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The same vector x⃗ can be written in terms of the basis vectors ei of the dual space V* as: xiei, why is this true? It's not true. The elements of the dual space are not vectors as we ordinarily conceive of them geometrically, e.g., directed line segments. Rather, they are (geometrically) a set of oriented surfaces with a density proportional to ...


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If $V$ is a (finite dimensional) vector space with no additional structure, then $V^*$ is a different vector space of the same dimension, hence isomorphic to $V$ --- but it would be a great mistake to think of $V$ and $V^*$ as "the same'' or even ``almost the same'', because there is no preferred isomorphism between them. In other words, you can pick a ...


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Here, the idea behind introducing the subspace $S$ and its orthogonal complement $S^\bot$ was to show that all vectors on the RHS of equation 6.5.1 and 6.5.2 form a vector space of dimension $n-1$. As you say, indeed, the zero vector $\mathbf{0}$ is a vector that we need, but you are wrong in saying that it is 'forbidden' because it appears in the subspace ...


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Here's a counterexample: Suppose two identical, ideal batteries (with zero internal resistance) are both connected in parallel across a single resistor; equivalently, replace one of the resistors in your diagram with a second, identical battery. Also assume the conducting wires are ideal (again, no resistance). Kirchhoff's laws in this case result in an ...


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The answer is not quite simple, to show this we need some graph theory and matrices. There is a beautiful document explaining this relation in detail http://www2.math.uu.se/~takis/L/Circuits/2000/handouts/graphsandckts/graphsandckts.pdf I think the "fundamental reason" of this is related with the fact that every loop have different variables, if we can ...


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It really is obvious if you understand how tensor products work. Essentially, your state has two indices instead of one, and a tensor product of operators means that the first operator acts on the first index, and the second operator acts on the second. The operator that act on the "other" index just goes along for the ride. If it helps, you can write the ...


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If you want to write a super-operator representing left- or right-multiplication, there is a distinct method which is simpler and more elegant. Let us define the left-multiplication superoperator by $$ \mathcal{L}(A)[\rho] = A\rho,$$ and the right-multiplication superoperator by $$ \mathcal{R}(A)[\rho] = \rho A.$$ It should be clear that these operations ...


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This is exactly analogous to the procedure for finding matrix elements of normal operators. Let's first recall how this works in the familiar case. You choose an orthonormal basis of vectors, say $|n\rangle$, with $n = 1,2,\ldots D$, where $D$ is the dimension of the Hilbert space, such that $\langle n\rvert m\rangle = \delta_{mn}$. Now the matrix elements ...


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The statement generalizes to all dimensions. Given a vector space $\mathbb{R}^n$ with the usual Euclidean metric, we represent a $k$-dimensional subplane spanned by the vectors $v_1,\dots,v_k$ with $v_1 \wedge v_2 \wedge \dots \wedge v_k$, where the wedge $\wedge$ is the antisymmetric linear product of the exterior algebra. Also, we have that the volume of ...



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