The textbook computes the time derivative of an expectation value as follows: ddt⟨Q⟩=ddt⟨Ψ|ˆQΨ⟩=⟨∂Ψ∂t|ˆQΨ⟩+⟨Ψ|∂ˆQ∂tΨ⟩+⟨Ψ|ˆQ∂Ψ∂t⟩
I guess the chain rule should be used, but the result should be the product of two derivatives instead of the sum.
Answer
Q is not a functional, but a linear operator. Since it is linear, there are no problems in using the chain rule. I will pretend there are no domain problems here, and that you can exchange limits and integrals as you wish (e.g. I'm supposing you can use the dominated convergence theorem), and obviously that each quantity is differentiable. ddt⟨ψ(t),Q(t)ψ(t)⟩=limh→01h(⟨ψ(t+h),Q(t+h)ψ(t+h)⟩−⟨ψ(t),Q(t)ψ(t)⟩)=limh→0⟨1h(ψ(t+h)−ψ(t)+ψ(t)),Q(t+h)ψ(t+h)⟩−1h⟨ψ(t),Q(t)ψ(t)⟩=⟨∂tψ(t),Q(t)ψ(t)⟩+limh→01h⟨ψ(t),Q(t+h)ψ(t+h)⟩−1h⟨ψ(t),Q(t)ψ(t)⟩=⟨∂tψ(t),Q(t)ψ(t)⟩+limh→0⟨ψ(t),1h(Q(t+h)−Q(t)+Q(t))ψ(t+h)⟩−1h⟨ψ(t),Q(t)ψ(t)⟩=⟨∂tψ(t),Q(t)ψ(t)⟩+⟨ψ(t),(∂tQ(t))ψ(t)⟩+limh→0⟨ψ(t),Q(t)1h(ψ(t+h)−ψ(t))⟩=⟨∂tψ(t),Q(t)ψ(t)⟩+⟨ψ(t),(∂tQ(t))ψ(t)⟩+⟨ψ(t),Q(t)∂tψ(t)⟩.
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