# Can the uncertainty principle be redefined for different standard deviations?

$$\sigma_x \cdot \sigma_p \ge {{\hbar} \over {2}}$$

Where the $\sigma$ is the standard deviation. What happens to the inequality if you use a different definition of $\sigma$. For instance what happens if $\sigma=\int |x-\mu| \cdot f(x) \ dx$ is used?

My guess is that some form of the uncertainty principle holds, but it may not be as pretty as the standard one.

• I don't understand what your alternative definition is supposed to be - you have to define the deviation of some operator $A$ for some state $\lvert \psi \rangle$. – ACuriousMind Jul 28 '15 at 16:42
• Actually it is not the statistical standard deviation in the uncertainty principle. It is about and interval delta(x) and an interval delta(p) which can be defined in any way the problem defines. An interval on the axis x and the axis p. hyperphysics.phy-astr.gsu.edu/hbase/uncer.html – anna v Jul 28 '15 at 16:44
• @annav: On the contrary, the uncertainty in the principle is precisely the standard deviation $\sqrt{\langle A^2\rangle -\langle A \rangle^2}$. – ACuriousMind Jul 28 '15 at 16:51
• @ACuriousMind I think you are confusing necessary and sufficient. A standard deviation is sufficient to define an interval, but not necessary. The heisenberg uncertainty can be applied to a larger set than standard deviations. – anna v Jul 28 '15 at 19:19
• @annav What does "defined in any way the problem defines" mean? I'm completely confused by that phrase. – joshphysics Jul 28 '15 at 19:55

Given two observables $A$ and $B$ such that $[A,B] = iC$, the most general form of the uncertainty principle is $$\Delta_\omega(A)\Delta_\omega(B)\geq\frac12|\omega(C)|,$$ where $\omega$ is any state of the algebra of observables. By the Riesz-Markov theorem, there is a regular probability measure such that $$\omega(f(A)) = \int_{\sigma(A)}f(\lambda)\ \text d\mu_\omega(\lambda).$$ Hence the standard deviation depends on the state the uncertainties are evaluated on.