New answers tagged

1

It is equivalent to ask the operators on different sites to commute or anticommute. Namely, there is always a so-called Klein transformation changing the commutation between different sites. If they anticommute one says they have natural commutation relations.


1

On the mere level of "second quantization" there is nothing wrong with fermionic operators commuting with other fermionic operators. They don't "know" that they are operators for "the same fermion" on different sites, so they could as well commute. But the deeper reason that fermionic operators on different sites anticommute is that they are just modes of ...


1

Restrictions can be imposed on the anomalous terms from general considerations even without fully solving the Feynman diagrams. In fact, the restrictions on the Schwinger term in the commutator given in the question are explicitly described in detail by Roman Jackiw in his review article: Field theoretic investigations in current algebra (section 2.2.). His ...


2

$$ [P_i,L_j]=\varepsilon_{jkl} ([P_i, X_k]P_l+X_k[P_i,P_l])=-\mathrm{i}\hbar \varepsilon_{jkl}\delta_{ki}P_l = -\mathrm{i} \varepsilon_{jil} P_l = \mathrm{i}\varepsilon_{ijl} P_l$$ Which shows that as expected $P_k$ is a vector. The lesson here is that you should use not use the same letter twice as an external index and a dummy index. Here $i, j$ are ...


1

I think the proof can actually work, but it needs to be formulated a bit better and it needs a bit more of explaining. Take $|\psi_n \rangle$ as a non-degenerate Eigenstate of $\hat{A}$. Then for all other Eigenvectors $|\psi_m \rangle$: $$\langle \psi_m | [\hat{A},\hat{B}]| \psi_n \rangle = (a_m - a_n)\langle \psi_m | \hat{B} | \psi_n \rangle = 0$$ So: $\...



Top 50 recent answers are included