# Why is the Heisenberg uncertainty principle stated the way it is?

I spent a long time being confused by the Heisenberg uncertainty principle in my quantum chemistry class.

It is frequently stated that the "position and momentum of a particle cannot be simultaneously known to arbitrary precision" (or any other observables $[A, B] \neq 0$).

This made no sense to me -- why can't you measure both of these? Is my instrument just going to stop working at a certain length scale? The Internet was of little help; Wikipedia describes it this way as well and gets into philosophical arguments on what "position" and "momentum" mean and whether they really exist (in my opinion, irrelevant nonsense that has no effect on our ability to predict things).

Eventually it was the equation itself that gave me the most insight:

$$\sigma_x \sigma_p \geq \frac{\hbar}{2}$$

Look at that – there's two standard deviations in there! It is impossible by definition to have a standard deviation of one measurement. It requires multiple measurements to have any meaning at all.

After some probing and asking around I figured out what this really means:

Multiple repeated measurements of identically prepared systems don't give identical results. The distribution of these results is limited by that formula.

Wow! So much clearer. Thus $\hat{r}(t)$ and $\hat{p}(t)$ can be known for the same values of $t$ to as much precision as your measuring equipment will allow. But if you repeat the experiment, you won't get identical data.

Why doesn't everyone just state it that way? I feel like that would eliminate many a student's confusion. (Unless, of course, I'm still missing something – feel free to enlighten me should that be the case).

EDIT: This post was at +1. Who downvoted me? I took a while to write out my question clearly and made sure it followed the guidelines on here.

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The uncertainty principle is often stated in the form you started with because it's the most accurate yet universal way to state it. In particular, your "version of the principle" directly contradicts the uncertainty principle and it is therefore wrong. No, $r(t)$ and $p(t)$ cannot have well-defined values at the same moment, not even in principle. It's because $r$ and $p$ are really identified with operators that satisfy $rp-pr=i\hbar$ and no actual numbers $r,p$ can obey this equation. It follows that $r(t)$ and $p(t)$ can't be equal to two particular numbers. – Luboš Motl Sep 3 '12 at 9:37
It is true that in general, repetitions of the same experiment - with the same initial state - will yield different, partly random outcomes (and quantum mechanics allows us to predict all the relevant probability distributions). But this observation isn't something that allows you to circumvent or deny the Heisenberg uncertainty principle; on the contrary, the randomness and "not strict reproducibility" of the experiments are consequences of the uncertainty principle. – Luboš Motl Sep 3 '12 at 9:40
The claim "one gets non-unique, fluctuating predictions for $x$" is the very same thing as saying that $\Delta x\gt 0$: the probabilistic distribution isn't peaked, it has a width. And the reason why $\Delta x\gt 0$ is that $\Delta p\lt \infty$ as well as $\Delta x\cdot \Delta p \geq \hbar/2$ (the latter is the uncertainty principle). One may also exchange $p$ and $x$ and say the thing in the opposite way. The Heisenberg inequality itself says that $p,x$ can't be both sharply determined at a given moment. – Luboš Motl Sep 3 '12 at 9:42
One more comment: the fact that $\Delta x\gt 0$ which means that there is an uncertainty in $x$ isn't associated with any measuring apparatus. It's the whole point of the uncertainty principle that it is a universal principle in physics that no measuring apparatus can overcome. – Luboš Motl Sep 3 '12 at 10:36
Thanks for the replies.  I believe I am still somewhat confused in terms of what this means for experiment.  Sorry for the really absurd example here, but: if you have a hypothetical digital "position" probe and a digital "momentum" probe and the position probe displays the number 4.5293029086467... then what will the momentum probe be displaying at the same time?  I mean, it can't just stop working once the position probe shows enough digits.  It's got to show some value, right? – Nick Sep 3 '12 at 15:49
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## 2 Answers

Nick, Don't be surprised that this is confusing. There are a lot of concepts intermixed in the discussion of the uncertainty principle that are frequently not clearly understood and are intertwined unintentionally.

Although one often sees that these are stated in statistical terms, the standard deviation does not directly require multiple observations of a sample to understand. Traditional statistics does rely upon repeated sampling in order to develop a standard deviation, however in quantum mechanics the idea is more closely associated with properties associated with the Fourier transform.

To understand the Fourier transform one must first understand what a Fourier series is. The hyperlink will take you to a discussion about the Fourier series as it relates to sound. Starting at about minute two you see a representation of a saw-tooth like wave form. When they show you in the video how the saw-tooth like wave has many components, those components are determined by performing a Fourier transform. In many cases, they transform time series functions into frequency functions (which is directly proportional to energy) but the transform is also applicable to situations where one is transforming position into momentum.

Essentially what happens, is that if one wants to have complete certainty in the value of momentum (or energy), one must look at the entire position (or time) spectrum. In other words, a definite position, when transformed into the momentum domain, requires the entire momentum domain. If one allows a little uncertainty in the position, one does not require the entire momentum domain.

This relationship can be well defined as it relates to Fourier Transforms. This is the real source of the uncertainty principle, and does not require a statistical interpretation to understand.

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No objection to most of this, but I would say that the observable consequences of having a particular wavefunction only become clear with the statistical analysis of a large set of measurements. That's where the statistical connection comes in. If you're just doing mathematical manipulations on the wavefunctions, then sure, no statistics need be involved. – David Zaslavsky♦ Sep 3 '12 at 22:20
Thanks David, definitely agree with the point! – Hal Swyers Sep 3 '12 at 23:12
I have given an example down to the bare basics of HUP. One particle, no statistics, in a magnetic field. Its circle can be predicted by construction with accuracy. The HUP says that that I cannot see in  a detector a more accurate position than it (the HUP)  allows, this means that the position I find will deviate from the predicted circle within HUP. I mean that it will be equally probable for the measured point to deviate at the edge of HUP as at the center of the interval. I wonder if anyone has done the experiment. – anna v Sep 4 '12 at 4:39

I will turn my comments into an answer.

Quantum mechanics is the underlying stratum of nature but it dominates the microcosm, of dimensions commensurable to hbar 4.135667516(91)×10^−15 eV·s . Very small numbers. The Heisenberg Uncertainty Principle (HUP) does not apply statistically, it applies on individual particles, be they elementary, molecules or phonons etc in solids, individual "particles".

We are not talking of accumulating statistics and statistical errors.Half of your misunderstanding comes because of those sigmas in the formula you found, since the sigma symbol is associated with the standard deviation.

The delta symbol is more appropriate because it has nothing to do with statistics, it is the mathematical range of the value: p +/- Delta(p). If one wants to constrain the momentum within this range, then the measurement of position is constrained by the Heisenberg uncertainty principle. So the better one knows momentum the worse one knows position, and vice versa.

Probes are macroscopic and used statistically . When we talk about quantum mechanical quantities we have to address single individual particles:

Take a single ionized molecule, put it in a magnetic field so that the circle it will make will give you the momentum p with designed accuracy delta(p). The HUP tells you that any position detector you devise for the same particle, for example an emulsion film, will only be able to give you a delta(x) for the same particle within the HUP constraint.

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In my experience this interpretation of the uncertainty principle (as a constraint instead of a standard deviation) has been responsible for more incorrect conclusions than any other aspect of it. Yes, it works, but only because the wavefunction of a constrained particle has a standard deviation of the same order as the size of the constraint. – David Zaslavsky♦ Sep 3 '12 at 22:23
@DavidZaslavsky My point is that when you call it a standard deviation people start looking at classical statistics, which defeats the point . It is not classical statistics that defines the delta but the commutators. – anna v Sep 4 '12 at 3:00
I'm not sure I see what you mean... the $\sigma$ in the uncertainty principle is nothing more than the standard deviation of the probability distribution on eigenvalues of the relevant operator. That's a (classical, I suppose) statistical quantity. – David Zaslavsky♦ Sep 4 '12 at 3:51
@DavidZaslavsky and do you think it has a gaussian distribution, as it would if it were a real sigma? Or is it an interval of undefined values for that variable? I am afraid that this is how the thinking about underlying classical explanations of QM start. – anna v Sep 4 '12 at 4:01
In fact, my textbook (Quantum Chemistry by McQuarrie) actually used the words "standard deviation" and "normal distribution". Maybe I'm not the only one who is confused. If authoritative sources all say different things, it's no wonder nobody grasps this concept. – Nick Sep 4 '12 at 4:31
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