# Quantum mechanics, Fourier transformation

Why do we use $$p=-i\hbar\frac{\partial}{\partial x}$$ in quantum physics? (I know the reason for $$i\hbar$$, quantization). Is this right to say we can't measure velocity and position of electrons at the same time, so we use mathematical method, Fourier transformation?

We define the quantum Poisson bracket for the operators corresponding to dynamical variables $$u$$ and $$v$$ as $${\{u,v\}}=uv-vu=i\hbar[u,v]$$ where $$[u,v]$$ is the classical Poisson bracket. For the simple case of momentum and position (considering one dimension for simplicity), we get $$xp-px=i\hbar.\tag{*}$$ We now define the differentiation operator $$\frac{d}{dx}$$ acting on a ket $$|\psi\rangle$$ as $$\frac{d}{dx}|\psi\rangle=|\frac{d\psi}{dx}\rangle.$$ Using this we get $$\frac{d}{dx}x-x\frac{d}{dx}=1.$$ From here we can see that $$p=-i\hbar\frac{d}{dx}$$ satisfies the commutation relation $$(*)$$. Now it is not necessary for us to take $$p$$ as defined above but with a bit more work it can be shown that choosing a suitable representation (basis) allows (forces) us to take $$p$$ as above.
Refer to section $$22$$ of Principle of Quantum Mechanics by Dirac ($$4^{th}$$ edition) for more details.
• You answer is true comes from the plane wave function for a free particle ,it is also similar to tangent vector (intrinsic) we take x as parameter ,using ih ,again we reach ih {\partial}{\partial x}$, but why we take it as p? however , I think there is a problem with these ,is there any Physical justification ? – John Jun 26 at 12:31 • @JohnChambers I think what you're asking, correct me if I'm wrong, is why we take$-i\hbar\frac{d}{dx}$to be$p$even if it satisfies the commutation relation. That is simply because we started with the classical Poisson bracket for$x$and$p$, i.e., we got the commutation relation for$x$and$p$from the classical Poisson bracket for$x$and$p$. – Anonymous_original Jun 26 at 12:57 • @JohnChambers And what you're asking physical justification for is not clear. – Anonymous_original Jun 26 at 12:59 • On suggestion,$\psi\rangle$has been changed to$|\psi\rangle$as it might cause confusion to some readers. However, it must be pointed that the former notation is correct and the$|\$ is unnecessary as Dirac showed. – Anonymous_original Jun 26 at 13:04