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8

Let's try the Baker–Campbell–Hausdorff formula in the form $$e^X Y e^{-X} = Y + [X, Y] + \frac{1}{2!}[X,[X,Y]] + \frac{1}{3!}[X,[X,[X,Y]]]\; + \;...$$ Take $X = ix_jA_{jk}p_k$ and $Y = x_i$. It is not the simplest case since the first commutator, $$[X, Y] = [ix_jA_{jk}p_k, x_i] = ix_jA_{jk}(-i\delta_{ki}) = x_jA_{ji}$$ does not commute with $X = ... 1 Comments to the question (v3): Eqs. (1) are part of the CCR for a scalar field, such as, e.g., a real or complex Klein-Gordon field, a Schrödinger field, etc. Eq. (2) refers to the Schrödinger field, which is a complex field, see e.g. this Phys.SE post. A real Schroedinger field does naively not make sense since e.g. the expected kinetic term$\propto ...

2

The point is that the equation of motion of the fields is different if referring to temporal derivatives. In relativistic field theory, it is a second-order one and you need two initial conditions i.e. $\pi$ and $\pi$ to solve it. Quantizing, and interpreting the Fourier coefficients of the initial conditions as creation and annihilation operators, ...

0

Two main points are.... Generally $\langle{x}|[X,P]|\alpha\rangle \not= \langle{x}|XP|\alpha\rangle-\langle{x}|PX|\alpha\rangle$ When $[X,P]=XP-PX$ is an well-defined operator in a Hilbert space, $H=L^2([a,b])$, space of square-integrable functions in $[a,b]$, the domain of definition of $[X,P]$ is a set of functions $|\alpha\rangle$ satisfying ...

0

I think you'd probably need to integrate that term by parts, lowering $k = p'/\hbar$ to $1$ while raising $\exp[i~k~(x - x')]~dk$ into $[-i\hbar/(x - x')]~\exp[i~k~(x - x')].$ The result you get for the middle integral is then $$-2\pi i\hbar ~ \frac{\delta(x - x')}{x - x'}.$$ If you hold off evaluating the integral further, the other integral will be ...

1

The Weyl system $\exp\left[\frac{i}{\hbar} Q \hat{p}\right]$ and $\exp\left[\frac{i}{\hbar}P\hat{q}\right]$ comprise two "presentation" elements of the Heisenberg group. To the extent $\hat{p}$ is a derivative with respect to position q, Q is the shift amount q in any function of it is translated by the action of $\exp\left[\frac{i}{\hbar} Q ... 0 Eq.(8.2.4) assumes the corrections in$\alpha$depend just on A, and not on any other quantities. The explicit form of the dependence is not really important at this point and it is left as$f(A)$. As for Eq,(8.2.12), start with $$U(\alpha +\delta\alpha) = U(\delta\alpha)U(\alpha)$$ and substitute the 1st of Eqs.(8.2.8) for$U(\delta\alpha)$, neglecting ... 2 The operator$\partial_i \partial_j$is symmetric by switching$i$and$j$. All tensors can be decomposed the following way :$T_{ab} = T_{[ab]} + T_{\{ab\}}$With$T_{\{ab\}}$the symmetrized tensor ($\frac{1}{2}(T_{ab} + T_{ba})$)and$T_{[ab]}$the antisymmetrized tensor ($\frac{1}{2}(T_{ab} - T_{ba})\$). Any antisymmetric tensor contracted with a ...

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