# Tag Info

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List (to be completed with more references and/or items, details of the relation to physics) there is a notion of positive energy representation (cf. Haag-Kastler axiom, "Spectrum" or "stability" condition) in which generators of translations can be choosen in the von Neumann algebra associated to the representation of the observables, but not necessarily ...

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In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics.

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The Hilbert space of quantum mechanics arises from considering irreducible representations of the C*-algebra of observables. When the C*-algebra of a physical system is commutative all its irreducible representations are one dimensional and therefore the corresponding Hilbert space is $\mathbb C$. Hence any classical physical system is associated to the ...

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Vectors (which kets are) don't have adjoints, they have duals. Whether the dual of $\lvert n_1,\dots,n_n\rangle$ is denoted by $\langle n_1,\dots,n_n\rvert$ or $\langle n_n,\dots,n_1 \rvert$ is entirely conventional.

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SUMMARY OF EDITED VERSION: You cannot place any conditions on $V(x)$ and $E$ that guarantee that solutions to the time-independent Schrödinger equation are normalizable, for something of a silly reason. Initial, partial answer: If the potential is bounded below by some value $V_\text{min}$, then a solution to the time-independent Schrödinger equation ...

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Note: There is a short summary at the bottom. This is actually also described in Nielsen&Chuang: You don't learn about general measurements, because they are completely equivalent to projective measurements + unitary time evolution + ancillary systems, all of which is described in your usual QM formalism. The Measurement Postulate Let's start from ...

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I can understand your confusion. Let me start by saying that this is the "correct" way of writing down what you want to have in physics notation: $$\langle e_i|\otimes \mathbf{1} \sum_j |e_j\rangle\otimes |w_j\rangle = \sum_j \delta_{ij} |w_j\rangle$$ Note the absence of the tensor product sign. Otherwise, you will never have this reduction of dimension, ...

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Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...

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(1) Answer is Yes in this example: Consider complete set of orthonormal energy eigenfunctions of "particle in a box problem (infinite potential well)" in quantum mechanics. Choose any 3 of them. If any one of these chosen 3 is orthogonal to remaining two, then these two are orthogonal to each other. (2) Answer is No in the example of cross product of ...

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In general no. If you think of $\left|a\right>$, $\left|b\right>$ and $\left|c\right>$ as vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$, you can say that $\vec{a} = \vec{b} \times \vec{c}$, so that $\vec{a}$ is orthogonal to both $\vec{b}$ and $\vec{c}$, but this doesn't imply that $\vec{b} \cdot \vec{c} = 0$.

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No. Just let $|b\rangle = |c \rangle$.

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Since $V(x)$ is bounded from above we have three possibilities. Either it oscillates at infinity with an upper bound, or it asymptotes to a constant $<E$ or it diverges to $-\infty$. Since we are interested in $x\rightarrow\infty$ we may average the oscillation in the first case to the mean, and if it diverges then we concern our selves with the leading ...

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The reason why it is often stressed that Hawking radiation is in a pure state, is that this is in apparent contradiction to the fact that Hawking radiation is also said to be thermal. The apparent contradiction is solved when one realizes that in a general curved spacetime there is no unique definition of the vacuum state and therefore the whole Hilbert ...

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