Using the Schrodinger eqn in finding the momentum operator

[Hamiltonian]

TL;DR

how can we use the Schrodinger equation while finding $\hat p$ when in fact we have already used $\hat p$(i.e. $\hat p ^2$ in the $\hat T$ term of the $\hat H$) in the Schrodinger equation?

I have read that the Schrödinger equation has no formal derivation we are simply applying the Hamiltonian operator on the wave function
$$\hat H = i\hbar \frac{\partial}{\partial t} = \hat T + \hat V$$
here we substitute $$\hat T = \frac{\hat p^2}{2m}$$ where $$\hat p = -i \hbar \frac{\partial}{\partial x}$$
but when we derive the equation for $\hat p$ we actually substitute $\frac{\partial \psi}{\partial t}$ and $\frac{\partial \psi *}{\partial t}$ from the Schrödinger equation.

$$< p> = m\frac{d<x>}{dt} = m\int_{-\infty}^{+\infty} x\frac{\partial (\psi^*\psi)}{\partial t}$$
$$<p> = m\int_{-\infty}^{+\infty} x[\psi^*\frac{\partial \psi}{\partial t}+\psi\frac{\partial \psi^*}{\partial t}] dx$$
here we substitute $\frac{\partial \psi}{\partial t}$ and $\frac{\partial \psi *}{\partial t}$ as
$$\frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2} -\frac{i}{\hbar} V\psi$$
$$\frac{\partial \psi^*}{\partial t} = \frac{-i\hbar}{2m}\frac{\partial^2 \psi^*}{\partial x^2} +\frac{i}{\hbar} V\psi^*$$
after some simplification we end up with
$$<p> = \int_{-\infty}^{+\infty} \psi^* (-i\hbar \frac{\partial}{\partial x})\psi dx$$
and then finally we get $$\hat p = -i\hbar \frac{\partial }{\partial x}$$
so I don't understand how we can use the Schrödinger equation while finding $\hat p$ when in fact we have already used $\hat p$(i.e. $\hat p ^2$ in the $\hat T$ term of the $\hat H$) in the Schrödinger equation?
this video does the derivation for the momentum operator

 

[PeroK]

What you appear to be showing is that$$m\frac{d\langle x \rangle}{dt} = \langle p \rangle$$ where $\hat p = -i\hbar \frac{\partial}{\partial x}$. And that justifies the original definition of $\hat p$.

 

[Hamiltonian]

What you appear to be showing is that$$m\frac{d\langle x \rangle}{dt} = \langle p \rangle$$ where $\hat p = -i\hbar \frac{\partial}{\partial x}$. And that justifies the original definition of $\hat p$.

how exactly do we find $\hat p$ without using the Schrödinger equation? by finding $\hat p$ I mean how do we arrive at $\hat p = -i\hbar \frac{\partial}{\partial x}$. I thought the only way of arriving at this would be by using $m\frac{d<x>}{dt} = <p>$ but when we use this approach we need to use the Schrödinger equation but the KE energy term in the Hamiltonian is already using $\hat p = -i\hbar \frac{\partial}{\partial x}$

 

[PeroK]

how exactly do we find $\hat p$ without using the Schrödinger equation? by finding $\hat p$ I mean how do we arrive at $\hat p = -i\hbar \frac{\partial}{\partial x}$. I thought the only way of arriving at this would be by using $m\frac{d<x>}{dt} = <p>$ but when we use this approach we need to use the Schrödinger equation but the KE energy term in the Hamiltonian is already using $\hat p = -i\hbar \frac{\partial}{\partial x}$

There are ways to motivate the definition of momentum. For example:

https://en.wikipedia.org/wiki/Momentum_operator#Origin_from_De_Broglie_plane_waves

And also on that page momentum as the generator of spatial translations (this is done in Sakurai's book).