# 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?

## 1 Answer

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.

• Simple and good answer. – gented Jun 26 at 9:11
• 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