I've been studying the postulates of QM and seeing how to derive important ideas from them. One thing that I haven't been able to derive from them, however, is the identity of the momentum operator.
For simplicity, I'm only thinking about no relativistic effects, no spin, no time-dependent potentials, and one spatial dimension. Also I'm assuming the position operator is simply multiplication by $x$, as in, I'm in position space. So the Hamiltonian operator is $ H = -\frac{\hbar^2}{2m}\nabla^2+V$.
I know that the momentum operator is $p = -i\hbar \frac{\partial}{\partial x}$.
But how do I get there from the postulates? I know that it makes sense, as it results in the Ehrenfest Theorem, the De Broglie wavelength hypothesis, the Heisenberg Uncertainty Principle (for $x$ and $p$), the momentum operator being the generator of the translation operator, and possibly many other desirable theorems, and correlations with classical momentum.
But none of these are postulates (at least, not in the various formalisms I encountered), so you can't derive $p = -i\hbar \frac{\partial}{\partial x}$ from them. Rather, they are consequences of it. You need to know the operator beforehand to see that they are correct. Yes, this is just semantics, but that is the core issue for me:
Regardless of how much sense it makes, is the identity $p = -i\hbar \frac{\partial}{\partial x}$ (under the assumptions I made) a Postulate, meaning that you can't derive it from other postulates, or can it in fact be obtained from them? And in the latter case, could you show me how?
Note: I know that there are many different and equivalent sets of postulates for QM. But in none that I saw did they name it as a postulate nor properly derived it.
Answer
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 $\langle x|\hat{P}|\phi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\phi\rangle$
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