Poisson bracket

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In classical mechanics, the time dependence of some function $f(p_i,q_i,t)\,$ is given by

$\frac{df}{dt} = \{f, H\}+ \frac{\partial f}{\partial t}\,$ ,

where

$\{f,g\} = \sum_{i=1}^{N} \left( \frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} - \frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}} \right)$ ,

is the Poisson bracket of $f\,$ and $g\,$ . Being of fundamental importance in classical mechanics, we would like to see what happens to the Poisson bracket when $f\,$ and $g\,$ are replaced by non-commuting operators.

1 Commutator as derivation
2 Classical limit
3 Coherent states
4 References
5 References

Commutator as derivation

First, let us note that the commutator has the same algebraic properties as a derivation acting on matrices. For instance, compare

$[A,BC] = [A,B]C + B[A,C] \,$ ,

with

$\frac{d}{ds} BC = \frac{dA}{ds}C + B\frac{dC}{ds} \,$ .

Let us therefore make an algebraic association between the two. Then

$\left[q, p\right]\,$	$= i \hbar\,$
	$= i \hbar \frac{\delta p}{\delta p}\,$ .
$\left[p, q\right]\,$	$= - i \hbar\,$
	$= - i \hbar \frac{\delta q}{\delta q}\,$ .

More generally,

$\left[q, A \right]\,$	$= i\hbar \frac{\delta A}{\delta p}\,$ .
$\left[p, A \right]\,$	$= - i\hbar \frac{\delta A}{\delta q}\,$ .

For example, $A = pq\,$ .

$\left[A, B \right]\,$	$=\left[pq, B\right]\,$
	$= p\left[ q, B\right] + \left[p, A\right]q \,$ .
	$= i \hbar \left( p\frac{\delta B}{\delta p} - \frac{\delta B}{\delta q} q \right)\,$ .
	$= i \hbar \left( \frac{\delta A}{\delta q} \frac{\delta B}{\delta p} - \frac{\delta B}{\delta q} \frac{\delta A}{\delta p}\right)\,$ .

Now, consider the commutator of two functions $F(q,p)\,$ , and $G(q,p)\,$ where $G = BC(...)\,$ and where $B,C,...\,$ stand in for either $p\,$ or $q\,$ . In other words, $G\,$ is some product of the operators $p\,$ and $q\,$ . Then

$\left[F, G\right]=\left[F, B\right]C(...) + B\left[F, C\right](...) + BC\left[F, (...)\right]\,$ .

Some commutators are of the form $[F,q]\to -i \hbar \tfrac{\delta F} {\delta q}\,$ while others are of the form $[F,p] \to i \hbar \tfrac{\delta F} {\delta p}\,$ . Therefore, let us adopt the notation that

$\left[F, G\right]=\frac{\delta G}{\delta q}_{\to \left[F,q\right]} + \frac{\delta G}{\delta p}_{\to \left[F,p\right]}\,$ ,

where the operator $\frac{\delta}{\delta q}_{\to\left[F,q\right]}\,$ instructs us to remove each instance of $q\,$ (as if differentiating), and replace it with $\left[F,q\right]\,$ . In other words,

$\frac{\delta G(p,q)}{\delta q}_{\to \left[F,q\right]} \equiv \left. \frac{d}{d \epsilon} G(p, q + \epsilon \left[F,q\right] ) \right|_{\epsilon \to 0}\,$ .

Then,

$\left[F, G\right] =i \hbar \left( \frac{\delta G}{\delta p}_{ \to \frac{\delta F}{\delta q} } - \frac{\delta G}{\delta q}_{\to \frac{\delta F}{\delta p} } \right)\,$ .

At this point we can use the linearity of the commutator to allow $G\,$ to be any function of $q\,$ and $p\,$ .

Classical limit

Note that, if $p\,$ and $q\,$ were commuting operators, then the placement of (for example) $\tfrac{\delta F}{\delta q}\,$ would be irrelevant, and we could write

$\left[F, G\right] \approx i \hbar \left( \frac{\delta F}{\delta q} \frac{\delta G}{\delta p} - \frac{\delta F}{\delta p} \frac{\delta G}{\delta q} \right)\,$ .

Since reordering the operators costs at least a factor of $\hbar\,$ , we could write

$\left[F, G\right] \approx i \hbar \left( \frac{\delta F}{\delta q} \frac{\delta G}{\delta p} - \frac{\delta F}{\delta p} \frac{\delta G}{\delta q} \right) + O(\hbar^2)\,$ .

Therefore, in the classical limit in which $\hbar \to 0\,$ , the Poisson bracket and commutator are equal up to a factor. In going from classical mechanics to quantum mechanics, we should therefore make the replacement