Question

According to the Heisenberg uncertainty principle if the position of a moving particle is known what other cannot be known?

Answer 1

Its velocity. However, the particle has to be microscopic in order for this phenomenon to be most applicable. For macroscopic objects, for most intents and purposes, you can observe both its position and velocity.

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You can prove the Heisenberg Uncertainty Principle using something called operators. The ones you use would be the position operator `hatx` and momentum operator `hatp`. Since momentum is proportional to velocity, it follows that if you know the position of a moving microscopic particle, you don't know its velocity to a good enough certainty.

These two operators are defined as follows:

`hatx[f(x)] = x*f(x)`
(left-multiply `x` by the function)

`hatp[f(x)] = (-ih)/(2pi)*d/(dx)[f(x)]`
(take the derivative of the function, then left-multiply by `(-ih)/(2pi)`)

These two operators represent observables---that is, events you can observe in real life. The position operator represents the position of an object, and the momentum operator represents its momentum.

What the Heisenberg Uncertainty Principle depends on is that these two operators mathematically commute. In other words, if you use these operators to operate on (affect) `f(x)`...

If `hatxhatp - hatphatx = 0`, then these operators commute, and thus you can observe both phenomena in real life.

Operators affect `f(x)` from right to left. Let us see how that turns out:

`hatxhatp[f(x)]`

`= x*(-ih)/(2pi)*d/(dx)[f(x)]`

`= color(green)(overbrace([(-ixh)/(2pi)d/(dx)])^(hatxhatp)[f(x)])`

`hatphatx[f(x)]`

`= (-ih)/(2pi)*d/(dx) [x*f(x)]`

`= (-ih)/(2pi) [x d/(dx)[f(x)]+ f(x)]`

`= (-ixh)/(2pi)d/(dx)[f(x)] + (-ih)/(2pi) [f(x)]`

`= (-ixh)/(2pi)d/(dx)[f(x)] -(ih)/(2pi) [f(x)]`

`= color(green)(overbrace([(-ixh)/(2pi)d/(dx) -(ih)/(2pi)])^(hatphatx)[f(x)])`

Now, comparing them:

`hatxhatp[f(x)] - hatphatx[f(x)] stackrel(?)(=) 0`

`[hatxhatp - hatphatx]f(x) stackrel(?)(=) 0`

`[hatxhatp - hatphatx] stackrel(?)(=) 0`

`[overbrace(cancel(-(ixh)/(2pi)d/(dx)))^(hatxhatp) - overbrace((cancel(-(ixh)/(2pi)d/(dx)) -(ih)/(2pi)))^(hatphatx)] stackrel(?)(=) 0`

`color(blue)([(ih)/(2pi)] ne 0)`

Therefore, these two operators do not commute, and you can say that for a moving microscopic particle, you cannot observe both of these at once---only one at a time.

Since momentum is proportional to velocity (`p = mv`), when momentum cannot be observed, neither can velocity.

(Velocity is more straightforward to observe, so we might sometimes say velocity instead of momentum.)