# Heisenberg Picture from $\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$

I have a question about the equation below:

$$\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$$

Is this equation valid in the Schrödinger picture, Heisenberg picture, or in some hybrid representation? For the Heisenberg picture (ie. $$A = t$$, $$B = H$$) I'm wondering why this equation does not include the total derivative of $$f$$, and if there is an even more general equation that includes it.

This equation looks like it is valid iff $$[A,[A,B]]=0$$which is necessary because it is the case when $$f(A)=A^2$$ and sufficient because you can then use it to attack the power series.

If that precondition holds in any picture then it is true in that picture. The partial derivative is incorrect in this syntax but could apply if the function secretly has other variables like time in it. Crucial to this whole enterprise is understanding that $$f$$ is an ordinary function $$C^\infty(\mathbb R \to \mathbb R)$$ being “lifted” to a function from operators to operators, and the derivative is happening on the un-lifted function. So this is expressing that, for example, because $$[\hat x, \hat p]=i\hbar$$ and $$[\hat x, i\hbar] = 0$$ that therefore $$[e^{ik\hat x}, \hat p] = (i\hbar)(i k) ~e^{ik\hat x}.$$ The precondition holds in both the Schrödinger and Heisenberg pictures so this expression holds in both of them. But we have lifted a function $$f(x) = e^{ikx}$$ into a function $$f(\hat x) = e^{ik\hat x}.$$ And the partial is just to remind us that we should not afterwards form something barbaric like $$\partial k/\partial \hat x$$ or so afterwards if $$k$$ is being assumed to be a constant above.

I do not think this “works” with $$t$$ as time without some strange treatment of time. Well, let me take that back: it works, it merely is not useful for any purpose under the sun. If $$A=t$$ then your victory in finding this expression $$f'(t)$$ is compensated by a great defeat in that you have formed the tautology $$0=0~f'(t)$$ because time is an ordinary parameter and not a quantum operator so $$f(t)$$ commutes with all operators. You still satisfy the precondition and the theorem holds: but it cannot give you any juicy details relating to $$f$$.

But anyway, this equation does not include a total derivative of $$f$$ because it might be confusing if $$f$$ has scalar parameters like time or a frequency.

Your equation is actually an identity of operators, assuming that

$$[A,[A,B]]=0$$

and if you replace $$A$$ by a real number known as $$t$$ you get $$0=0$$ since $$t$$ is a c-number.

The Heinseberg equation is not an identity but an equation with content. It tells how the operators in your theory involve on time $$t$$. Note that time is a real number, a c-number, that parametrize the time evolution. The equation is given by:

$$\frac{d A(t)}{dt} = [H, A(t)]$$

where I am assuming that the system is closed. This is not an identify, it fixes $$A(t)$$ in terms of $$A(0)$$ and $$H$$.