Power Series For Function of Operators

[ghotra]

Hi, I'm looking for a general power series for a function of F of n operators. As normal, the operators do not necessarily commute.

My first guess was:

F(x,p) = \sum_{i=0}^\infty \sum_{j=0}^\infty a_{ij} x^i p^j + b_{ij}p^i x^j


However, I don't think this is correct as it is possible to have operators between x and p.
So then I thought that this might be the correct expansion:

F(x,p) = \sum_{i=0}^\infty \sum_{j=0}^\infty a_{ij} x^i b_{ij} p^j + c_{ij} p^i d_{ij} x^j


Obviously, I'm just guessing here and could use some help. Why do I care about this? I am trying to derive a formula for

[F(x_1,....,x_n), G]

Suppose I know how G commutes with each of the operators x_i. I want to get an expansion for G commuting with some function F of the x_i operators. I'll work on the details of that, but first I need some help with the power series of F.

[dextercioby]

There exists a simple rule, if you'd be applying your discussion to quantum mechanics.

Daniel.

[ghotra]

So here is what I am actually trying to do. I have:


[P^j,\phi_r(x)] = -i \hbar \frac{\partial\phi_r(x)}{\partial x_j}


and


[P^j,\pi_r(x)] = -i \hbar \frac{\partial\pi_r(x)}{\partial x_j}


For a function F\left(\phi_r(x),\pi_r(x)\right), I need to show the following:


[P^j,F\left(\phi_r(x),\pi_r(x)\right)] = -i \hbar \frac{\partial}{\partial x_j}F\left(\phi_r(x),\pi_r(x)\right)


I was thinking of considering the various commutator relations:


[P^j,\phi^n_r(x)]


but since the operators don't commute, there are (too) many possible combinations to consider. I would be interested in knowing this trick you speak of.

Thanks.

[dextercioby]

Use the Poisson bracket and Dirac's rule giving the canonical quantization.

Daniel.

[ghotra]

Could someone spell this out for me? I have convinced myself that there is no pretty way to write a power series for a function of operators (that do not necessarily commute). It seems like you'd have a sum of an infinite product...each term in the product with their own index...so you are summing over an infinite number of indices.