# Commutation relation

1. Aug 18, 2006

### kakarukeys

sometimes I see [\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2)
what does the last term O(\hbar^2) mean?

$$x=y$$

Last edited: Aug 18, 2006
2. Aug 20, 2006

### kakarukeys

the latex is not working....

3. Aug 20, 2006

### Hurkyl

Staff Emeritus
It's asymptotic notation. Since you're probably in a context where everything's analytic... O(x^2) is essentially just shorthand for "possibly plus some more terms involving powers of x with exponent at least 2".

4. Aug 20, 2006

Staff Emeritus

It actually means "what I have ignored doesn't increase toward infinity any faster than the quadratic function y = x^2." The big O notation sets a bound on the rate of increase of the function as the argument increases without bound. In practice, as physicists use it, it means "All power series stop at the linear term"

5. Aug 20, 2006

### Hurkyl

Staff Emeritus
AFAIK, the O-notation is being used in the opposite direction here. They're interested in what happens as the argument goes to zero, not when it goes to infinity!

So, in this context, O(x^2) is hiding something that goes to zero quadratically (or faster!) as x -> 0... and not to hide something that goes to infinity quadratically (or slower) as x -> infinity, as you would expect in other contexts.

6. Aug 22, 2006

### kakarukeys

I think I understand the notation....
But undergraduate physics only taught me [x,p] = i hbar
there are no other terms behind, looks like the equation is suggesting there are some terms behind, and [x,p] = i hbar is only a first order approximation??

7. Aug 22, 2006

### joelperr

Since you have provided no context in which [x,p] = i*hbar appears, the most likely answer is that it is exact, since this is a basic Poisson Bracket commutation relation in QM that is seen in quite a few places.

8. Aug 23, 2006

### kakarukeys

finally latex is back:
here's the equation that's puzzling me:
$$[\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2)$$
without $$O(\hbar^2)$$, the equation is just the usual canonical quantization recipe. what is that term for?

Last edited: Aug 23, 2006
9. Aug 23, 2006

Staff Emeritus

Usually this comes up where they are showing the commutation relationship applies in the context of something they are expanding in a power series, and then they ignore all terms except the first to get the CR. It would really help if instead of saying "sometimes I find" this expression, you would give some specific citation that people could examine.

10. Sep 11, 2006

### kakarukeys

11. Sep 13, 2006

### samalkhaiat

Last edited: Sep 13, 2006