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2

As everyone points out commutation is not a shorthand for equivalence, as, given your relations, the Jacobi identity, [A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0 dictates that when the first two terms vanish, the third must too, so that B must commute with [C,A], non vanishing in general, as remarked repeatedly. Lie algebra commutators do, nevertheless, parameterize ...


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I'm sorry I first misinterpreted your braces as anticommutators and not as set inclusions! First absorb the constants into the normalizations of H and B so the (1,1) entry of each is now 1. The common eigenvectors of them both are then the transposes of (1,0,0), (0,1,1), and (0,1,-1); the eigenvalues of each are, under H: 1,-1, -1 ; and under B: 1, 1, -1, ...


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Operators form a complete set if specifying eigenvalues of all of them already determines a state uniquely (up to normalization and phase). Since they commute, they will have simultaneous eigenspaces. Eigenstate is uniquely determined if all these eigenspaces are one dimensional.


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Yes, their outer product is defined as you said. Further, the product of operators is given by $$ (A \otimes 1_B)(F \otimes 1_B) = (AF) \otimes (1_B1_B) = (AF) \otimes 1_B $$ Therefore, $$ [A \otimes B, F \otimes 1_B] = [A,F] \otimes [B,I_B] = [A,F]\otimes {\bf 0} = 0 $$


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I) For the record, here is the operator calculation that OP wants to avoid. The benefit of the calculation is that the operators are not sandwich with any bra/ket representation, and hence we do not have to worry about whether the bra/ket representation is faithful. Let us put $\hbar=1$ for simplicity. The starting point is the CCR $$ [x^i, ...


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Take for example $q$ a vector field: $$ q^a=A^\mu $$ where $a=\mu$ is a vector index. The conjugate momentum is $$ \frac{\partial\mathcal L}{\partial A^\mu} $$ and, as it is an upper index in the denominator, it makes sense to write it as $\pi_\mu$. Also, you can use the definition of vectors and covectors to prove that $\pi_\mu$ transforms as a covector, ...


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When you write $n(x)=\psi^\dagger(x)\psi(x)$, you basically claim that $n(x)$ is a composite operator. Such naively defined composite operators in QFT suffer from UV-divergences, and this is essentially what you observe. In order to have a well-defined $n(x)$, you need to renormalize it, and the standard approach for free theories is to take a normal-ordered ...


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You are not missing anything: the commutator $[\psi(0),\psi^\dagger(0)]$ is ill defined. This is related to the fact that operators are actually distributions, not functions of $x$, so taking $x=0$ is meaningless.


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So after much deliberation, I think I have found the answer: $$ \hat{a }^\dagger \hat{a} = \hat{N} = \hat{a }^\dagger[\hat{a},\hat{a }^\dagger]\hat{a}$$ $$ = \hat{a }^\dagger(\hat{a} \hat{a }^\dagger - \hat{a }^\dagger\hat{a})\hat{a}$$ as $\hat{a}^2 = 0$, we have: $$ \hat{a }^\dagger \hat{a} = \hat{N} = \hat{a }^\dagger \hat{a}\hat{a }^\dagger \hat{a} = ...


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In spinor space, the fermionic fields $\psi$ are vectors, for example in the 4x4 representation of the $\gamma$-matrices: $\psi^\dagger = \begin{pmatrix} \psi_1^* & \psi_2^* & \psi_3^* & \psi_4^* \end{pmatrix}^{T}$ and $\psi=\begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3\\ \psi_4 ...


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Expectation values of constants or numbers are just those constants or numbers. The expectation value of the anti commutator of $\hat x$ and $\hat p$, that is, $\langle\{\hat x,\ \hat p\}\rangle$, for the Harmonic Oscillator, or coherent states of the Harmonic Oscillator, is equal to $0$. $$\langle\{\hat x,\ \hat p\}\rangle = \langle\hat x \hat p + \hat p ...



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