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I believe this is a simple question, however I cannot find it explained anywhere what the term:

"Impose canonical commutation relations" means.

If I have a classical equation, and I wish to quantise it, I will first promote the classical coordinates to operators, nothing more than notation change. What is the process to then impose the commutation relations for the operators that I have promoted the coordinates to?

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  • $\begingroup$ Thank you, however I think I am still missing something. I do not see how this will change my equation. For example, if I have a classical Hamiltonian for the harmonic oscillator, and I want to quantise it, how will the Hamiltonian change other than putting hats onto the coordinates, and is the method the same for any Hamiltonian? $\endgroup$
    – Sean
    Commented Dec 5, 2015 at 10:44
  • $\begingroup$ Related: physics.stackexchange.com/q/19770/2451 , physics.stackexchange.com/q/46988/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Dec 5, 2015 at 11:06

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Not sure if I understood your question totally. But when you need to change a classical entity to its quantum counterpart you need to make sure that the symmetry is observed if they are not commuting. For instance:

$xp$---change to quantum mechanical operators----> $\frac{1}{2}(\hat x\hat p+\hat p\hat x)$

or

$x^2p$---change to quantum mechanical operators----> $\frac{1}{4}(\hat x^2\hat p+2\hat x\hat p\hat x+\hat p\hat x^2)$

where $x$ and $p$ are position and momentum operators, respectively.

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  • $\begingroup$ That is absolutely what I was unsure about, thank you. So in my equation, ensuring symmetry is observed for the non commuting operators I will end up with a longer equation as you have suggested above? Because xp and px are equivalent, as are 1/2(x'p'+p'x') and 1/2(p'x'+x'p')? --(with primes for hats) $\endgroup$
    – Sean
    Commented Dec 5, 2015 at 11:15
  • $\begingroup$ Yes. Replace your classical operators with the quantum ones the way that I mentioned. You can be more specific if you like then I can help you. $\endgroup$
    – Amin
    Commented Dec 5, 2015 at 11:18
  • $\begingroup$ That is okay, thank you, that does make sense to me now. $\endgroup$
    – Sean
    Commented Dec 5, 2015 at 11:20
  • $\begingroup$ If you found the answer correct please mark it so people with same problem can rely on that :) $\endgroup$
    – Amin
    Commented Dec 5, 2015 at 11:23
  • $\begingroup$ Unfortunately I need 15 rating to do so,but if I get it, I will do! $\endgroup$
    – Sean
    Commented Dec 5, 2015 at 11:42

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