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sometimes I see [\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2)

what does the last term O(\hbar^2) mean?

[tex]x=y[/tex]

what does the last term O(\hbar^2) mean?

[tex]x=y[/tex]

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what does the last term O(\hbar^2) mean?

[tex]x=y[/tex]

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the latex is not working....

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Hurkyl

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selfAdjoint

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Hurkyl said:O(x^2)is essentially just shorthand for "possibly plus some more terms involving powers ofxwith 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"

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Hurkyl

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So, in this context, O(x^2) is hiding something that goes to zero quadratically (or faster!) as x -> 0... and

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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??

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finally latex is back:

here's the equation that's puzzling me:

[tex][\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2)[/tex]

without [tex]O(\hbar^2)[/tex], the equation is just the usual canonical quantization recipe. what is that term for?

here's the equation that's puzzling me:

[tex][\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2)[/tex]

without [tex]O(\hbar^2)[/tex], the equation is just the usual canonical quantization recipe. what is that term for?

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selfAdjoint

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kakarukeys said:finally latex is back:

here's the equation that's puzzling me:

[tex][\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2)[/tex]

without [tex]O(\hbar^2)[/tex], 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.

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samalkhaiat

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kakarukeys said:Read post#7 in the thread "transition from poisson brackets to ..." if you follow my derivation, you will be able to see where the [tex]O(\hbar^{2})[/tex] term comes from.

O(something) stands for "Of Order Of" that "something".

sam

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