# Why can't derivative notation of Hamiltonian and momentum prove $[H, p] = 0$?

We already know that in quantum mechanics, hamiltonian and momentum don't commute with each other in general sense.

However, if we regard them as generator of wave function like,

$$H \to i\hbar\frac{\partial}{\partial t}$$

$$P \to -i\hbar\frac{\partial}{\partial x }$$

then

$$HP = \hbar^2\frac{\partial^2}{\partial t\partial x}$$

$$PH = \hbar^2\frac{\partial^2}{\partial x\partial t}$$

so that these are same operator (two derivatives commute) and it seems $$[H, P] = 0$$

this must not be true, but I don't know where it goes wrong. Could you point the contradiction in this argument?

The short answer is $$\hat{H} = i \hbar \frac{\partial}{\partial t}$$ is false because, it is not implied by Schrödinger's equation unless you require for the Schrödinger's equation to be true for all wave functions.

This is an understandable pitfall with Schrödinger's equation. Let's look at the (non-relativistic) time dependent Schrödinger's equation.

$$\hat{H} \left| \psi(t) \right> = i \hbar \frac{\partial}{\partial t} \left| \psi(t) \right>$$

One could come to the conclusion that $$\hat{H} = i \hbar \frac{\partial}{\partial t}$$, but this is incorrect and would make Schrödinger's equation useless. Here's why...

First let's subtract the manipulate the express via subtraction and factoring to get

$$(\hat{H} - i \hbar \frac{\partial}{\partial t}) \left| \psi(t) \right> = 0$$

If $$\hat{H} = i \hbar \frac{\partial}{\partial t}$$ is indeed true, we see for any $$\left| \psi(t) \right>$$, Schrödinger's equation is satisfied, so any wave equation is allowed!

To get use from Schrödinger's equation, one should see this a condition $$\left| \psi(t) \right>$$ must satisfy, given a system with it's own Hamiltonian. Every system has it's own hamiltonian so you have to work on a case by case basis to see if $$\hat{H}$$ does commute with $$\hat{P}$$.