Zoran Skoda
exp(xF G)

Let F=F(d/dx),G=G(d/dx)F = F(d/d x), G = G(d/d x) be polynomial differential operators. The commutator [F,x]=:D(F)[F,x] =: D(F) is the derivative of FF with respect to its argument, namely d/dxd/d x.

For convenience, let us introduce a parameter λ\lambda. We want to polarize the operator exp(λxF+G)exp(\lambda x F + G).

Let A s,lA_{s,l}, be a double sequence defined as follows: A s,l=0A_{s,l} = 0 unless 0sl0\leq s\leq l, A s,s=F sA_{s,s} = F^s (ss-th power) and the basic recursion

A s,l+1=F(D(A s,l)+A s1,l)+A s,lG A_{s,l+1} = F (D(A_{s,l}) + A_{s-1,l}) + A_{s,l} G

holds.

Theorem. exp(λ(xF+G))= s=0 x sα ss!Cexp(\lambda (x F + G)) = \sum_{s=0}^\infty \frac{x^s \alpha^s}{s!} C where C= l=0 λ lA 0,ll!C = \sum_{l = 0}^\infty \frac{\lambda^l A_{0,l}}{l!} and α=1C l=0 λ lA 1,ll!\alpha = \frac{1}{C}\sum_{l = 0}^\infty \frac{\lambda^l A_{1,l}}{l!}.

Sketch of the proof. We can write (xF+G) l= s=0 lx sB s,l(x F + G)^l = \sum_{s=0}^l x^s B_{s,l}. Clearly, B(s,s)=F sB(s,s) = F^s. Now

(xF+G) l+1= s=0 lx sB s,l(xF+G) (x F + G)^{l+1} = \sum_{s=0}^l x^s B_{s,l} (x F + G)

Commuting B s,lB_{s,l} and xx we get

= s=0 l(x s+1B s,lF+x sD(B s,l)+x sB s,lG), = \sum_{s=0}^l (x^{s+1} B_{s,l} F + x^s D (B_{s,l})+ x^s B_{s,l} G),

and after the renaming of the indices a comparison shows that the recursion relation for AA is satisfied by BB. Extending BB by zero to other values of s,ls,l (and recalling the initial conditions) we see that A=BA=B and

exp(λ(xF+G))= l=0 λ ll!(xF+G) l= s,l=0 λ ll!x sA s,l exp(\lambda (x F + G)) = \sum_{l=0}^\infty \frac{\lambda^l}{l!} (x F + G)^l = \sum_{s,l=0}^\infty\frac{\lambda^l}{l!} x^s A_{s,l}

It is clear that Cα s/s!= lλ lA l,sl!C \alpha^s/s! = \sum_{l} \lambda^l \frac{A_{l,s}}{l!} stands along x sx^s in the polarized form of
exp(λ(xF+G))exp(\lambda (x F + G)). Then

s(Cα)(Cα s1/s!)=C(Cα s/s!) s (C\alpha) (C \alpha^{s-1}/s!) = C (C\alpha^s/s!)

translates into

s r 1+r 2=lA 1,r 1r 1!A s1,r 2r 2!= l 1+l 2=lA 0,l 1l 1!A s,l 2l 2! s\sum_{r_1+r_2 = l} \frac{A_{1,r_1}}{r_1!}\frac{A_{s-1,r_2}}{r_2!} = \sum_{l_1+l_2 = l} \frac{A_{0,l_1}}{l_1!}\frac{A_{s,l_2}}{l_2!}

If this is true for some ll let us prove it for l+1l+1. As this is true identically, we can apply DD to the identity, multiply by FF and apply the Leibniz rule for DD, obtaining

s r 1+r 2=lFD(A 1,r 1)A s1,r 2+FA 1,r 1D(A s1,r 2)r 1!r 2!= s \sum_{r_1+r_2 = l} \frac{F D(A_{1,r_1}) A_{s-1,r_2} + F A_{1,r_1} D(A_{s-1,r_2})}{r_1! r_2!} =
= l 1+l 2=lFD(A 0.l 1)A s,l 2+FA 0.l 1D(A s,l 2)l 1!l 2! = \sum_{l_1+l_2 = l} \frac{F D(A_{0.l_1}) A_{s,l_2}+ F A_{0.l_1} D(A_{s,l_2})}{l_1! l_2!}

hence

s r 1+r 2=l(A 1,r 1+1FA 0,r 1A 1,r 1G)A s1,r 2+A 1,r 1(A s1,r 2+1FA s2,r 2A s1,r 2G)r 1!r 2!= s\sum_{r_1+r_2 = l} \frac{(A_{1,r_1+1}- F A_{0,r_1} - A _{1,r_1} G) A_{s-1,r_2} + A_{1,r_1} (A_{s-1,r_2+1}- F A_{s-2,r_2} - A_{s-1,r_2} G)}{r_1! r_2!} =
= l 1+l 2=l(A 0.l 1+1A 1.l 1)A s,l 2+A 0.l 1(A s,l 2+1FA s1,l 2A s,l 2G)l 1!l 2! = \sum_{l_1+l_2 = l} \frac{(A_{0.l_1+1} -A_{1.l_1}) A_{s,l_2}+ A_{0.l_1} (A_{s,l_2+1}- F A_{s-1,l_2} - A_{s,l_2} G)}{l_1! l_2!}

By the induction hypothesis for ll, all the terms containing GG clearly cancel. Similarly, the terms with GG: sF 0,r 1F s1,r 2s F_{0,r_1} F_{s-1,r_2} splits into the sum of (s1)F 0,r 1F s1,r 2(s-1)F_{0,r_1} F_{s-1,r_2} and F 0,r 1F s1,r 2F_{0,r_1} F_{s-1,r_2}; the first cancels with the corresponding term on the LHS by hypothesis and the other term with a term which is identical after renaming indices. The remaining terms multiply by l!l!. One is left with the identity

s r 1+r 2=ll!r 1!r 2!(A 1,r 1+1A s1,r 2+A 1,r 1A s1,r 2+1)= l 1+l 2=ll!l 1!l 2!(A 0,l 1+1A s,l 2+A 0,l 1A s,l 2+1) s \sum_{r_1+r_2 = l} \frac{l!}{r_1! r_2!} (A_{1,r_1+1} A_{s-1,r_2} + A_{1,r_1} A_{s-1,r_2+1}) = \sum_{l_1+l_2 = l} \frac{l!}{l_1! l_2!} (A_{0,l_1+1} A_{s,l_2} + A_{0,l_1} A_{s,l_2+1})

Rename the indices so that the two summands on each side have identical labels:

s r 1+r 2=l+1(l!(r 11)!r 2!+l!r 1!(r 21)!)A 1,r 1A s1,r 2= l 1+l 2=l+1(l!(l 11)!l 2!+l!l 1!(l 21)!)A 0,l 1A s,l 2 s \sum_{r_1+r_2 = l+1} \left(\frac{l!}{(r_1-1)! r_2!} +\frac{l!}{r_1! (r_2-1)!}\right) A_{1,r_1} A_{s-1,r_2} = \sum_{l_1+l_2 = l+1} \left(\frac{l!}{(l_1-1)! l_2!} +\frac{l!}{l_1! (l_2-1)!}\right) A_{0,l_1} A_{s,l_2}

hence by the Pascal triangle we get the identity we needed to prove.

Last revised on May 12, 2014 at 07:40:09. See the history of this page for a list of all contributions to it.