Zero-point energy amplitude calculation

Question

On this page

https://www.miniphysics.com/simple-harmonic-oscillator.html

It is stated that for a linear restoring force of $F = -k \Delta x$, the total energy is

$ E = K + U $

or rather

$ \\ E = \frac{(\Delta p)^2}{2m} + \frac{1}{2}m\omega^2(\Delta x)^2 $

Then, the uncertainty principle is applied to say that $\Delta x\Delta p \approx \hbar/2$ and this is then used to reduce the expression for the energy to be solely in terms of $\Delta x$.

From that point,

$ \frac{d E}{d(\Delta x)} = 0 $

is calculated to show that, in general, assuming the uncertainty principle, it must follow that the Zero-Point energy amplitude is

$ \Delta x = \sqrt{\frac{\hbar}{2m\omega}}. $

Why is it valid to assume that the $\Delta p$ in the expression for the energy and the $\Delta p$ in the uncertainty principal are interchangeable? Doesn't the $\Delta p$ in the uncertainty principal stand for the standard deviation, while the $\Delta p$ that appears in the expression stand for instantaneous momentum?

Answer 1

a) The instantaneous momentum is represented by $dp$ and not $\Delta$p.Its more universally accepted , unless explicitly mentioned its just a matter of choice.

$\Delta$p is a finite change.

b) The author tells that the forces between atoms in a solid can be approximated by this kind of interaction and thats why $\Delta$p is used as uncertainty in the momentum.

c)The $\Delta$x here is actually the uncertainty in position in which the particle is most likely to be found even when the temperature is taken to be very low is what the author says. Now consider the third law of thermodynamics which states that at absolute zero temperature, "which is the lowest possible temperature", the atoms or molecules arrange themselves in perfect positions ,with no motion at all. However we just showed that there is this zero point energy which is omnipresent which never allows any particles to settle completely which by the way is due to Heisenberg's Uncertainty principle.