# Tag Info

0

While the accepted answer is very clear, I'll write an operator proof. The $\hat{p^2}$ in $\hat{H}$ commutes with $\hat{\mathbb{P}}$ (the parity operator). So, to show that $\hat{H}$ and $\hat{\mathbb{P}}$ commute, we have to show this: $[\hat{V},\hat{\mathbb{P}}]=0$ Note that since $V(x)$ is a symmetric function i.e. even function, it is an eigenfunction ...

1

I believe that you must study (not simply reading but trying to prove) that beautiful story of angular momentum quantization starting from the non-commutativity of the coordinate and momentum of a particle along any axis. The following is a starting point (found in any introductory book on Quantum Mechanics). In Classical Mechanics the angular momentum ...

2

Short answer: yes A map $A: V \to V$ on a $\mathbb{K}$-vectorspace $V$ is linear if $\forall v_1, v_2 \in V$ and $\forall \lambda \in \mathbb{K}$ we have: $$A(\lambda v_1 + v_2)= \lambda A(v_1)+A(v_2)$$ If we have two such maps $A,B$ then $[A,B]=AB-BA$ is again linear because, well AB is linear: $$(AB)(\lambda v_1+v_2)=A(B(\lambda v_1 + v_2))=A(\lambda ... 0 Elaborating on ACruiosMind's comment, assume that the matrices A and B are defined the following way:$$A=\begin{pmatrix} 1 & 2 \\ 5 & 4 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$Notice that the eigenvectors of A are$$\begin{pmatrix} 1 \\ 5/2 \end{pmatrix} \quad \text {and} \quad ...

0

Although $[ \hat{L_{x}}, \hat{L_{y}}] \phi_{l, m_{l}} = 0$, $\phi_{l, m_{l}}$ is neither $\hat{L_{x}}$'s nor $\hat{L_{y}}$'s eigenstate.

1

Commuting matrices We take two Hermitian matrices A and B that commute. Since they are Hermitian and act on the same Hilbert space they have a common (orthonormal) basis of eigenvectors. In general all Hermitian matrices have the same amount of lineary independent eigenvectors since they all have a orthonormal basis of eigenvectors. Commuting observables ...

4

As noted by Sebastian, $[Q,H]$ will be anti-Hermitian and therefore generally not an observable (except in the trivial case). However $i[Q,H]$ is an important observable. This corresponds to the classical Poisson bracket which can be see in the following formula, $$\frac{d \langle Q\rangle}{d t} = \frac{i}{\hbar} \langle [H,Q]\rangle + \langle ... 1 Do A and B have the same amount of eigenvalues? They don't have to, S_z and the identity commute. The former has two eigenvalues. The latter has one. Other examples exist too, such as the operator, H=p^2/2m+e/r which has an infinite number of eigenvalues but can commute with operators with a finite number of eigenvalues such as both operators ... 6 [A, B] for two observables A and B is an observable if, and only if, A and B commute. Proof:$$ [A, B]^\dagger = (AB)^\dagger - (BA)^\dagger = B^\dagger A^\dagger - A^\dagger B^\dagger = BA - AB = -[A, B]. Note: An observable is any Hermitian operator. The commutator of two Hermitian operators is anti-Hermitian, as the proof shows. $0$ is an ...

Top 50 recent answers are included