19.1 OPERATOR FORMALISM
defining series, we have
f(A)=2∑∞m=0(−1)mA^2 m+1
(2m+1)!∑∞n=0(−1)nA^2 n
(2n)!.Writingm+nasrand replacingnbys, we have
f(A)=2∑∞r=0A^2 r+1(r
∑s=0(−1)r−s
(2r− 2 s+1)!(−1)s
(2s)!)=2∑∞r=0(−1)rcrA^2 r+1,where
cr=∑rs=01
(2r− 2 s+ 1)! (2s)!=1
(2r+1)!∑rs=02 r+1C
2 s.By adding the binomial expansions of 2^2 r+1=(1+1)^2 r+1and 0 = (1−1)^2 r+1,it
can easily be shown that
22 r+1=2∑rs=02 r+1C 2 s ⇒ cr=^22 r(2r+1)!.It then follows that
2sinAcosA=2∑∞r=0(−1)rA^2 r+1 22 r
(2r+1)!=∑∞r=0(−1)r(2A)^2 r+1
(2r+1)!= sin 2A,a not unexpected result.
However, if two (or more) linear operators that do not commute are involved,combining functions of them is more complicated and the results less intuitively
obvious. We take as a particular case the product of two exponential functions
and, even then, take the simplified case in which each linear operator commutes
with their commutator (so that we may use the results from the previous worked
example).
IfAandBare two linear operators that both commute with their commutator, show thatexp(A)exp(B)=exp(A+B+^12 [A, B]).We first find the commutator ofAand expλB,whereλis a scalar quantity introduced for