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Heisenberg imcertainty principle (get it!)

  1. I dont understand why it is so hard to find the exact position and velocity of orbiting electron. And also, why would we want to know it, if it is always moving? im not trying to disprove it or anything, so dont make fun of me, i am an uneducated peon
  2. jcsd
  3. whats your idea of finding both the exact position and velocity of the electron?

    how do you find the location of your keyboard? shine light on it (lamp or diffuse sunlight through the window) what happens if you try do the same for something tiny like an electron ?
  4. you should do that then, just prove the heisenberg uncertainty therom but just looking at the orbiting electrons with your own eyes
  5. when you look at your keyboard, the photons of light that 'interact' with it (and ten hit our eyes and give us what we percieve as colour) are much much smaller than the keyboard.
    but if we are trying to see an electron using a photon..
    the wavelength ("size") of an photon is alot larger than that of an electron, what implications does this have?
  6. For complete understanding of Uncertainity Principle , check out :


    TOPIC#2 on the above site is Heisenberg's..
    Last edited by a moderator: Jun 14, 2005
  7. ZapperZ

    ZapperZ 29,892
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    I would not recommend the site listed above. It perpetuates the misconception of the HUP that I have written about[1], that it is the uncertainty in a single measurement. I have seen this mistake repeated several times within the past week on here.

    The HUP is NOT about the uncertainty in INSTRUMENTATION or measurement. One can easily verify this by looking at HOW we measure certain quantities. It is silly for Heisenberg to know about technological advances in the future and how much more accurate we can measure things. This is NOT what the HUP is describing. The HUP is NOT describing how well we know about the quantities in a single measurement. I can make as precise of a measurement of the position and momentum of an electron as arbitrary as I want simultaneously, limited to the technology I have on hand. I can make improvements in my accuracy of one without affecting the accuracy of measurement of the other.

    What the HUP is telling you is the difference between a classical system and a quantum system. In a classical system, if you have a set of identical initial condition, and you measure ONE observable, and then you measure another observable, you will continue to get the SAME value of that 2nd observable everytime you measure the same value of that 1st observable. The more accurate you measure the 1st observable, the more accuract you can predict the value of the 2nd observable the next time you want to do such a measurement. The only limitation to how accurate you can determine these observable is the limitation to your measuring instruments. But these limitations do NOT scale like the HUP. You don't make one worse as the other one becomes better, because these are technical issues and are not related to one another.

    On the other hand, in a quantum system, under the IDENTICAL initial conditions, even if you measure a series of identical values for the 1st observable, the 2nd observable may NOT yield the identical result each time. In fact, as you narrow down the uncertainty of the 1st observable, the 2nd observable may start showing wildly different values as you do this REPEATEDLY. Therefore, unlike the classical system, your ability to know and predict what is going to be the outcome of the 2nd observable goes progressively WORSE as you improve your knowledge about the 1st observable!

    Again, it has NOTHING to do with the uncertainty in a SINGLE measurement! It doesn't mean that if you measure with utmost accuracy the position of an electron, that that electron momentum is "spread out" all over the place. This is wrong! I can STILL make an accurate determination of that electron's momentum - only my instrument will limit my accuracy of determining that. However, my ability to know what its momentum is going to be the NEXT time I measure it under the idential situation is what is dictated by the HUP!


    [1] [11-15-2004 09:26 AM] - Misconception of the Heisenberg Uncertainty Principle
    Last edited: Jun 14, 2005
  8. time is also an observable so why do we always take Δt= 0 while
    Δx*Δp >= hbar/2 ?
  9. ZapperZ

    ZapperZ 29,892
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    Who is this "we"?

    You will note that in typical school problems, you often work with a solution to the time-INDEPENDENT solution. It doesn't mean that the uncertainty in the time period is zero. It means that for that case, it is irrelevant since the description does not contain any time dynamics.

  10. I've a simple answer, hope u won't consider it naive...

    Most of the working theories right now weren't based on solid proofs but because we needed them, and some observations needed explanations...Theorists make theories, without confitmations, but we assume they r correct untill they r proven wrong, technology gives us the chance to make sure that we r on the right track, we can't yet find the electron and the uncertainty principle is what really works right now...

    See u when they find out something strong that would give us the opportunity to observe electrons..
  11. ZapperZ

    ZapperZ 29,892
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    What are "solid proofs"? Would the statment that says "IT WORKS" be considered as "solid proof"? How about if I point to you your modern electronics? Would that be considered as "solid proof"?

    Most people forget that the HUP is a CONSEQUENCE, not the origin, of quantum mechanics! To find a problem in the HUP is to find a problem in QM. And unless people also forgot about the centenial year of QM in 1999, let me remind you that there was an almost universal acclaimed by physicists that QM is THE most successful theory SO FAR in the history of human civilization!

    This means that if you think QM does not have "solid proofs", then other parts of physics suffer from even a worse level of lacking of solid proofs!

    Please keep in mind that NO part of physics is considered to be accepted and valid until there are sufficient experimental/empirical agreement! In fact, many theories and ideas originally came out of unexpected experimental observation in the first place!

    There is no lacking of "solid proofs" for QM, and the HUP. Why this is even brought up here, I have no idea.

  12. "i" was thinking these as you said , until i see the word 'simultaneous' for measurements in the defn of HUP. if we say nearly simultaneous then i think there will be no problem(i hope so...)

    Also there are different explanations for HUP, and maybe this the problem about understanding it. at first i was thinking i get it , but it didnt take a long time for me to confuse (because i am a beginner only)
    Zz 's article is very helpful but if anyone can send a sketch of proof for HUP it will be more clear

  13. ZapperZ

    ZapperZ 29,892
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    I don't understand. You want a "sketch of proof" for the HUP? What is this?

    I think every student has either done, or seen the diffraction from a single slit. To me, this is a VERY clear example of the HUP! It just happens that we typically use wave description of light to account for such effects. But with the photon picture, the identical diffraction pattern can be directly obtained and the spreading is a direct consequence of the HUP!

    More? The deBoer effect that is very pronounced in noble gasses is a direct consequence of the HUP. This leads to a correction to the internal energy (and thus, the specific heat capacity) of the gasses at very low temperatures. Only via taking into account such corrections can one obtain the experimental values!

    But I think people pay waaaaay too much attention at disecting the consequences and forgetting the principles that CAUSE such consequences. The fact that this came out of "First Quantization" principle of QM that is based on [A,B] operations of two non-commuting observables is less understood by many who do not understand the formalism of QM. This is a crucial part of elementary QM with which a whole slew of consequences are built upon!

  14. what i wonder is the mathematical part :

    [A,B]= C --> ΔA * ΔB >= |<C>|/2

    is it just a thm about standart deviation?
    Last edited: Jun 14, 2005
  15. The second part answers to the first part of your post.

    In QM, we may define the position of a particle. We may also define the momentum of a particle. However momentum is not the velocity of the particle. Many people in this forum always mix the momentum with the velocity (due to their equality for average values: i.e. Erhenfest Theorem) and tend to make incorrect deductions.
    QM tells one thing: we cannot associate a classical path to a particle, hence we cannot define a couple (position, velocity) to the particle.
    The HUP property applied to the (position, momentum) observables just highligh this fact: we cannot find a particle where both position and momentum have "defined" values equal to their mean values (i.e. through the Erhenfest Theorem, they have a classical path if is the case).

  16. Yes.

  17. What sort of technology do we have to achieve this?
  18. my (rather naive) understanding of it is that in a classical system:

    [tex]xp - px = 0[/tex]

    but accord to HUP:

    [tex]xp - px \neq 0 [/tex]

    so that measuring the position and then measuring the momentum, is not the same as measuring the momentum and then measuring the position. infact, they will always differ by [tex]\frac{ih}{2\pi}[/tex]

    position and momentum are not seen as real values, but as non-commutative operators. you can derive the schrodinger wave equation in a straightforward manner from this.

    (this is one interpretation).
  19. No, that's not correct. The HUP doesn't says something about only one simultaneoulsy measurment. It says something about a serie of simultaneously measurements (always the same conditions). You see?
    If you make 1000 simultaneoulsy measurments (position and momentum) and you measure the position at each experiment exactly then you will get at each measurement of momentum a completely different value.
  20. Umm, how can momentum be equal to velocity ? Ehrenfest theorem is about equality of average QM momentum, which is defined as [tex]-i \hbar \vec \nabla[/tex] and classical momentum [tex]m \vec v[/tex]. Or generally it shows that average values of QM operators are equal to corresponding quantities in classical mechanics (I guess that's what you had in mind).


    HUP actually does say [tex]\left< (\Delta p_x)^2 \right> \left< (\Delta x)^2 \right> \neq 0[/tex]. What you wrote, are commutation relations for operators, not a HUP (however, it's used in derivation of HUP). And you cannot get deltas by taking ONE particle. Let's say you have an instrument which determines position and momenutum up to 10 decimal places [in some units]. And let's say measurement of momentum of electron always gives one value, eg. 1.0000000001 [in some units]. Then you measure it's position 1st time and get let's say 0.0000000044 [some units]. A set of repeated measurement of position (with momentum set to 1.0000000001) will get you just random results, like 50.3243243212, 0, 13.1313131313, -400000.0000000001, etc. (but each of these measurements will have high precision, and that's the one that depends on the instrument itself!). However, calulating the average of the square of deltas of random numbers like that (you see, I HAVE to have more than one measurement to do average!), gives us a big number. This number appears in HUP.

    When I let particle momentums have some distribution of let's say between 1.0 and 1.1 , the numbers I get for position will not be so random - they will have a peak value around some number. If I don't control momentum at all, but let particles pass a very small hole (that way I'm controling position) and measure momentum afterwards, I will get random results.

    Hope this helps! :smile:
  21. Ok, let's explain the "due to their equality for average values" in my previous post.
    <P>=m.d<X>/dt= m<V> if no em field (case H=p^2/2m+V(q)).
    => implictly assuming m=1 units, we have <P>=<V> QED.

    (I thought it was clear enough, but your post showed I was wrong with this assumption, now I hope it is clearer).

    However, for a given relation V=dX/dt on operators (e.g. V=P/m), we have the eigenvalue relation v=dx/dt iff [V,X]=0 (if the operators are sufficently "gentle").

    If V and X have not the same eigenbasis (in other words they do not commute: [V,X]=/=0) => the relation v=dx/dt is no more valid for the eigenvalues => we cannot associate a classical path to a particle.

    HUP just reflects this fundamental property of operators.

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