The General Uncertainty Principle of 2 Operators is given by
$$(\Delta X)^2\,(\Delta Y)^2 \geq\left(\dfrac{1}{2}\big\langle\{X,Y\}\big\rangle – \big\langle X\big\rangle \big\langle Y\big\rangle\right)^2+\left(\dfrac{1}{2\,i}\big\langle[X,Y]\big\rangle\right)^2$$
Now I witnessed a lot of people asking why omitting the anti-commutator that is to say:
$$(\Delta X)^2\,(\Delta Y)^2 \geq \left|\dfrac{1}{2\,i}\big\langle[X,Y]\big\rangle\right|$$
It oftentimes had been argued the other vanishing terms really have no other meaning, but what about calculating it explicitly e.g. for $X = \hat{x},\quad Y = \hat{p}?$ There I receive in position space:
$$(\Delta \hat{x})^2\,(\Delta \hat{y})^2 \geq \left(\dfrac{1}{2}\Big\langle \dfrac{2}{\hbar\,i}\,x\,\partial_x+\dfrac{\hbar}{i}\Big\rangle-\big\langle x\big\rangle\big\langle \hbar/i\,\partial_x\rangle\right)^2 – \dfrac{1}{4}\left(\Big\langle\dfrac{h}{i}\Big\rangle\right)^2$$
$\underline{\text{I really do not see how this expression is supposed to be larger than the minimal Heisenberg estimate:}}$
$$(\Delta X)^2\,(\Delta Y)^2 \geq \left|\dfrac{1}{2\,i}\big\langle[X,Y]\big\rangle\right| = \dfrac{\hbar}{2}$$
Also what does it mean having unmatched derivative $(\partial_x)$ in the unpleasant large formula?
Best Answer
The correct generalized uncertainty relation is
$$(\Delta X)^2\,(\Delta Y)^2 \geq\left|\dfrac{1}{2}\big\langle\{\hat X,\hat Y\}\big\rangle - \big\langle \hat X\big\rangle \big\langle \hat Y\big\rangle\right|^2+\left|\dfrac{1}{2\,i}\big\langle[\hat X,\hat Y]\big\rangle\right|^2.$$
Note the absolute values: both terms are non-negative. The weaker uncertainty relation follows immediately from the fact that the first term is non-negative:
$$(\Delta X)^2\,(\Delta Y)^2 \geq\left|\dfrac{1}{2}\big\langle\{\hat X,\hat Y\}\big\rangle - \big\langle \hat X\big\rangle \big\langle \hat Y\big\rangle\right|^2+\left|\dfrac{1}{2\,i}\big\langle[\hat X,\hat Y]\big\rangle\right|^2 \geq \left|\dfrac{1}{2\,i}\big\langle[\hat X,\hat Y]\big\rangle\right|^2.$$