Question

Why is the Heisenberg uncertainty principle not significant when describing macroscopic object behavior?

Answer 1

The basic idea is that the smaller an object gets, the more quantum mechanical it gets. That is, it's less able to be described by Newtonian mechanics. Whenever we can describe stuff using something like forces and momentum and be quite sure about it, it's when the object is observable. You can't really observe an electron whizzing around, and you can't catch a runaway proton in a net. So now, I guess it's time to define an observable.

The following are the quantum mechanical observables:
Position
Momentum
Potential Energy
Kinetic Energy
Hamiltonian (total energy)
Angular Momentum

They each have their own operators, such as momentum being `(-ih)/(2pi)d/(dx)` or the Hamiltonian being `-h^2/(8pi^2m)delta^2/(deltax^2)` for a one-dimensional inescapable boundary with infinitely tall walls (Particle in a "Box").

When these operators are used on each other, and you can have them commute, you can observe both corresponding observables at once. The quantum mechanics description of the Heisenberg Uncertainty Principle is as follows (paraphrased):

If and only if `[hatx,hatp] = hatxhatp - hatphatx = 0`, both position and momentum can be observed at the same time. Otherwise, if the certainty in one is good, the uncertainty in the other is too large to provide good enough assurance.

Let's see how that works out. The position operator is just when you multiply by `x`. The momentum operator is, as stated above, `(-ih)/(2pi)d/(dx)`, which means you take the derivative and then multiply by `(-ih)/(2pi)`. Let's see why they don't commute:

`x[(-ih)/(2pi)d/(dx)] - [(-ih)/(2pi)d/(dx)]x = 0?`

Operate on x by taking its first derivative, multiplying by `(ih)/(2pi)`, and changing `-(-u)` to `+u`.
`cancel(x[(-ih)/(2pi)d/(dx)]1) + (ih)/(2pi) = 0?`

Oh, look at that! The derivative of 1 is 0! So you know what, `x*(-ih)/(2pi)*0 = 0`.

And we know that can't be equal to 0.
`(ih)/(2pi) != 0`

So, that means position and momentum do not commute. But, this is only an issue with something like an electron (so, a fermion) because:
- Electrons are indistinguishable between each other
- Electrons are tiny and very light
- Electrons can tunnel
- Electrons act like waves AND particles

The larger the object is, the more sure we can be that it obeys the standard laws of physics, so the Heisenberg Uncertainty Principle only applies to those things that we can't readily observe.