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1

Right, in general you're not going to see a straightforward equivalence there. We can use Dirac notation with $\hat P_b = |b\rangle\langle b|$ to see that $\langle \hat A \hat B \rangle = \sum_{a,b} a~b~\langle \psi | a \rangle~\langle a | b \rangle~\langle b | \psi \rangle$ and even inserting an identity matrix for $b$ (call it $b'$) gives: $$ ...


0

When I was asking this question, I didn't understand the relation between the commutativity of two operators and their eigenspaces: If an operator $A$ commutates with another operator $B$, then $A$ leaves the eigenspaces of $B$ invariant: $$ B\psi = \epsilon\psi \implies BA\psi = AB\psi = \epsilon A\psi $$ But this does not imply that $\psi$ is an ...


6

OP is essentially pondering if commutativity is a transitive relation, ie. if three normal operators$^{1}$ $A$, $B$, and $C$ satisfies $$ [A,B]~=~0\quad \wedge\quad [B,C]~=~0 \quad\stackrel{?}{\Rightarrow} \quad [A,C]~=~0 .\tag{T}$$ The answer is No, but OP argues via the existence of a common basis of eigenvectors for two commuting normal operators that ...


2

The $L_i$ has many eigenspaces corresponding to many eigenvalues. Each of those eigenspaces is also an eigenspace of the Casimir operator. So they share common eigenspaces in the sense that there are eigenspaces that are eigen to both. But they don't share them in the sense that they are the same. Look at the hydrogen atom. There are energy eigenspaces and ...


4

Two operators may be simultaneously diagonalized if and only if they commute. As you can see, $L_z$ commutes neither with $L_x$ nor with $L_y$ – and not with any other linear combinations different from a multiple of $L_z$ – so there's no way to diagonalize two different components of $L_i$ at all. However, $L_z$ (and similarly other components) commutes ...


3

The eigenspaces of the quadratic Casimir $L^2 = L_x^2+L_y^2+L_z^2$ of the Lie algebra of infinitesimal rotations $\mathfrak{so}(3)$ are precisely the irreducible representations of $\mathfrak{so}(3)$ - we usually label a representation by its highest weight $l$, which is in this case just a number telling you what the largest possible value for any of the ...


1

Comments to the question (v2): It seems that the question does not explain how a 'Hamiltonian' $H$ differs from a self-adjoint operator $A$ (presumably bounded from below). This would make OP's question a duplicate of the linked Phys.SE post. Perhaps a 'Hamiltonian' $H$ is also supposed to generate 'time'-evolution for some distinguished parameter $t$, ...


1

This is pretty niche notation, and it is indeed not defined in the paper, but the name "vector-coupled product" does seem to be used by a few people beyond Varga and Suzuki. In essence, $$ [\mathcal Y_{l_1}(\mathbf x_1)\mathcal Y_{l_2}(\mathbf x_2)]_{LM} $$ is a coupled wavefunction with total angular momentum $L$ that's made up of the single-particle ...


0

It is so, because Dyson formula was designed for the interaction picture in Quantum Mechanics. Interaction picture may seem strange, especially in comparison with the 'natural' Heisenberg picture, but has been proven useful, especially in perturbative computations. The basic idea is to split the total Hamiltonian into two parts: $$ H = H _0 + H _{int} $$ ...


0

Comments to the questions (v2): The Hamiltonian $H(t)$ in QFT often depends on time $t$, e.g. if there are interactions in the interaction picture, or if there are external sources. In particular, Hamiltonians at different times need not commute. For an explicit example see e.g. my Phys.SE answer here. Well, if the Hamiltonian $H(t_1)=H(t_2)$ is the same ...



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