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Notice that in $[\partial_xO(x),O(x')]$, the partial derivative only acts on $x$, not on $x'$. So we can pull the partial derivative operator out of the bracket and get $\partial_x[O(x),O(x')]$.


0

I came here before asking myself the same question and, as I figured it out, I'd like to answer just for the record. Respecting the former answer, with plain and usual commutators (e.g. see this question) in the Schrödinger picture with locality and causality, if $X(\xi)=\sum_i\left(\alpha_i(\xi){U}+b_i(\xi){V}\right)$, the condition ...


1

As was said in the comments, my attempt to a solution is to decompose $\hat{x}$ as a linear combination $\alpha \hat{A} + \beta \hat{B}$. This seems to be general, because if $\hat{p}$ is a linear combination of $\hat{A}$ and $\hat{B}$, then the position will be too, since it's the inverse Fourier transform of $\hat{p}$ and such transformation is linear. I'd ...


1

Lie algebras are not a group w.r.t. to the commutator (the Lie bracket). The first reason is that the commutator is not associative. Another is that they almost always lack an identity element, since the identity matrix is, for example, not in $\mathfrak{su}(2)$, and Schur's lemma would, in the fundamental representation, guarantee that only multiples of ...


3

Short answer: complexifications facilitate representation theory. In physics, we typically want to find representations of a Lie algebra $\mathfrak g$, and often times determining the representations of its complexification $\mathfrak g_\mathbb C$ is easier. Moreover, we have the following theorem (see ref 1. Proposition 4.6) which tells us that ...


2

From a mathematical perspective, to develop Lie algebra representation theory most efficiently, we need the field $\mathbb{F}$ of the Lie algebra to be algebraically closed. See e.g. Ref. 1, where this assumption is used already in the beginning of Chapter II. The situation for Lie algebras is similar to when we in linear algebra try to diagonalize, say, a ...


9

Qmechanic explained a way in which something with the word "commutator" in it doesn't vanish when applied to two of the same operator. However, I feel it is necessary to point out that plain commutators, as seen in a quantum mechanics course, really, honestly, always, and without fail satisfy $[Q,Q] = 0$ for any operator $Q$. This is because $[A,B]$ is ...


8

I) Yes, they are probably referring to that a Grassmann-odd operator needs not (super)commute with itself. Take e.g. the 1st order Grassmann-odd differential operator $$\tag{1} D~:=~\frac{d}{d\theta}+ \theta\frac{d}{dt}. $$ In eq. (1) $t$ is a Grassmann-even variable and $\theta$ is a Grassmann-odd variable, which (super)commute $$\tag{2} ...


2

According to the topic of deformation quantization, the first few entries in the dictionary between $$\tag{0} \text{Quantum Mechanics}\quad\longleftrightarrow\quad\text{Classical Mechanics}$$ read $$\tag{1} \text{Operator}\quad\hat{f}\quad\longleftrightarrow\quad\text{Function/Symbol}\quad f,$$ $$\tag{2} \text{Composition}\quad\hat{f}\circ\hat{g} ...


3

Choose the momentum representation, $$x_i = i \hbar \frac{\partial}{\partial p_i}$$ distribute $i \hbar$ and act the commutator on vector $\psi$, $$[x_i, F(\mathbf p)] \psi = i \hbar \left(\frac{\partial}{\partial p_i}(F(\mathbf{p}) \space \psi) -F(\mathbf p) \frac{\partial }{\partial p_i} \psi \right)$$ and apply the product rule: $$= i \hbar ...


3

The commutation of two variables, in some cases, can be related to Poisson Bracket via $$ \left[\hat A,\,\hat B\right]=i\hbar\left\{\hat A,\,\hat B\right\} $$ Thus, $$ \left[\hat A,\,\hat B\right]=i\hbar\sum_i\left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)\tag{1} $$ ...


3

As @Qmechanic pointed out in a comment, we are free to use any operator representation. In momentum space, $\hat{\bf x} = + i \hbar \ \partial/\partial {\bf p} $ and $\hat{\bf p} = {\bf p}$, so $$ \begin{eqnarray} \left[\hat{x}_i,F\left(\hat{\bf p}\right)\right] &=& \left[i \hbar \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] \\ ...



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