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## Understanding the Commutator

 Quote by Elekko That's a perfect explanation. Thank you very much!
You're welcome. Thanks for letting me know that you read it and understood it.

Here's a slightly different version that doesn't even use the d/dx notation. I will not include all the information about domains and stuff this time. Here I denotes the function that takes x to x. To avoid confusion, the identity operator will be denoted by 1 instead of I. We define two operators, Q and D.

Qf=If (This means that for all real numbers x, Qf(x)=(If)(x)=I(x)f(x)=xf(x), so the Q we have defined here is the position operator).
Df=f' (We define the momentum operator by P=-iD, so D=iP).

\begin{align}
DQf(x) &=(If)'(x)=I'(x)f(x)+I(x)f'(x) =f(x)+xf'(x)=f(x)+QDf(x) =(f+QDf)(x) \\
DQf &=f+QDf=(1+QD)f\\
DQ &= 1+QD\\
\left[Q,P\right] &= \left[Q,-iD\right]=-i\left[Q,D\right]=-i(-1)=i.
\end{align}

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