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If $[A,B]=0$ then there is a unitary transformation $U_{A,B}$ that diagonalises both $A$ and $B$ simultaneously. This transformation depends on the pair of commuting operators $(A,B)$, so that for a different pair there could be a different unitary. Assume that all the eigenvalues of $H$ have multiplicity 1. Then there exists a unique (up to a phase factor) ...


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If $H$ commutes with $A_1$, then it will indeed share an eigenbasis with it. Your mistake is in supposing that it will share the same eigenbasis with both $A_i$s. Examples are easy to provide: On the trivial side, if $H=E_0\mathbf 1$ is trivial, then it shares an eigenbasis with $A_1=x$ and it shares an eigenbasis $A_2=p$, but it cannot share an ...


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For bosons: We put ourselves in a suitable common domain, i.e. the finite particle vectors. Then $$\bigl(a^*(f)\Psi\bigr)_n(X_n)=\frac{1}{\sqrt{n}}\sum_{j=1}^n f(x_j)\Psi_{n-1}(X_n\setminus{x_j})\\ \bigl(a(f)\Psi\bigr)_n(X_n)=\sqrt{n+1}\int \bar{f}(x)\Psi_{n+1}(x,X_n)dx$$ Hence by definition $$\bigl(a(g)a^*(f)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n+1}\int ...


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For normal elements in a C*-algebra you can do continuous functional calculus, that is, if $a$ is a normal operator, then $f(a)$ is well-defined for any $f\in C(\sigma(a))$. Since $\sigma(a)$ is always compact you can use Stone-Weierstrass to write $f$ as a uniform limit of polynomials in one complex variable and its complex conjugate. Hence you can verify ...


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Perhaps the easiest way to see that there should be a Grassmann sign factor $(-1)^{|A| |B|}$ in the definition of time ordering $$\tag{1} {\cal T} \left\{ A(t_A) B(t_B)\right\} ~:=~ \theta(t_A-t_B) A(t_A) B(t_B) + (-1)^{|A| |B|} \theta(t_B-t_A) B(t_B) A(t_A), \qquad $$ is to go to the classical limit $\hbar\to 0$. Here $|A|$ denotes the Grassmann parity, ...


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Let's calculate $[L_x L_y, L_z]$. I'm going to use the property $[AB,C]=A[B,C]+[A,C]B$. If you apply the property to our case, you obtain $L_x[L_y,L_z]+[L_x,L_z]L_y$. Now you can substitute the value of the commutators and find the correct answer. Note that the quantum commutation relations are pretty similar to vector product in cartesian coordinates. For ...


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commuting operators are any two operators which can be applied to a function in any order without altering the outcome


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The reason why the commutation relations between a field and its conjugate at equal times are of the form $$ \left[\phi(t,\textbf{x}),\pi(t,\textbf{y})\right]=i\hbar\,\delta^{(3)}(\textbf{x}-\textbf{y}) $$ is only to mirror and copy the canonical hamiltonian commutation relations $[q_i,p_j]=i\hbar\,\delta_{ij}$. No causality is involved, rather it is somehow ...


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One way to define spacelike separation in special relativity is that any two events are spacelike separated if and only if there exists a reference frame in which the two events have the same time coordinate. So yes, if $x^0 = y^0$ the separation is spacelike. Alternatively you can work from the definition where two events are spacelike separated if (and ...



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