2
votes
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 ...
- 15.7k
2
votes
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 ...
- 1,709
2
votes
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
...
- 24k
2
votes
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, ...
- 66k
2
votes
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.
...
- 35k
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 ...
- 2,153
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^\...
- 24k
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 ...
- 507
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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