How do you prove that momentum and position cannot be observed simultaneously as per Heisenberg's Uncertainty Principle?
1 Answer
There is a quantum mechanical proof of this involving operators. Just a heads-up that it is difficult, so if you have any questions on it, please ask. I've written it out below.
Physical chemists have used operators to express the Heisenberg Uncertainty Principle.
An operator must correspond directly to an observable; that is, it must describe a real, physical property or event.
Therefore, these two operators for position and momentum respectively describe the position and momentum (and thus velocity) of a quantum mechanical particle such as an electron:
#hatx[f(x)] = x*f(x)#
(left-multiply the function by#x# )
#hatp[f(x)] = (-ih)/(2pi)*d/(dx)[f(x)]#
(take the derivative of the function, then left-multiply by#(-ih)/(2pi)# )
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)
If
Operators affect
#= x*(-ih)/(2pi)*d/(dx)[f(x)]#
#= color(green)(overbrace([(-ixh)/(2pi)d/(dx)])^(hatxhatp)[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)#
This result means that they do NOT commute, which means that the position and momentum (and thus velocity) of a quantum mechanical particle such as an electron cannot be observed simultaneously to a reasonable level of certainty.
