- #1
KFC
- 488
- 4
Hi all,
I am reading an article about uncertainty principle. If we consider a Gaussian wave packet which standard deviation of momentum ##\sigma_p##. The uncertainty principle states that the multiplication of variance of x and variance of p is larger or equal to half ##\hbar##
##\Delta x\Delta p \geq \dfrac{\hbar}{2}##
I think ##\Delta x## has the unit of meter, ##\Delta p## has the unit of kg.meter/second, so the multiplication of them give the same unit of ##\hbar##.
But if we have the Gaussian wave packet, the standard deviation of ##\sigma_x## and ##\sigma_p## should have the unit of meter and kg.meter/second. But reading the expression of variance for Gaussian given by standard deviation
##
\Delta x = \sigma_x^2, \Delta p = \sigma_p^2
##
So the unit for ##\Delta x \Delta p ## becomes kg*meter^4/second^2 ? I am confusing what mistakes I made here.
I am reading an article about uncertainty principle. If we consider a Gaussian wave packet which standard deviation of momentum ##\sigma_p##. The uncertainty principle states that the multiplication of variance of x and variance of p is larger or equal to half ##\hbar##
##\Delta x\Delta p \geq \dfrac{\hbar}{2}##
I think ##\Delta x## has the unit of meter, ##\Delta p## has the unit of kg.meter/second, so the multiplication of them give the same unit of ##\hbar##.
But if we have the Gaussian wave packet, the standard deviation of ##\sigma_x## and ##\sigma_p## should have the unit of meter and kg.meter/second. But reading the expression of variance for Gaussian given by standard deviation
##
\Delta x = \sigma_x^2, \Delta p = \sigma_p^2
##
So the unit for ##\Delta x \Delta p ## becomes kg*meter^4/second^2 ? I am confusing what mistakes I made here.