The Heisenberg Uncertainty Principle is derived for two operators $\hat A$ and $\hat B$ as $$\Delta \hat A\ \Delta \hat B \geq \dfrac{1}{2}|\langle[\hat A, \hat B] \rangle|$$ where $\Delta$ denotes the standard deviation in a variable.

Taking position and momentum as $\hat A$ and $\hat B$ respectively, we get- $$\Delta \hat x\ \Delta \hat p_x\geq \hbar/2$$

Now, mathematically what this means is that if we prepare a large number of states $\ | \Psi \rangle$ and perform measurements of the position and momentum on them one by one, the RMS value of deviation from the mean for both $x$ and $p_x$ will show an inverse relationship with each other.

How then does this lead to a restriction on the individually measured values of position and momentum? How do we make claims such as the particle is restricted to this box so it can't have a zero momentum and so forth?

  • $\begingroup$ Related: physics.stackexchange.com/q/285290 $\endgroup$ Apr 21, 2021 at 16:33
  • $\begingroup$ The obligatory link to the video by Grant Sanderson The more general uncertainty principle, beyond quantum For the key point of that video see the following answer to the question How can I accurately state the uncertainty principle? $\endgroup$
    – Cleonis
    Apr 21, 2021 at 18:50
  • $\begingroup$ Related: physics.stackexchange.com/q/534614/195139 $\endgroup$
    – Sandejo
    Apr 22, 2021 at 3:47
  • $\begingroup$ It doesn't lead to any restrictions on the range of possible values you can obtain in a measurement (except when one of the standard deviations is zero). For example, if a particle is restricted to a box (which means you can predict beforehand that the result of a $\hat{x}$ measurement will be within two hard bounds), a measurement of $\hat{p}$ can still yield the value $0$. Partially related: physics.stackexchange.com/questions/599980/… $\endgroup$
    – pglpm
    Apr 22, 2021 at 8:53
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    $\begingroup$ @SarthakGirdhar From the context I assume we're speaking about one measurement on one system at some specific time, not repeated measurements on the same system (or measurements on a collection of identically prepared systems). So stationarity is not relevant in this particular case, if I'm not mistaken. $\endgroup$
    – pglpm
    Apr 22, 2021 at 9:08

5 Answers 5


While uncertainties can be interpreted as statements about ensembles of identically-prepared states, that's not how they are defined. You take a state $\psi$ and then the uncertainty $\sigma_\psi(A) = \sqrt{\langle A^2\rangle - \langle A\rangle^2}$ for any observable $A$ is simply a property of that state.

A state that has a definite momentum of zero would have to be an eigenstate of momentum. You can straightforwardly show that $\sigma_\psi(A) = 0$ for an eigenstate of $A$. So any state that has a well-defined $\sigma_\psi(x)$ cannot be an eigenstate of momentum, since the uncertainty principle implies $\sigma_\psi(p)\neq 0$. Note that this argument shows that in general you cannot have eigen"states" of position or momentum if both $\sigma_\psi(x)$ and $\sigma_\psi(p)$ exist.

"It can't have zero momentum" is not supposed to mean it's impossible to measure 0 momentum - just that it's not an eigenstate where the only possible result for momentum is zero (or indeed any other value of momentum).

In somewhat sloppy phrasing, "It can't have zero momentum" might also refer to the expectation value of the magnitude of momentum: If $\sigma_\psi(p)\neq 0$, then since $\sigma_\psi(p)\leq \sqrt{\langle p^2\rangle}$ we have that $\langle p^2\rangle$ is also non-zero.

  • $\begingroup$ "You take a state ψ and then the uncertainty ... for any observable A is simply a property of that state." Is it fair to say that it is gosh darn curious that things we can observe have this property? My instinct is that that particular "coincidence" is at the root of a great deal of confusion about QM. $\endgroup$
    – Cort Ammon
    Apr 23, 2021 at 2:28
  • $\begingroup$ @CortAmmon I'd say it's just another manifestation of standard QM being probabilistic - quantum states encode probability distributions, and so of course they also encode the standard deviations of these probability distributions. We're not used to thinking about single "definite" states in probabilistic terms, and indeed this is at the root of both confusion and the desire of many who first encounter it to find some underlying "deterministic" theory that removes this weirdness. $\endgroup$
    – ACuriousMind
    Apr 23, 2021 at 8:15
  • $\begingroup$ @CortAmmon it's actually a consequence of certain notable observables (time, position) being fourier transforms of other notable observables (energy, momentum) and of a mathematical truth that applies to such pairs of observables in general. $\endgroup$ Apr 23, 2021 at 14:33
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    $\begingroup$ @JohnDvorak Time is not an observable (otherwise energy could not be bounded from below, see physics.stackexchange.com/a/268344/50583), and the $\Delta t$ in the "energy-time uncertainty principle" likely does not mean what you think it means, see physics.stackexchange.com/a/53804/50583 $\endgroup$
    – ACuriousMind
    Apr 23, 2021 at 15:27
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    $\begingroup$ @LoneAcademic If you have a new question, you should make a new post asking that question. The comments of an answer is not the place to ask new questions. $\endgroup$ Apr 27, 2021 at 13:52

(This first section refers to v1 of the post)

Now, mathematically what this means is that if we prepare a large number of states $|Ψ⟩$ and perform measurements of the position and momentum on them (not doing >1 measurement on any prepared system)...

First mistake. For example, measuring $X$ on one state gives you a new state, so the resulting $\Delta p$ you get from subsequent momentum measurements will not be related (by the HUP) to the previous $\Delta x$ you obtained.

The HUP is a statement about measurements of $X$ or $P$ separately on a large number of similar states. So, take a large number of similar states and measure $X$ for each one to get $\Delta x$. Take another larger number of similar states and measure their momentum to get $\Delta p$. The HUP says that no matter what state you started with, you will always find $\Delta x\cdot\Delta p$ to be no smaller than $\hbar/2$.

...the RMS value of deviation from the mean for both will show an inverse relationship with each other.

Second mistake. If you have multiple similar states and do the above steps you will get a single $\Delta x$ and a single $\Delta p$. There is no inverse relationship to be found.

However, let's say we were to then take a different state and do the same procedure, and let's say $\Delta x$ is now smaller than before. We still cannot say anything about the new $\Delta p$. $\Delta p$ could increase or decrease; all we would know is that $\Delta x\cdot\Delta p$ can be no smaller than $\hbar/2$. However, if the previous $\Delta x\cdot\Delta p$ was exactly equal to $\hbar/2$, then we could guarantee a larger $\Delta p$ because the HUP must hold.

How then does this lead to a restriction on the individually measured values of position and momentum?

It doesn't. The HUP does not say anything about individual measurements.

How do we make claims such as the particle is restricted to this box so it can't have a zero momentum and so forth?

You can certainly make a $0$ momentum measurement. That doesn't violate the HUP. The HUP would be violated if $\Delta p=0$, i.e., if every momentum measurement of our large collection of similar states gave us the same value every single time.

To address the title then:

Why can't the Uncertainty Principle be broken for individual measurements if it is a statistical law?

As stated above, the HUP does not say anything about individual measurements. It just puts a bound on the product of the "spreads" of two types of measurements.

  • $\begingroup$ I think there has been a confusion in my wordings in the question. I will edit it. But what I mean is- Take a large number of states $| \Psi \rangle$ and perform $\Delta x$ and $\Delta p$ calculation on them, while performing the experiment only on $|\Psi \rangle$ at a time. After that you will get an inverse relationship between the two. Okay sure. But now, how does this tell us that you can't find a particle at rest in a box, since zero point energy is often attributed to the uncertainty principle. $\endgroup$
    – Physiker
    Apr 21, 2021 at 17:00
  • $\begingroup$ @SarthakGirdhar I have addressed your concerns in my answer. The zero point energy being due to the HUP is not correct though. That is just a property of certain quantum mechanical systems in general. $\endgroup$ Apr 21, 2021 at 17:02
  • $\begingroup$ I have edited me question, please check if it makes sense now. $\endgroup$
    – Physiker
    Apr 21, 2021 at 17:06
  • $\begingroup$ @SarthakGirdhar Thanks. My answer is still valid and addresses your concerns $\endgroup$ Apr 21, 2021 at 17:07
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    $\begingroup$ @SarthakGirdhar There is a lot of pop-sci descriptions of the HUP that are not rigorous in their description. As for the zero-point energy, that is just a property of quantum systems in general. They do not arise from the HUP. They arise from the system only being able to take on certain energy values, and the smallest of those not being $0$. $\endgroup$ Apr 21, 2021 at 17:29

All measurements in QM are implied to be the ensemble measurements, i.e., they are done on many particles/systems "prepared" in the same way. A measurement perturbs the system, so it can never be done on the same system twice. Measuring two non-commuting operators on the same system would require two measurements, so it cannot be done on one system. But consecutive measurements may indeed violate the uncertainty principle, before we collect enough data for reliable statistical inference.

It is necessary to distinguish between the mathematical and the sample means. Thus, mathematically we define the mean and the variance of a quantity as $$ \langle A\rangle = \int dx \psi^*(x)\hat{A}\psi(x),\\ \langle (\Delta A)^2\rangle = \int dx \psi^*(x)(\hat{A}-\langle A\rangle)^2\psi(x) $$ The uncertainty relation applied to such mathematical quantities is a rigorous mathematical statement, allowing no ambiguity.

In experiment however we will have measurements $\mathbf{A}=(A_1, A_2, ..., A_N)$ and calculate sample averages $$ \overline{x} = \frac{1}{N}\sum_{i=1}^Nx_i,\\ var(x) = \overline{(x -\overline{x})^2} = \frac{1}{N-1}\sum_{i=1}^N(x_i-\overline{x})^2 $$ These averages are themselves random quantities, which approach the mathematical averages only in the limit $N\rightarrow\infty$ (Note factor $1/(N-1)$ in the definition of the invariance - if it were $1/N$, the estimate would be biased, i.e., it would never approach the theoretical value). Thus, for finite $N$ it is quite possible to have a situation where the sample variances do not satisfy the uncertainty relation.

To summarize: the physical meaning of the uncertainty relation is statistical. More specifically:

  • it is valid only for an ensemble of measurements
  • it cannot be used for a single measurement (moreover, two non-commuting quantities cannot be measured simultaneously)
  • 1
    $\begingroup$ What then does measuring position and momentum simultaneously mean? $\endgroup$
    – Physiker
    Apr 21, 2021 at 16:38
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    $\begingroup$ They cannot be measured simultaneously. $\endgroup$
    – Roger V.
    Apr 21, 2021 at 16:52
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    $\begingroup$ Don't we say casually that the Uncertainty Principle says that measuring position and momentum simultaneously to arbitrary accuracy is impossible. Does that refer to the ensemble average? $\endgroup$
    – Physiker
    Apr 21, 2021 at 16:54
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    $\begingroup$ But consecutive measurements may indeed violate the uncertainty principle... I am not sure about what this actually means. Are you just saying with limited data calculated $\Delta x$ and $\Delta p$ will have substantial errors? $\endgroup$ Apr 21, 2021 at 16:57
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    $\begingroup$ @BioPhysicist this is precisely what I meant. $\endgroup$
    – Roger V.
    Apr 21, 2021 at 18:56

Quantum mechanics cannot be reduced to the HUP. There are some basic axioms of QM and the HUP follows from them. The relevant axioms here are:

  1. Quantum states are vectors $\psi$ in a Hilbert space $\mathbf{H}$.
  2. Observables are operators $A: \mathbf{H} \rightarrow\mathbf{H}$
  3. The measurement outcomes can only be the eigenvalues of the corresponding operator.
  4. The measurement outcome is random (that means unknowable in general). The probability of measuring the value $a$ is given by the absolute square of the overlab $\langle \phi_a |\psi \rangle$ of the quantum state and the eigenvector of $A$, corresponding to the eigenvalue $a$

Therefore the value of an observable for a given state is only known, if the quantum state is an eigenstate of that observable (for example immediately after a measurement). It follows that we can know the value of two observables simultaneously only if the state is an eigenstate of both observables at once. This is only possible if the two observables commute.


How do we make claims such as the particle is restricted to this box so it can't have a zero momentum and so forth?

Have you seen this claim made? It is at best shorthand for something else, and at worst just wrong. First, one should talk of the expected value or measured value of a particle, rather than speaking of it just having a value. The expected value of the momentum of a particle confined to a box absolutely can be zero; if it's confined to a box, and the box isn't moving, then overall expected value has to be zero (there are states where the expected value varies over time, but the average is still zero).

The probability of the measured value of its momentum being zero is zero (that is, $p(\text{momentum}=0)=0$), but that's just because the probability of it being any particular value is zero. The probability density for its momentum will, for normal states, be highest at zero.

Another correct statement would be that the momentum can't be restricted to zero.


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