What Does O(\hbar^2) Mean in Commutation Relations?

In summary, the conversation discussed the meaning of the last term O(\hbar^2) in the equation [\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2), which is used in asymptotic notation and represents a bound on the rate of increase of the function. This term is typically used in power series expansions and is often ignored in the context of canonical quantization. The discussion also included a specific citation on the topic for further examination.
  • #1
kakarukeys
190
0
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]
 
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  • #2
the latex is not working...
 
  • #3
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
Hurkyl said:
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:
 
  • #5
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
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
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
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?
 
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  • #9
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.
 
  • #11
kakarukeys said:
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 [tex]O(\hbar^{2})[/tex] term comes from.
O(something) stands for "Of Order Of" that "something".
sam
 
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1. What is the significance of the symbol \hbar in commutation relations?

The symbol \hbar, also known as "h-bar", is the reduced Planck's constant in quantum mechanics. It is a fundamental constant that relates to the discrete nature of quantum systems and is used to determine the uncertainty in measuring a particle's position and momentum simultaneously.

2. How is \hbar^2 related to commutation relations?

In quantum mechanics, commutation relations describe how two physical quantities, such as position and momentum, behave when measured simultaneously. \hbar^2 is often seen in these equations and represents the square of the reduced Planck's constant. It is used to calculate the uncertainty in the measurement of these quantities.

3. What does O(\hbar^2) mean in commutation relations?

O(\hbar^2) is known as the "order of magnitude" of \hbar^2 in commutation relations. This means that \hbar^2 is being used as a small quantity compared to other terms in the equation. It represents the level of precision in calculating the uncertainty between two quantities.

4. Why is \hbar^2 often used in commutation relations?

Since \hbar^2 is a small quantity, it is often used in commutation relations to represent the uncertainty in measuring two quantities simultaneously. This is because, in quantum mechanics, the more precisely one quantity is measured, the less precisely the other can be measured. \hbar^2 helps to quantify this uncertainty.

5. How does \hbar^2 affect the behavior of quantum systems?

The value of \hbar^2 can affect the behavior of quantum systems in terms of the uncertainty principle. The smaller the value of \hbar^2, the more precise the measurements of two quantities can be, but this also means that the uncertainty in the third quantity will increase. This is a fundamental concept in quantum mechanics that helps to explain the probabilistic nature of quantum systems.

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