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) <br />
- \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
