# Tag Info

### Commuting operators and their physical interpretation in QM

I'm studying Quantum Mechanics for the first time at the moment and I have a few questions in mind. can we interpret an operator as an act of "measuring", No, not an "act," in ...
Accepted

### Commutator of annichilation and creation operators

Yes, it is true. Boson creation operators only have non-zero commutators with annihilation operators of the same type, and Fermion creation operators only have non-zero anticommutators with Fermion ...

### What went wrong in the following calculation of $\langle p'|[x,p]|p'\rangle$?

The issue you have (as mentioned in comments and @J. Murray) is that none of the things you are evaluating are defined. We instead calculate $\langle p'| [ X , P ] | p \rangle$. This is manipulated as ...
Accepted

### What went wrong in the following calculation of $\langle p'|[x,p]|p'\rangle$?

The wrong step was in assuming that those expressions are well-defined in the first place. The "generalized" bras/kets $|p\rangle$ and $|x\rangle$ are not true members of the Hilbert space, ...
Accepted

### Commuting operators and their physical interpretation in QM

can we interpret an operator as an act of "measuring", Yes, there is a one-to-one mapping between mathematical operators (more precisely: self-adjoint operators) and physical observables. ...
1 vote
Accepted

### Commutator of gamma matrices with scalar product of 4-vectors

If you write $[\gamma\cdot p,\gamma\cdot q] = [\gamma^\mu p_{\mu}, \gamma^\nu q_{\nu}]$ then you can treat $p_{\mu}$ and $q_{\nu}$ as just numbers, i.e. they are "scalars" with respect to ...
1 vote
Accepted

### Strange definition of the fermion number operator in Polchinski

Recall that $$\{ \psi_r^\mu , \psi_s^\nu \} = \eta^{\mu\nu} \delta_{r,-s}. \tag{1}$$ Let's start by trying your definition, \begin{equation} \begin{split} [ F' , \psi_r^\mu ] &= \sum_s [ \psi_s^\...
1 vote

### Operators depending on the same independent variable but commuting between them

Actually, if you restrict attention to the finite-dimensional version of the problem, it is the other way around: two self-adjoint operators commute if and only if the are polynomials of the same ...
1 vote

### Commuting operators and their physical interpretation in QM

It is not the case that all operators represent measurements. A measurement is an interaction that produces a record that can be copied with arbitrarily high accuracy. This requirement roughly ...

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