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).

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    $\begingroup$ 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. $\endgroup$ – Luboš Motl Sep 3 '12 at 9:37
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    $\begingroup$ 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. $\endgroup$ – Luboš Motl Sep 3 '12 at 9:40
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Sep 3 '12 at 9:42
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Sep 3 '12 at 10:36
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    $\begingroup$ @Nick: Listen carefully to what Lubos is saying. Most quantum states are not eigenstates of either the position or momentum operator, and no quantum state is an eigenstate of both. (You can easily check this.) So at the moment when you measure the position (meaning that you've put the particle into an eigenstate of the position operator) it has no well-defined momentum (i.e. it cannot be in an eigenstate of momentum). There is no such thing as a "position probe" and a "momentum probe" that can be used simultaneously. If there were, the corresponding operators would commute. $\endgroup$ – WillO Feb 7 '14 at 6:16

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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    $\begingroup$ 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. $\endgroup$ – David Z Sep 3 '12 at 22:20
  • $\begingroup$ 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. $\endgroup$ – anna v Sep 4 '12 at 4:39
  • $\begingroup$ @annav What is the physical meaning that the position of the particle would be spread out once it is measured by the detector? $\endgroup$ – Antonios Sarikas Dec 13 '19 at 22:08
  • $\begingroup$ When one measures something, there is always an experimental error. If one makes many measurements one can have a distribution of experimental errors. The uncertainty principle says that there are variables coupled quantum mechanically, in this case momentum and position, that introduce an inherent deviation in the "error" curve,. from the simple statistical errors. No matter how accurate one makes the instrument there will be Δx associate with it, dependent on the accuracy Δp of the momentum measurement. $\endgroup$ – anna v Dec 14 '19 at 4:50

The uncertainty principle of quantum mechanics is neither a statement about Fourier transforms, nor is it a statement about the "precision" of measurements as such (although the experimental measurement of the standard deviation is of course only accessible by repeated experiments).

For any quantum state $\lvert \psi \rangle$ and any observable $A$, we define the quantum mechanical expectation value as $$ \langle A \rangle_\psi := \langle \psi \vert A \vert \psi \rangle $$ because this is indeed the expectation value when you perform the (strong) measurement of $A$ on an ensemble of identically prepared states $\lvert \psi \rangle$.

The standard deviation of the observable for this particular state is then $$ \sigma_A(\psi) = \sqrt{\langle A^2 \rangle_\psi - \langle A \rangle_\psi^2}$$ and the reason we say that non-zero standard deviation means that $\lvert \psi \rangle$ doesn't have a well-defined value for $A$ is that "having a well-defined value" means that $\lvert \psi\rangle$ is an eigenstate of $A$, and this directly implies $\sigma_A(\psi) = 0$.

Now, for two observables $A,B$, one may derive the general uncertainty relation $$ \sigma_A(\psi)\sigma_B(\psi)\geq\frac{1}{2}\lvert\langle [A,B] \rangle_\psi\rvert$$ and the dependence of this on the state $\psi$ is crucial. Even if two observables do not commute, the expectation value of their commutator might still be zero on some states, i.e. they may have some common eigenstates, just not a basis consisting of those. In particular, non-zero expectation of the commutator implies that both standard devitations are non-zero, so the state we are examining is not an eigenstate of $A$ or $B$, i.e. it doesn't have a "well-defined value" for $A$ or $B$.

The reason this is not discussed for the particular pair of observables $x$ and $p$ is that their commutator is a multiple of the identity whose expectation value does not depend on the state. Additionally, yes, for $x$ and $p$ this coincides with the product of the width of a function of $x$ and the width of its Fourier transform, but this is not true in general - not all observables are Fourier related (in fact, $x$ and $p$ are pretty much the only ones - and the uncertainty relation is much more general than this.

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  • $\begingroup$ Any two operators whose commutator is a number times the identity would necessarily be related by the Fourier transform? Or the other way around, does being Fourier transform conjugates of each other imply anything of general value about their uncertainty relation? $\endgroup$ – Dvij D.C. Sep 9 '18 at 15:24
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    $\begingroup$ @DvijMankad Both. There is nothing special about $x$ and $p$ - you have the Fourier relation between all canonically conjugate variables (canonically conjugate means that the classical Poisson bracket resp. the quantum commutator of the variables is a multiple of the identity). Likewise, variables that are "Fourier conjugate" are also canonically conjugate. $\endgroup$ – ACuriousMind Sep 9 '18 at 16:25

Classical mechanics assumes that the state of a particle can be described, with arbitrary precision, by its position and momentum. That is in classical mechanics the state of the particle is, mathematically speaking, a phase vector. (And, the physical laws governing the particle's motion can be thought of as a trajectory in this vector space.)

But quantum mechanics uses a completely different notion of the state of a particle.

In QM, the 'state' is a weighted sum of a set of probability distribution functions. Each function in the set describes a possible configuration of the system, and the weights represent the probability that the system is in that configuration. To qualify as an allowed function, each function must be a possible solution to the wave equation (which is the physical law that defines the whole system.)

These functions are called the eigenfunctions of the wave equation. Now, a particular state can be described in more that one way as the weighted sum of more that one possible set of eigenfunctions, in the same way that a particular given three-space vector can be described differently using different co-ordinates and different basis vectors.

But, the particular set of eigenfunctions that locates the position of a particle in space precisely (dirac spikes) contains no information about momentum, and similarly the set of eigenfunctions that describe momentum precisely (non-localized waveforms) don't specify the particle's location in any way.

Other sets of eigenfunctions between these two extremes might be chosen as a basis to describe the state, but the total product of the information about location and momentum is always capped.

The uncertainty principle is a consequence of how 'state' is defined in quantum mechanics, and applies to any description of any particle. It really has nothing to do with measurement at all (or the statistical properties of aggregates of particles.)

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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 $2\pi\hbar\;\; 4.135667516(91)×10^{−15}~\text{eV·s}$ . 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: $p \pm \Delta p.$ 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.

Edit after thinking on comments:

At the mathematical level the Heisenberg Uncertainty Principle ( HUP) is directly relate to the commutation relations. In turn this operator relationship imposed on the wave function will give a value which will depend on the probability function given by the wave-function squared.

Now in classical mechanics we have a very specific probability distribution which is called the Gaussian and which is the distribution of the probability of errors in a measured value, if the error is random, not systematic. When one uses the symbol sigma one is immediately referring to a Gaussian probability distribution.

The probability distribution from a particular wave-function is generally not a Gaussian. It can be a wave solution, it can be an exponential decay and depending on the complexity any one of a large number of functions which cannot in any sense be approximated by the behaviour of the Gaussian over the interval of the variables of $x$ or $p.$

Therefore the mathematical use of $\Delta x\; \&\; \Delta p$, an interval on the variable, is a better description of the truth if one enters into solutions of equations.

At the same time it should be stressed that the HUP is an envelope of the result possible from detailed solutions, without going to the trouble of solving equations and imposing boundary conditions. Thus $\Delta x\; \&\; \Delta p$ are a much more accurate description of the probability nature of QM. For example, given a central $x$ , the HUP says that without detailed solutions, the probability of finding the particle in a point within the interval $\Delta x$ is unknown (within the bound). A Gaussian would give higher probability for the central value.

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    $\begingroup$ 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. $\endgroup$ – David Z Sep 3 '12 at 22:23
  • $\begingroup$ @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. $\endgroup$ – anna v Sep 4 '12 at 3:00
  • $\begingroup$ 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. $\endgroup$ – David Z Sep 4 '12 at 3:51
  • $\begingroup$ @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. $\endgroup$ – anna v Sep 4 '12 at 4:01
  • $\begingroup$ 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. $\endgroup$ – Nick Sep 4 '12 at 4:31

The particle is described by a wave that is spread out in position and momentum space. It is not possible to produce a wavepacket that does not respect the HUP. You can make such wavefunctions undergo interference and when it's undergoing interference the square amplitudes don't respect the calculus of probability, for an example see


As a result it doesn't make sense to say that all the wavefunction is doing is representing a probability distribution. Since all measurement instruments are made of particles that also respect the HUP there are no measuring instruments that can record both position and momentum with arbitrary accuracy. The fact that things around you don't look spread out is due to the fact that they are spread out on a scale that is too small for you to see in everyday life, and to more complicated issues like decoherence.

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