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Suppose I know the time evolution of an operator is given by $\dot{\hat{O}} = \frac{i}{\hbar}[\hat{H}(t), \hat{O}(t)]$. Now I want to look at a function $\hat{f}(\hat{O}$, and I want to know the time evolution of the "values" of this function. I assume that I can expand $\hat{f}$ in a taylor series: Since $\hat{O}$ has an Unitary time evolution $U$, I can write: $$\hat{f}(\hat{O}(t)) = \Sigma_i c_i \hat{O}^i(t) = \\ \Sigma_i c_i (\hat{U}_{t, t_0} \hat{O}(t_0) \hat{U}_{t_0, t})^i = \\ \Sigma_i c_i \hat{U}_{t, t_0} \hat{O}(t_0)^i \hat{U}_{t_0, t} = \\ \hat{U}_{t, t_0} \hat{f}(\hat{O}(t_0)) \hat{U}_{t_0, t} $$ Then computing the time derivative will yield: $$ \frac{d}{dt} \hat{f}(\hat{O}(t)) = \frac{i}{\hbar}[\hat{H}, \hat{f}(\hat{O}(t))] $$

However, I fail to arrive at this same result, if I try to compute the derivative directly: $$ \frac{d}{dt} \hat{f}(\hat{O}(t)) = \Sigma_{j=1} c_j j \hat{O}^{j-1}(t) \dot{\hat{O}}\\ = \Sigma_{j=1} c_j j \hat{O}^{j-1}(t) \frac{i}{\hbar}[\hat{H}(t), \hat{O}(t)] $$ To arrive at the same result, I would have to permute the Hamiltonian $\hat{H}$ through the whole product $(\hat{O}(t))^{j-1}$. I'm stuck here. Am I making a mistake somewhere?

Quantumwhisp
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2 Answers2

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Derivative of product of operator valued functions is :

$$\frac{d}{dt}(A(t)B(t))=(\frac{d}{dt}A(t))B(t)+A(t)(\frac{d}{dt}B(t)).$$

You can proove it using standard definition of derivative.

You have missed this non commutativity of derivative of an operator with operator itself in writing down the derivative after Taylor expansion. Once you fix it you can see that both expressions for time evolution of function of an operator will agree.

Sunyam
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What user Sunyam said about a non-commutative Leibniz rule is precisely right. Therefore OP's last step for the $j$'th power should be

$$\frac{d}{dt} O^j~\stackrel{\text{Leibniz}}{=}~\sum_{i=0}^{j-1} O^i \dot{O} O^{j-1-i}~=~\frac{i}{\hbar} \sum_{i=0}^{j-1} O^i [H,O] O^{j-1-i}~=~\frac{i}{\hbar} [H,O^j] .$$

Qmechanic
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