Heisenberg is said to have inferred the uncertainty relation from $[x,p] =-i\hbar\neq 0.$ This gave rise to some questions about the details of commutator operations in the QM/classical contexts (restricting attention to p,x).

  1. For the classical* commutator, we have that \begin{align} \left(\frac{\mathrm d}{\mathrm dx}x -x \frac{\mathrm d}{\mathrm dx}\right)\psi (x) & = \psi(x) + xD_x \psi(x)-xD_x \psi(x) \\ & = \psi(x),\text{ or}\\ D_x\cdot x-x\cdot D_x&=1\hspace{10mm} \tag A \end{align}

    So my first question is, isn't it also true that the classical commutator $[p,x]\neq 0$?

  2. My second question is:

    whether we can therefore infer uncertainty from the classical non-zero commutator as well.

    Here is my reasoning. In the classical setting we have

    $$( \Delta x\Delta p)^2 =\frac{1}{4}\left[4\frac{\int \psi^* x^2\psi \mathrm dx \int\frac{d\psi^*}{\mathrm dx}\frac{d\psi}{\mathrm dx}\mathrm dx}{\left[\int\psi^*\psi \mathrm dx\right]^2}\right]$$

    Via Cauchy-Schwarz this gives the uncertainty relation since the quantity in brackets on the right is always $\geq 1,$ that is,

    $$\left[\int \psi^*\psi ~\mathrm dx\right]^2 < \left[ \int \psi^* x^2 \psi~ \mathrm dx \int \frac{d \psi^*}{dx}\frac{d\psi}{\mathrm dx} ~dx\right] $$

    But the left hand side is

    $$\left[\int \psi^*\psi~ \mathrm dx \right]^2 = \lvert\lvert \psi \rvert\rvert _2^4 = \lvert\lvert[p,x]_\psi\rvert\rvert_2^4 $$

    where $[p,x]_\psi$ is the commutator applied to $\psi.$

Edit: the part above is not crucial to the argument. Below is more to the point.

But we only need to look at a typical statement of the uncertainty relation--this one is adapted from Folland and Sitarum, The Uncertainty Principle: A Mathematical Survey at p.210. For $\psi$ in R and any $a,b$ in R, $\lvert\lvert\psi \rvert\rvert_2=1,$ the last equality is mine--

$$ \int (x-a)^2 |\psi|^2 \mathrm dx \int (p-b)^2 \hat{|\psi|^2} \mathrm dp \geq \frac{1}{16 \pi^2} \lvert \lvert \psi\rvert\rvert_2^4 = \frac{1}{16\pi^2}\lvert\lvert(px-xp)(\psi) \rvert\rvert_2^4$$ So because the commutator was $[p,x] = \psi \neq 0$ we knew without more that an uncertainty relation existed...?

  1. If (1) and (2) are more or less correct, then:

    restricting attention to ground state in QM, for these conjugates, the major distinction between the classical and QM setting is not whether the commutator is zero or not, but the scale in QM due to $\hbar$...is that correct?

  2. Finally:

    what is the right way to express $(\mathrm A)$ from (1) above in "bra-ket" notation?

*By "classical" I mean from Fourier analysis in, say, $L^2,$ in which $\Delta x $ is a well-defined standard deviation.

  • $\begingroup$ Related/possible duplicates (I'm not quite sure what you're asking): physics.stackexchange.com/q/197821/50583, physics.stackexchange.com/q/47458/50583 $\endgroup$
    – ACuriousMind
    Apr 22, 2017 at 20:00
  • $\begingroup$ I think my answer to the first question I linked answers that - the QM uncertainty relation is a much more general consequence of the non-commutativity of operators. $\endgroup$
    – ACuriousMind
    Apr 22, 2017 at 20:20
  • $\begingroup$ Part of the delicate nature of the discussion is that the constants are quite important. Crawford ("Waves; Berkeley physics course vol. 3." (1968)) gives $\Delta k\Delta x\ge 2\pi $ from Fourier analysis of a pulse. There is also work by Arthurs and Kelly ("BSTJ briefs: On the simultaneous measurement of a pair of conjugate observables." The Bell System Technical Journal 44.4 (1965): 725-729.) which gives a kind of "classical limit" to the quantum uncertainty relation. See also work by Ozawa v.g. arxiv.org/pdf/quant-ph/0207121.pdf. Getting the constants right is the tricky part. $\endgroup$ Apr 24, 2017 at 19:43
  • 1
    $\begingroup$ There is more by Uffink and Maassen on the interpretation and difference between the quantum and classical versions (seeEq.(14) in here stanford.library.sydney.edu.au/entries/qt-uncertainty for example of the kinds of delicate discussions involved.) $\endgroup$ Apr 24, 2017 at 19:53
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    $\begingroup$ It does actually: this is more or less the contents of Robertson's derivation of the uncertainty relation. But your question is implicitly deeper IMO because there is some "classical" uncertainty, which is deeply related to the way detectors are coupled to quantum states, as per Arthurs and Kelly (a nice paper... also Stig Stenholm has nice work on this) $\endgroup$ Apr 25, 2017 at 4:08

2 Answers 2


Classically, momentum has nothing to do with the spatial derivative $d/dx$ - that's a purely quantum-mechanical relation. So if by "$p$" you mean momentum, then your point 1 is invalid. If by "$p$" you instead simply mean the spatial derivative operator $d/dx$, then yes, you can derive a purely mathematical uncertainty relation for the Laplace transform, but it has no physical content, and in particular has nothing to do with momentum.

  • $\begingroup$ @daniel Responding to your comments in order: (1) The Fourier transform relates the $x$ basis and the eigenbasis of $-i\, d/dx$. Your "classical momentum operator" (in quotes because I'm not sure what that means) doesn't seem to have an $i$, so the Fourier transform gets modified to the very similar Laplace transform (which is basically just the Fourier transform without the $i$). (2) Momentum is certainly a valid quantity in classical mechanics, it just doesn't have anything to with the spatial derivative operator. It's not that your question is "not realistic" so much as ill-posed. $\endgroup$
    – tparker
    Apr 25, 2017 at 7:29
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    $\begingroup$ (3) Any two operators $\hat{A}$ and $\hat{B}$ with a nonzero commutator have an uncertainty relation $\Delta A \ \Delta B \geq (1/2) \left| \left \langle [ \hat{A}, \hat{B} ] \right \rangle \right|$, whether or not they are related by Fourier transform (this was pointed out by Howard Robertson in 1929). See here for a proof. In the case of position and momentum, this reduces to the Heisenberg uncertainty relation. $\endgroup$
    – tparker
    Apr 25, 2017 at 7:35
  • $\begingroup$ The question was bad all around, considering the comment above is really what I was looking for. Yes, I left out the i. $\endgroup$
    – daniel
    Apr 25, 2017 at 16:00
  • $\begingroup$ Yes, and the 2 points will ease the sting of down-votes... $\endgroup$
    – daniel
    Apr 26, 2017 at 4:13

I think this somewhat misses the point. Your 'classical uncertainty principle' holds for classical waves, such as light or sound, and it was known long before the discovery of quantum mechanics.

The new feature of quantum mechanics is that things we considered to be particles classically have wavelike properties, with their momentum related to their wavenumber by $p = \hbar k$. Only by having this relation do you get a bound on position and momentum uncertainty.

  • $\begingroup$ I understand both of the above. tparker's answer at least addresses the question. If the premise of the question is wrong (in the way he points out), then the question is just a minor math point about commutators--does a non-zero commutator imply an uncertainty relation. $\endgroup$
    – daniel
    Apr 25, 2017 at 3:41
  • $\begingroup$ How can a question miss the point? I'm curious about manipulations with commutators. Heisenberg is said to have inferred an uncertainty relation from a nonzero commutator. How did he see that? In some purely abstract sense, a nonzero commutator implies uncertainty. Is is wrong to ask about it? $\endgroup$
    – daniel
    Apr 25, 2017 at 4:00
  • $\begingroup$ @daniel Absolutely, there's a mathematical link between commutators and uncertainty, and I wrote about a simple example here. A more general example is here. I said your question missed the point because it seemed you were saying that quantum mechanics was literally the same thing as classical wave mechanics. $\endgroup$
    – knzhou
    Apr 25, 2017 at 4:04
  • $\begingroup$ A. Khurasia addressed part (3) and I saw that the answer to the question was "no." I should have left that part out as it only muddied the waters. I deleted my comment in response to save space... $\endgroup$
    – daniel
    Apr 25, 2017 at 4:15

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