Is it possible to measure simultaneously the x and y positions of a particle exactly?
1 Answer
Yes. The position and momentum of a quantum mechanical particle are known as "conjugate variables", which means if you exactly measure the position of this particle, you know the momentum with much less certainty.
But there is no reason why the position of a particle would be a conjugate variable with itself. We live in three dimensions, don't we?
The Heisenberg Uncertainty Principle may be defined for the
One way to show that they can be observed simultaneously is to check the commutativity of the position operators for the
The position operators are fairly simple.
#color(green)(hatxf(x) = x*f(x))#
#color(green)(hatyf(y) = y*f(y))#
That is, the operation is simply to multiply the position by the function that describes the particle. We'll also introduce a simple zero operator, which we'll use as well:
#color(green)(hat0f(y) = 0*f(y) = 0)#
What we want to do is evaluate the following expression:
#hatxhatyf(x,y) stackrel(?" ")(=) hatyhatxf(x,y)#
We can do that by checking the following relationship:
#\mathbf([hatx,haty] = hatxhaty - hatyhatx stackrel(?" ")(=) 0)#
If this turns out to be true, then we can observe these two positions simultaneously to any precision. For this, we write:
#[hatx,haty] stackrel(?" ")(=) 0#
#[hatxhaty - hatyhatx]f(x,y) stackrel(?" ")(=) hat0[f(x,y)]#
#hatxhaty[f(x,y)] - hatyhatx[f(x,y)] stackrel(?" ")(=) 0*f(x,y)#
#hatx[y*f(x,y)] - haty[x*f(x,y)] stackrel(?" ")(=) 0#
#xy*f(x,y) - yx*f(x,y) stackrel(?" ")(=) 0#
#xy*f(x,y) - xy*f(x,y) stackrel(?" ")(=) 0#
#[xy - xy]*f(x,y) stackrel(?" ")(=) 0#
#0*f(x,y) stackrel(?" ")(=) 0#
#color(blue)(0 = 0)#
Thus, these operators commute, and we can observe the
