Commutators and physical interpretation

[vsage]

An interesting question was posed, and since I have many problems of this type I'll just make the question general:

Suppose you have operations A and B, if [A, B] != 0, then what can you conclude about a simultaneous measurement of A and B? For example, if A was momentum in the x direction's operator and B was the position operator, what does a nonzero commutator mean?

The best answer I can guess is that depending on which operator is done first, the result will be different and therefore a simultaneous operation to get the information from both operators is impossible, or more accurately gives physically meaningless or incorrect results. Thoughts?

 

[xman]

Since the commutator of measurable quantities commutes, then we are able to measure both quantities simultaneously. On the other hand, when the commutator between measurables does not commute we are unable to measure both with any certainty. The famous Heisenberg Uncertainty Principle (HUP) is exactly this statement regarding position and momentum. If I remember correctly, the result of the HUP is actually just a mathematical fact, but has profound physical significance.

 

[Sojourner01]

xman is correct. The commutator defines the uncertainty in the two measurements; specifically, the product of the uncertainties of each measurement. If the commutator is zero, the uncertainty of each variable is zero and so both can be measured with complete certainty simultaneously.

If the commutator is not zero (say as in your example), the uncertainty in position can be expressed as the commutator divided by some arbitrary uncertainty in momentum. If you give the momentum an uncertainty of zero, the uncertainty of position becomes infinite and so you have no idea where the particle is; and vice versa.

 

[vsage]

Duely noted. I sort of understood the answer but was unsure of how to articulate it correctly. Thanks for the insights.