# Uncertainty Principle: Commutators [duplicate]

How are commutators the mathematical basis for uncertainty principle? What makes one say that commutators imply uncertainty principle or vice-versa?

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• Do you mean to say that non-commuting observables imply the uncertainty principle? Because the mere existence of commutators does not imply anything like that. – probably_someone Aug 2 '17 at 7:43
• Yes I am asking whether non-commuting observable imply uncertainty principle or uncertainty principle implies non commuting observable? Which one is correct? – Khushal Aug 2 '17 at 7:46
• What do you want to assume? Some people start with assuming that non-commuting observables exist and then derive the uncertainty principle, and others start with the uncertainty principle and derive the existence of non-commuting observables. – probably_someone Aug 2 '17 at 7:49
• Possible duplicates: physics.stackexchange.com/q/24116/2451 , physics.stackexchange.com/q/186746/2451 , physics.stackexchange.com/q/43821/2451 and links therein. – Qmechanic Aug 2 '17 at 8:00

Consider a commutator relation $[\hat{A},\hat{B}]=\hat{C}$. This commutator relation is preserved if you take $\hat{A} \mapsto \hat{A} - <\hat{A}>,\hat{B} \mapsto \hat{B} - <\hat{B}>$ with expectation value denoted by $<...>$.
for Operators $\hat{A},\hat{B},\hat{C}$. Then you can act a norm on it and obtain
$||\hat{C}||=||\hat{A}\hat{B}-\hat{B}\hat{A}||\le ||\hat{A} \hat{B}||+||\hat{B}\hat{A}||$ (triangle inequality).
Now you can set this norm to supremum norm and obtain $||\hat{A}\hat{B}||\le||\hat{A}||||\hat{B}||$.
Set e.g. $\hat{A} = \hat{p},\hat{B}=\hat{x},C=-i\hbar$ and you will get the uncertainty relation for this variables.