Different forms of Heisenberg Uncertainty Principle The Heisenberg Uncertainty Principle is often written in two forms:
$$\Delta x \Delta p \geq \frac{\hbar}{2} $$
and
$$\sigma_x \sigma_p \geq \frac{\hbar}{2}. $$
Are these two equivalent? I've been told they are, but it doesn't make sense to me. For instance, in an infinite square well, $\Delta x = $ the width of the well. Whereas $\sigma_x$ is equal to the standard deviation of the wave function.
If they are equivalent, then $\Delta x = \sigma_x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} $ meaning we can find the value of $\langle x^2\rangle  $ using:
$\langle x^2\rangle = (\Delta x)^2 + \langle x\rangle^2  $
Does this make sense? Or have I been misled about the equivalence of the two forms of the HUP?
 A: The statistical interpretation of quantum mechanics tells us that the "best" that we can know a priori (i.e. before carrying out a measurement, an experiment) from a theoretical study of a physical system is, in general, a range of possible values. Since you have a range of possibilities, the way naturally opens up for a statistical analysis: you have a distribution of values ​​characterized by an average value and a dispersion, $\sigma$, around it.The product of the two $\sigma$ associated with the distribuitions of two coniugate osservables can't go below the value indicated in the HUP.
If we instead carry out an experiment and successive measurements of two conjugate quantities, "each time returning the system to the $\Psi$ preceding the measurements", A and B, we obtain different values ​​characterized by uncertainties $\Delta A$ and $\Delta B$ whose product will have an upper limit.
As De Broglie said, we are therefore dealing with pre-measurement (in the first case) and post-measurement (in the second) uncertainty relations.
For instance, the infinite square well centered in the origin, the particle can occupy all positions between -L/2 and +L/2: so the average value is x=0 and the dispersion is L/2. Or, if you performe a large number of measurements, you'll obtain  the uncertainty, $\Delta x$, is L/2 for the mean value x=0.
I hope I was helpful.
A: Most general form of uncertainty principle is :
$$ \boxed{\sigma _{A}\sigma _{B}\geq \left|{\frac {1}{2i}}\langle [{A},{B}]\rangle \right|} $$
Where $A,B$ is Hermitian operators which do not commute, namely their commutator :
$$ [A,B]={A}{B}-{B}{A} \neq 0 $$
If that's the case, then it's impossible to measure variables $A,B$ simultaneously.
Now let's try to find out commutator $[x,p]$, for that we need to solve expression $$xp~\psi(x) -px~\psi(x)$$
Substituting quantum mechanical momentum operator and re-arranging terms gives :
$$ i\hbar\frac{\partial}{\partial x} \left(x\psi(x)\right) - xi\hbar\frac{\partial}{\partial x}\psi(x) $$
Using multiplication rule for first term gives :
$$ i\hbar\psi(x) + xi\hbar\frac{\partial}{\partial x}\psi(x) - xi\hbar\frac{\partial}{\partial x}\psi(x) $$
Noticing that the last two terms cancels each other, so we get final relationship:
$$ [x,p]\psi(x) = i\hbar~\psi(x)$$
This means that our position-momentum commutator is :
$$ [x,p]=i\hbar $$
Substituting resulting commutator back into general uncertainty principle form we get:
$$ \sigma _{x}\sigma _{p}\geq \frac{\hbar}{2} $$
