The uncertainty principle is NOT about "disturbance through measurement". It's a fundamental ingredient of nature. Moreover, "the" uncertainty principle doesn't only apply to position and momentum. It applies to any two observable quantities whose corresponding operators don't commute.
jmlfa, are you saying that DrPhysicsA is incorrect in his explanation in this video? It certainly seems to me that he is describing uncertainty as a result of "disturbance through measurement"? How do you show that uncertainty is a fundamental property of nature?
@Dan Hillman. I am a layman but I agree with you. I think what some people don't understand is the difference between the macro realm and the quantum realm. Within classical physics, throw a piece of paper at someone when impact occurs the change in person's position and movement remain unchanged. A speeding bullet on the other hand will cause a change. In other words, in classical physics there is a great deal of variation for size, mass etc. However, in the quantum realm those differences are narrowed down and it becomes a matter of wavelength and frequency. At least, that's a very simplistic way of looking at it.
Thanks, I seriously doubted the correctness of this explanation, but as a layman, I thought maybe I am the one misunderstanding something here. Should not be attempting to explain something before you fully understand it yourself imo.
In the fifth part of my video on special relativity I derive the formula E^2 = p^2c^2 + m^2c&4. When m=0 as is the case with photons that E=pc. So p=E/c = hf/c =h/wavelength
uncertainty principle applies to any situation where U x V = K. Just consider: If the product is constant then increasing one factor must decrease the other. Nothing to do with interference due to measurement. Another example is speed of light = C (a constant) = Frequency X Wavelength. Here again decreasing one increase the other.
I love your video's. Fantastic job at being direct,and simple. In college I always would seek out Professors like you. In the maths, and sciences too many profs would not be able to be organized, and explain the material simply. Great job!
love this series because I'm in year 9 so as you can imagine school isn't much help when it comes to particle physics, but that's where this series comes in. Again, great help!
That is right. To get the photon to interact with the electron you need the light to be of low wavelength, comparable to the size of the electron. But low wavelength = high frequency = high energy (E=hf). So high energy photon will kick the electron away from where it was. But actually I think all these explanations are just illustrative of a quantum mechanical phenomena which can't properly be understand using classical analogies.
That is a property of waves. You can see it with water waves. If the waves go through a slit which is broadly the same width as the wavelength the wave spreads out. It was Huygens who suggested that every point on a wave front becomes a source of a spherical wave. It is the combined effect of all these secondary waves which causes the wave to propagate but it is also the reason that light spreads out when it passes thro a slit.
It seems to me that this video is describing the Observer Effect and not the Uncertainty Principle. The Uncertainty Principle is a fundamental reality of matter not a limitation of how humans can measure particles. As I understand it, whether a particle is being observed or not, the location and momentum are not defined (as opposed to we can't measure them) beyond a certain level of precision (h/4π). And this is what the uncertainty principle is actually describing.
If you look up Heisenberg's uncertainty principle in Wikipedia it will indeed say that Δx Δp is of the order of h. But further on it says that Kennard in 1927 first proved the modern inequality σx σp ≽ h bar/2 - where σx σp are the standard deviations of position and momentum. Actually the exact form doesn't really matter since its an inequality. All we really need to know is that you cannot measure both position and momentum with absolute certainty at the same time.
A Little error at 7:19...I believe. : the small distance he defines as lambda must be in fact lambda/2, half a wavelength in order for the two waves to cancel out. This error needs to be corrected through the end of the explanation resulting in the correct formula h/2 = dx.dp
Thanks. Although the electron's mass and charge are known, its size is unknown because it is not known whether it is a point like particle with no spacial dimensions or extremely small. Its not because of the Uncertainty Principle.
I think you are describing what is known as the Compton effect. A high energy photon hits an electron and scatters with a higher wavelength. The difference in wavelength enables change in energy and momentum of electron.
It is true that this video is more about the observer effect. For the quantum mechanical explanation you need to look at the videos in the quantum mechanics playlist, And in particular the series of seven videos which I produced last year.
We assume that light travelling from the source to the two slits remains in phase. The light arriving at the top slit is then diffracted at a particular angle. The light arriving at the bottom split is diffracted at the same angle but it will have further to travel to reach the Observer. If that additional distance which is calculated by dropping the perpendicular to which you refer happens to be a complete number of wavelengths then the light will remain in phase.
the measurement shortcoming is a lame explanation. real explanation is much simpler and based on complementarity what is complementarity? if the product of 2 variables is a constant e.g A x B = a constant then A and B show complementarity coz if you increase or decrease one the other must change in the opposite direction since their product is a constant. here deltaX x deltaP = h/2 which is a constant hence the 2 variables are complementary. QED
@ThrowHerAway Thanks for watching. I think you'll find it is greater than or equal to. If it were less than, then it could be equal to 0 and then there would be no uncertainty. Planck's constant (h) over 4pi is the same as Planck's reduced constant (h bar) over 2 - since h bar = h/2pi
the right relativistic relation is E^2 = Eo^2 + (p.c)^2, with Eo as the rest energy of the considered particle which for photon is 0, Eo photon = 0. this reduces the eq. to E = pc for photon. this prevents the wrong assumption of mass m = E/c^2 for photon, i think. tnx
@1996sagark Thanks. Have thought about a board, but difficult to get camera in the right place so that I don't block the board. The paper seems to work quite well.
You could write on glass with a phosphorescent pen illuminated by a side light & then use a mirror to reverse the writing. ruclips.net/video/FYwXOLU4TKk/видео.html&ab_channel=JulesWhite
Correct me if i am wrong about light (particle) propagate as a wave. I see light as water current. A faucet if we reduce the opening, it will drop as a droplet. Now if we enlarge the opening, these droplets become as a steady current. Light as we know is really both as particles and wave, it just depends how we approach them.
Space is a division of solidity into tenuity. Also a multiplication of volume at the expense of gravitational potential. Matter in violent motion simulates rest and balance through violent motion. In the future Heisenbergs Uncertainty Principle will be the required time cycle for quanta to exist. The interval of the current moment reaching its threshold when C is canceled out by its anti+C collasping and reforming+/-eXploding into this enomous collage of what we experience as the current moment
*Transcript* 0:00 Hello. Today, we're going to look at Heisenberg's uncertainty principle which generally relates to things at the atomic & subatomic level. [BECBE writes:] ΔxΔp ≥ ℏ/2 [BECBE calls: the Δ symbol 'delter' the: ℏ symbol 'h bar' the: θ symbol 'theeter' & the: λ symbol 'lamder'] [& of course, these are spelt: delta, aitch bar, theta, lambda] 0:10 Heisenberg's uncertainty principle is usually written in this form: Δx times Δp ≥ ℏ over 2 0:24 What does that mean? 0:25 Well, Δx talks about the uncertainty in position. 0:30 If you want to know, for example, where an electron is, how clear can you be? 0:35 How precise can you be about where it is? 0:38 Δp talks about the uncertainty of that electron's momentum - in other words, its velocity & its direction. 0:47 & the idea is that the uncertainty in position multiplied by uncertainty of momentum is greater than ℏ over 2 0:57 So, you cannot precisely define position & momentum at the same time. 1:04 Here is, let's say, an electron. 1:06 That electron has a certain position, x & it has a certain momentum, p but we cannot measure both of them at the same time. [BECBE draws a dot] 1:16 The reason for this is that electrons, even atoms, are very small compared with anything that you're going to use to measure them. 1:27 For example, let's say we use light to look at an electron. 1:31 Well, here's a light wave [BECBE draws a wave that's about 10 times taller than the dot] which goes right past the electron without even noticing it. 1:38 In fact, I haven't drawn this to scale at all because a light wave has a wavelength of approximately 5 x 10ᐨ⁷ metres. 1:50 Whereas, even if this were an atom, it would be approximately 10ᐨ¹⁰ metres. 1:57 If it were a proton, it would be 10ᐨ¹⁵ metres. 2:04 & if it were an electron, it would be something like 10ᐨ¹⁸ metres. 2:11 In other words, even an atom is a thousand smaller than the wavelength of this light. 2:19 A proton would be about a hundred *million* times smaller than the wavelength of this light. 2:25 So, any visible light will just go straight past the electron or the proton or the atom without noticing it. 2:30 If you want to actually *see* something of the size of an atom or a proton or an electron, you have to use some form of radiation whose wavelength is broadly comparable to the size of the thing you are trying to see. 2:48 & therein lies the problem: because the smaller the wavelength, the larger the momentum. [BECBE writes:] p = h/λ 2:57 It's given by a formula that says that the momentum of a wave = h (Planck's constant) divided by λ 3:06 & so you can see that as λ gets smaller, p gets larger. 3:11 Where does this formula come from? 3:13 Well, we can derive it in a simplistic way: we take the famous formula of Einstein: E = mc² [BECBE writes:] p = mv 3:21 & we say that momentum is classically given by mass times velocity. 3:28 Now, when we're talking about electromagnetic radiation like light, photons - which are the constituent parts of that light, don't have a mass, [BECBE writes:] m = E/c² 3:37 but we can say that the mass of a photon is kind of equivalent to E over c² from... [BECBE points at: E = mc² ] ...this formula here: mass is E over c² [BECBE writes:] p = Ec/c² 3:46 & so, you can say that the momentum is the mass E/c² times the velocity which is c [BECBE writes:] p = Ec/c² = E/c & that gives you E over c [BECBE writes:] E = hf 3:59 But you'll know that energy of an electromagnetic wave is hf, Planck's constant times the frequency of the radiation. 4:10 & that is known as the packet of energy - the photon packet of energy - the quantised energy. [BECBE writes:] p = hf/c 4:16 & so now we can say that p, the momentum is E which is hf divided by c [BECBE writes:] p = hf/c = h/λ but c over f is λ & so we derive the formula that we started with: p = h/λ [BECBE draws a dot & a horizontal arrow touching it] 4:35 & now we can see the problem that if we have an electron (or a proton or whatever) & we send in very low wavelength electromagnetic radiation in order to be able to *see* this proton, it's going to have such a high momentum, that although the wave may well reflect & we can detect it, it will... [BECBE energetically draws a 2nd arrow from the dot towards the top right of the screen] ...give a huge *kick* to this electron or proton & send it scurrying away. 5:06 It's rather like hitting a billiard ball with a super fast billiard ball that knocks it for 6 & so you can't tell what its position & its momentum are. 5:15 We can perhaps explain this by reference to the so-called 'single slit experiment'. 5:21 This is where you take a single slit - very narrow & you pass ordinary light through it. 5:29 & what happens is that the light beam comes through the slit but as it goes through the slit, it spreads out, onto a screen. [BECBE draws a -boob- bell shape] 5:37 & what you find is that the intensity of the light on the screen, goes something like that. [BECBE points at the top & bottom of the bell shape. The bell shape almost touches the screen at these points.] 5:44 In other words, at these points it's virtually 0. [BECBE's pen is touching the middle of the bell shape] Up here it's very high. 5:49 & it doesn't get very much bigger thereafter. [What BECBE does now with his pen is beyond my ability to describe.] 5:52 So it's kind of spread out across the screen & the spread is at an angle θ 6:01 Why does it do that? 6:03 Well, if we magnify the slit we can see what's actually happening here. 6:08 Here is the slit of dimension d 6:11 & here is the wave that is going up to reach this point here - in other words, the point where there's no light at all. 6:18 Now, if we drop a perpendicular here... 6:24 This angle, as we've said is θ 6:30 That's that angle there. 6:31 & by geometry, this angle here is also θ 6:39 Now, what will cause these 2 waves (when viewed by my eye here) to cancel out. 6:49 By 'cancel out' I mean that 1 wave will look like this, 6:54 & the other wave will look like this, 6:57 & when they are superimposed, they will simply produce a flat line which is the no light that you get here. 7:06 Well, the answer is that they have to be completely out of phase. 7:11 & how can that happen? 7:13 Well, that happens if this distance here - from here to here is λ 7:19 because that's the difference between the phase of this wave & this wave. 7:24 Why is that the case? 7:25 Well, let's blow this up again. 7:27 Here is that famous triangle. This is d. This is this gap here. 7:32 This is λ. That's there. 7:35 & take the point halfway. 7:37 Now, this is λ/2 7:43 & this wave & this wave are going to be ½ a wavelength apart. 7:48 & so there are going to be in precisely this position of superimposing. 7:52 So, the light there, will cancel the light there. 7:56 & the light there, will cancel the light there. 8:00 Because they're all λ/2 apart. 8:03 & similarly every position here, will cancel a position here. 8:07 Every point there, will cancel there. 8:11 Every point there, will cancel there. 8:12 So, *all* the points in the upper ½, cancel all the point in the lower ½ & produce this minimum here. 8:23 & you can see from this diagram here that λ = d sin θ 8:33 So, if you know the wavelength of the light, you know what the angle is where you're going to get a minimum. 8:40 You can do the same thing - instead of using light, you can do the same thing with electrons.
I am confused on why you can use E=mc^2 to find the momentum of a photon? Are we considering variable mass because if we are not then E=mc^2 is only applicable for rest energy? I just used the E^2 = (mc^2) ^2 + (pc)^2, plugged in m=0 and solved for p.
7:14 I don't get why you say the answer's λ. I see why wave has to be 180° out of of phase to cancel out the other wave. But why can't you have really long wavelengths? Then it wouldn't be λ Edit: Ok, I think he made 2 errors. 1) He meant λ/2 not λ Before the slit, it's coherent laser light therefore the top & bottom rays are in phase. After the slit, they have to quickly change from 1) being in phase to being 180° out of of phase. At 6:22 he draws the adjacent side. When the light reaches the adjacent side it has to be 180 deg out of phase. No further changes are possible. Therefore the bottom ray must have travelled λ/2 2) He says the middle ray is 180° out of phase with the top ray. But he had already established that the top & bottom rays are 180° out of phase: 6:39 "Now, what will cause these 2 waves (when viewed by my eye here) to cancel out. By 'cancel out' I mean that 1 wave will look like this & the other wave will look like this." If we treat 6:39 as false then everything else works fine: the top & bottom rays are in phase at the slit & in phase at the adjacent side & the answer's λ & the middle ray's 180° out of phase like he says
Unnecessary complex explanation. Simply put plancks h/2pi= k Joules x sec = k momentum x distance = k mv x (x). Since the units of planck's constant are mv x (x) it is obvious that if delta (x) decreases mv must increase and vice versa since their product is a constant = k. Also since momentum is mv and if m is tiny then changes in mv must be largely due to v. In very massive objects with large m (like a baseball) the change in momentum is due to only tiny, imperceptible changes in v since the k in pianck's h is so tiny. QED.
To expand on the previous comment, check out the much touted original experiment. They proved fringing, and nothing else. Check it out (hint: It's not on utube)
Not my joke but worth posting on the matter: Heisenberg and Schrodinger are speeding down the highway when they are pulled over by a cop. The cop strolls up to the window as Heisenberg roles it down. The cop asks, "Sir do you know how fast you were going?" Heisenberg's, "No, but I know where I was." The cop decides to search the car. He opens the trunk and says, "Sir do you know you have a dead cat?" Schrodinger say, "I do now!"
This is not the Heisenberg uncertainty principle. The principle is due to the wave nature of matter. I can illustrate it this way. Suppose you have a wave with well defined momentum. Due to de Broglie, you can determine its frequency which is also uniquely defined. So you can represent that state as a sine wave. But where is the wave then? Well it's everywhere. You have complete uncertainty about you position. It's just functional analysis. Nothing to do with this, it's absurd.
A stationary photon has no mass. This is known as the rest mass, which is 0. However, photons do have a mass when moving (momentum) and contain a quantum 'package' of energy. This is due to the way in which the electric and magnetic field oscillate, following the photon's creation. A photon wouldn't exist if it wasn't moving and didn't have energy. Basically, photons do have mass in the form of the quantum energy level they have, which differs depending on the frequency.
I have a question. You used a lightwave to measure the electron. But if we use the wave-particle duality model light is a particle or pocket of a wave (photon), so how come you represented it with a wave? Also according my very limited understanding of QED electrons and protons interchange photons. I have always understood Heisenberg's uncertainty principle as if you use a photon to try to measure the position of an electron that photon will interact with the electron thus changing its position.
of *course* photons have a mass...all energy has an associated mass (that's the meaning of e=mc^2). you are thinking of the concept of "rest mass" which, since a single photons is *never* at rest in the reference frame of any observer, does not exist. thus, the mass of the photon is not "kind of equivalent" to e/c^2, it is precisely equivalent...but the better derivation comes via the relativistic e-p equation e^2-p^2c^2=m_0^2c^4, where rest mass m_0 can be set explicitly to 0.
My belief is that at the current level of science, we are not able to 'measure' the size and mass of some 'particles'. another reason could be that our understanding of 'particle-wave duality' is incomplete to explain exactly what particle or wave is. you must understand that our science has always been to quantify all matter with existing SI units.
I think Bruce is correct about the interpretation of the uncertainty principle. A particle in principle does not have an exact position and momentum. This is independent of whether we can meaure it or not. The debate about this interpretation is still going on today.
here you actually speak about Heisenberg measurment-disturbance relations and not about uncertainty principle...it's not your fault, that misconception is something most of people learn in school...these two are nto the same, you can check wikipedia on uncertainty relations...
If E^2=(mc^2)^2+(pc)^2 then why does it cancel to E=mc^2 implying that p=0 and therefore m=0 if the particle (massless and speed c) has momentum of h/ lamda (sorry no key for that symbol). Basically why does the mass=0 in p=mv but not mc^2.
err, i mean rest mass does not exist for a single photon, or rest mass = 0 for any system consisting solely of a single photon or collection of photons with the same exact same momentum.
Utterly wrong. The position and the momentium of a particle are Fourier transforms, one of the other. That is the uncertanity. That's true if nobody is measuring anything.
so in quantum mechanics waves can have a momentum,but in classical mechanics,they cant? because they have no mass and since momentum=mass x velocity,momentum should be 0?
I'm confused because in my workbook as well as in wikipedia, I found that (delta p)(delta x) is greater than planck's constant h. Whithout being devided by 4π.
The reason we can not measure (know) velocity and momentum of an electronat at the same time is not because of size of it and uncertanity of measurement but because of the uncertain nature of the electron.
If we cannot measure momentum and position at the same time, why not just minimize the time between the measurement of one and the measurement of the other such that we approximate simitanaeity?
3:23 I understand you seek to simplify, but I think it would still have been better if you had used : E² = p²c² + m²c^4 obviously giving E = pc for m=0 Although unlike "E=mc²", the above formula may not be known by "everybody on earth", it would still allow you to avoid "confusions" such as : -(remark mentionned by someone else in one of the comments below) : using on photons a formula that would normally represent "rest energy" -(my personal remark) : using CLASSICAL momentum on something going at the speed of Light !.. Your way of deriving it obviously works, and thus can be seen as a "nice trick to find it back", but it doesn't seem rigorous at all :/ ...Nice video though :)
Heisenberg's Uncertainty Principle is soo uncertain , even I don't have the certainty, that is actually 100 % certain, ..you know ,..that a particle ( we were talking about particles aren't we ? ) ..is were it's supposed to be , even If a I look at it with certainty, cause if I look ..or try to measure it , it will move......somewhere.............uncertain.....end of history :P
For the single slit experiment, wouldn't you get similar interference for integral multiples of lambda? This would imply that there would be fainter fringes extending on either side of the main band. Or am I missing something?
How do we know that we need to send out a wave to 'try' locate the electron, what do we expect when the wave hits the electron ? is this anything related to the compton effect ?
Starting at 06:20 you say that light coming from the two edges of the slit meet up at some specific point on the screen where they will interfere destructively because of the phase difference. What I do not understand is when you draw that perpendicular line and assume that starting from the perpendicular line the the distances from the edges to that specific point on the screen are the same. But one of the distances is the hypotenuse the other is not. What am I missing here?
The equation E^2=(mc^2)^2+(pc)^2 is the complete equation but it does not cancel to E=MC^2. The E=MC^2 we usually talk about is just an approximation. The reason is because in first part we have (mc^2)^2, this basically result in c^4 and we get a ridiculously large number compare to (pc)^2, which only has c^2. Even if we omit (pc)^2 the result won't have significant changes in most cases. Thus we can approximate the equation to E^2=(mc^2)^2 and the ^2 cancel each other and we get E=MC^2.
How can you know that they aren't defined? If they get defined as soon as you measure them or if you would just measure a defined position and momentum, that should give the same test result, wouldn't it?
Even Heisenberg used the Observer Effect to demonstrate the Uncertainty Principle and made it easier to confuse both. DrPhysicsA pointed out it is a fundamental property of wave matter in the end of the 2nd part.
You are right, the video try to explain the uncertainty principle from a more classical view, i.e, from the Copenhagen interpretation. As you have just mentioned the Uncertainty Principle is a fundamental description of an objective reality, and this is not so easy to understand.
Best video I learned more here than any other video I've ever watched.... the wave size matching the object size hit home..ha' I am a slow learner not trained in math this illustration was easy for me to understand
Glad to find explanations I can follow. Sufficient to prompt questions in my mind. An example you use is an electron of size ~10^-18m. Is this size is real? Or is it due to the uncertainty in position of a point like particle?
Hi, how to you record this video, i mean how you position camera or phone over the paper you write? this is the most engaging way to give a lesson on the web.
Just fucked up my physics exam by misunderstanding 2 questions. It's nice to come here and learn something to feel like i'm not useless. I never knew the geometrics of the huygens-fresnel principle. I just knew the formula.
good to know about how many dimentions we are talking about. the HUP only works for a 4D environment or could it be true for as many other dymentions we can have? Is it ok for Kaluza-Klein 5 dimentions architecture? Does it works for a Ricci tensor configuration?
For understanding HUP at the kaluza klein level , you might should first take into account scale at which you're talking about the strings existing , you've first got to sort the dimensionality of strings as string theory is the only present theory giving a viable explanation of kaluza klein theory
The uncertainty principle is NOT about "disturbance through measurement". It's a fundamental ingredient of nature. Moreover, "the" uncertainty principle doesn't only apply to position and momentum. It applies to any two observable quantities whose corresponding operators don't commute.
jmlfa, are you saying that DrPhysicsA is incorrect in his explanation in this video? It certainly seems to me that he is describing uncertainty as a result of "disturbance through measurement"? How do you show that uncertainty is a fundamental property of nature?
@Dan Hillman. I am a layman but I agree with you. I think what some people don't understand is the difference between the macro realm and the quantum realm. Within classical physics, throw a piece of paper at someone when impact occurs the change in person's position and movement remain unchanged. A speeding bullet on the other hand will cause a change. In other words, in classical physics there is a great deal of variation for size, mass etc. However, in the quantum realm those differences are narrowed down and it becomes a matter of wavelength and frequency. At least, that's a very simplistic way of looking at it.
Thanks, I seriously doubted the correctness of this explanation, but as a layman, I thought maybe I am the one misunderstanding something here. Should not be attempting to explain something before you fully understand it yourself imo.
That is straight.
It's Heisenberg's uncertainty principle..and it's about position and momentum. We can't measure both simultaneously with 100% accuracy.
In the fifth part of my video on special relativity I derive the formula E^2 = p^2c^2 + m^2c&4. When m=0 as is the case with photons that E=pc. So p=E/c = hf/c =h/wavelength
Tripod on table with camera pointing vertically down onto a sheet of A3 paper. I am also branching out into using a white board.
uncertainty principle applies to any situation where U x V = K.
Just consider: If the product is constant then increasing one factor must decrease the other.
Nothing to do with interference due to measurement.
Another example is speed of light = C (a constant) = Frequency X Wavelength. Here again decreasing one increase the other.
does that mean you can't measure it's wavelenght and frecuency at the same time?
I love your video's. Fantastic job at being direct,and simple. In college I always would seek out Professors like you. In the maths, and sciences too many profs would not be able to be organized, and explain the material simply. Great job!
love this series because I'm in year 9 so as you can imagine school isn't much help when it comes to particle physics, but that's where this series comes in. Again, great help!
That is right. To get the photon to interact with the electron you need the light to be of low wavelength, comparable to the size of the electron. But low wavelength = high frequency = high energy (E=hf). So high energy photon will kick the electron away from where it was. But actually I think all these explanations are just illustrative of a quantum mechanical phenomena which can't properly be understand using classical analogies.
That is a property of waves. You can see it with water waves. If the waves go through a slit which is broadly the same width as the wavelength the wave spreads out. It was Huygens who suggested that every point on a wave front becomes a source of a spherical wave. It is the combined effect of all these secondary waves which causes the wave to propagate but it is also the reason that light spreads out when it passes thro a slit.
It seems to me that this video is describing the Observer Effect and not the Uncertainty Principle. The Uncertainty Principle is a fundamental reality of matter not a limitation of how humans can measure particles. As I understand it, whether a particle is being observed or not, the location and momentum are not defined (as opposed to we can't measure them) beyond a certain level of precision (h/4π). And this is what the uncertainty principle is actually describing.
If you look up Heisenberg's uncertainty principle in Wikipedia it will indeed say that Δx Δp is of the order of h. But further on it says that Kennard in 1927 first proved the modern inequality σx σp ≽ h bar/2 - where σx σp are the standard deviations of position and momentum. Actually the exact form doesn't really matter since its an inequality. All we really need to know is that you cannot measure both position and momentum with absolute certainty at the same time.
A Little error at 7:19...I believe. : the small distance he defines as lambda must be in fact lambda/2, half a wavelength in order for the two waves to cancel out. This error needs to be corrected through the end of the explanation resulting in the correct formula h/2 = dx.dp
Thanks. Although the electron's mass and charge are known, its size is unknown because it is not known whether it is a point like particle with no spacial dimensions or extremely small. Its not because of the Uncertainty Principle.
I think you are describing what is known as the Compton effect. A high energy photon hits an electron and scatters with a higher wavelength. The difference in wavelength enables change in energy and momentum of electron.
hello, could i clarify if the video describes the uncertainty principle, or the observer effect? Thank you :)
It is true that this video is more about the observer effect. For the quantum mechanical explanation you need to look at the videos in the quantum mechanics playlist, And in particular the series of seven videos which I produced last year.
Once you use two sources you cannot be certain that both sources will be in phase. It only works with a single source.
We assume that light travelling from the source to the two slits remains in phase. The light arriving at the top slit is then diffracted at a particular angle. The light arriving at the bottom split is diffracted at the same angle but it will have further to travel to reach the Observer. If that additional distance which is calculated by dropping the perpendicular to which you refer happens to be a complete number of wavelengths then the light will remain in phase.
Yes you are right. You do get fringes on either side of the main central fringe. But they are much fainter than for the double slit experiment.
It could be related to the Compton effect, and I shall be doing a video on that, about absorption of electromagnetic radiation, very shortly.
You are right of course. The subtitle is unclear: It was supposed to say that h is Planck's constant but I can see that it appears to be h bar.
the measurement shortcoming is a lame explanation.
real explanation is much simpler and based on complementarity
what is complementarity? if the product of 2 variables is a constant e.g A x B = a constant then A and B show complementarity coz if you increase or decrease one the other must change in the opposite direction since their product is a constant.
here deltaX x deltaP = h/2 which is a constant hence the 2 variables are complementary. QED
OK but why does Δx complement Δp ?
Pressed play, heard accent, subscribed. Well played good sir.
@ThrowHerAway Thanks for watching. I think you'll find it is greater than or equal to. If it were less than, then it could be equal to 0 and then there would be no uncertainty. Planck's constant (h) over 4pi is the same as Planck's reduced constant (h bar) over 2 - since h bar = h/2pi
As long as you are talking about massless waves. A water wave, for example, clearly has classical momentum
the right relativistic relation is E^2 = Eo^2 + (p.c)^2, with Eo as the rest energy of the considered particle which for photon is 0, Eo photon = 0. this reduces the eq. to E = pc for photon. this prevents the wrong assumption of mass m = E/c^2 for photon, i think. tnx
Partly. But you might like to look at my videos on Atomic Physics which give a little more info.
@1996sagark Thanks. Have thought about a board, but difficult to get camera in the right place so that I don't block the board. The paper seems to work quite well.
You could write on glass with a phosphorescent pen illuminated by a side light & then use a mirror to reverse the writing. ruclips.net/video/FYwXOLU4TKk/видео.html&ab_channel=JulesWhite
I have seen lots of videos about this topic, but this is definitely the best, very well explained. Thanks DrPhysicsA. Keep the good work.
Correct me if i am wrong about light (particle) propagate as a wave. I see light as water current. A faucet if we reduce the opening, it will drop as a droplet. Now if we enlarge the opening, these droplets become as a steady current. Light as we know is really both as particles and wave, it just depends how we approach them.
Space is a division of solidity into tenuity. Also a multiplication of volume at the expense of gravitational potential.
Matter in violent motion simulates rest and balance through violent motion.
In the future Heisenbergs Uncertainty Principle will be the required time cycle for quanta to exist. The interval of the current moment reaching its threshold when C is canceled out by its anti+C collasping and reforming+/-eXploding into this enomous collage of what we experience as the current moment
Awesome !! Don't know why but only your Explanations is Easy to understand for me ? Very Great :)
*Transcript*
0:00
Hello. Today, we're going to look at Heisenberg's uncertainty principle which generally relates to things at the atomic & subatomic level.
[BECBE writes:]
ΔxΔp ≥ ℏ/2
[BECBE calls:
the Δ symbol 'delter'
the: ℏ symbol 'h bar'
the: θ symbol 'theeter'
& the: λ symbol 'lamder']
[& of course, these are spelt: delta, aitch bar, theta, lambda]
0:10
Heisenberg's uncertainty principle is usually written in this form: Δx times Δp ≥ ℏ over 2
0:24
What does that mean?
0:25
Well, Δx talks about the uncertainty in position.
0:30
If you want to know, for example, where an electron is, how clear can you be?
0:35
How precise can you be about where it is?
0:38
Δp talks about the uncertainty of that electron's momentum - in other words, its velocity & its direction.
0:47
& the idea is that the uncertainty in position multiplied by uncertainty of momentum is greater than ℏ over 2
0:57
So, you cannot precisely define position & momentum at the same time.
1:04
Here is, let's say, an electron.
1:06
That electron has a certain position, x & it has a certain momentum, p but we cannot measure both of them at the same time.
[BECBE draws a dot]
1:16
The reason for this is that electrons, even atoms, are very small compared with anything that you're going to use to measure them.
1:27
For example, let's say we use light to look at an electron.
1:31
Well, here's a light wave
[BECBE draws a wave that's about 10 times taller than the dot]
which goes right past the electron without even noticing it.
1:38
In fact, I haven't drawn this to scale at all because a light wave has a wavelength of approximately 5 x 10ᐨ⁷ metres.
1:50
Whereas, even if this were an atom, it would be approximately 10ᐨ¹⁰ metres.
1:57
If it were a proton, it would be 10ᐨ¹⁵ metres.
2:04
& if it were an electron, it would be something like 10ᐨ¹⁸ metres.
2:11
In other words, even an atom is a thousand smaller than the wavelength of this light.
2:19
A proton would be about a hundred *million* times smaller than the wavelength of this light.
2:25
So, any visible light will just go straight past the electron or the proton or the atom without noticing it.
2:30
If you want to actually *see* something of the size of an atom or a proton or an electron, you have to use some form of radiation whose wavelength is broadly comparable to the size of the thing you are trying to see.
2:48
& therein lies the problem: because the smaller the wavelength, the larger the momentum.
[BECBE writes:]
p = h/λ
2:57
It's given by a formula that says that the momentum of a wave = h (Planck's constant) divided by λ
3:06
& so you can see that as λ gets smaller, p gets larger.
3:11
Where does this formula come from?
3:13
Well, we can derive it in a simplistic way: we take the famous formula of Einstein: E = mc²
[BECBE writes:]
p = mv
3:21
& we say that momentum is classically given by mass times velocity.
3:28
Now, when we're talking about electromagnetic radiation like light, photons - which are the constituent parts of that light, don't have a mass,
[BECBE writes:]
m = E/c²
3:37
but we can say that the mass of a photon is kind of equivalent to E over c² from...
[BECBE points at: E = mc² ]
...this formula here: mass is E over c²
[BECBE writes:]
p = Ec/c²
3:46
& so, you can say that the momentum is the mass E/c² times the velocity which is c
[BECBE writes:]
p = Ec/c² = E/c
& that gives you E over c
[BECBE writes:]
E = hf
3:59
But you'll know that energy of an electromagnetic wave is hf, Planck's constant times the frequency of the radiation.
4:10
& that is known as the packet of energy - the photon packet of energy - the quantised energy.
[BECBE writes:]
p = hf/c
4:16
& so now we can say that p, the momentum is E which is hf divided by c
[BECBE writes:]
p = hf/c = h/λ
but c over f is λ & so we derive the formula that we started with: p = h/λ
[BECBE draws a dot & a horizontal arrow touching it]
4:35
& now we can see the problem that if we have an electron (or a proton or whatever) & we send in very low wavelength electromagnetic radiation in order to be able to *see* this proton, it's going to have such a high momentum, that although the wave may well reflect & we can detect it, it will...
[BECBE energetically draws a 2nd arrow from the dot towards the top right of the screen]
...give a huge *kick* to this electron or proton & send it scurrying away.
5:06
It's rather like hitting a billiard ball with a super fast billiard ball that knocks it for 6 & so you can't tell what its position & its momentum are.
5:15
We can perhaps explain this by reference to the so-called 'single slit experiment'.
5:21
This is where you take a single slit - very narrow & you pass ordinary light through it.
5:29
& what happens is that the light beam comes through the slit but as it goes through the slit, it spreads out, onto a screen.
[BECBE draws a -boob- bell shape]
5:37
& what you find is that the intensity of the light on the screen, goes something like that.
[BECBE points at the top & bottom of the bell shape. The bell shape almost touches the screen at these points.]
5:44
In other words, at these points it's virtually 0.
[BECBE's pen is touching the middle of the bell shape]
Up here it's very high.
5:49
& it doesn't get very much bigger thereafter.
[What BECBE does now with his pen is beyond my ability to describe.]
5:52
So it's kind of spread out across the screen & the spread is at an angle θ
6:01
Why does it do that?
6:03
Well, if we magnify the slit we can see what's actually happening here.
6:08
Here is the slit of dimension d
6:11
& here is the wave that is going up to reach this point here - in other words, the point where there's no light at all.
6:18
Now, if we drop a perpendicular here...
6:24
This angle, as we've said is θ
6:30
That's that angle there.
6:31
& by geometry, this angle here is also θ
6:39
Now, what will cause these 2 waves (when viewed by my eye here) to cancel out.
6:49
By 'cancel out' I mean that 1 wave will look like this,
6:54
& the other wave will look like this,
6:57
& when they are superimposed, they will simply produce a flat line which is the no light that you get here.
7:06
Well, the answer is that they have to be completely out of phase.
7:11
& how can that happen?
7:13
Well, that happens if this distance here - from here to here is λ
7:19
because that's the difference between the phase of this wave & this wave.
7:24
Why is that the case?
7:25
Well, let's blow this up again.
7:27
Here is that famous triangle. This is d. This is this gap here.
7:32
This is λ. That's there.
7:35
& take the point halfway.
7:37
Now, this is λ/2
7:43
& this wave & this wave are going to be ½ a wavelength apart.
7:48
& so there are going to be in precisely this position of superimposing.
7:52
So, the light there, will cancel the light there.
7:56
& the light there, will cancel the light there.
8:00
Because they're all λ/2 apart.
8:03
& similarly every position here, will cancel a position here.
8:07
Every point there, will cancel there.
8:11
Every point there, will cancel there.
8:12
So, *all* the points in the upper ½, cancel all the point in the lower ½ & produce this minimum here.
8:23
& you can see from this diagram here that λ = d sin θ
8:33
So, if you know the wavelength of the light, you know what the angle is where you're going to get a minimum.
8:40
You can do the same thing - instead of using light, you can do the same thing with electrons.
I am confused on why you can use E=mc^2 to find the momentum of a photon? Are we considering variable mass because if we are not then E=mc^2 is only applicable for rest energy?
I just used the E^2 = (mc^2) ^2 + (pc)^2, plugged in m=0 and solved for p.
7:14 I don't get why you say the answer's λ. I see why wave has to be 180° out of of phase to cancel out the other wave. But why can't you have really long wavelengths? Then it wouldn't be λ
Edit: Ok, I think he made 2 errors. 1) He meant λ/2 not λ
Before the slit, it's coherent laser light therefore the top & bottom rays are in phase.
After the slit, they have to quickly change from 1) being in phase to being 180° out of of phase.
At 6:22 he draws the adjacent side. When the light reaches the adjacent side it has to be 180 deg out of phase. No further changes are possible. Therefore the bottom ray must have travelled λ/2
2) He says the middle ray is 180° out of phase with the top ray.
But he had already established that the top & bottom rays are 180° out of phase:
6:39 "Now, what will cause these 2 waves (when viewed by my eye here) to cancel out. By 'cancel out' I mean that 1 wave will look like this & the other wave will look like this."
If we treat 6:39 as false then everything else works fine: the top & bottom rays are in phase at the slit & in phase at the adjacent side & the answer's λ & the middle ray's 180° out of phase like he says
Unnecessary complex explanation.
Simply put plancks h/2pi= k Joules x sec = k momentum x distance = k mv x (x). Since the units of planck's constant are mv x (x) it is obvious that if delta (x) decreases mv must increase and vice versa since their product is a constant = k. Also since momentum is mv and if m is tiny then changes in mv must be largely due to v. In very massive objects with large m (like a baseball) the change in momentum is due to only tiny, imperceptible changes in v since the k in pianck's h is so tiny. QED.
To expand on the previous comment, check out the much touted original experiment. They proved fringing, and nothing else. Check it out (hint: It's not on utube)
Not my joke but worth posting on the matter:
Heisenberg and Schrodinger are speeding down the highway when they are pulled over by a cop. The cop strolls up to the window as Heisenberg roles it down.
The cop asks, "Sir do you know how fast you were going?"
Heisenberg's, "No, but I know where I was."
The cop decides to search the car.
He opens the trunk and says, "Sir do you know you have a dead cat?"
Schrodinger say, "I do now!"
This is not the Heisenberg uncertainty principle. The principle is due to the wave nature of matter. I can illustrate it this way. Suppose you have a wave with well defined momentum. Due to de Broglie, you can determine its frequency which is also uniquely defined. So you can represent that state as a sine wave. But where is the wave then? Well it's everywhere. You have complete uncertainty about you position. It's just functional analysis. Nothing to do with this, it's absurd.
A stationary photon has no mass. This is known as the rest mass, which is 0. However, photons do have a mass when moving (momentum) and contain a quantum 'package' of energy. This is due to the way in which the electric and magnetic field oscillate, following the photon's creation. A photon wouldn't exist if it wasn't moving and didn't have energy.
Basically, photons do have mass in the form of the quantum energy level they have, which differs depending on the frequency.
I have a question. You used a lightwave to measure the electron. But if we use the wave-particle duality model light is a particle or pocket of a wave (photon), so how come you represented it with a wave? Also according my very limited understanding of QED electrons and protons interchange photons. I have always understood Heisenberg's uncertainty principle as if you use a photon to try to measure the position of an electron that photon will interact with the electron thus changing its position.
of *course* photons have a mass...all energy has an associated mass (that's the meaning of e=mc^2). you are thinking of the concept of "rest mass" which, since a single photons is *never* at rest in the reference frame of any observer, does not exist.
thus, the mass of the photon is not "kind of equivalent" to e/c^2, it is precisely equivalent...but the better derivation comes via the relativistic e-p equation e^2-p^2c^2=m_0^2c^4, where rest mass m_0 can be set explicitly to 0.
My belief is that at the current level of science, we are not able to 'measure' the size and mass of some 'particles'. another reason could be that our understanding of 'particle-wave duality' is incomplete to explain exactly what particle or wave is. you must understand that our science has always been to quantify all matter with existing SI units.
And what would happen if 2 sources were used, one source half the wave length of the other?
I think Bruce is correct about the interpretation of the uncertainty principle. A particle in principle does not have an exact position and momentum. This is independent of whether we can meaure it or not. The debate about this interpretation is still going on today.
here you actually speak about Heisenberg measurment-disturbance relations and not about uncertainty principle...it's not your fault, that misconception is something most of people learn in school...these two are nto the same, you can check wikipedia on uncertainty relations...
What would happen if you correlated the value of Lambda, to to the mass of the object? And then just plugged that into the math?
If E^2=(mc^2)^2+(pc)^2 then why does it cancel to E=mc^2 implying that p=0 and therefore m=0 if the particle (massless and speed c) has momentum of h/ lamda (sorry no key for that symbol). Basically why does the mass=0 in p=mv but not mc^2.
err, i mean rest mass does not exist for a single photon, or rest mass = 0 for any system consisting solely of a single photon or collection of photons with the same exact same momentum.
Utterly wrong. The position and the momentium of a particle are Fourier transforms, one of the other. That is the uncertanity. That's true if nobody is measuring anything.
so in quantum mechanics waves can have a momentum,but in classical mechanics,they cant? because they have no mass and since momentum=mass x velocity,momentum should be 0?
I'm confused🙇
∆x • ∆p (> or =) h
∆x•∆p (> or =) h/4π
Which is correct???
Hbar is also referred to as Planck's Reduced Constant
I'm confused because in my workbook as well as in wikipedia, I found that (delta p)(delta x) is greater than planck's constant h. Whithout being devided by 4π.
hi may I ask, can we not use the difference of light wavelengths to calculate the momentum of the electron? due to conservation of momentum
what happened to C=lambda x nu.... with the constant as like.. 2.996x10 to the -8th.... -dies- i'm gonna die in my chem test..
yes, otherwise you would not have any momentum, and you wouldn't transfer the momentum, which is needed to push it away.
I'm sure you will have to take a Quantum Chemistry course in your undergrad years. Your prof will explain it in details.
LOOOOL HEISENBERG, Walter White is that you behind the camera
P.S google Walter White if you dont get it
how do I get the uncertainty in velocity if I have 15 m as the uncertainty in position and a mass of 2025 kg
I certainly dont have the knowledge to understand this yet I feel like if I did then I would already know this
Thank you so much for taking the time to do these.
The reason we can not measure (know) velocity and momentum of an electronat at the same time is not because of size of it and uncertanity of measurement but because of the uncertain nature of the electron.
Breaking Bad brought me here haha
Im learning chemistry right now. Does this relate to the positioning of electons within an atom?
wait i thought the equation was (delta x)(delta p) is less than or equal to planck's constant/4pie
If we cannot measure momentum and position at the same time, why not just minimize the time between the measurement of one and the measurement of the other such that we approximate simitanaeity?
3:23 I understand you seek to simplify, but I think it would still have been better if you had used :
E² = p²c² + m²c^4
obviously giving E = pc for m=0
Although unlike "E=mc²", the above formula may not be known by "everybody on earth", it would still allow you to avoid "confusions" such as :
-(remark mentionned by someone else in one of the comments below) : using on photons a formula that would normally represent "rest energy"
-(my personal remark) : using CLASSICAL momentum on something going at the speed of Light !..
Your way of deriving it obviously works, and thus can be seen as a "nice trick to find it back", but it doesn't seem rigorous at all :/
...Nice video though :)
Heisenberg's Uncertainty Principle is soo uncertain , even I don't have the certainty, that is actually 100 % certain, ..you know ,..that a particle ( we were talking about particles aren't we ? ) ..is were it's supposed to be , even If a I look at it with certainty, cause if I look ..or try to measure it , it will move......somewhere.............uncertain.....end of history :P
Does anyone use solid light as a measurement tool? And, if not, why?
Great lecture
Beautifully explained!
but why does the light bend when it passes through the slit in the first place?
For the single slit experiment, wouldn't you get similar interference for integral multiples of lambda? This would imply that there would be fainter fringes extending on either side of the main band. Or am I missing something?
How do we know that we need to send out a wave to 'try' locate the electron, what do we expect when the wave hits the electron ? is this anything related to the compton effect ?
Does the light (or electrons) passing through a slit roughly form a bell curve, and is it subject to probability?
I have observed some of your lectures.They are really great.Tribute to you and your attitude of spreading knowledge.
Why didn't my engineering school explain it this way. Now it makes perfect sense.
how can a wave a momentum when it doesnt have a mass?
Starting at 06:20 you say that light coming from the two edges of the slit meet up at some specific point on the screen where they will interfere destructively because of the phase difference. What I do not understand is when you draw that perpendicular line and assume that starting from the perpendicular line the the distances from the edges to that specific point on the screen are the same. But one of the distances is the hypotenuse the other is not. What am I missing here?
The equation E^2=(mc^2)^2+(pc)^2 is the complete equation but it does not cancel to E=MC^2. The E=MC^2 we usually talk about is just an approximation.
The reason is because in first part we have (mc^2)^2, this basically result in c^4 and we get a ridiculously large number compare to (pc)^2, which only has c^2.
Even if we omit (pc)^2 the result won't have significant changes in most cases. Thus we can approximate the equation to E^2=(mc^2)^2 and the ^2 cancel each other and we get E=MC^2.
How can you know that they aren't defined? If they get defined as soon as you measure them or if you would just measure a defined position and momentum, that should give the same test result, wouldn't it?
Even Heisenberg used the Observer Effect to demonstrate the Uncertainty Principle and made it easier to confuse both.
DrPhysicsA pointed out it is a fundamental property of wave matter in the end of the 2nd part.
You are right, the video try to explain the uncertainty principle from a more classical view, i.e, from the Copenhagen interpretation. As you have just mentioned the Uncertainty Principle is a fundamental description of an objective reality, and this is not so easy to understand.
Best video I learned more here than any other video I've ever watched.... the wave size matching the object size hit home..ha' I am a slow learner not trained in math this illustration was easy for me to understand
Yet, if we assume based on Higgs, that there is no mass for the particle (boson) (or it is mass-less), how does that effect the math?
Glad to find explanations I can follow. Sufficient to prompt questions in my mind. An example you use is an electron of size ~10^-18m. Is this size is real? Or is it due to the uncertainty in position of a point like particle?
you are right, most of people mix these two.
I don't understand why a higher wave length can make light pass through the electron instead of hitting it
Great Video
But I will suggest you to make vids on a board
It will save paper
Btw great explaination
:)
Continue with your vids :D
Hi, how to you record this video, i mean how you position camera or phone over the paper you write? this is the most engaging way to give a lesson on the web.
I think it would clarify the definition of delta x by explaining 'how' we measure: we measure a position as x +- delta x and momentum as p +-delta p.
Shite, you sound like Richard Dawkins.
Just fucked up my physics exam by misunderstanding 2 questions. It's nice to come here and learn something to feel like i'm not useless. I never knew the geometrics of the huygens-fresnel principle. I just knew the formula.
Hey lelouch !
Well hello there.
You speak like we know what amplitude of the light wave means, we dont'!?
amplitude is how high the wave is
But if you wanted to use light to measure an electron, why not use solid light as a measurement tool?
Thank you Dr physics, I'm a freshman in high school and I love your physics videos. Keep up the good work.
THANKSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
if you need support, just put down a thing where we can donate we would love too :)
It could be better if you took the slit's axis in the x direction. 😆
directions weren't clear enough...got my girlfriend stuck in a microwave
So much maths :( Can you teach me maths so I can pursue my dream? :(Thank you for a good video
good to know about how many dimentions we are talking about. the HUP only works for a 4D environment or could it be true for as many other dymentions we can have? Is it ok for Kaluza-Klein 5 dimentions architecture? Does it works for a Ricci tensor configuration?
For understanding HUP at the kaluza klein level , you might should first take into account scale at which you're talking about the strings existing , you've first got to sort the dimensionality of strings as string theory is the only present theory giving a viable explanation of kaluza klein theory