quantum mechanics - Momentum as Generator of Translations

I understand from some studies in mathematics, that the generator of translations is given by the operator $\frac{d}{dx}$.

Similarly, I know from quantum mechanics that the momentum operator is $-i\hbar\frac{d}{dx}$.

Therefore, we can see that the momentum operator is the generator of translations, multiplied by $-i\hbar$.

I however, am interested in whether an argument can be made along the lines of "since $\frac{d}{dx}$ is the generator of translations, then the momentum operator must be proportional to $\frac{d}{dx}$". If you could outline such an argument, I believe this will help me understand the physical connection between the generator of translations and the momentum operator in quantum mechanics.

Answer

In the position representation, the matrix elements (wavefunction) of a momentum eigenstate are $$\langle x | p\rangle = \psi_p(x) = e^{ipx}$$ The wavefunction shifted by a constant finite translation $a$ is $$\psi(x+a)$$ Now the momentum operator is the thing which, acting on the momentum eigenstates, returns the value of the momentum in these states, this is clearly $-i\frac{d}{dx}$.

For our momentum eigenstate, if I spatially shift it by an infinitesimal amount $\epsilon$, it becomes $$\psi(x+\epsilon) = e^{ip(x+\epsilon)} = e^{ip\epsilon}e^{ipx} = (1+i\epsilon p + ...)e^{ipx}$$ i.e. the shift modifies it by an expansion in its momentum value. But if I Taylor expand $\psi(x+\epsilon)$, I get $$\psi(x+\epsilon)= \psi(x)+\epsilon \frac{d}{dx} \psi(x)+... = \psi(x)+i\epsilon(-i\frac{d}{dx})\psi(x)+...$$ So this is consistent since the infinitesimal spatial shift operator $-i\frac{d}{dx}$ is precisely the operator which is pulling out the momentum eigenvalue.