# Uncertainty principle for two compatible observables. When do we get the equality and the inequality?

For two compatible observables A and B i.e. if $$[A, B]=0$$, the uncertainty principle says that $$(\Delta A)_\psi(\Delta B)_\psi\geq 0$$ in any state $$|\psi\rangle$$ where $$(\Delta A)_\psi=(\langle \psi|A^2|\psi\rangle-(\langle \psi|A|\psi\rangle)^2)^{1/2}$$ . I know that these uncertainties have nothing to do with the precision of measurement. It is however not clear to me when will we get equality and when inequality?

• If you look at the derivation of the general uncertainty principle, you should be able to tell which terms are neglected there to obtain this inequality from a more complicated equality, and so this will be an equality when these neglected terms are zero. Do you have some difficulty with doing this on your own? Mar 27 at 18:02
• Consider $A=B$, when would $(\Delta A)_\psi$ be non-zero? Mar 27 at 18:04
• I think, there exist some states $\psi$ for which either $(\Delta A)_\psi$ or $(\Delta B)_\psi$ or both are zero. In those states, the uncertainty product is zero. Am I right? Mar 27 at 18:11
• @DvijD.C. When $\psi$ is not an eigenstate of $A$. Also, see my comment above. Mar 27 at 18:16

The Heisenberg uncertainty relation looks like $$(\Delta A)^2(\Delta B)^2\geq \frac{1}{4}\langle \psi|[\hat{A},\hat{B}]_+|\psi\rangle^2+\frac{\hbar^2}{4}$$ The above inequality becomes (if you look for the whole derivation) equality only if

• $$\hat{A}|\psi\rangle =c\hat{B}|\psi\rangle$$
• $$\langle \psi|[\hat{A},\hat{B}]_+|\psi\rangle =0$$

where $$\hat{A}=A-\langle A\rangle$$ and $$\hat{B}=B-\langle B\rangle$$.