# Tag Info

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Here, $V_0(k_2)$ could have been replaced by $e^{k_2 \alpha_{-1}} e^{-k_2 \alpha_1}$ while the factors from $\alpha_{\pm n}$ for $n\gt 1$ could have been neglected because $\alpha_n$ annihilates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.

Top 50 recent answers are included