commutation relation

[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

[kakarukeys]

the latex is not working....

[Hurkyl]

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".

[selfAdjoint]

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".


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":rolleyes:

[Hurkyl]

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.

[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??

[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.

[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?

[selfAdjoint]

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?


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.

[kakarukeys]

here is one:
http://arxiv.org/pdf/quant-ph/9606031
page 16
could you explain what it means?

[samalkhaiat]

[QUOTE]here is one:
http://arxiv.org/pdf/quant-ph/9606031
page 16
could you explain what it means?

Read post#7 in the thread "transition from poisson brackets to ..." if you follow my derivation, you will be able to see where the O(\hbar^{2}) term comes from.
O(something) stands for "Of Order Of" that "something".
sam