About the Heisenberg uncertainty principle

[Another]

heisenberg uncertainty principle

$Δx Δp ≥ ħ$

where

$Δx = \sqrt{<\hat{x}^2>-<\hat{x}>^2}$
$Δp = \sqrt{<\hat{p}^2>-<\hat{p}>^2}$

I don't know. Why $Δx$ equal to $\sqrt{<\hat{x}^2>-<\hat{x}>^2}$ and $Δp$ equal to $\sqrt{<\hat{p}^2>-<\hat{p}>^2}$

What can I find out about this keyword ? I want to see the proof.

 

[kuruman]

$\Delta x$ is just the root mean square (RMS) value of the difference between the average value $<x>$ and some $x$. Same for $\Delta p$.
Mathematically $(\Delta x)^2=\left< (<x>-x)^2 \right>$
Expanding the square on the right side,
$\left< (<x>-x)^2 \right>=<x>^2-2<<x>x>+<x^2>=<x>^2-2<x><x>+<x^2>=<x^2>-<x>^2$
Therefore $\Delta x=\sqrt{<x^2>-<x>^2}$.

Note: The angular brackets imply averaging or expectation value.

 

[ZapperZ]

heisenberg uncertainty principle

$Δx Δp ≥ ħ$

where

$Δx = \sqrt{<\hat{x}^2>-<\hat{x}>^2}$
$Δp = \sqrt{<\hat{p}^2>-<\hat{p}>^2}$

I don't know. Why $Δx$ equal to $\sqrt{<\hat{x}^2>-<\hat{x}>^2}$ and $Δp$ equal to $\sqrt{<\hat{p}^2>-<\hat{p}>^2}$

What can I find out about this keyword ? I want to see the proof.

I'm not sure what you mean by wanting to see the "proof". This is the definition of "standard deviation" in field of statistics. Show those expressions to a statistician who knows nothing about QM, and he/she should be able to tell you what they are.

Zz.