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When we have a normal operator like momentum or Hamiltonian we know it acts over a wavefunction as $\hat{H}\psi$ or $\hat p\psi$. … But when we talk about Symmetry Operators, everywhere I see it being operated along with its inverse like $U\psi U^{-1}$ where U is any of Symmetry operator. …

I understand that the momentum operator is Hermitian (thanks to this proof), as demonstrated by verifying the inner product relation. … The $L_x$ operator can be expressed as: $$ L_x = y p_z - z p_y, $$ where $p_z = -i\hbar \frac{\partial}{\partial z}$ and $p_y = -i\hbar \frac{\partial}{\partial y}$ are the momentum operators in the …

Let's have two non commuting operators $\hat A$ & $\hat B$ corresponding to the physical quantities $A$ & $B$. Also let's assume we are given a wave function $\Psi(\vec{r})$. … $ So we would have measured both the position and the momentum of the particle with an arbitrary precision even though the position and momentum operators don't commute ($[\hat x, \hat p]\ne\hat 0$). …

If we are given two operators $\hat A $ & $\hat B$ corresponding to the physical quantities $A$ & $B$, then for a given wave function $\Psi$ we know that the average values of those quantites over all … So I imagine that it's a way to measure how "spread out" are the quantities for this exact wave function. …

In case of crystal, $\hat p$ does not commute with Hamiltonian, but the translation operator with lattice vector $\mathbf R$ commutes with Hamiltonian. … $$[\hat H, \hat T(\mathbf R)]=0$$ Also by Bloch's Theorem, there exists quantum number $\mathbf k$ (which looks so similiar with the wave number $\mathbf k$) that satisfies $$\psi_\mathbf k(\mathbf r+\ …

it is also quite clear the solution can be expressed by exponentiating the right hand side: $$\Psi = e^{\frac{-i}{\hbar} H t}\Psi_0$$ where $U=e^{\frac{-i}{\hbar} H t}$ is the time evolution unitary operator … solution can be written as the matrix resulting from exponentiating the right hand side acting on some initial state $\Psi_0$: $$\Psi = e^{\frac{\pmb{+}i}{\hbar} L_z\theta}\Psi_0$$ therefore, the rotation operator …

EDIT:- I recently noticed an issue with the phenomenon of wave function collapse . … After collapse the wavefunction collapses to a single eigenstate of its respective observable like position or momentum or something else. …

I can now represent position eigenvector in momentum space(that is in the form of sum of momentum eigenvectors) and vice versa (fourier transforms of each other ). …

What I mean is, is this about the case where the state is in the subspace of the Hilbert space, where the energy is strictly at most $E$, i.e. in the image of the projection operator $$1_{(-\infty,E]}( … same section: "because the system’s Compton length must be smaller than its size R in order that the very notion of size be well defined" which seems to suggest to me that maybe the support of the wavefunction …

Why are particles represented by plane waves instead of spherical waves? Quantum field theory is based on the theory of plane waves, which simplifies particles into a quasi plane wave. … For example, in quantum field theory, particles generated by the generation operator, particles annihilated by the annihilation operator, or particles excited at a certain point are actually plane waves …

Thus, when we want to prove that the momentum operator $\hat{p_x}$ is Hermitian, we must show that $$\int_{-\infty}^\infty f^* \hat{p_x}g dx = \left(\int_{-\infty}^\infty g^* \hat{p_x} fdx\right)^*$$ because … For example, in the time-dependent Schrodinger equation (in one dimension) $$-\frac{\hbar}{i}\frac{\partial \psi(x,t)}{\partial t} = \hat{H}\psi(x,t)$$ the Hamiltonian operator acts on a wavefunction that …

at some momentum $k = \frac{2\pi}{N}n$, here $n = 1\cdots N$ be the $n$-th wave mode. … But instead I got the exact same Hamiltonian back from momentum to real space, and it's just the number operator in each site. I tried for 3 sites and got the same thing. …

The Hamiltonian in terms of these operators is $$\hat{H}=2(\hat{b}^\dagger\hat{b}+\frac{1}{2})$$ which is just the harmonic oscillator. … Which of the bosonic operators should I use to create coherent states? Why are coherent states used to represent the Landau level eigenstates? …

I'm learning the basics of quantum mechanics from Binney and Skinner's book, and I'm trying to do a very basic exercise (2.5d), yet am struggling. The exercise is to calculate $\langle p^2 \rangle$. S …

I'm confused with something regarding the rigorous mathematical definition of an arbitrary operator in the position/momentum basis and how it connects to the phase-space representation (https://en.wikipedia.org … that $<x|\hat A(\hat x) | \psi(t)> = A(x)\psi(x,t)$ which I intuitively understand because said operator and the position operator $\hat x$ share the position basis as an eigenbasis. …