Time dependence of the momentum operator for a free particle

I was studying Modern Quantum Mechanics by Sakurai, and at the page 85, it is given the analysis of a free particle.

There, the author assumes that Hamiltonian is $$\hat H = \frac{ \hat p ^2}{ 2m},$$

and do things like $$\frac{\partial p}{\partial t} = [p, H] = 0$$ etc. and "derives" that $$\frac{\partial p}{\partial t}(t) = 0$$.

However, there are a couple of contradictory statements that I see there, and all of them resolves if one assumes that $$\hat p$$ is time-independent, but the author does not state that, but rather claims to derive that. So is there any such implicit assumption on the time dependence of $$\hat p$$ (or at least an assumption that $$[p(t), p(t')] = 0$$) in the Hamiltonian of a free particle ? If not, the following argument invalidates all of the analysis given in the book.

First of all, the relation $$\frac{\partial \hat A_H}{\partial t} (t) = \frac{1}{i \hbar } [\hat A_H(t), \hat H(t_0)] + \left( \frac{\partial \hat A}{\partial t} (t) \right)_H$$ holds only for operators defined in Heisenberg's picture, i.e if you give us any operator $$\hat A(t)$$, then we define a new operator $$\hat A_h := U^\dagger (t, t_0) \hat A(t) U(t, t_0)$$, and the above relation holds only for such operator. Therefore, unless we explicitly define $$\hat H = \frac{\hat p_H^2}{ 2m},$$ we we cannot derive the same result $$\frac{\partial p_H}{\partial t} = [p_H, H] = 0$$ because otherwise (i.e if $$H = \frac{\hat p^2 }{2m }$$, where $$\hat p$$ is the momentum operator in Schrödinger's picture)

we have $$\frac{\partial p_H}{\partial t}(t) = [U^\dagger (t,t_0)p(t)U(t,t_0), H(t_0)] + \left ( \frac{\partial p_S}{ \partial t} (t) \right),$$ and at this point, we need to know whether $$\hat p$$ is time-dependent or not, because that will determine the time-dependence of $$\hat H$$, hence the operator $$U(t,t_0)$$.

At this point, if we make an assumption that $$\hat p$$ is time-dependent, but $$[p(t), p(t')] = 0$$, then one can easily show that $$\hat p$$ and $$\hat U$$ commutes, and we get $$\frac{\partial p_H}{\partial t}(t) = \frac{\partial p_S}{\partial t}(t)$$, or if we assume that $$p_S$$ is time-independent, everything plays nicely.

• @AaronStevens Where does that relation come from ? The book does not mention it ,as far as I can remember. – onurcanbektas Jan 25 at 14:14
• @AaronStevens Ok, let me first ask, when you write $\hat p$, do you mean the momentum operator in Schrödinger's picture, or Heisenberg's picture ? – onurcanbektas Jan 25 at 14:24
• If the commutator is $0$ then you end up with $\frac{\text d p_H}{\text d t}=\frac{\partial p_H}{\partial t}$. I don't understand the question – Aaron Stevens Jan 25 at 14:24
• @AaronStevens Yes, but before saying that, you need to show that the commutator is zero, but that what I'm rejecting because the equation $\frac{\partial \hat A_H}{\partial t} (t) = \frac{1}{i \hbar } [\hat A_H(t), \hat H(t_0)] + \left( \frac{\partial \hat A}{\partial t} (t) \right)_H$ is derived for operators in Heisenberg's picture only ,so you cannot put $\hat p$ instead of $\hat A$, where $\hat p$ is the momentum operator in Schrödinger's picture. – onurcanbektas Jan 25 at 14:26
• Then work in the Heisenberg picture. $H=\frac{p^2}{2m}$ is true in all picture is it not? – Aaron Stevens Jan 25 at 14:28