Quantum Mechanics: Uncertainty and Commutation relation

[Xyius]

I am stuck on one part of my Quantum Mechanics HW. Above the question it says "Try and answer the following question." So I can only assume that he isn't looking for something incredibly detailed. (Ill explain why after the question is given.)

Homework Statement

What is the meaning of the uncertainty relation [itex]\Delta \overline{x}\overline{p}≤\hbar /2[/itex] associated with the commutation relation [itex][\overline{x},\overline{p}]=i \hbar[/itex] (The bars over x and p are supposed to be hats, as in a unit vector but I do not know how to do them with Latex.)

The Attempt at a Solution

At this point in time we only had one class and the whole first chapter of the book doesn't delve much into either one of these relations. It says that
[tex][\overline{A},\overline{B}]=AB-BA[/tex]
That is really all I know about this relation. There is nothing in the lecture notes and nothing about these relations until Chapter 3. Can anyone lead me in the right direction?

 

[zhermes]

The relationship between the uncertainty relation and the commutation relation given is somewhat complex; but I think they're asking more about the uncertainty relation itself---which (happens) to be associated with the given commutation relation.

 

[Simon Bridge]

I think you mean: [itex]\Delta x \Delta p_x \leq \hbar/2\pi[/itex] in terms of [itex][\hat{x},\hat{p}_x]=i\hbar[/itex] and I think he means the hats to indicate operators.

The 1D x-p commutation looks like:[tex]\int_\infty \psi^\star (\hat{x}\hat{p}_x-\hat{p}_x\hat{x})\psi.dx = i\hbar[/tex][tex]\hat{x}=x \; ; \qquad \hat{p}_x=-i\hbar\frac{d}{dx}[/tex]

The question is about what the uncertainty relation means - which you can presumably describe OK ... but they want to to describe it in terms of the commutation. The commutation implies Heisenberg's uncertainty - you have to find out how and write it down.

[aside]

The bars over x and p are supposed to be hats, as in a unit vector but I do not know how to do them with Latex

[itex]\text{\hat{a}} \rightarrow \hat{a}[/itex]

LaTeX reference - get one of the pdf's (bottom of page).
For PF you only need the chapters on math typesetting, but hey: it's all good.
There's a PFLaTeX guide someplace around here but I keep losing it.

 

[Xyius]

I do not know what commutation means though. This question is throwing me for a loop because this was not discussed at all in our lecture. And there is very little on it in the first chapter of the book. There is a proof that relates the generalized uncertainty principle with the commutation relation but how can I describe what anything means if I do not know what commutation means? I know it is defined as AB-BA, but I do not know the physical significance of this definition and how it relates the the uncertainty principle. :(

 

[Xyius]

Wait! I think I may be getting somewhere! (just did some more reading and something clicked!)
So for two physical quantities to be simultaneously observable then [A,B]=0. But since [itex][A,B]=i\hbar[/itex] this implies only one quantity to be observable. Relating this to the uncertainty principle, this means that either x OR p can be observable not both.

Am I close?

 

[Simon Bridge]

Close - they are both observable, but the quantities are related in such a way that the uncertainties of the measurements are inversely related. They are not independently observable. The commutator describes and quantifies this.

 

[conquest]

Exactly you can actually see that quite clearly. Picture applying an operator to a wavefunction as measuring it for the property that belongs to that operator (x is measuring location, p is measuring momentum). The fact that the commutator is not zero means ,measuring first momentum and then location does not lead to the same results as first measuring location and then momentum.

The commutator gives you a description of the difference which then gives rise to the uncertainty relation. So be prepared there is not just one uncertainty relation all non-commuting operators give rise to one.

 

[Xyius]

Cool! Thanks everyone!