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1) What is the difference between these two momentum operators: $\frac{\hbar}{i}\frac{\partial}{\partial x}$ and $-i\hbar\frac{\partial}{\partial x}$? How are these two operators the same?

My textbook says that $\frac{\hbar}{i}\frac{\partial}{\partial x}$ is the mathematical operator acting on $\Psi$ that produces the $x$ component of the momentum.

2) What is an operator? By operator, do they mean like addition, subtration, differentiation? Things of that nature? So if I take the partial with respect to $x$ of $\Psi$ and then multiply that whole thing by $\hbar / i$ then I should get the $x$ component of the momentum?

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    $\begingroup$ Notice that $1/i= -i$, so... $\endgroup$ – Valter Moretti Mar 18 '14 at 18:35
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    $\begingroup$ @DWade64: As V. Moretti says, for the first part of your question, there is no difference between $\frac{\hbar}{i}\frac{\partial}{\partial x}$ and $-i\hbar\frac{\partial}{\partial x}$. $\endgroup$ – DumpsterDoofus Mar 18 '14 at 18:36
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There's no difference between those two operators you wrote, since $\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i.$

In QM, an operator is something that when acting on a state returns another state. So if $\hat{A}$ is an operator and $|\psi\rangle$ is a state, the quantity $\hat{A}|\psi\rangle$ is another state, which you could relabel with $|\phi\rangle \equiv \hat{A}|\psi\rangle.$ If this Dirac notation looks unfamiliar, think of it as $\hat{A}$ acting on $\Psi_1(x,t)$ producing another state $\Psi_2(x,t)=\hat{A}\Psi_1(x,t)$ where $\Psi_1$ and $\Psi_2$ are just different states or wave functions.

When you act on a state with your operator, you don't really "get" momentum; you get another state. However, for particular states, it is possible to extract what a measurement of momentum would yield once you know what the resulting state is, but that's another question. The term eigenstate may help you in your discovery. But very briefly, if $\hat{A}\Psi_1=a\Psi_1$ (note that its the same state on both sides!), one can interpret the number $a$ as a physical observable if $\hat{A}$ meets certain requirements. This is probably what your textbook refers to.

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It's simply because $i^{-1} = -i$

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