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I know the equation for the momentum operator, but what exactly is the momentum operator? … It's bizarre to me that taking the derivative of the wave function, which is an operator, should return something that isn't just a function. …

I'm studying quantum mechanics and I have a question about the momentum operator. … Edit: The domain on which I mean the momentum operator to act is the set of quantum states. …

Consider momentum operator representation in position space. … .\ Now consider the matrix representation of the momentum operator. …

In the following stack exchange post Momentum Operator in curved spacetime (QFT) a general expression for the momentum operator is given for a Riemannian manifold $(M,g)$. … To make sense out of this, we should think of the wave function as a section of a complex line bundle, which is associated to the frame bundle $LM$ of the manifold $(M,g)$. …

From a video lecture on quantum mechanics at MIT OCW I found that you didn't need to know Schrödinger's equation to know the momentum operator which is $\frac{\hbar}{i}\frac{\partial}{\partial x}$. … Can it be derived besides using the wave function I mentioned? …

Using the momentum operator $ \hat{p}\rightarrow-i\hbar\frac{d}{dx} $ and the translation operator $ e^{-i\hat{p}a/\hbar}\psi(x)=\psi(x-a) $, how to I go about showing that the eigenfunctions of the momentum … operator are plane waves? …

So my question is: For an operator to yield an observable, the wave function should be square integrable and thus belong to a Hilbert space (a complete inner-product space). … So if the eigenfunctions of the momentum operator do not belong to a Hilbert space, how is the operator Hermitian? …

Is it still called a wave function? If so, what is the meaning of the value $\hat{p}\psi(x, t)$ for a given $x$? … operators are what they are, or how to even interpret them momentum or energy being a function of position. …

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? …

I am just wondering how we actually derive the momentum operator in quantum mechanics. … Secondly, we derive this operator for only free particle. How do we know this will work for a general wavefunction? …

One thing that I haven't been able to derive from them, however, is the identity of the momentum operator. … I know that it makes sense, as it results in the Ehrenfest Theorem, the De Broglie wavelength hypothesis, the Heisenberg Uncertainty Principle (for $x$ and $p$), the momentum operator being the generator …

If eigenfunction of momentum operator is $e^{-x^3}$, then calculate its eigenvalue. …

In a previous question How to get the position operator in the momentum representation from knowing the momentum operator in the position representation? … \end{align}$$ In the above expressions, the $ p $ is a wavefunction in momentum space but $ \hat p$ is an operator in $x$ i.e $\frac{\hbar}{i}\frac{\partial}{\partial x} $, so can it act on momentum …

I have two related questions on the representation of the momentum operator in the position basis. … The action of the momentum operator on a wave function is to derive it: $$\hat{p} \psi(x)=-i\hbar\frac{\partial\psi(x)}{\partial x}$$ (1) Is it ok to conclude from this that: $$\langle x | \hat{p} | …

I was reading a book on theoretical quantum mechanics and the authors introduced the (orbital) angular momentum operator as the operator that generates rotations around an (arbitrary) axis. … They then proceeded by examining how a unitary operator corresponding to a rotation acts on the wave function and eventually related the two using the formula: \begin{align} \psi\left(e^{-\vartheta\boldsymbol …