# Derive Heisenberg Uncertainty from General Uncertainty Principle

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?

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:

$$(\Delta X)^2\,(\Delta Y)^2 \geq\left|\dfrac{1}{2}\big\langle\{\hat X,\hat Y\}\big\rangle - \big\langle \hat X\big\rangle \big\langle \hat Y\big\rangle\right|^2+\left|\dfrac{1}{2\,i}\big\langle[\hat X,\hat Y]\big\rangle\right|^2 \geq \left|\dfrac{1}{2\,i}\big\langle[\hat X,\hat Y]\big\rangle\right|^2.$$

• yes the absolute value is essential… Jun 2, 2022 at 22:04
• Alright I am left with another confusion: why in Heisenberg uncertainty $$|\dfrac{1}{2\,i}\langle[x,p]\rangle|^2 > |\dfrac{1}{2\,i}\langle[x,p]\rangle|$$ while $$\dfrac{\hbar^2}{4}<\dfrac{\hbar}{2}$$ ?
– Leon
Jun 4, 2022 at 22:18
• @Leon You have an error in your question. The Heisenberg uncertainty relation is $$\Delta X\Delta Y \geq \left|\dfrac{1}{2\,i}\big\langle[\hat X,\hat Y]\big\rangle\right|.$$ No squares on the left hand side. It also makes no sense to compare $\hbar^2/4$ and $\hbar/2$: they have different dimensions.
– Puk
Jun 5, 2022 at 0:24