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Please forgive me if I'm wrong, as I have no formal physics training (apart from some in high school and personal reading), but there's something about Heisenberg's Uncertainty Principle that strikes me as quite obvious, and I find it strange that nobody thought about it before quantum mechanics development began, and still most people and texts explain it in quantum mechanics terms (such as citing wave/particle dualism, or the observer effect)... while actually it should appear blatantly obvious in classical mechanics too, at least regarding the position and momentum variables, due to the very definition of speed.

As everyone knows, the speed of an object is the variation of its position over an interval of time; in order to measure an object's speed, you need at least two measurements of its position at different times, and as much as you can minimize this time interval, this would always create an uncertainty on the object's position; even if the object was exactly in the same place at both times, and even if the time was a single nanosecond, this still wouldn't guarantee its speed is exactly zero, as it could have moved in the meantime.

If you, on the contrary, reduce the time interval to exactly zero and only measure the object's position at a specific time, you will know very precisely where the object is, but you will never be able to know where it came from and where it's going to, thus you will have no information at all about its speed.

So, shouldn't the inability to exactly measure the position and speed (and thus the momentum) of an object derive directly from the very definition of speed?

This line of reasoning could also be generalized to any couple of variables of which one is defined as a variation of the other over time; thus, the general principle should be:

You can't misure with complete accuracy both $x$ and $\frac{\Delta x}{\Delta t}$

For any possible two points in time, there will always be a (however small) time interval between them, and during that interval the value of $x$ could have changed in any way that the two consecutive measurements couldn't possibly show. Thus, there will always be a (however low) uncertainty for every physical quantity if you try to misure both its value and its variation over time. This is what should have been obvious from the beginning even in classical mechanics, yet nobody seem to have tought about it until the same conclusion was reached in quantum mechanics, for completely different reasons...

Massimo
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    Related: http://physics.stackexchange.com/q/24068/2451 and links therein. – Qmechanic Mar 10 '14 at 14:02
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    @Qmechanic, thanks but I'm not talking about limits in our ability to measure things here. I just think the very definition of speed implies that you can't exactly measure it and position at the same time. – Massimo Mar 10 '14 at 14:06
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    @Massimo: so it doesn't seem normal to you that one can measure both the value of a curve $y = f(x)$ AND its tangent at its tangent? Is it because the problem is always presented as if we were doing a single measurement? What if one imagined simply sampling positions at regular time intervals such that Shannon's sampling theorem applies and THEN measure position and velocity at any point of the reconstructed trajectory with arbitrary accuracy? Just in case it is nor clear: even that strategy doesn't lead to statistically null variances in position and momentum. – gatsu Mar 10 '14 at 15:41
  • I think that reducing the time interval between measurements gives you a better value for both position AND velocity. After all, if you want to find the slope of a curve, you want to send $\Delta t\rightarrow 0$. – Jahan Claes Nov 04 '15 at 18:20
  • Of course you'll get a better (= more precise) measurement by reducing the time interval; but you can't reduce it to zero, otherwise you'll have no information about speed. Hence the intrinsic (however low) uncertainity of measuring both position and speed. – Massimo Nov 04 '15 at 20:00
  • Nothing in your post appears to have anything to do with the uncertainty principle. In particular, it appears to apply to classical mechanics as well as to quantum mechanics, and there is no uncertainty principle in classical mechanics. – WillO Nov 04 '15 at 23:57
  • @Massimo Yes, but this uncertainty can be made arbitrarily small in your formulation, just by making $\Delta t$ sufficiently small. The uncertainty principle, in contrast, gives a HARD lower limit. – Jahan Claes Nov 05 '15 at 04:27
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    @Massimo, I've had exactly this thought since the first time I heard about the uncertainty principle. At the very least, if there is no connection to these two different notions of uncertainty, it is a suspicious coincidence! – Tom B. Feb 09 '18 at 19:03

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The uncertainty principle says something a deeper than "it is impossible to measure both position and momentum to arbitrary accuracy". It says

1) The accuracy is precisely limited by $\Delta x \Delta p > \hbar/2$.

2) In fact, this is not a limit of our measuring procedure, but a limit of reality. If something has well-defined position, it does not have a well defined momentum, and vice-versa. In other words, it's not that we can't precisely measure the exact momentum; if an object has a well-defined position, it does not have a well-defined momentum.

Your argument doesn't really get at the real content of the uncertainty relationship, which is that there is no such thing as a particle with well-defined momentum and position.

Jahan Claes
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  • While this is indeed correct, I never said that the problem was in measuring both variables at the same time; I said, instead, something very similar to what you are saying: that the very definition of speed ceases to make sense if you reduce to zero the time interval between two measurements, and thus speed and position can't be both precisely determined at the same time, regardless of the accuracy of measuring instruments. – Massimo Nov 04 '15 at 18:59
  • What I was arguing, however, was that this should be obvious by the very definition of speed: if the speed is completely defined then the position isn't, and vice versa. – Massimo Nov 04 '15 at 19:01
  • @Massimo, I don't think that's true. Classically, we can measure both position and momentum arbitrarily well just by shining high-wavelength, low-energy light on an object, and deduce both the position and the speed. – Jahan Claes Nov 04 '15 at 19:39
  • @Massimo How about this: Let's say I classically measure the position of an object at $t=0$ to some arbitrary precision. Then I measure the position at $t=\Delta t$. I can then find BOTH the approximate velocity (by taking $\Delta x/\Delta t$) and the exact position (to some high precision). By decreasing $\Delta t$, I'm INCREASING the precision of my velocity measurement, and leaving my position measurement just as precise. – Jahan Claes Nov 04 '15 at 19:42
  • @Massimo this is slightly separate from my original point, but classically, velocity is only defined in the limit $\Delta t\rightarrow 0$. – Jahan Claes Nov 04 '15 at 19:53
  • @Jahan Claes - This is a nice answer, except for the opening sentence. I suggest changing "a little deeper" to "a lot deeper". It's your answer, so I won't make that change. – David Hammen Nov 04 '15 at 19:54
  • @Jahan Claes - Please re-read my question. As soon as we measure the position at two different times, we will know the object's speed, but we will not know its exact position anymore, because we'll be only able to guess what heppened between the two measurements. The only way to know for sure the position of an object is to take a single measurement, but this will of course not be enough to know its speed. Thus, either we precisely know the position, or we precisely know the speed; we can never exactly know both. – Massimo Nov 04 '15 at 19:56
  • But, again: what I was trying to say in my question is, this should be blatantly obvious even in classical mechanics. – Massimo Nov 04 '15 at 19:58
  • @Massimo - Stop rejecting calculus. You do not need to measure position at two different times to measure velocity. Radar guns, for example, don't measure a car's velocity by two different measurements of position. – David Hammen Nov 04 '15 at 19:59
  • @DavidHammen, I'm not rejecting calculus at all; but calculus only applies if we have a well defined function to describe how the position varies over time. If we are measuring an object's position and speed to actually understand how it moves, we can't assume it follows a a known differentiable function. – Massimo Nov 04 '15 at 20:05
  • @Massimo you know the position at time $\Delta t$ EXACTLY. And by shrinking $\Delta t$ you get better and better momentum as well! – Jahan Claes Nov 05 '15 at 02:22
  • @Massimo Even in your original formulation, we get both position and velocity more accurately if we make $\Delta t$ smaller. – Jahan Claes Nov 05 '15 at 03:38
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The uncertainty principle doesn't say anything about simultaneous measurements of a particle - that's just a myth which originated from Heisenberg's interpretation of it.

Let us first describe the basis of quantum physics and let's start with the most innocent looking object: the quantum state. We can see a quantum state as a prescription to prepare a system. It's a number of steps how to prepare your system (e.g. how to build a one-electron source, how to set it up in a vacuum, how to install magnetic fields and apply them to produce an electron with specific spin direction). In the literature, you'll hear something like "this electron is in the state..." - this is just an extreme short cut for saying: in our theory, we can define objects such as electrons and they have properties and we have an experimental procedure that produces results that seems to work exactly as we would predict if it were an electron of our theory.

The second step is a measurement. From classical physics, it is pretty clear what a position measurement is and a momentum measurement is only slightly more difficult. In quantum physics, we already have to be much more careful, but let's suppose we know what it means to measure position and momentum (for position, we can for example take a number of detectors in an array and when one detector makes "click", this tells us the position of the instance of the state we created). This implies that we have another set of rules that gives us some classical output to read of a screen and that we call "momentum" of the state or "position".

Now, suppose we want to have a very special state that is as localized as possible, i.e.: we prepare a state (or better: we perform the procedure defined by the state) over and over and each time we measure the position. Like this, we get a distribution of the positions - if our preparation is not very accurate, the variance of that distribution will be large, if it is quite accurate, it will be small. We go on and change the state (i.e. refine the procedure) such that ultimately, whenever we create an instance of the state and we measure its position, every time the same detector clicks. Let's assume we can achieve arbitrary accuracy, i.e. the detector is really small and it's really always the same detector that clicks.

And now, instead of measuring the position, let's measure the momentum of our perfectly localized state. What happens? We'll get some result, but when we redo the experiment, it'll be completely different. We do the preparation & measurement over and over again and the probability distribution for momentum we receive will have a huge variance.

Heisenberg's uncertainty relation tells us that regardless of what we do, this is the picture we must obtain. We can try to change the preparation procedure as much as we want and assuming we could work with arbitrary precision, we cannot define a procedure were both the position and the momentum density we obtain have a small variance.

This is contrary to classical mechanics: Let's suppose I want to do the experiment where I try to see whether two stones of different masses fall differently. Galileo supposedly did this in Pisa by throwing two stones from the tower. His "state" was: I take two stones and hold them directly next to each other. Then I let them fall. But this means that at the beginning of the experiment, he knew both the position (next to each other) and the velocity (zero) perfectly - otherwise the conclusion wouldn't make sense. Heisenberg's uncertainty relation tells us that this is not possible in quantum mechanics - it doesn't even really make sense to ask the question.

Martin
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  • By the way, Galilean parabole is very nice to consider energy time indetermination. – arivero Apr 14 '16 at 19:21
  • "And now, instead of measuring the position, let's measure the momentum of our perfectly localized state. What happens? We'll get some result, but when we redo the experiment, it'll be completely different." So what? Each measurement will result in a precise momentum at a precise position. – Tom B. Feb 09 '18 at 19:10
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In order to measure an object's speed, you need at least two measurements of its position at different times.

This is not the case. The radar guns used by police to determine if you are exceeding the speed limit do not use position measurements. They instead measure the frequency difference between the outgoing and reflected signals. No position measurements are required.

Conceptually, one could use a measurement from a radar gun to simultaneously determine position and velocity: Measure the frequency shift between the outgoing and incoming signals and measure the time difference between transmission and reception. The first measurement yields velocity, the second, position.

Shouldn't the Uncertainty Principle be intuitively obvious, at least when talking about the position and momentum of an object?

No, it's not. Classically, one could make measurement devices that simultaneously measure position and velocity to any desired degree of precision. I gave an example above. Even if two position measurements are used, simply making those position measurements ever more precise lets one use two position measurements closely-spaced in time to measure both position and velocity to any desired degree of precision. There are no limits to the precision of measurements of canonically conjugate variables in classical mechanics.

The uncertainty principle says that this is not possible. There are limits, specifically $\Delta x \Delta p>\hbar/2$. This is not just a limitation on measurement devices. It is much deeper than that. The uncertainty principle is a fundamental limitation of reality, as opposed to a minor constraint on measurement devices.

David Hammen
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Shouldn't the Uncertainty Principle be intuitively obvious, at least when talking about the position and momentum of an object?

No, not necessarily.

in order to measure an object's speed, you need at least two measurements of its position at different times,

A police radar gun can be used to measure the speed of a object at a single point in time.

It can also be used to measure its position in space at the same time.

Using Heisenberg's Uncertainty Principle:

$$ \sigma_x \sigma_\rho \geq \frac{\hbar}{2} $$

Which leaves a minimum accuracy of speed of a one tonne car measured to $1~\text{nm}$ accuracy at $5 \times 10^{-29}~\text{ms}^{-1}$. So classically for all intents and purposes one can measure a car to an arbitrary accuracy of both position and momentum at any point in time without invoking HUP.

This is made even easier when you assume that the measurement of position does not affect its position or momentum, which classically is true for the car, so you can measure them separately in any order.

user288447
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  • Well, of course you can measure both quantities at the same time, but only up to a certain degree of precision, however high. Never exactly. – Massimo Mar 10 '14 at 14:24
  • Of course, this becomes practically irrelevant for large (i.e. bigger than atoms) objects, just like most quantum physics. But it's the principle that counts. – Massimo Mar 10 '14 at 14:32
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    True, (your first comment) but that is the measurement error and it will exist for all measurements, the limit is not given classically except by the precision of the instruments. The HUP tells us that in the microcosm no matter how precisely we measure there exist pairs of variables, called conjugate, which are connected through the HUP indeterminancy, not measurement error. – anna v Mar 10 '14 at 14:33
  • I'm not talking about measuremt errors here. I'm talking about how speed is defined, as the variation of position over time; which implies that either you precisely know the position and thus can't know how it varies, or you precisely know its variation and thus you can't know its exact value. – Massimo Mar 10 '14 at 14:37
  • Of course what's intuitively obvious depends on your intuition, but it should at least be obvious that a delta-function looks nothing like a sine wave, and nearly obvious that it looks nothing like the sum of any small number of sine waves. So in that sense, yes, the qualitative outline of the uncertainty principle was certainly clear all along. It takes a little more work to quantify it. – WillO Mar 10 '14 at 14:39
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    @Massimo I assume you don't know calculus, because calculus works on the very principle you're trying to deny. Instantaneous speed is a concept in classical mechanics and is completely defined. One can exactly know what an object's velocity is at a given time if one knows exactly how the object is moving around that time. In quantum mechanics the object does not have a definite velocity, which is very different. – Robert Mastragostino Mar 10 '14 at 18:25
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    A radar gun can measure the speed of a object at a single point in time. Not true. A radar gun measures the Doppler shift of the reflected wave. That is, it measures the frequency of the reflected wave. But you can't define the frequency of a wave at a single instant in time. I'm not going to attempt the math, but I'm pretty sure that a radar gun effectively tells you some weighted average of the target speed over some finite interval of time. – Solomon Slow Nov 04 '15 at 17:58
  • @jameslarge - In theory, one could make a single photon radar gun and "simply" simultaneously measure the energy of the returning photon and the time at which it arrives. The energy measurement yields the car's velocity and the time measurement yields the car's position. Except it's not that simple. Energy and time are canonically conjugate variables. – David Hammen Nov 04 '15 at 19:34
  • @DavidHammen touché. :-) – Solomon Slow Nov 04 '15 at 19:41
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"So, shouldn't the inability to exactly measure the position and speed (and thus the momentum) of an object derive directly from the very definition of speed?"

Yes, if you are talking about instantaneous speed. There is no speed at a point because the definition requires two points.

No, if you are talking about average speed, which unsurprisingly is what everyone means when they are talking about speed in the macroscopic regime.

gregsan
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  • And thus, the very fact that speed needs to be measured as an average between two points explicitly forbids measuring it to an arbitrary level of precision; and this limit becomes increasingly relevant the smaller the objects and quantities involved become. – Massimo Mar 10 '14 at 14:45
  • Really, this limitation should have immediately become self-evident as soon as $\Delta t$ was used as a denominator... – Massimo Mar 10 '14 at 14:47
  • There is no speed at a point because the definition requires two points Except, if you know calculus, then you can define the speed at an instant in time. Assuming you know the position of the object as an function of time, then the derivative of the position function w.r.t. time gives you the instantaneous speed function. If the position function is analytic, then the instantaneous speed will also be analytic. – Solomon Slow Nov 04 '15 at 18:05
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Shouldn't the Uncertainty Principle be intuitively obvious, at least when talking about the position and momentum of an object?

No. You can not talk about the uncertainty principle without associating a wave to a particle according to the de Broigle hypothesis.

"A function (a wave) and its Fourier transform cannot both be sharply localized in time and frequency, respectively." A short pulse around moment $t_0$ appears like a vertical line on the screen of an oscilloscope but if you display its spectrum (calculated by the same oscilloscope if it is a smart one) you will remark that the spectrum covers a large part of the screen spreading over a broad range of frequencies.

If you do not associate a wave to a particle or to a macroscopic body the uncertainty principle does not make sense.

See also the following explanation:

Mathematically, in wave mechanics, the uncertainty relation between position and momentum arises because the expressions of the wavefunction in the two corresponding orthonormal bases in Hilbert space are Fourier transforms of one another (i.e., position and momentum are conjugate variables). A nonzero function and its Fourier transform cannot both be sharply localized. A similar tradeoff between the variances of Fourier conjugates arises in all systems underlain by Fourier analysis, for example in sound waves: A pure tone is a sharp spike at a single frequency, while its Fourier transform gives the shape of the sound wave in the time domain, which is a completely delocalized sine wave. In quantum mechanics, the two key points are that the position of the particle takes the form of a matter wave, and momentum is its Fourier conjugate, assured by the de Broglie relation p = ħk, where k is the wavenumber. Source

Energizer777
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Yes the argument that you can not measure the speed at single point is correct, but if you take a distance 'dx' and then measure the time 'dt', you will measure the momentum absolutely and position can be determined with the uncertainty of dx. Now you may divide dx by n and measure the time dt again you will get absolute momentum and even more precise position.

But how small you can go. classically you may make n tends to infinite and you will have no uncertainty in both momentum and position.

If you try to derive the uncertainty relation classically you can derive it but you can not define the lower limit, as soon as you apply the quantum considerations (wave particle duality) you can set the lower limit such that

$\Delta x$ $\Delta p$> $\hbar$/2.

hsinghal
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