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1

Well, this is a problem of linear algebra substantially. The basic idea is that a square matrix $A$ is invertible if and only if $detA\neq 0$. Then, if the matrix is diagonalizable, since the determinant is invariant under coordinate transformations $A'=C^{-1}AC^1$, when you compute the determinant you get that, you get $detA=\lambda_1\dots\lambda_n$. (Note ...


4

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, \begin{equation} x'^{i}=x^{i}+\epsilon x^{i} \ . \end{equation} The change in the coords is, \begin{equation} \frac{d x^{i}}{d\epsilon}=x^{i} \end{equation} In the Hamiltonian ...


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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 ...


3

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 | #2 \right>}$ $\newcommand{bok}[3]{\left< #1| #2 |#3\right>}$ It means basically that all of the energy eigenstates has zero energy eigenvalue. Ups... Let $\left| \psi \right>$ be a normalized energy eigenstate with energy ...


0

In matrix form we have this: $\begin{array}{l} H \otimes H = \frac{1}{2}\left( {\begin{array}{*{20}{c}} 1&1&1&1\\ 1&{ - 1}&1&{ - 1}\\ 1&1&{ - 1}&{ - 1}\\ 1&{ - 1}&{ - 1}&1 \end{array}} \right)\\ CNOT = \left( {\begin{array}{*{20}{c}} 1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\\ ...


3

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 ...


2

Looking at the Hilbert spaces themselves, we indeed find the puzzling equality $$\mathbb{C}^2\times\mathbb{C}^2 = \mathbb{C}^2\otimes\mathbb{C}^2 = \mathbb{C}^4$$ so the tensor product of the qubit spaces is equal to the pairs of non-entangled states. Or so it would seem. The superficial equality is wrong in the physical context, because the isomorphism ...



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