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## New answers tagged commutator

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In classical Hamiltonian mechanics one has the generalized coordinates $q^{i}(t)$ and momenta $p_{i}(t)$. The Poisson bracket is defined as, $$[F,G]_{PB}=\frac{\partial F}{\partial q^{k}}\frac{\partial G}{\partial p_{k}}-\frac{\partial F}{\partial p_{k}}\frac{\partial G}{\partial q^{k}}\ .$$ Using the q's and p's in place of $F$ and $G$ one has the ...

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The canonical commutation relations are postulates (i.e. you can't derive them) from which you can derive the commutation relations of the annihilation and creation operators. However, for the electromagnetic field this is quite complicated and if I were you I would start by looking at the scalar Klein-Gordon field. The postulates come from the ...

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As operators, you have $[L_{\vec e}, L_{\vec f}] = i L_{\vec e \wedge \vec f}$ (in units $\hbar=1$)

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In fact, you have $\{\psi^a(x), \bar \psi_b(y)\} = 0$, as an operator, for a space-like interval $(x-y)^2 <0$ (stricly speaking, this is a distribution, for instance, at $x_0=y_0$, this is the distribution $\delta^a_b \delta^3(\vec x-\vec y))$, together with relations $\{\psi^a(x), \psi_b(y)\} =\{\bar \psi^a(x), \bar \psi_b(y)\} = 0$ Now, if you look ...

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You have already got "practical" answers, so I intend to answer form another point of view. There is a quite famous theorem due to Stone and von Neumann, later improved by Mackay, and finally by Dixmier and Nelson, roughly speaking establishing the following result within the most elementary version. (Another version of the theorem focuses on the unitary ...

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It is possible to say something more precise than Martin's answer (that is correct however). The key-point is that self-adjoint operators are closed operators. An operator $A: D(A) \to H$, with $D(A) \subset H$ a linear subspace of the Hilbert space $H$ is said to be closed if, for every sequence of vectors $f_n\in D(A)$ such that (1) f_n \to f \in ... 4 Firstly, note that they postulate those commutation relations in the beginning of section 3.5 in order to show that they are wrong, which they demonstrate in the ensuing pages. The ultimate point is to show that one needs to impose anti-commutation relations on fermionic fields. In fact, the correct relations are postulated in equation 3.96; \begin{align} ... 1 Yes. Also note that in the momentum representation,x = i\hbar \frac{d}{dp}$, which is what your commutation relation proved as a special case. You could use this shortcut right off the bat. 2 My reading of the electron photon experiment is that intrinsic uncertainty enters the problem by limiting the resolving power of the photon. In other words, the electron is along for the ride, and perhaps historically it was chosen because it is such a simple system. But the recoil of the electron seems to confuse the issue. Instead of a free electron, we ... 2 Yes, the experiment is oversimplified, because the uncertainty principle is not about "disturbance through measurement". Although that's what Heisenberg said (one of the things he said), it turned out you can't interpret it that way in a very rigorous sense. Whether there is something like "disturbance through measurement" that gives rise to an uncertainty ... 3 Since I'm not an expert on spectral theory, this will only be a partial answer, however, I believe that this question, is mathematically much more involved than you think. First of all, let's review the finite dimensional case: We have two Hermitian matrices$A,B\in\mathcal{M}_d\$ and they commute if and only if their spectral projections commute, i.e. they ...

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