A few details about commutators in QM and Fourier analysis

Question

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...?

3. 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?

4. Finally:

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

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

Answer 1

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.

Answer 2

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.