# Transition from Poisson brackets to commutors?

#### dantypas

Hi to everyone.

I am a new member in this forum. I was wondering if there is a rigorous proof
on to how one passes from Poisson brackets to commutor relations in QM. Any help on that would be appreciated.

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#### jtbell

Mentor
As far as I know, the rule that you can make the transition from CM to QM by replacing Poission brackets with commutators can be viewed as a postulate, from which you can derive QM (maybe with additional assumptions/postulates, I'm not very familiar with this). The "proof" is experimental, that is, the physical predictions of QM agree with experiment, at least to date.

Or are you looking to go the other way, from QM as derived from other postulates to the fact that commutators correspond to Poisson brackets in CM?

#### lalbatros

dantypas,

In principle, it should be possible to proof it in the reverse direction.
In the limit hbar->0 , the operators & commutators formalism should translate in Poisson brackets.

It is well known that the Schrödinger equation can lead to the classical least action principle of CM and the Lagragian formulation. This simple fact garantees that the proof will work. However, I have never seen a proof QM->CM focusing on the {} formulation.

Michel

#### samalkhaiat

lalbatros said:
However, I have never seen a proof QM->CM focusing on the {} formulation.
I can show you how to do it, if you want to.

sam

#### Careful

dantypas said:
Hi to everyone.

I am a new member in this forum. I was wondering if there is a rigorous proof
on to how one passes from Poisson brackets to commutor relations in QM. Any help on that would be appreciated.
Hi, it is rather well known that there is no unique way to do this: the troubles being collectively named as normal ordering problems and anomalies. Usually one chooses x,p as basic variables, quantizes the corresponding poisson brackets (which comes down to working in the Schrodinger representation) and builds up the representation from there.

#### lalbatros

samalkhaiat said:
lalbatros said:
I can show you how to do it, if you want to.

sam
Yes I am interrested.
If possible, give us just a hint about how to do it.
Would it start by the evoluation of the density operator?

A good opportunity to fill a gap for me,

Michel

#### samalkhaiat

lalbatros said:
samalkhaiat said:
Yes I am interrested.
If possible, give us just a hint about how to do it.
It is lengthy and requires some tricks.

Would it start by the evoluation of the density operator?
NO, it is rather general. Here we go;

take the matrix element of commutator of two arbitrary operators in the coordinate representation;

$$\left[a,b\right]_{xx'} = \int dx'' [a(x,x'')b(x'',x') - b(x,x'')a(x'',x')]$$

change the arguments of all matrix elements as follow;

$$a(x,x'') = a\left( \frac{x+x''}{2}, x-x'' \right)$$
$$b(x'',x') = b\left( \frac{x''+x'}{2}, x''-x' \right)$$
....
....
etc.

now, expand the RH-sides in terms of momentum eigenfunctions;

$$a(x,x'') = \frac{1}{2\pi \hbar} \int dp e^{ip(x-x'')/ \hbar} a\left[ \frac{x+x''}{2} ,p\right]$$
$$b(x'',x') = \frac{1}{2\pi \hbar} \int dp' e^{ip'(x''-x')/ \hbar} b\left[ \frac{x''+x'}{2} ,p'\right]$$

In the semiclassical limit we take the differences between the coordinates to be small (wave packets).
Thus, we expand all Fourier amplitudes on the RHS in power series in
$$\delta x = x - x''$$ and $$\delta x'' = x'' - x'$$

keeping linear terms;

$$a(x,x'') = \frac{1}{2\pi \hbar} \int dp e^{ip\delta x/ \hbar} \left[ a(x,p) + \frac{1}{2} \delta x \partial_{x}a(x,p) + .. \right]$$
$$b(x'',x') = \frac{1}{2\pi \hbar} \int dp' e^{ip'\delta x''/ \hbar} \left[ b(x',p') - \frac{1}{2} \delta x'' \partial_{x'}b(x',p') + .. \right]$$

(note that these equations give the (semi)classical limit
$$x'' \rightarrow x$$
of any matrix element:
$$c(x' \rightarrow x) = \frac{1}{2\pi \hbar} \int dp e^{ip(x-x')/ \hbar} c(x,p)$$
later we use this for c = [ , ])

Let us go back and integrate the linear terms by parts

$$a(x,x'') = \frac{1}{2\pi \hbar} \int dp e^{ip\delta x/ \hbar} \left[ a(x,p) - \frac{\hbar}{2i} \partial_{x}\partial_{p}a(x,p) + O(\hbar^2) \right]$$
$$b(x'',x') = \frac{1}{2\pi \hbar} \int dp' e^{ip'\delta x''/ \hbar} \left[ b(x',p') + \frac{\hbar}{2i} \partial_{x'}\partial_{p'}b(x',p') + O(\hbar^2) \right]$$

Now, take a coffee break, then substitute these and other expansions of the matrix elements into the commutator and use the properties of Dirac's delta

$$\left[\hat{a},\hat{b}\right]_{xx'} = \frac{1}{2\pi \hbar} \int dp e^{ip(x-x')/ \hbar} \left[a(x,p)b(x',p) - a(x',p)b(x,p) - \frac{\hbar}{2i} \left( b(x',p)\partial_{x}\partial_{p}a - a(x,p)\partial_{x'}\partial_{p}b + b(x,p)\partial_{x'}\partial_{p}a - a(x',p)\partial_{x}\partial_{p}b \right) \right]$$

Now we integrate by parts all the terms in (...) and pass to the limit by putting $$x' \rightarrow x$$ everywhere except the exponential.
For example:
$$\int dp e^{ip(x-x')/ \hbar} \partial_{x}\partial_{p}ab(x',p) \rightarrow - \int dp e^{ip(x-x')/ \hbar} \partial_{x}a \partial_{p}b$$

Thus the commutator in the semiclassical limit

$$\left[\hat{a},\hat{b}\right]_{x'\rightarrow x} = \frac{1}{2\pi \hbar} \int dp e^{ip(x-x')/ \hbar} \frac{\hbar}{i} [\partial_{p}a\partial_{x}b - \partial_{x}a\partial_{p}b]$$

We almost there.Now the semi-classical limit of any matrix element
$$[\hat{a},\hat{b}]_{x'\rightarrow x} = \frac{1}{2\pi \hbar} \int dp e^{ip(x-x')/ \hbar} \left{[\hat{a},\hat{b}](x,p) + O(\hbar^{2}) \right}$$
By comparing Fourier amplitudes in the above two equations you get it
$$Lim_{\hbar \rightarrow 0} \frac{i}{\hbar} [a,b] = [a,b]_{PB}$$

A good opportunity to fill a gap for me,
It is a gap in all textbooks.

regards

sam

Last edited:

#### lalbatros

Thanks a lot samalkhaiat.
Quite a hard Latex job!

With my best regards,

Michel

#### samalkhaiat

lalbatros said:
Thanks a lot samalkhaiat.
Quite a hard Latex job!

With my best regards,
Playing with equations is my game.

regards

sam

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