# The Heisenberg Uncertainty Principle **URL:** \ **Category:** Factual Questions\ **Created:** [June 17, 2003, 10:17pm UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562 "2003-06-17T22:17:24Z")\ **Posts on this page:** 8\ **Page:** 2
**Author:** ![Donut](https://avatars.discourse-cdn.com/v4/letter/d/e19adc/32.png) [@Donut](https://boards.straightdope.com/u/Donut)\ **Post date:** [June 18, 2003, 3:34am UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/21 "2003-06-18T03:34:55Z")
I’m sorry for the rollyeyes (and the discount comment); at the time it seemed to me you were being intentionally disingenuous. I was, in retrospect, wrong about that. The reason I thought you were being disingenous is that the sentence you left you out (with no indication that you cut anything) does make a difference, at least for those who are trying to figure out what Feynman is talking about without having the book in front of them. However, I just noticed that I left out a sentence also, so that hardly leaves me in any position to judge. Anyway, what Feynman is talking about is sending a stream of particles through a slit from very far away, so all particles hitting the slit can be considered to have no vertical momentum (or at least an arbitrarily small vertical momentum). Any particles which happen to pass through the slit must have a certain vertical position, which we can make arbitrarily precise by making the slit smaller. So it may seem, on the face of it, that a particle passing through the slit has its position and momentum measured to arbitrary precision. The problem is that passing through the slit is a measurement (it tells us the vertical position) so it effects the momentum. Thus, as soon as we know the position to arbitrary precision we no longer know the momentum. The proof of this lies in the interesting diffraction pattern we get when particles pass through a slit. At no time in the experiment will we know both the position and momentum of the particle to arbitrary precision at the same time. I’m not going to try to explain what exactly Feynman is trying to say; I find the paragraph confusingly worded. ---
**Author:** ![Sqwert](https://avatars.discourse-cdn.com/v4/letter/s/a9adbd/32.png) [@Sqwert](https://boards.straightdope.com/u/Sqwert)\ **Post date:** [June 18, 2003, 3:41am UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/22 "2003-06-18T03:41:59Z")
Hi, I’m not quite sure exactly how you are proposing to defeat the HUP, but what I think you are suggesting is the following: (If I am mistaken, please tell me where I misunderstood your proposed experiment - or better yet, go into more detail with it) ASSUMPTION: You are planning on bouncing a photon off of a screen, and determining the following: (1) The position of the photon at the time of the impact as wherever the screen is marked. (2) The magnitude of the momentum of the photon at the time of the impact can be defined by knowing the wavelength of the photon used. (3) The direction of the momentum of the photon at the time of the impact can be extrapolated by measuring the angle of incidence after the deflection. If this is your plan, here is why I think it will not work. First off for those that don’t know, the minimum value for the HUP is a very tiny number indeed [h-bar] = 1.055 x 10^-34 Js. Some things to keep in mind that may cause you to miss your less than H-bar target. : MOMENTUM PROBLEMS : If the screen you use is an actual physical screen, then it will be made up of constantly moving molecules. The idea that the screen acts as a perfectly smooth perfectly flat surface would be a wrong assumption. Theoretically if you could achieve the temperature of the screen to be absolute zero, then the molecules, would no longer be moving. A second theoretical possibility is you might be able to create some perfectly flat electro-magnetic field as your “screen”, but then the impact wouldn’t leave a mark - thus completely removing the possibility of knowing the position the photon hit the screen at the time of impact. : POSITION PROBLEMS : Secondly, what will you use to measure the exact position of the impact between photon and screen. Even if you could track it to a specific molecule on the screen, you still have an accuracy problem. Because h-bar is very small, the precision you are expecting [1.055 x 10^-34] is much smaller even than the diameter of an atom [1x10^-8 cm.] ---
**Author:** ![Sacroiliac](https://avatars.discourse-cdn.com/v4/letter/s/ecb155/32.png) [@Sacroiliac](https://boards.straightdope.com/u/Sacroiliac)\ **Post date:** [July 2, 2003, 9:31pm UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/23 "2003-07-02T21:31:09Z")
From _Introduction to Quantum Mechanics_ by David J. Griffiths. > [@](#): > > Please understand what the uncertainty principle means: Like position measurements, momentum measurements yield precise answers—the “spread” here refers to the fact that measurements on identical systems do not yield consistent results. > > _Underlining mine._ From \* Understanding Quantum Physics\* by Michael Morrison. > [@](#): > > On the Heisenberg Uncertainty Principle > > To determine the accuracy of our results we would perform such measurements not just on one particle but on a large number of identical particles, all in the same state. Each individual measurement yields a value of x and a value for p[sub]x[/sub] (subject to experimental uncertainties). But the results of _different_ measurements are not the same even though the systems are identical. Understand that I’m talking about uncertainties due to the HUP not uncertainties due to experimental uncertainties ---
**Author:** ![AskNott](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/asknott/32/5790_2.png) [@AskNott](https://boards.straightdope.com/u/AskNott)\ **Post date:** [July 2, 2003, 10:32pm UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/24 "2003-07-02T22:32:11Z")
I have years of experience in uncertainty, but I’m grateful that it’s somebody else’s job to understand and explain Heisenberg’s Principle. My hat is off to you folks. ---
**Author:** ![Fuji\_Kitakyusho](https://avatars.discourse-cdn.com/v4/letter/f/f0a364/32.png) [@Fuji\_Kitakyusho](https://boards.straightdope.com/u/Fuji_Kitakyusho)\ **Post date:** [July 2, 2003, 11:18pm UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/25 "2003-07-02T23:18:34Z")
How can we be certain that Heisenburg was right? 😉 Grafitti: “Heisenburg may have been here.” ---
**Author:** ![Don\_Roberto](https://avatars.discourse-cdn.com/v4/letter/d/e47774/32.png) [@Don\_Roberto](https://boards.straightdope.com/u/Don_Roberto)\ **Post date:** [July 2, 2003, 11:32pm UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/26 "2003-07-02T23:32:37Z")
Sacroiliac, I’m not entirely sure what your point is, but here’s something to think about: Have you considered the fact that your measurement device is also subject to the Uncertainty Principle? In order to determine the position of your destroyed photon when it hit the screen, as well as its momentum, you must precisely determine the position and momentum of the screen as well. There’s no getting around the HUP. From my perspective, the HUP is fundamentally about measurement, so it doesn’t make sense to talk about it without regards to measurement. ---
**Author:** ![Joe\_Random](https://avatars.discourse-cdn.com/v4/letter/j/e8c25b/32.png) [@Joe\_Random](https://boards.straightdope.com/u/Joe_Random)\ **Post date:** [July 2, 2003, 11:44pm UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/27 "2003-07-02T23:44:10Z")
> [@](#): > > \*Originally posted by Sacroiliac \* > \*\*After the particle has reacted with the screen both the position and momentum are known to arbitrary accuracy. \*\* You seem to be conflating “known” with “measured”. You cannot simultaneously _measure_ position and momentum with arbitrary accuracy. I assume that someone smarter than me could come up with a way of _calculating_ one of those values with arbitrary accuracy _after the fact_, but that’s not the same thing. ---
**Author:** ![Sacroiliac](https://avatars.discourse-cdn.com/v4/letter/s/ecb155/32.png) [@Sacroiliac](https://boards.straightdope.com/u/Sacroiliac)\ **Post date:** [July 3, 2003, 1:24am UTC](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562/28 "2003-07-03T01:24:44Z")
_”I’m not entirely sure what your point is”_ More importantly what is Dr. Griffiths and Dr. Morrisons point? What I think they’re saying is if you perform a _ **destructive** _ measurement on a quantum particle your measurement apparatus will yield specific numbers for the particles position and momentum (say 3 for position and 6 for momentum). And these values are correct _within the experimental accuracy_ of your measurement system. Who’s to say they aren’t? And if you say they’re aren’t correct then tell me what the correct values are. The HUP doesn’t rear its ugly head until you perform this experiment on an ensemble of identically prepared particles. Then you’ll see the uncertainty relation. [Previous page](https://boards.straightdope.com/t/the-heisenberg-uncertainty-principle/182562.md?page=1)