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At that stage one is fairly comfortable with the concept of wavefunctions and with the Schrödinger equation, and has had some limited exposure to operators. … One common case, for example, is to explain that some operators commute and that this means the corresponding observables are 'compatible' and that there exists a mutual eigenbasis; the commutation relation …

Why do we exclude half-integer values of the orbital angular momentum? It's clear for me that an angular momentum operator can only have integer values or half-integer values. … Of course, when we do the experiments we confirm that a scalar wavefunction and integer spherical harmonics are enough to describe everything. …

It is easy to prove that this is equivalent to $[U,H] = 0$ for all $U$ representing the transformation, where $H$ is the Hamiltonian operator. … Presumably after quantization these can be associated to the Lie algebra generators, acting on wavefunctions on the manifold. Does this sound right to anyone? …

For this quantum mechanical system, is there an operator that corresponds to "position" the same way that there are operators corresponding to angular momentum and energy? … My first guess for what the operator would be is $$ \hat \Phi \big[ \psi(\phi) \big] = \phi \, \psi(\phi) $$ which would be analogous to the position operator for the "particle in a box" system, but I …

In wave-function theories we have differential operators (e.g., $-i\nabla$), but these aren't dynamical, so there is a clear distinction between an operator and a differential operator. … \ \text{operator theory} & (\mathrm{QM.1}) & (\mathrm{QFT.1}) \\ \text{wave-function theory} & (\mathrm{QM.2}) & \quad ?? …

When we define an operator in physics, e.g. the momentum operator as $\hat{p} = i \frac{d}{dx}$, it is said this operator acts on the wave functions. … Why do we call $\hat{p}$ momentum operator and not momentum functional? …

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 …

We all we told that to some extent it makes sense to treat the particle as a Gaussian wave packet. … After the measurement, the wave function collapses and forgets about both 'separate' Gaussians. …

($|\alpha\rangle$ is a Gaussian wave packet.) … I think my question boils down to: Does the operator $\hat{p}$ act on the basis kets $|x\rangle$ or on their coefficients? …

.$$ However, in class I've also seen operators applied to scalar-valued functions, such as the momentum operator in position space: $$P = - i \hbar \frac{\partial}{\partial x},\quad P \psi(x) = - i \hbar … On the Wikipedia article "Operator (physics)", under "Linear operators in wave mechanics", I found the following: $$A \psi(x) = A \langle x | \psi \rangle = \langle x | A | \psi \rangle$$ However, that …

Can these functions be described as eigenfunctions of $d$-dimensional angular momentum operators, analogous to $L^2$ and $L_z$? If so, what are the eigenvalues? … Is there a reasonable closed form for the $d$-dimensional hydrogenic stationary-state wavefunctions? If not, is there a reasonable (asymptotic) approximation formula? …

I'm making examples of wave functions to incorporate in a QM exam. … I came up with the following wave function, which gives me some troubles: $$\psi(x,0) = \begin{cases} A(a-x), & -a \leq x \leq a,\\ 0& \text{otherwise}. …

Wavefunction fundamentalism. All knowable information about a system is encoded in its wavefunction (ignoring phase and normalization). Unitary evolution of the wavefunction. … The wavefunction evolves over time in a deterministic and unitary manner. Observables. Any observable is represented by a Hermitian operator. Inner product. …

For example, we know that the eigenfunctions of the momentum operator (in 1D for simplicity) $$\hat p = -i \hbar \frac{\partial}{\partial x}$$ are plane waves: $$\psi_p(x) = A e^{ipx/\hbar}$$ These eigenfunctions … If we try to apply the cited postulate to the momentum operator, we would therefore incur in a contradiction: the system cannot jump into an eigenstate of the momentum operator, because such an eigenstate …

My layman understanding of the Uncertainty Principle is that you can't determine the both the position and momentum of a particle at the same point in time, because measuring one variable changes the other …