Is the Momentum Operator a Postulate?

I cannot totally agree with @dmckee.

First it is totally wrong to write something such as: $$\hat{P} = -i\hbar\partial /\partial x.$$ The correct way is to write: $$\langle x|\hat{P}|\phi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\phi\rangle,$$ and it should be interpreted as the momentum operator in spatial representation.

Derivations:

The physical meaning behind momentum is that: 1. It is the conserved quantity corresponding to spatial translation symmetry. 2. Because of 1, the momentum operator (Hermitian) is the generator of the spatial translation operator (unitary).

In terms of equations:

Define the spatial translation operator $D(a)$ s.t. $$C|x+a \rangle = D(a)|x \rangle,$$ and: $$D(a) = e^{-ia\hat{p}/\hbar}$$

I assume you have no problem deriving this.

Please note that this only depends on the quantization condition $[x,p] = i\hbar$, which is one of the postulates of quantum mechanics.

Take an arbitrary state $|\phi\rangle$ and apply $D(a)$ on it: $$D(a)|\phi\rangle = \int D(a)|\phi\rangle |x\rangle \langle x|dx$$ Change of variable, RHS = $$\int C|x\rangle \langle x-a|\phi\rangle dx$$

Take $a\to 0$, plug in to RHS: $$\phi(x-a) = \phi(x) - a\frac{\partial}{\partial x}\phi(x)$$ and to LHS: $$D(a) = 1-ia\hat{p}/\hbar$$

you can recover $\langle x|\hat{P}|\phi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\phi\rangle$

There is no derivation, but there is a heuristic argument.

Assume that it is 1926 and Derby has just challenged us to show him the wave equation that goes with de Broglie "waves" (as he did challenge Schrödinger). That means that we are working on a wave equation. The solutions should be of the form (in one dimension) $$ \Psi(x,t) = A e ^{i(kx - \omega t)} \,$$ where $k = 2 \pi / \lambda$ is the wave-number and $\omega = 2 \pi / T$ is the angular frequency.

We also want \begin{align*} p &= h/\lambda = \hbar k\\ E &= h f = \hbar \omega \end{align*} to agree with de Broglie and Plank's ad hoc assumptions that are working.

We could notice (as I presume that Schrödinger did) that the spatial and temporal derivatives that usually appear in a wave equation will give us factors of $k$ and $\omega$ respectively (with some inconvenient factors of $i$ hanging around, but we just have to live with that.). That is, we've just decided to go with \begin{align*} p &\mapsto \frac{\hbar}{i}\frac{\partial}{\partial x} \\ E &\mapsto -\frac{\hbar}{i}\frac{\partial}{\partial t} \\ \end{align*}

From there it is just a matter of saying that for a particle moving in a potential $V$ the total energy (Hamiltonian in many cases) is \begin{align*} E &= T + V \\ &= \frac{p^2}{2m} + V \;, \end{align*} Seeing this as one derivative with respect to time and two derivatives with respect to space and then fixing up the constants, we can arrive at $$ \left[-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)\right] \Psi(x,t) = i\hbar\frac{\partial}{\partial t} \Psi(x,t) \tag{TDSE}$$

I should re-iterate that this is in no way a proof. It's a kind of extended plausibility argument. And one that rather strains the suspension of disbelief except that it works.

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I have a more carefully constructed version of this argument that I give to my modern physics students and variations can be found in many place that predate my version.

The fact that $\hat{P} \to -i \hbar\, \partial / \partial x $ in the position basis is neither derivable nor a postulate, because it's not always true. The canonical commutation relation $[\hat{X}, \hat{P}] = i \hbar \hat{I}$ is generally taken as a postulate, but even if you choose the representation $\hat{X} \to x$ in the position basis, then the CCR allows infinitely many representations of $\hat{P}$ of the form $\hat{P} \to (-i \hbar\, \partial / \partial x) + f(x)$ for any function $f(x)$. The choice of representation corresponds to a gauge choice for the wavefunction, and does not affect any physically observable quantities. See Exercise 7.4.9 on pgs. 213-214 of Shankar for further discussion.