Newton's expansion for non-commutative quantities

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SUMMARY

The discussion focuses on the extension of Newton's binomial expansion to non-commutative quantities, specifically when two operators, x and y, do not commute, and their commutation relation is given by [x,y]=c, with [x,c]=[y,c]=0. Participants explore the implications of non-commutativity on the expansion formula, noting that unlike the commutative case, terms cannot be simply ordered as x^i y^j. Special cases for n=2 and n=3 are suggested as potential starting points for deriving a generalized formula.

PREREQUISITES
  • Understanding of commutator notation in quantum mechanics, specifically [x,y].
  • Familiarity with Newton's binomial theorem for commutative quantities.
  • Basic knowledge of operator algebra and its properties.
  • Experience with mathematical proofs and derivations in abstract algebra.
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  • Research the implications of non-commutative algebra in quantum mechanics.
  • Study the derivation of binomial expansions in non-commutative settings.
  • Explore the mathematical framework of operator theory and its applications.
  • Investigate specific cases of non-commutative expansions for n=2 and n=3.
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Mathematicians, physicists, and researchers in quantum mechanics or operator theory who are interested in advanced algebraic structures and their applications in non-commutative settings.

ShayanJ
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You probably know that for two commutative quantities x and y,we have:
[itex](x+y)^n=\sum_{r=0}^n \left( \begin{array}{c} n \\ r \end{array} \right) x^{n-r} y^r[/itex]
Now I want to know is there a similar formula for the case when x and y don't commute and we have [itex][x,y]=c[/itex] and [itex][x,c]=[y,c]=0[/itex]?
Thanks
 
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In the case of commutative types, the terms can be ordered as x^i y^j, as you showed, because x and y commute.
In general, you will have all sorts of ordering, like xxy, xyx, yxx.
Maybe you can work out the special cases n=2 and n=3, to guess an answer.
 

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