The left hand side of the equality is thus e A a e − − A . We can now make use of the general equality e A B e − − A = ∑ ∑ k = 0 ∞ ∞ 1 k ! [ A , [ A , … … , [ A ⏟ ⏟ k times , B ] … … ] ] , which holds true for any pair of operators A and B . To compute e A a e − − A thus reduces to the problem of computing the repeated commutators between A and a . As can be readily verified, we have [ A , a ] = 1 2 [ z a † † 2 − − z ∗ ∗ a 2 , a ] = z 2 [ a † † 2 , a ] = − − z a † † , [ A , a † † ] = 1 2 [ z a † † 2 − − z ∗ ∗ a 2 , a † † ] = − − z ∗ ∗ 2 [ a 2 , a † † ] = − − z ∗ ∗ a . Using these equalities, we obtain