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1

$$[x^2,p^2]=x[x,p^2]+[x,p^2]x=x[x,p]p+xp[x,p]+[x,p]px+p[x,p]x=i\hbar(2xp+2px)=2iā€Œā€‹\hbar[x,p]_+=4i\hbar xp +2\hbar^2\; ;$$ using the fact that $[x,p]=i\hbar$ and $[AB,C]=A[B,C]+[A,C]B$ .


2

The uncertainty product is bounded from below by the expectation value of the commutator of the relevant observables. If $A$ and $B$ are any two observables, then the generalized Heisenberg uncertainty relation reads as $$ \sigma_A\sigma_B \geq \frac{1}{2}\vert \langle[A,B]\rangle\vert .$$ For the case of position - linear momentum pair, the commutator is ...


3

The Heisenberg uncertainty principle in the most general form $$\Delta_\omega(A)\Delta_\omega(B)\geq\frac12|\omega([A,B])|$$ depends on the state $\omega$ on which it is evaluated. In the special case of the canonical commutation relations $[q,p]= i\hbar I$, $\omega(I)=1$ for any state and therefore the RHS reduces to a constant. For more general commutators ...


1

The $i$ in the formula of commutator is necessary in order to make the quantities hermitian. Indeed, the commutator of two hermitian operators is NOT hermitian, but anti-hermitian. \begin{equation} \left(\left[x_i , p_j \right] \right)^\dagger = - \left[x_i^\dagger, p_j \dagger\right] = - \left[x_i , p_j \right] \ . \end{equation} In a certain sense - it ...


1

It's hard to answer matters of convention, but here's my take: the traditional way of writing them keeps everything about the operators on the left, and it's more useful to know during calculations. If I know the commutator, then when the commutator turns up I know to just replace with it with $i \hbar$. If I knew your formula, there's an extra mental step ...



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