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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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  • Have you tried the links and references at the end of the wikipedia article? Otherwise a general internet search? – sammy gerbil Jul 26 '16 at 15:05
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    I'm voting to close this question as off-topic because of insufficient research effort. – sammy gerbil Jul 26 '16 at 15:05
  • @sammygerbil Please look at the related question, which explains clearly that these doubts have come up based on answers to the linked questions. I have edited the question to clarify this part. Happy to provide more details if necessary. – texmex Jul 27 '16 at 02:25

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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  • Given that $p=-i\hbar\partial/\partial x$, then, by the rules of calculus, they have to satisfy that relationship. Or, are you asking why $p$ needs to be $-i\hbar\partial/\partial x$? – John1024 Jul 26 '16 at 01:42
  • Thanks for this clarification. Please note, I have just started learning quantum mechanics through self study one month back. Hence my level is very basic at best. Can you please elaborate how / why you go from step one to two in your equations. – texmex Jul 26 '16 at 01:46
  • Yes indeed; Why does $p$ need to be $-i\hbar {\partial}/{\partial x}$ Thanks for your patient clarifications. – texmex Jul 26 '16 at 01:54
  • @user249613 Have you studied DeBroglie's relationship between wavelength and momentum? – John1024 Jul 26 '16 at 01:56
  • Not in detail (Will certainly look through it; If you have any good source, please do point out); but if I understand the uncertainty principle correctly any two operators that are the fourier transforms of one another satisfy this conjugate relationship. So this applies to operators other than position $x$ and momentum $p$. Could you please clarify how / why and under what conditions any two operators satisfy this relationship? – texmex Jul 26 '16 at 01:58
  • $p=-i\hbar\partial/\partial x$ is just a restatement of DeBroglie's relation in calculus form. So, to understand why $p$ has that value, a review of the evidence that led DeBroglie to this relation would be very helpful. – John1024 Jul 26 '16 at 02:19
  • Thanks for your patient answers. I will look through this in more detail. But please note, as I understand it, the uncertainty priciple can apply to any operators related by a fourier transform. Could you point out a source with details steps and other relevant materials? – texmex Jul 26 '16 at 02:22
  • @user249613 As far as the commutation relationship is concerned, the fourier transform does not tell you anymore than freshman calculus tells you. If you are finding the calculus challenging, I would leave the fourier transforms until later. – John1024 Jul 26 '16 at 02:25
  • Please note, I have no issues with the calculus. But I fail to see how the $\hbar$ gets introduced by doing fourier transforms of two operators. – texmex Jul 26 '16 at 03:09
  • OK. Mathematically, the uncertainty principle relates position and wavenumber (or time and frequency or ...). $\hbar$ only enters when the De Broglie relation is used to convert wavenumber to momentum. – John1024 Jul 26 '16 at 03:15