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From this related question,

Rigorous Mathematical Proof of the Uncertainty Principle from First Principles

The key assumption to derive the uncertainty principle seems to be the relationship between canonical conjugate operators, $x , p $ such that,

$$ [x,p]=iℏ.$$

I have seen at this link (https://en.wikipedia.org/wiki/Canonical_commutation_relation) that this means the two operators are fourier transforms of each other.

1)

Could someone please point out a good source for more intuition and detailed steps on these canonical conjugate operators, showing their fourier transforms and why they need to satisfy this commutator relationship?

2)

Are there any other properties two operators need to satisfy so that they can be shown to be governed by the Uncertainty Principle?

EDIT:

Please note, the related question links in other questions that provide proofs for the uncertainty principle. But, after having looked at many of the mathematical proofs (some long and many short versions) for the uncertainty principle. It becomes clear that the uncertainty principle is the result of the assumption regarding the relationship between canonical conjugate operators.

Position and Momentum satisfy this relation because of the DeBroglie experiments. But how can this be generalized to a larger set of conjugate operators (as the wikipedia article seems to imply)? Any pointers would be appreciated.

The mathematics is fairly straightfoward (as it usually is when the rules applied are clearly stated) but the reason why conjugate operators need to satify this relation is not clear (As stated on the wikipedia link).

I admit I lack a deeper understanding of quantum mechanics (or of most things); But happy to delete this if it adds to the noise instead of adding clarify

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texmex
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1 Answers1

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  1. To understand the commutation relationship, let's just apply the operator to a wave function and use some basic calculus:

    $$(xp - px)\Psi = -i\hbar \bigg(x\frac{\partial\Psi}{\partial x} - \frac{\partial (x\Psi)}{\partial{x}}\bigg) = -i\hbar \bigg(x\frac{\partial\Psi}{\partial x} - \Psi -x\frac{\partial \Psi}{\partial{x}}\bigg) = i\hbar \Psi$$

    The left-hand side above is just the commutation operator. The next step is to substitute in $p=-i\hbar\partial/\partial x$. The step after that is just applying the usual calculus rules for a derivative of a product. The last stop is just the result of canceling out identical terms.

  2. Time and energy also obey an uncertainty relationship. (Whether "time" is a proper QM operator is a discussion I will leave to someone else.)

John1024
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