# Time-derivative of an operator

Would I be right in thinking that the operator $$\hat O'(t)$$ is different from the operator $$D\hat O(t)$$ where $D={d\over dt}$, since when acting on a function $f$, the second corresponds to $$\hat O'(t)f+\hat O(t){d\over dt}f$$?

Suppose $\hat O(t)$ is an operator acting on a space of functions $f=f(x,t)$.

I wish to find an antisymmetric operator $\hat P$ such that $$\hat O'(t)=P\hat O$$. As commented above, I don't think $P={d\over dt}$ would work. Can anyone help me out? Thanks.

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It is not clear to me that is the derivative of an operator. I don't know if there is a standard definition, what is the definition in your context? –  tst Nov 8 '12 at 17:20