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3

Classically, angular momentum is only conserved in a central potential by considering the torque (correct me if I am wrong). In quantum mechanics, it is also true, isn't it? In QM, an operator is conserved iff it commutes with $H$, because $$ i\dot {\mathcal O}=[H,\mathcal O] $$ Therefore, the angular momentum is conserved iff it commutes with $H$. As ...


0

No, it's not true and a simple counter example suffices. Let $A = \vec L^2$, the square of the angular momentum operator, and $B = L_z$, the z component of the angular momentum operator. $[\vec L ^ 2, L_z] = 0$, but $[\vec L , L_z] != 0$ because the components of the angular momentum operator do not commute with each other.


3

It's not true. For example take $$ A = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \qquad\qquad\qquad B = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} $$ you have $A^2=0$ so that $[A^2,B]=0$, but $$ [A,B] = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. $$ You must add some condition, for example if you know also $$ [A,[A,B]]=0 $$ ...


3

Similarly to AccidentalFourierTransform I am not sure to understand well your issue. However there is a crucial missed point in your argument, usually absent in many textbooks on these topics. It is true that decomposing $H$ as $H= \hbar\omega( a^\dagger a + \frac{1}{2})$ and taking the relations (following from CCR) $[a, a^\dagger] =I$ into account one ...


3

Well, I'm not sure I understood your question so I'm going to write what I think and let's see if it's useful :-) The algebra $[a,a^\dagger]=1$ is all you need to diagonalise $H$, but this is because what $H$ looks like: $$ H=\omega a^\dagger a $$ The important observables, namely $H,P,X$, can be written as polynomials in $a,a^\dagger$: \begin{aligned} ...


2

What's special about $H$ is that it generates time translations. This means that operators evolve, by postulate, through Heisenberg Equations of motion $$ i\frac{\mathrm d}{\mathrm dt}\mathcal O(t)=[H,\mathcal O] $$ modulo an explicit time dependence. Therefore, if an operator commutes with the Hamiltonian, $[H,\mathcal O]=0$ we automatically conclude that ...


1

Here, $V_0(k_2)$ could have been replaced by $e^{k_2 \alpha_{-1}} e^{-k_2 \alpha_1}$ while the factors from $\alpha_{\pm n}$ for $n\gt 1$ could have been neglected because $\alpha_n$ annihilates everything that appears in the matrix element on the right side from $V_0$ (because it ultimately annihilates $|0\rangle$), and similarly for $\alpha_{-n}$ that ...


4

Heisenberg's uncertainty principle is in fact not a principle but a consequence of the operator formalism of QM. If we associate to the operator $X$ the standard deviation $$\Delta_X = \sqrt{ \langle{X^2}\rangle -\langle X \rangle^2}$$ it can be then shown that, given two operators $A,B$ $$\Delta_A \Delta_B \geq \frac{1}{2} \left| \langle ...


2

As everyone points out commutation is not a shorthand for equivalence, as, given your relations, the Jacobi identity, [A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0 dictates that when the first two terms vanish, the third must too, so that B must commute with [C,A], non vanishing in general, as remarked repeatedly. Lie algebra commutators do, nevertheless, parameterize ...


1

I'm sorry I first misinterpreted your braces as anticommutators and not as set inclusions! First absorb the constants into the normalizations of H and B so the (1,1) entry of each is now 1. The common eigenvectors of them both are then the transposes of (1,0,0), (0,1,1), and (0,1,-1); the eigenvalues of each are, under H: 1,-1, -1 ; and under B: 1, 1, -1, ...


1

Operators form a complete set if specifying eigenvalues of all of them already determines a state uniquely (up to normalization and phase). Since they commute, they will have simultaneous eigenspaces. Eigenstate is uniquely determined if all these eigenspaces are one dimensional.



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