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10

Does this mean that the operator $\hat O$ (an observable) is special in some way? I believe it means there is no such $\hat O$. If $\hat O$ corresponds to an observable, we require the eigenvalues to be real. Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$: $$\hat O |o\rangle = o |o\rangle$$ Now consider the following $$\hat O ... 6 If |\phi⟩ and |\psi⟩ are linearly independent, then it is always possible to assign them to the column vectors$$ |\phi⟩\mapsto\begin{pmatrix}1\\0\end{pmatrix} \text{ and } |\psi⟩\mapsto\begin{pmatrix}0\\1\end{pmatrix}, $$but if they're not orthogonal you're obviously going to need to work harder on the representation of the inner product in this basis. ... 6 First of all, the equation$$ $$A\otimes B=A\otimes \mathbb{1}+\mathbb{1}\otimes B,$$ $$is a claim about an identity, and this claim is incorrect. Note that for 1\times 1 matrices, the matrices are numbers and the equation above reduces to$$ a\cdot b = a\cdot 1 + 1 \cdot b$$which is clearly wrong because the addition (the right ... 6 It's not possible to derive the orbital angular momentum L = r \times p from the \mathfrak{so}(3) commutation relations alone, since the spin operator S also fulfills the same commutation relations, but certainly is different from r \times p. 6 I have read somewhere that commutation relations of the form $$[a_i,b_j]=\epsilon_{ijk} c_k$$ admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is:$$ ...

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In QM, a real valued observable $A$ is mathematically represented by a projector valued measure over $\mathbb{R}$, $P^{(A)}$, i.e., if $E$ is a Borel subset of the real line, then $P^{(A)}(E)$ is a projector representing the proposition "the outcome of measuring $A$ falls in $E$". In principle, that's all you need for representing, mathematically, ...

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Suppose the space-time group includes dilatations which expand or contract space. Points in space $x^{i}\in V_{3}$ transform under a small dilatation $\epsilon$ near the identity as, $$x'^{i}=x^{i}+\epsilon x^{i} \ .$$ The change in the coords is, $$\frac{d x^{i}}{d\epsilon}=x^{i}$$ In the Hamiltonian ...

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In this case $p_0$ is just a number. It's the amount of phase that you add to the wave function. $\text{e}^{ip_0 x/\hbar}$ is not a translation operator. It's just multiplying the wave function by a complex number of norm $1$. In you previous question you had $\hat{p}$ which is an operator. It acts on the wave function by differentiating it.

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It is invertible iff its determinate doesn't vanish $$\det([H]_B) \ne 0$$ Note that this property of the determinate is invariant under a change of basis since: \begin{align} \det(S^{-1} \cdot [H]_B \cdot S) & = \det(S^{-1}) \cdot \det([H]_B) \cdot \det(S) = \frac{1}{\det(S)} \cdot \det([H]_B) \cdot \det(S) \\ & = \det([H]_B) \end{align} with ...

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Adding to Lubos' answer, let me address this part more specificially: I am very confused about this - actually - I'm guessing that transformations that can be written $A⊗B$ are a subset of all possible transformations with all 16 coefficients free. This is correct. First some notation: Let $H_1$ and $H_2$ be two (finite-dimensional) Hilbert spaces for ...

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The time-ordered product is not an operation on operators. It is an operation on time-dependent operators. Let $\cal{O}$ denote the algebra of linear operators on a Hilbert space $\cal{H}$. If $A, B: [0, T] \to \cal{O}$ are two time-dependent operators, we can define an operator $$AB: [0,T]^2 \to \mathcal{O}, \ \ (AB)(t_1, t_2) = A(t_1) B(t_2)$$ which ...

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Given a densely defined pre-closed operator $T$ on a Hilbert space $H$, you can define its transpose (more properly called the adjoint) $T^*$ by requiring it to be the operator with the property that $$(\eta,T\psi)=(T^*\eta,\psi)$$ for any $\eta$ in the domain of $T^*$ (which is dense) and $\psi$ in the domain of $T$. Using Dirac's notation we can rewrite ...

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It was an incorrect statement, as it is explained here by Griffiths himself. Anyways, the mathematical explanation is straightforward: given a self-adjoint operator $A$ with domain $D(A)$, any sufficiently regular real function $f(A)$ of it (and the square is perfectly ok for $-\Delta=p^2$) is self-adjoint on some domain $D(f(A))$ by the spectral theorem ...

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Without specifying the domains of the involved operators all the discussion has not much sense. Let me say that, if $A :D(A) \to H$ with $D(A)\subset H$ a linear dense subspace of the Hilbert space $H$, $A$ is self-adjoint if $D(A)=D(A^\dagger)$ and $A=A^\dagger$. Notice that consequently (I stress that the converse is false) self-adjointness implies ...

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Let me assume $\hbar = 1$ We have the following expression for the commutator $$[\nabla^2, L_z] = -[p^2, L_z] = -[p_x^2 + p_y^2 + p_z^2, L_z]$$ Now using the fact that the action of the commutator is linear we can write $$[p_x^2 + p_y^2 + p_z^2, L_z] = [p_x^2, L_z] + [p_y^2, L_z] + [p_z^2, L_z]$$ Now I'm going to use the property  [AC, B] = A[C, B] + ...

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I think I'm essentially supposed to show... Why do you think this? Is this a homework question? If so then it should be tagged as such. If this is not a homework question and you are interested to read an explanation involving no explicit equations then consider the following: The operator $\vec L$ is the generator of rotations. Therefore, any ...

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$\frac{dU}{ds}+\frac{dU^\dagger}{ds}=0$ implies that $\frac{dU}{ds}$ is anti-selfadjoint: set $A = \frac{dU}{ds}$, then what you have is $A + A^\dagger = 0$. Therefore by setting $A = iK$ you have that $K$ is a self-adjoint operator.

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