Justify the value of the product of the uncertainty in position and momentum

The Heisenberg uncertainty principle states that: $$\Delta x\Delta p_x\ge\frac12\hbar$$ But more generally for operators $A$ and $B$: $$\Delta A\Delta B=\frac12|\langle[A,B]\rangle|$$ However, $[x,p_x]=i\hbar$. With this information, I don't see how to get from the second equation to the first.

• What you think you'll get upon this trivial substitution if not $1/2 \hbar$? What's your result? Sep 13, 2015 at 13:06

The commutator $[\hat{A}, \hat{B}]$ of two operators $\hat{A}$ and $\hat{B}$ is also an operator, so when using the Robertson uncertainty relation we have to take the expectation value of the commutator first and then take the absolute value of it.
But for $\hat{x}$ and $\hat{p}$ the commutator $[\hat{x}, \hat{p}]$ is simply a constant $\mathrm{i} \hbar$, so that taking the expectation value is trivial: $\langle [\hat{x}, \hat{p}] \rangle = \mathrm{i} \hbar$. Then we take the absolute value of this result which is $\hbar$ and divide it by two to get $\frac{1}{2} \hbar$ as expected.
• I think that perhaps I am being stupid but why is the absolute value of $i\hbar$ equal to $\hbar$? Sep 13, 2015 at 13:28
• The absolute value of a complex number goes as the square root of (the complex number multiplied by the complex conjugate of the complex number), or $\hbar = \sqrt{i\hbar\cdot -i\hbar}$ in this case. Sep 13, 2015 at 13:53
• @RobChem, the absolute value of $\mathrm{i} \hbar$ is $\hbar$ by the very definition of the absolute value of a complex number. Sep 13, 2015 at 13:56