How to Expand Noncommuting Variables in a Formal Power Series?

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Homework Help Overview

The problem involves expanding a function of noncommuting variables, specifically f(a, a^\dagger), in a formal power series while adhering to the commutation relation [a, a^\dagger] = 1.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the challenge of expanding f(a, a^\dagger) due to the noncommuting nature of the variables. There are attempts to express the function in terms of a power series and derivatives, with some questioning how to incorporate the noncommuting aspect into their expressions.

Discussion Status

The discussion is active, with participants exploring different approaches, including starting with simpler cases and using induction. Some guidance has been provided regarding the use of derivatives and the structure of the series expansion.

Contextual Notes

Participants are navigating the complexities introduced by the noncommuting variables and are considering how this affects their standard methods of expansion and differentiation.

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Homework Statement


Need to show that [a,f(a,a^\dagger]=\frac{\partial f}{\partial a^\dagger}

Homework Equations


[a,a^\dagger]=1

The Attempt at a Solution


Need to expand f(a,a^\dagger) in a formal power series. However I don´t know how to do it if the variables don´t commute.
 
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##f(a,a^+)= b_0+ b_1 a+b_2 a^+ b_3 a^2 +b_4 aa^+ +b_5 a^+a+b_6 (a^+)^2+\ldots##
So the general term is some product of a and ##a^+## in arbitrary order.
 
Thanks for answering, but how would it be in terms of derivatives? Normally It would be:

f=\sum_{m,n} \frac{a^m a^{\dagger m}}{n!m!}\frac{\partial^{n+m} f}{\partial a^n \partial a^{\dagger m}}

But in this case I guess I have to take into account that they're noncommuting.
 
Start with a simpler case, i.e., ##f = (a^\dagger)^n## only. Use induction (on ##n##) to show that the desired formula holds. Once you understand the induction method for this problem, you'll probably work it out for more general ##f## more easily.
 
Thanks, that did the trick!
 

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