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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?

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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.

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  • $\begingroup$ Simple and good answer. $\endgroup$ – gented Jun 26 at 9:11
  • $\begingroup$ 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 ? $\endgroup$ – John Jun 26 at 12:31
  • $\begingroup$ @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 $p$. $\endgroup$ – Anonymous_original Jun 26 at 12:57
  • $\begingroup$ @JohnChambers And what you're asking physical justification for is not clear. $\endgroup$ – Anonymous_original Jun 26 at 12:59
  • $\begingroup$ 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. $\endgroup$ – Anonymous_original Jun 26 at 13:04

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