668 Trotter theorem
(
PˆhQˆh− 1)
ψ=h(
Aˆ+Bˆ
)
ψ+O(h), (C.7)whereO(h) denotes any vectorφsuch that
lim
h→ 0||φ||
h= 0. (C.8)
In other words,||x||goes to 0 faster thanhdoes. Note that we can also write
(
Rˆh− 1)
ψ=h(
Aˆ+Bˆ
)
ψ+O(h). (C.9)Consequently, (
PˆhQˆh−Rˆh)
ψ=O(h). (C.10)Now leth=t/n. We need to show that
∣
∣
∣∣
∣
∣
[(
PˆhQˆh)n
−Rˆhn]
ψ∣
∣
∣
∣
∣
∣→^0 (C.11)
asn→∞.
To see how this limit can be demonstrated, consider first the casen= 2. It is
straightforward to show that
(
PˆhQˆh) 2
−Rˆ^2 h=(
PˆhQˆh−Rˆh)
Rˆh+PˆhQˆh(
PˆhQˆh−Rˆh)
. (C.12)
Likewise, forn= 3, a little algebra reveals that
(
PˆhQˆh) 3
−Rˆ^3 h=(
PˆhQˆh−Rˆh)
Rˆ^2 h+PˆhQˆh(
PˆhQˆh−Rˆh)
Rˆh+(
PˆhQˆh) 2 (
PˆhQˆh−Rˆh)
. (C.13)
Generally, therefore,
(
PˆhQˆh)n
−Rˆnh=(
PˆhQˆh−Rˆh)
Rˆ(n−1)h+PˆhQˆh(
PˆhQˆh−Rˆh)
Rˆ(n−2)h+···+(
PˆhQˆh)n− 1 (
PˆhQˆh−Rˆh)
. (C.14)
We now let the operators on the left and right sides of eqn. (C.14) act onψand take
the norm. This yields