compacted waveform because to create an electron from the field energy needs to jump to a very high and excited state to produce the particle. But none the less the particle is (if I am not mistaken) point of extremely dense packet of energy in a highly excited state....and by point mean it is expressed at a particular coordinate in space/time from the observers frame of reference. This is the picture of any particle as I understand it. So, since the particle has a wave/particle duality, I was asking if it can be inferred by the wave function alone without collapse of the wave?!?
8 лет назад
Jonathan Owens Well, your ear does interact with the wave - if it didn't interact with the wave, it basically means that the sound didn't reach your ear, in which case you won't hear anything. The momentum of the wave has to be imparted in part to the sensing hairs in the inner ear before you hear it. You can't detect anything without interacting with it. It is as simple as that.
We can pretty easily measure the location and momentum of, for example, an automobile. Up to what size scale is the Uncertainty principle relevant. electrons, hadrons, atoms, molecules, etc.? And why?
When you use low accuracy you can do both. When you measure with the accuracy of let's say a billionth of a meter, you need a photon of that wavelength, which means you're putting a lot of energy into the system and disturbing it
So if I wanted to measure that position of an automobile with the same accuracy as used in measuring the position of an electron, I would not be able to measure the automobile's momentum with the same accuracy used, or desired in measuring the momentum of an electron. The difference being that for the automobile, that type of measurement is not useful or practical anyway. . In other words the Uncertainty Principle applies equally on the macro scale, but it is not relevant because at that scale, that very fine level of accuracy is not needed or desired. Is that what you mean?
The way to get around it is to measure the particle's gravitational field. Not the most practical way of doing it, but at least you don't have to hit it with a photon.
If its not possible to measure a quantum encryption and therefore interpret the message without the other party knowing the message had been disturbed. What’s to stop people from measuring every packet of data that passes and prevent anyone from using quantum cryptography?
@@HeyImLucious Using boiling water steam to loosen the adhesive properties of glue in order to open envelope without a trace (or without easily noticeable trace) and easily glue it shut again is a really old technique (espionage technique one could say), and with traditional mail falling out of fashion, it is guaranteed to be lost to time in near future
Cant you somehow measure the electric field the electron gives off at varying points to pinpoint the electrons location? Or would measuring that field somehow interact with that particle, or is the field even measurable, or is there really a field, is there really even an electron? quantum physics is weird.
No there's no field, there are just photons and virtual photons (google it). We just call it a field when there's a lot of photons, just taking the average of a lot of particles. It's all particles, and you have to hit them to find out anything about them
Why don't you just detect the electron via exclusion? Here is a 3x3 imaginary wall and each number is the size of an electron. 123 456 789 You know that there is one electron somewhere in this wall. You send a photon through all other numbers except the number 5. If you receive same amount of photons on the other side of the wall, then you know that you have not hit the electron, thus have not affected it with your measurement and you also know its location (5). Manage to do this twice (lets say leaving out number 3) and you also know the speed and direction on one axis. Add another axis to the experiment and you can then plot the direction and speed of the electron after successfully not hitting it with photons.
I believe you would get interference on the photons, because the photons are acting like waves. Perhaps you could send a polarized photon so that it doesnt interfere, not quite sure how interference works.
Well if you measure it from every direction with millions of weak measurements, it should average to around not moving the particle at all, but also not moving it, since every measurement averaged moves it exactly the same.
how do you make sure it's the same intensity? There's an uncertainty in the energy of whatever it is you want to use and so there's an uncertainty in the disturbance.
You can't create a photon with an exact energy and timing either (if you could, then you'd know the position and momentum of that particle). So it is impossible to create the photons to do what you're proposing. If you want to generate a photon of a very precise frequency, you don't know exactly when it will be emitted, so you don't know exactly when it will hit the electron.
Nope, because you don't know where to aim it until AFTER you've shot it with the photon. But then you don't know where it's at. Thus you can't aim a second one at it.
+Mortizul, in order to do that, you would need to know where the particle is so you can hit it from each side. But in order to know where it is... you have to measure where it is.
but the particles dont have a well defined position nor momentum to begin with, so how does weak measurement measure a position more exact than the particle actually has??? .
The uncertainty principle is what dictates "the particles don't have a well defined position nor momentum". What they're saying is, that may not be strictly true in all cases. They're saying Heisenberg might have been wrong.
I still feel like it can be done...Maybe we can figure out a way to measure everywhere the particle isn't.
8 лет назад+1
If we don't know where it is in the first place, hitting everywhere but the point the particle is in seems like a very unlikely option. Also, particles tend to move, so we would have to measure *everywhere* around the particle at one instant in order to know its position... no, it can't be done.
In order to do that you'd have to know with certainty that the particle was within a particular well defined volume of space. You can't even do that because you can't put particles in tiny boxes. See Quantum Tunneling.
Why can’t you hold the electron with tweezers? In which case, you know its location and you know that its momentum is exactly 0. This would be classical physics. So are you saying that quantum systems can never be “held” between your fingers? This makes quantum mechanics literally intangible to any possible human senses and the brain because you have no idea what the quantum system even is. You can never hold it in your hand, because you’d then know it’s position and momentum simultaneously
If you try to constrain a particle's position like that, then its momentum becomes huge - it begins to move very, very fast. Essentially, you won't be able to squeeze it down to being in one place and not moving, as it will start to push against your "tweezers" with enormous force. You have to think about what it means to "hold" something. Think about how that works at the atomic level.
You measure something and note its value, then measure it again and again. Then you take the difference between each measurement and find it's relative changing ratio. you can now log the changing ratio according to the number of time you measured it and then find the real value of what you are measuring. Why don't you do this?
The uncertainty principle doesn't really cause the uncertainty of the future. The future is uncertain because it can't causally affect the present, whereas the past can... at least, that's the case on the macroscopic level. The Schrödinger equation _is_ t-symmetric, but statistically speaking, things end up being time irreversible in the end... not really sure of the specifics of how the second law of thermodynamics can be gotten to from a quantum mechanics starting point though.
Ben Rowe We can think of classical physics as explaining processes over a period of time and quantum physics as explaining ‘time’ itself as a physical process. Photon energy cascades down forming greater degrees of freedom for entropy and chaos. These photon oscillations or vibrations do not just form greater degrees freedom for the aging process of decay. They also form greater degrees of freedom for the evolution and diversity of life that has lead to the creativity of art, poetry and music!
Quantum Atom Theory an artist theory on the physics of 'time' that doesn't really explain how the uncertainty principle has anything to do with the uncertainty of the future... it's good that you're enthusiastic about quantum mechanics, but you're just kinda jumping around right now.
why wouldn't you just ask god whre the electron is, not the weather god though, he has a lot of particles to take care of and he doesn't care for thunderstorms
E=MC^2 leaves out an important part: E^2 = (MC^2)^2 + (PC)^2 where P is momentum. If momentum is zero, it is at rest and (PC)^2 is zero, so E^2 = (MC^2)^2 reduces to E=MC^2 which represents rest mass. A photon has zero mass, so (MC^2)^2 resolves to zero and you get the reduced energy equation for a photon: E=PC.
Because people who do not like Donald Trump running for office and they disagree with his VP candidate so they disturb the electrons en mass or they just vote for the other person!!!!
Couldn't you just put an electron in an electric field with equally powerful magnets on all sides meaning that the electron would be equally repelled by all of them and thus forced to stop. It would have no momentum and you would know its position.
Unless you could start in perfectly the middle, like down to the planck length, it would oscillate, because it would be closer to one side, accelerate away from it, pass the middle, and repeat. That's assuming it's in a vacuum.
I'm thinking of two problems: -you couldn't tell if all the magnets are exerting the exact same magnetic field -you couldn't tell if they are positioned precisely correct But its hypothetically plausible
The strength of the force between an electrical charge and a magnetic field depends on how quickly they are moving relative to each other. My guess is that if you set up your experiment, the electron would orbit in a tiny circle, or maybe bounce along an axis, but never actually stop.
This is actually a really good question and I hope someone who really knows would be able to tell you why. I think the key here is that you don't actually *know* its position. Just because you think the electron is in a particular place, doesn't actually mean it's standing still. Actually by just examining the uncertainty relation equation, you know that there is no such thing as a completely still particle because if: delta position * delta momentum >= h bar / 2 then if delta momentum is zero, which you assumed, you see that h bar (the planck constant) would have to be zero to make the equation satisfy. This tells us that there is no such thing as a completely stationary particle, so that electron must still be moving in one way or another, which would also mean it's position is highly non-localized, meaning the particle is almost acting like a foggy interference than a precise particle. Do research on something called "standing matter waves". You also were describing a real device called a "Penning trap", so I would look into that as well :) Please tell me what you find, I am by no means a physicist and I would be just as intrigued as you are to find the real answer (if this wasn't right).
What you have basically described is a pendulum. The electron won't sit nicely in the centre, since when it is in the centre it had momentum (I mean, it had to move there to start with right?). So that momentum will take it out of the exact centre, so it will feel a force due to the net magnetic field, in the same way a pendulum feels a net force when it isn't at the bottom of its swing. The electron will then rock back and forward, creating uncertainty in position and momentum all the time.
There are security proofs of the quantum key distribution protocol that follows from the postulates of quantum mechanics. What you're implying about the entire field of quantum cryptography being in the balance based on some debate about weak measurements is simply not correct. When claiming an entire field is about to come crashing down, please contact someone familiar with the field first.
4:04 CANNOT BELIEVE YOU BOTHERED TO DECODE THIS
impressive
false.
Forget about Weak Measurement... has anyone tested if the Brady Feather will work?
RimstarOrg Yes, it is known as quantum tickling.
I'm almost afraid to ask how much quantum tickling would cost downtown.....(but I did anyway haha)
Jonathan Owens abou' tree fiddy.
And you can use the sharp end of the feather to dig a quantum tunnel.
Heisenberg Compensator
I think about weak measurement like this: we've got a hockey puck on the ice, of unknown dimension; if we poke it we gain some information about its diameter at a point, but it moves/slides by an unknown amount, confounding our efforts to learn the diameter by poking it again on the opposite side. BUT: if we poke it SOFTLY ENOUGH, it won't slide at all. Now it may be difficult to gain the same amount of information from our gentlest of pokes as from our hard poke, but we can do this over and over, and take averages. That's the idea, but by the time you get don't propagating the error my money says you're right back at Heisenberg's limit
as much as I love EVERYONE from sixty symbols, Merrifield provides the most clarity and seems the most intelligent.
I got an information pack from the University of Nottingham today, I was very impressed with 'The Ruler of the Universe' that came with it!
we want the new video!!!! :)
uncertainty principle is insamnely interesting :)
I would love if you did a video on how weak measurement works!
I should first point out that I'm a biochemistry student and my knowledge of physics is fairly limited so I apologize in advance if I get some details wrong.
But seeing how all of these particles have some mass, and gravity propagates in a wave-like manner, is it possible to measure the gravitational field around an isolated particle so carefully that you can establish the exact center or origin? Or, if that's not possible right now, is it at least plausible? From what I understand, gravitons haven't been confirmed but if they were, could the detection of gravitons be used to pinpoint the position of a particle with mass? The graviton would be emitted by the particle you're trying to measure, as opposed to a photon that hits the particle and is then detected.
Mmm, doesn't sound too bad to me... Still, you'd be capable of doing that only with nonzero-mass particles, which are (mostly) uninteresting ;/
I really don't think that's possible at all, for starters gravity is incredibly weak, so weak that the gravitational force created by a particle is so immeasurably small it can easily be cancelled out by its surroundings at the quantum scale all the way to the planetary scale, if gravitons are confirmed and depending on how they are detected the issue would be trying to figure out if they originate from the particle or from the background noise, I think it would be like trying to measure the temperature of a one millimetre squared patch of the suns photosphere to the nearest micro-kelvin.
I think i got it.. particle interaction is like making subtle waves of known wavelength to create a picture of defined position.. but i can also see looping itself.. it's the first time that i really have the grasp on what "uncertainty" really means.. thanks
laying in bed watching sixty symbols videos is the best thing
The graphics are great hah. Another great video Brady.
The trick to not disturbing the particle would be to detecting where the particle isn't. Sort of a template field. Greater resolution would be derived by narrowing how close you get to the particle without interacting with it.
But as soon as you run into it, the measurement is useless, and you still wouldn't be gaining any information about the particle itself.
You can't precisely bound the location of where the particle can be like that. See Quantum Tunneling. You could in fact check all of the locations inside the box and not find the particle because it can be slightly outside the box.
Hi Brady, could you ask this question for Prof Michael?
If it was possible, if the particle was to be measured by 2 photons hitting the particle at the same time on exactly symmetrical sides instead of only 1 photon wouldn't this neglect the disturbance on the particle and there for would be able to do the measurements correctly?
how would you know that you're hitting it symmetrically without knowing where it is and the momentum first?
This, unfortunately, hits the relativity laws, because you're dealing with very high speeds. There's no simultaneity when something travels at a near-C speed, so the electron would happen to.... wwell, it's goose-damn'd complicated.
Very good question, I'd like to hear the answer to that aswell.
There's actually another uncertainty principle which says you can't measure a photons energy and the time it arrives at a location perfectly. So you'd run into the same problem.
***** That only applies if you're trying to measure a free photon's energy/location. If you create a photon with an exact amount of energy at an exact moment of time, you can use that photon to "see" another particle's position or momentum.
I don't think it is a good idea to talk about the observer effect and Heisenberg uncertainty principle in the same video without making it clear those are two different things that have for a very long time been talked about interchangeably. the former is debateable enough to be attracting research into novel ways of measurement while the latter is pretty much mathematical fact from fourier analysis and the deBroglie relation.
Maybe the Heisenberg uncertainty principle is "uncertain" for us, 'cause we are too "big" to see small particles without perturbate their states. If we are small like particles we can see how they really "want" to move and where, and the world would be predictable... anyway right or no, I think that uncertainty and unpredictability is better than being an oracle.
@@itsiwhatitsi No, thats the point. To our best knowledge its not because we cant build a better microscope.
Just ask O'brien to calibrate the Heisenberg compensator.
Could you try this weak measurement with particles that have been chilled to very very low temperatures.
As we know that the momentum is going to be very low, we could assume that it is zero but this is physics so I guess we will still have the momentum before to consider. But the momentum could just become negligible.
Hence we can measure the disturbance and hence you know the momentum from the disturbance and the position from the measurement.
I'm no quantum physicist, just a guy doing a Maths Degree, who also did physics at A-level and watches videos online about it.
I think you need to sum all of the disturbances to say how much you disturb the system. Each individual disturbance isn't much, but if you are using a lot of disturbances, to create an average, then you also need to add all the disturbances. Perhaps Heisenberg's principal can be modified slightly to account for averaging of values?
I still have some trouble understanding the problem of this meta measurement he is talking about at 1:27, which is trying to determine how much the particle has been disturbed by measuring.
Here are my thoughts on this. I'm making some assumptions on things of which I'm not sure if/how they work, please correct me wherever I'm wrong:
In the example, a particle is measured by a single photon. When the photon is fired at the particle, we know the momentum/wavelength of this photon, obviously. After hitting the particle, getting deflected and hitting the lens/detector, the photon is measured. When the photon hit the particle, it disturbed it by transferring some of its momentum to the particle, which means, the photon lost momentum and therefore changed its wavelength. Why can't we infer from measuring the photon after the collision how much we disturbed the particle? After all, we know how much of the momentum was given to the particle (change in wavelength) and we even know at which angle the photon was deflected, because we know the angle at which the photon hit the detector.
TL;DR: We know the initial momentum and the initial angle of the photon in the beginning. We can measure the momentum and the angle of the photon after the collision. Where is the problem with the measurement of the particle?
I am not a physicist, but what would happen if you put a particle in a strong magnetic field which would guarantee that a particle stays in a region (or maybe at a fixed point) and you would try to measure it. Would there still be a problem (maybe because of the applied forces or would the particle oscillate?) or does it no longer makes sense to measure it?
Couldnt you messure the gravitational pull of the electron? It wouldnt interfere with the electron but still give information about the position. You simply have to make two messurements in a short time.
+Iasonas Zotos The method of measurement can't involve nearby mass, as then you are changing the position of the electron. Maybe send a photon and measure the change of direction from the electron ?
Also, we would need a theory of quantum gravity for that to work. We don't have that yet.
Can anyone answer me this....you have a lazer you can precisely aim at a target, shooting individual photons of a known wavelength. Wouldn't you know each photon's speed, energy, and when it arrives at a given point? So you'd know both its speed and position along its path?
Nope. The more accurately you know the photon's frequency, the less accurately you know when it left the laser (the less accurately you know its position).
Thats is James May's shirt.
Thanks to Mike and Brady.I thought quantum mechanics was just a collection of mathematical incantations.
Ive just learnt about this today from a fellow youtuber and ive got to say im excited and im thinking on ways around Heisenberg's Principle. Like does light make a diffraction pattern traveling through a fluid, and will single photons still make that pattern. How can one record a photons path without a disturbance to that photon. I dont buy photons being in multiple places at the same time.
I wonder if that 3rd video was ever made? This is so cool
If you had a negatively charged spherical shell and put one electron inside it, would that electron not end up right at the centre after a while? Then we would know its exact position (centre of the sphere) and speed (which would be zero) ?
so when comes the real weak measurement video ?
it may not be correct to assume that a particle even HAS a precise position or
momentum at the same time
what if you hit a particle you are measuring with more than one photon at the same time in a manner that they both cancel out the momentum imparted onto the measured object 8) E.g., if you hit a billiard ball with two other balls with the same energy on the same axis but on opposite sides, the middle ball won't move. Sounds impossibly precise to do, but if you got stupid lucky, could you?
That's assuming the particle is stationary though, which it probably isn't...
In order to create two photons like that, you'd have to be able to precisely specify the position and the energy of both of the photons. You can't do that for the very same reason.
Brady, did you make two videos from one interview just to inflate your publishing numbers compared to your podcast co-host? (Jk love the RUclips channels and Hello Internet)
Are you not still using Photons to disturb the -e ?
What are they using to measure in weak measurement?
If you had some insanely accurate detectors, couldn't you detect the gravitational field of the electron as it passes by and use that to determine both it's position and vector (and from that you can determine it's momentum)? I know that this isn't currently possible to do, but it's the only thing I can think of that doesn't involve bouncing a photon off the electron first. So would that work?
My question is if we took that electron and did something like throw it in a particle accelerator, couldn't we control both its speed and location? I know its not the same thing is measuring both the speed and location of a random particle but doesn't that count somehow?
Can't wait for the weak measurement video.
The thing with quantum cryptography I don't understand is why doesn't the interceptor just create a new message with the contents they have seen and pass that on? The receiver will then get a copy of the original, which as far as they know is the original, that looks completely fine. Is there something I'm missing here?
because when you intercept, you dont receive the original message, you receive a disturbed copy of it and so you arent actually able to view it
I'm by no means an expert on the matter but I believe that because of the No-cloning theorem you can never create an identical copy of a unknown quantum state. "In physics, the no-cloning theorem is a no-go theorem of quantum mechanics that forbids the creation of identical copies of an arbitrary unknown quantum state."
en.wikipedia.org/wiki/No_cloning
There is this theorem, the "No-cloning theorem" which says that you cannot copy quantum states. So, you can't copy the message, read one copy and send the other copy to the receiver.
And if you read the message, some bytes will be randomly wrong, so you'll send the wrong bytes too, and the receiver then reads out the message again and adds more errors. Therefore, the receiver can with some statistics notice more errors than there should be.
The idea is good, but in practice, it cannot be done. There are two ways quantum cryptography is accomplished (that i am aware of). One way it so send a message via conventional means but encoded with a quantum algorithm. You would not be able to replicate it as it is inherently impregnated with a unique quantum signature that only exists as the message is initially sent or decoded. The other is by quantum entanglement, where I have a quantum bit in my lab and my correspondence has the matching twin in his lab. When I 'flip' my bit his will spontaneously 'flip' as well. You would not be able to intercept the message without access to either bit.
Because the act of seeing the message will change the message, interceptor can't get original message in the first place.
How about this: what if you surround the area that the electron is in in some sort of electron detection grid. Then when your photon disturbed the location of the electron you'll be able to detect where that electron tries to escape the detection grid. Then using the image you've gotten and the time it took to hit the grid wouldn't that tell you its position?
If Heisenbergs uncertainty principle isn't correct, would that mean a particles groundstate energy could be 0? I was always taught that the reason a particle's minimum energy is greater than 0 is due to uncertainty.
Are electrons ticklish?
what about measuring the momentum of the disturbing instrument (photon) subtracted from the total momentum of the disturbed particle?
Couldn't you send an electron across and have stuff that detects electric charges to know the momentum and position at the same time? does merely detecting electric charge influence its momentum? You're not shooting anything at it that way...
It sounds WAY too obvious, but still curious
What if it was possible to get around Heisenberg's uncertainty principle? I mean, what would be possible to do knowing precisely the position and the momentum of an electron?
Watched a few videos on Heisenberg’s Microscope.
Why can’t you measure the electron with a single photon to see the position, why can’t you have multiple lenses 90* to measure the momentum.
Example. 1 photon bumps the electron. Then bombard the electron with 100 photons from 2/3 angled lenses to measure the momentum?
You mention at the end the fact that you could make 5000 measurements to get subtle information, this makes more sense to me.
You also mentioned that it does not violate the Heisenberg uncertainty principle. Is this because it’s using the principle but “outsmarting” it by using smaller energy to measure the system.
I think about it like the electron is a bowling ball and the measuring particle is a bb. If you hit a bowling bowl with a bb it’s unlikely to effect its position or momentum. If I hit a bowling ball through 5000 walls of bb’s I’m likely to gain some level of information from the bowling ball.
Do I understand this correctly?
Perhaps in the future, when we get to use (if ever) the particles we can't measure, produce or handle, like gravitons, neutrinos or dark matter, there could be a way to measure them.
Any sort of measurement has an intrinsic error due to the device itself. For example, when you put a thermometer into a hot sample, the thermometer will read a value lower than what the sample is because the room-temperature thermometer will have lowered the temperature of the sample. Even something as simple as placing a ruler against an object to measure its length is flawed, because the mass of the ruler causes it to gravitationally attract the object and distort its shape ever so slightly.
Has nothing to do with what is being discussed in the video. They are talking about fundamental limits in our ability to measure due to the Heisenberg uncertainty principle. And a ruler is not going to affect the length of any object due to gravity, that is just wrong.
+Meta Tron I never stated that I was referring to the Uncertainty principle. I was mentioning an additional source of error from any measuring device, regardless of whether or not it emits particles to do the detecting. And all objects with mass slightly distort the shapes of masses around it, because no object is perfectly rigid. There is a differential of force from all gravitating bodies. The ruler attracts the near side of an object more stronger than the far side, ever so slightly, and this creates a shear which can distort the shape of an object. The effect is incredible small but it does occur.
+Anthony Khodanian the gravitational force is so weak that it is negligible when doing everyday QM. It only becomes apparent when dealing with weird QM mechanical/massive gravitational objects aka a black hole. The measuring device itself is part of the entire QM system when measuring whatever particle you please, thus is also subject to the uncertainty principle. This is why you can not eliminate all the noise out of a QM system. So your statement about gravity affecting the QM system is meaningless unless it is in the extreme circumstances where gravity is actually strong enough to influence the systems. Once you do your QM corrections for Gravity you see that the force of gravity is negligible in most cases.
Is the zero sum game which he talked about at the end solved with the average ? Think of an pascals triangle, as you get lower the chances of you being in the center increase.
another continuation please
To me it always sounds like Uncertainty merely exists because we humans can't measure a particle properly.
But I always hear that in a system there is even an uncertainty in every system without a human interacting with it.
What's up with what? Do all particles interact with each other on their own so that they cause their own uncertainty?
Could you send a particle through a nano-tube and measure the path the particle took by how it interacted with the atoms of the tube along the way? You should be able to deduct both position and speed.
If you confine a particle to a small area so you know pretty well where it is, its momentum becomes huge.
How about if you catch the particle? You'd know where the particle is, obviously, and you'd know it's momentum by the force of the impact as you caught it?
Same force, photon interaction. It's electromagnetism that makes matter hard, when you press a finger against the table its the electrons repelling eachother via exchangeing photons that keeps your finger from going through the table..
In my ignorance I had a thought. If a particle is a wave doesn't that wave cause an interference similar to the wave(s) that are transmitted from a speaker. If the wave(s) are in a range that can be "heard" or if you have a device to interpret the wave and convert it to a wave range that can be heard (or seen as in an oscilloscope)...can you also do the same for a particle? Basically listening to the wave as it interferes with what is around and use this to measure? Instead of bouncing a photon off it. Admittedly, I am only a little read on this subject anf have come to grasp a small glimpse, thanks in no small part to "Sixty Symbols!"
No that's not how physics works.
Nothing can be detected without being interacted with. The sound waves you speak of is the air being pushed and pulled by the speaker membrane, only to interact with your ear after a while. The membrane pushing the air is the same as the photon scattering off the electron in the video.
Any interaction or "interference" as you mention will affect the particle we are trying to measure, putting us right back at the issue the video is all about.
You both are missing my point. Ok 1st, imagine the pond ripple, you can infer the mass of whatever is dropped into the water by the ripples.
Now imagine if you can't see the whole picture of those same ripples but just, let's say 20% +/- one can still infer an approximation with fairly high accuracy! 2nd imagine yourself listening to music, you don't need to see or interact with the sound coming out from the speaker to infer what is playing nor how loud it is AND to top it off without interacting/disturbing the soundwaves you could almost "see" the walls of a room because the sound is bouncing off the walls...finally
3rd, since a particle is an excitation of a field i.e. an electron is an excitation of the field or could also be considered a very dense or compacted waveform because to create an electron from the field energy needs to jump to a very high and excited state to produce the particle. But none the less the particle is (if I am not mistaken) point of extremely dense packet of energy in a highly excited state....and by point mean it is expressed at a particular coordinate in space/time from the observers frame of reference.
This is the picture of any particle as I understand it.
So, since the particle has a wave/particle duality, I was asking if it can be inferred by the wave function alone without collapse of the wave?!?
Jonathan Owens
Well, your ear does interact with the wave - if it didn't interact with the wave, it basically means that the sound didn't reach your ear, in which case you won't hear anything. The momentum of the wave has to be imparted in part to the sensing hairs in the inner ear before you hear it.
You can't detect anything without interacting with it. It is as simple as that.
I lack the words to express what I am talking about, but I'm trying to explain detection via inference and not direct detection!
Pretty cool channel for us science freaks :3 You got a sub!
Why couldn't you make a formula in regards to where the photon enters the lens and where the electron was?
What color is Professor Merrifield's shirt?
We can pretty easily measure the location and momentum of, for example, an automobile. Up to what size scale is the Uncertainty principle relevant. electrons, hadrons, atoms, molecules, etc.? And why?
The uncertainty principle is part of quantum mechanics and is fundamental to everything at the atomic scale and smaller.
When you use low accuracy you can do both. When you measure with the accuracy of let's say a billionth of a meter, you need a photon of that wavelength, which means you're putting a lot of energy into the system and disturbing it
Hjembrent Kent photon have very specific wavelengths and you can not generate a photon that does not have an "allowed wavelength"
So if I wanted to measure that position of an automobile with the same accuracy as used in measuring the position of an electron, I would not be able to measure the automobile's momentum with the same accuracy used, or desired in measuring the momentum of an electron. The difference being that for the automobile, that type of measurement is not useful or practical anyway. . In other words the Uncertainty Principle applies equally on the macro scale, but it is not relevant because at that scale, that very fine level of accuracy is not needed or desired. Is that what you mean?
The way to get around it is to measure the particle's gravitational field. Not the most practical way of doing it, but at least you don't have to hit it with a photon.
If its not possible to measure a quantum encryption and therefore interpret the message without the other party knowing the message had been disturbed. What’s to stop people from measuring every packet of data that passes and prevent anyone from using quantum cryptography?
By using a feather?
why is everything on this channel muted for me? every other youtube video i can see but on this channel sixty symbols i cant...
"You can tell wether someone steamed the envelope open"
Wondering when in future this saying will cause so much confusion
Well, I'm mildy confused, so does today count?
@@HeyImLucious
Using boiling water steam to loosen the adhesive properties of glue in order to open envelope without a trace (or without easily noticeable trace) and easily glue it shut again is a really old technique (espionage technique one could say), and with traditional mail falling out of fashion, it is guaranteed to be lost to time in near future
@@HeyImLucious and, it seems that it might count
Awesome.
Cant you somehow measure the electric field the electron gives off at varying points to pinpoint the electrons location? Or would measuring that field somehow interact with that particle, or is the field even measurable, or is there really a field, is there really even an electron? quantum physics is weird.
No there's no field, there are just photons and virtual photons (google it). We just call it a field when there's a lot of photons, just taking the average of a lot of particles. It's all particles, and you have to hit them to find out anything about them
Why don't you just detect the electron via exclusion? Here is a 3x3 imaginary wall and each number is the size of an electron.
123
456
789
You know that there is one electron somewhere in this wall. You send a photon through all other numbers except the number 5. If you receive same amount of photons on the other side of the wall, then you know that you have not hit the electron, thus have not affected it with your measurement and you also know its location (5). Manage to do this twice (lets say leaving out number 3) and you also know the speed and direction on one axis. Add another axis to the experiment and you can then plot the direction and speed of the electron after successfully not hitting it with photons.
+myowntubeusername Well, the electron's wave function would not have collapsed, so the electron's position would still be uncertain.
I believe you would get interference on the photons, because the photons are acting like waves. Perhaps you could send a polarized photon so that it doesnt interfere, not quite sure how interference works.
Well if you measure it from every direction with millions of weak measurements, it should average to around not moving the particle at all, but also not moving it, since every measurement averaged moves it exactly the same.
Couldn't the message be disturbed twice with the same intensity from opposing directions thus reading the message without leaving a trace?
no you can't disturb things from the exactly opposite direction.
how do you make sure it's the same intensity? There's an uncertainty in the energy of whatever it is you want to use and so there's an uncertainty in the disturbance.
I don't know, that's why I'm asking. :)
i meant it as rhetorical, sorry. The point is you can't make sure it's the same intensity in the first place to hit it the same way from both sides.
Why can't you use two photons of equal wavelength on exact opposite sides to cancel out each others effect on the electron being measured?
You can't create a photon with an exact energy and timing either (if you could, then you'd know the position and momentum of that particle). So it is impossible to create the photons to do what you're proposing. If you want to generate a photon of a very precise frequency, you don't know exactly when it will be emitted, so you don't know exactly when it will hit the electron.
So something is needed that will have practically zero interaction, like a Neutrino?
Cant you aim the photon at a precise point so you know how much the particle is gonna bounce away?
Nope, because you don't know where to aim it until AFTER you've shot it with the photon. But then you don't know where it's at. Thus you can't aim a second one at it.
I love this guy (too:) Favorite part: 4:44
What about Computer Simulation?
Surely if two photons hit the particle from opposite sides at the same angle, the particle will stay in the same place?
+Mortizul, in order to do that, you would need to know where the particle is so you can hit it from each side. But in order to know where it is... you have to measure where it is.
If you put that much energy into a particle there's no telling what it would do, might even turn into a different particle
What did they say right at the beginning? This is killing me
but the particles dont have a well defined position nor momentum to begin with, so how does weak measurement measure a position more exact than the particle actually has??? .
The uncertainty principle is what dictates "the particles don't have a well defined position nor momentum". What they're saying is, that may not be strictly true in all cases. They're saying Heisenberg might have been wrong.
His sweater is nearly the same color as "the dress", the brown is darker than I remember.
I still feel like it can be done...Maybe we can figure out a way to measure everywhere the particle isn't.
If we don't know where it is in the first place, hitting everywhere but the point the particle is in seems like a very unlikely option. Also, particles tend to move, so we would have to measure *everywhere* around the particle at one instant in order to know its position... no, it can't be done.
That doesn’t give you information about the particle though.
In order to do that you'd have to know with certainty that the particle was within a particular well defined volume of space. You can't even do that because you can't put particles in tiny boxes. See Quantum Tunneling.
This reminds me of the "Heisenberg compensator" from Star Trek. :))
Why can’t you hold the electron with tweezers? In which case, you know its location and you know that its momentum is exactly 0. This would be classical physics. So are you saying that quantum systems can never be “held” between your fingers? This makes quantum mechanics literally intangible to any possible human senses and the brain because you have no idea what the quantum system even is. You can never hold it in your hand, because you’d then know it’s position and momentum simultaneously
If you try to constrain a particle's position like that, then its momentum becomes huge - it begins to move very, very fast. Essentially, you won't be able to squeeze it down to being in one place and not moving, as it will start to push against your "tweezers" with enormous force. You have to think about what it means to "hold" something. Think about how that works at the atomic level.
How about using magnets?...
Can you do a programme on 'quantum erasure' ? thanks
You measure something and note its value, then measure it again and again. Then you take the difference between each measurement and find it's relative changing ratio. you can now log the changing ratio according to the number of time you measured it and then find the real value of what you are measuring. Why don't you do this?
Because each time you measure it you send the particle off in a completely new random direction.
I wonder what the encrypted message says
"Xsnnmg iypuyey bmf imgvyjyt gm tyxmty gvuh"
=
"CANNOT BELIEVE YOU BOTHERED TO DECODE THIS"
rumkin.com/tools/cipher/cryptogram-solver.php
4:04 cannot believe you bothered to decode this
It's very simple.
Everything has a defined p and x, but every time you observe it, you'll end up with uncertainty in your measurements.
We seem to have an interactive process with the future always being uncertain because of Heisenberg's Uncertainty Principle!
The uncertainty principle doesn't really cause the uncertainty of the future. The future is uncertain because it can't causally affect the present, whereas the past can... at least, that's the case on the macroscopic level. The Schrödinger equation _is_ t-symmetric, but statistically speaking, things end up being time irreversible in the end... not really sure of the specifics of how the second law of thermodynamics can be gotten to from a quantum mechanics starting point though.
Ben Rowe We can think of classical physics as explaining processes over a period of time and quantum physics as explaining ‘time’ itself as a physical process. Photon energy cascades down forming greater degrees of freedom for entropy and chaos. These photon oscillations or vibrations do not just form greater degrees freedom for the aging process of decay. They also form greater degrees of freedom for the evolution and diversity of life that has lead to the creativity of art, poetry and music!
I'm going to guess that it will create an impossibility in predicting the weather a month after the evidence is gathered.
Quantum Atom Theory an artist theory on the physics of 'time' that doesn't really explain how the uncertainty principle has anything to do with the uncertainty of the future... it's good that you're enthusiastic about quantum mechanics, but you're just kinda jumping around right now.
I think this is obvious
THIRST! ...FOR KNOWLEDGE!
Draw three cards and discard two, unless you discard an artifact
This guy says a whole lot without actually saying anything. It’s like he doesn’t want to honestly answer Brady’s question.
Or perhaps you just didn't understand what he said.
why wouldn't you just ask god whre the electron is, not the weather god though, he has a lot of particles to take care of and he doesn't care for thunderstorms
If photons are massless, why do they disturb massed electrons?
+Matt McAlister it has momentum, which it can transfer without mass.
Momentum is the movement of mass though
+Matt McAlister nope not necessarily
E=MC^2 leaves out an important part: E^2 = (MC^2)^2 + (PC)^2 where P is momentum. If momentum is zero, it is at rest and (PC)^2 is zero, so E^2 = (MC^2)^2 reduces to E=MC^2 which represents rest mass. A photon has zero mass, so (MC^2)^2 resolves to zero and you get the reduced energy equation for a photon: E=PC.
Because people who do not like Donald Trump running for office and they disagree with his VP candidate so they disturb the electrons en mass or they just vote for the other person!!!!
he's not explaining weak measurements - they are measured with the 2nd vector from the future!!
Try not to measure where it is but measure where it isn't.
I hope Heisenberg is happy in heaven
Schrödingers cat will only sit on Heisenbergs lap.
Couldn't you just put an electron in an electric field with equally powerful magnets on all sides meaning that the electron would be equally repelled by all of them and thus forced to stop. It would have no momentum and you would know its position.
Unless you could start in perfectly the middle, like down to the planck length, it would oscillate, because it would be closer to one side, accelerate away from it, pass the middle, and repeat. That's assuming it's in a vacuum.
I'm thinking of two problems:
-you couldn't tell if all the magnets are exerting the exact same magnetic field
-you couldn't tell if they are positioned precisely correct
But its hypothetically plausible
The strength of the force between an electrical charge and a magnetic field depends on how quickly they are moving relative to each other. My guess is that if you set up your experiment, the electron would orbit in a tiny circle, or maybe bounce along an axis, but never actually stop.
This is actually a really good question and I hope someone who really knows would be able to tell you why. I think the key here is that you don't actually *know* its position. Just because you think the electron is in a particular place, doesn't actually mean it's standing still.
Actually by just examining the uncertainty relation equation, you know that there is no such thing as a completely still particle because if:
delta position * delta momentum >= h bar / 2
then if delta momentum is zero, which you assumed, you see that h bar (the planck constant) would have to be zero to make the equation satisfy. This tells us that there is no such thing as a completely stationary particle, so that electron must still be moving in one way or another, which would also mean it's position is highly non-localized, meaning the particle is almost acting like a foggy interference than a precise particle.
Do research on something called "standing matter waves". You also were describing a real device called a "Penning trap", so I would look into that as well :) Please tell me what you find, I am by no means a physicist and I would be just as intrigued as you are to find the real answer (if this wasn't right).
What you have basically described is a pendulum. The electron won't sit nicely in the centre, since when it is in the centre it had momentum (I mean, it had to move there to start with right?). So that momentum will take it out of the exact centre, so it will feel a force due to the net magnetic field, in the same way a pendulum feels a net force when it isn't at the bottom of its swing. The electron will then rock back and forward, creating uncertainty in position and momentum all the time.
What color is his shirt????!!!!
Saemj Baumgartner gold and tartan.
Oh my.
+Saemj Baumgartner Schrodinger's calico! What other color WOULD a scientist wear?!?
😑
Black + Blue.
0:24 Didn't we recently figure out how to know both?
Google 'Quantum lines of desire'.
What happens when you move the dial to 11?
jk
There are security proofs of the quantum key distribution protocol that follows from the postulates of quantum mechanics. What you're implying about the entire field of quantum cryptography being in the balance based on some debate about weak measurements is simply not correct.
When claiming an entire field is about to come crashing down, please contact someone familiar with the field first.