In Heisenberg's Uncertainty Principle, is the uncertainty in the position a 2 dimensional, 3 dimensional, or 4 dimensional?

1 Answer
Apr 27, 2016

The easiest version of the Heisenberg Uncertainty Principle demonstrates the principle in one dimension, but it can be "expanded" to #n# dimensions, really.

"ONE-DIMENSIONAL VERSION"

The "one-dimensional version" is presented in the #x# direction as

#color(blue)(sigma_(vecx)sigma_(vecp_x) >= (h)/(4pi)),#

where:

  • #h# is Planck's constant, #6.626xx10^(-34) "J"cdot"s"#.
  • #sigma_(vecx)# is the uncertainty in the position in the #x# direction.
  • #sigma_(vecp_x)# is the uncertainty in the momentum in the #x# direction.

But this can be expanded and generalized to all three dimensions #x,y,z#. We would simply have:

#color(blue)(sigma_(vecx)sigma_(vecp_x) >= (h)/(4pi))#

#color(blue)(sigma_(vecy)sigma_(vecp_y) >= (h)/(4pi))#

#color(blue)(sigma_(vecz)sigma_(vecp_z) >= (h)/(4pi))#

GENERAL FORMULATION OF HEISENBERG UNCERTAINTY PRINCIPLE

Actually, a general formulation of the Heisenberg Uncertainty Principle to determine the capacity to observe two observables simultaneously is (Physical Chemistry: A Molecular Approach, McQuarrie):

#\mathbf([hatA,hatB] -= hatAhatB - hatBhatA stackrel(?)(=) 0)#

A quantum mechanical observable is the eigenvalue (a "fixed point") that corresponds to a given operator. So, it is what you observe in real life, with certainty that it is what you think it is, for each given time that you observe it.

If the anticommutation relation #[hatA,hatB]# is NOT equal to #\mathbf(0)#, then it is said that they do NOT commute. That is, the Heisenberg Uncertainty Principle DOES apply and you CANNOT observe the observables #A# and #B# simultaneously.

THE POSITION AND MOMENTUM OPERATORS

If we let #hatA# be #hatx# and #hatB# be #p#, then we define the following operators from quantum mechanics:

#color(green)(hatxf(x) = x[f(x)])#
(multiply by #x#)

#color(green)(hatpf(x) = -(ih)/(2pi)d/(dx)[f(x)])#
(multiply by #(-ih)/(2pi)# and take the derivative of #f(x)#; #i# is #sqrt(-1)#.)

...if and only if #f(x)# is an eigenfunction---that is, if the operation of #hatA# gives some eigenvalue #lambda# such that #hatAf(x) = lambdaf(x)#. (For example, #e^x# is an eigenfunction.)

In the above equations, #hatx# is the position operator corresponding to the position observable #vecx#, and #hatp# is the momentum operator corresponding to the momentum observable #vecp#.

AN ANTICOMMUTATION RELATION IS NONZERO FOR TWO OBSERVABLES THAT CANNOT BE OBSERVED SIMULTANEOUSLY

To perform the anticommutation relation to check if #hatxhatp = hatphatx#, we do the following:

#color(blue)([hatxhatp - hatphatx]f(x) stackrel(?)(=) 0)#

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

#= x*(-ih)/(2pi)d/(dx)f(x) - (-ih)/(2pi)d/(dx)*xf(x)#
(distribute)

#= (-ixh)/(2pi)d/(dx)f(x) - (-ih)/(2pi)[xd/(dx)f(x) + f(x)]#
(product rule)

#= (-ixh)/(2pi)d/(dx)f(x) - (-ih)/(2pi)[xd/(dx) + 1]f(x)#
(factor)

#= (-ixh)/(2pi)d/(dx)f(x) - [(-ih)/(2pi) + (-ihx)/(2pi)d/(dx)]f(x),#
(distribute)

#= [cancel((-ixh)/(2pi)d/(dx)) - ((-ih)/(2pi) + cancel((-ihx)/(2pi)d/(dx)))]f(x),#
(cancel)

#= color(blue)((ih)/(2pi))#

but #color(red)(0 ne (ih)/(2pi))#, so the position and momentum operators do NOT commute.

That means they are subject to the Heisenberg Uncertainty Principle and cannot be observed simultaneously. In other words, knowing one to maximum certainty implies knowing the other to maximum uncertainty.

Furthermore, the position operators for the #vecy# and #vecz# direction are similarly defined as:

#hatyf(y) = y[f(y)]#

#hatzf(z) = z[f(z)]#

All position operators do is multiply by the given eigenfunction #f#, so these #haty# and #hatz# operators, and similar ones for any given direction, really, function identically in the above proof.

At this point, we can then say that the Heisenberg Uncertainty Principle for position and momentum applies in any dimension.