# Does $[A , B] \not = 0$ necessarily mean $\Delta A \Delta B \ge k$ where $k \not = 0$?

For example : $$[\hat{x},\hat{p}] = i \hbar \hat{I}$$ and $$\Delta x \Delta p \ge \hbar/2$$ but in case of number states $$|n \rangle$$ $$[\hat{C},\hat{n}] = i \hat{S}\\ \Delta C \Delta n \ge 0$$ where $$\hat{C}=\frac{1}{2} [\hat{E}+\hat{E^\dagger}] \\ \hat{S}=\frac{1}{2i} [\hat{E}-\hat{E^\dagger}]\\ \hat{E}=\bigg( \frac{1}{\sqrt{\hat{a} \hat{a^\dagger}}} \bigg) \hat{a}\\ \hat{E^\dagger}=\hat{a}\bigg( \frac{1}{\sqrt{\hat{a} \hat{a^\dagger}}}\bigg)$$

and $$\hat{a^\dagger} (\hat{a})$$ are the creation(annhilation) operator of a Quantum Harmonic Oscillator. The C(S) are cos(sin) observables of quantum phase of a Quantised EM Field. The $$\hat{E}(\hat{E^\dagger})$$ are Susskind–Glogower (SG) phase operators. [Ref : GerryKnight Quantum Optics Ch. 2 Sec 7.]

When is an uncertainty product greater than zero and what does it signifiy?

• – Cosmas Zachos May 29 at 15:04
• The general uncertainty principle works for a state if the given state is in the domain of the commutator of the operators. I'm not very familiar with the SG phase operators but usual angular operators are such that their commutator with their conjugate operator is not defined for all states. – Dvij D.C. May 29 at 16:19