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I've heard from a lot of people that the reason momentum and position have an uncertainty relation is because of the Fourier Transform. But is this in any way the case?

If it were I would expect all conjugate quantities related by the canonical commutation relation to be Fourier transforms, but I don't think is true. in fact, I'm not sure how to write the eigenstates of an arbitrary operator as a function that I could Fourier transform.

So if it is the case that it is just, somehow, a coincidence that momentum and position are Fourier transforms of each other, which also coincidentally implies an uncertainty relation, could someone elaborate on this "coincidence"?

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2 Answers 2

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When you have a bunch of interrelated phenomena in physics, trying to figure out which one is the "reason" for the other ones is often just a recipe for confusion. Different people will start from different postulates, so they will disagree on which results are trivial and which aren't, but hopefully everyone agrees on what's true. In a first course on quantum mechanics, a teacher might define the momentum-basis wavefunction to simply be the Fourier transform of the position-basis wavefunction, from which one can derive the uncertainty relation. In a more advanced course, they might generalize the concept of the momentum basis to something more abstract or physical.

In the standard presentation of QM, the defining relationship between the position and momentum operators is postulated to be the canonical commutation relation $[x, p] = i \hbar$. From this we can derive the Heisenberg Uncertainty Principle from the Schrodinger uncertainty relation

$$ \sigma_A \sigma_B \geq \frac{1}{2} \left| \langle [A, B] \rangle \right|$$

without needing to go into any particular basis. So the uncertainty principle is far more general than just applying to operators that obey the canonical commutation relations or are Fourier transforms of each other. One could reasonably say that the nontrivial commutation relation is the "reason" for the uncertainty relation. But for the special case of canonically conjugate operators, thinking about the Fourier transform can make the uncertainty relation more intuitively clear than the abstract mathematical result above can.

Regarding your question about whether "all conjugate quantities related by the canonical commutation relation [are] Fourier transforms" - yes, there is a sense in which this is true. For any two canonically conjugate operators $A$ and $B$ which act on a Hilbert space $\{ | u \rangle \}$ that can be indexed by a continuous parameter $u$ (which could physically correspond to position, or momentum, or something else), one possible representation of the commutation relation in the $u$ basis is $A \rightarrow u,\ B \rightarrow -i \hbar \frac{d}{du}$. In this representation, the expressions for any wavefunction in the $A$- and $B$- bases will be Fourier transforms of each other. However, other representations are possible, which lead to physically equivalent results but wavefunctions that are not Fourier transforms of each other. (E.g. another representation would be $A \rightarrow u,\ B \rightarrow f(u) - i \hbar \frac{d}{du}$ for any function $f(u)$). In particular, the uncertainty principle holds in every representation, since it follows directly from the commutation relation. Transforming between different representations of the canonical commutation relations is an example of a "gauge transformation."

The Stone-von Neumann theorem very roughly says that for any operators $A$ and $B$ satisfying the canonical commutation relation, you can get away with using the standard representation $A \rightarrow u,\ B \rightarrow -i \hbar \frac{d}{du}$ without loss of generality. (More precisely, it says that any representation of the exponentiated canonical commutation relation on a sufficiently smooth Hilbert space is unitarily equivalent to the standard representation, so any other representation basically just describes the same physics in a different coordinate system.)

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Yes, this is not the "true" reason. The direct reason is that they are not commuting operators. See the Robertson-Schrodinger Relation.

You end up getting that the product of the uncertainties is bounded by 1/2 of the absolute value of the failure to commute. In the case of position and momentum this is $\hbar/2$.

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