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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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$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\o}{\mathbf 1}$ Physicists are lazy people we all are! When you see something like $S_{1z}+S_{2z}$ you should really think of the following: $$S_{1z}+S_{2z} \equiv S_{1z} \otimes \mathbf 1+ \mathbf 1 \otimes S_{2z}$$ Since you get tired of writing it over and over you just shorten it by an addition ...


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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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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 ...


2

A state $\psi$ corresponds to an energy $E$ if: $$H\psi = E\psi$$ Clearly, if there is a state $\psi = \sum_i c_i \psi_i$ where $H\psi_i = E_0\psi_i\ \forall i$, then $$H\psi = \sum_i c_i H\psi_i = \sum_i c_i E_0\psi_i = E_0\psi$$ A linear combination of states with the same energy value again has the same energy value. Now consider $\psi = c_1\psi_1 + ...


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It's convention to normalize the wave function, but this is purely something done to make calculations easier. States multiplied by a c number(complex number) are completely equivalent. This is why we choose to normalize states (magnitude 1). It makes the calculations for things like the transitions probability between states easier. If you include your ...


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The coefficients $a_k$ quantity the projection of the state $|\psi\rangle$ onto the $k^\mathrm{th}$ basis state. So if you measure in that basis you would expect $|a_k|^2$ of the time to measure state $|\phi_k\rangle$. If you change the basis you will need to recalculate the projection coefficients. If you are in a non-orthogonal basis the projections are ...


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$\renewcommand{ket}[1]{|#1\rangle}$ As others have pointed out, you can go whole hog and solve the characteristic equation $\text{det}(\hat{Q} - \lambda \hat{I})=0$ and find repeated solutions for $\lambda$. However, there is a simpler, more physically intuitive way to hunt for degeneracy: look for symmetry. When an operator $\hat{Q}$ has a symmetry there ...


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As Phoenix87 said, you need to solve the asociated eigenvalue problem. $\lambda$ is an eigenvalue if and only if $\hat{Q}|\Psi>=\lambda|\Psi>$. If you have a matrix expression for the operator $\hat{Q}$, then the usual way to solve the problem is writing the above equation as Phoenix87 did, writing everything in the left side: ...


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Assuming that your operator has a spectrum consisting of isolated points you can look for all the independent solutions of the eigenvalue equation $$(Q-\lambda I)\xi = 0$$ Let these solutions generate a vector space $V_\lambda$ and then compute the dimension of $V_\lambda$. If it is greater than 1 then the eigenvalue $\lambda$ is degenerate.


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I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace ...


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If you want to experimentally create a qubit, you need some actualization. One example is the z-component of the spin of a spin-1/2 particle such as an electron. There are two independent states which can be denoted $\left(\matrix{ 1 \\ 0}\right)$ and $\left(\matrix{ 0 \\ 1}\right)$ and each of which can be produced by orienting a stern-gerlach device in ...


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My understanding of this limited, but this might help (too long for a comment): The state space is spanned by the set of simultaneous eigenstates of the Hamiltonian, $ \hat L^2$, and $ L_z $. In fact, they form an orthonormal basis of a Hilbert space $ H $ which is the state space. Out of convenience, we denote the eigenstates by the quantum numbers, ...


2

For orbital angular momentum, indeed, $L = x\times p$ even as a quantum operator, see this question. When writing a ket $\lvert l,m \rangle$, this is meant to live in the $2l+1$-dimensional space $\mathcal{H}_l = \mathbb{C}^{2l+1}$ on which the representation of the angular momentum algebra labelled by $l$ exists ($m$ is the eigenvalue of the ket for ...


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For a normalised linear combination of (orthogonal) states like this, the probability of measuring one of them is the absolute square of the coefficient in the combination: If $$\Psi = a_1\psi_1+a_2\psi_2+...$$ where $|a_1|^2+|a_2|^2+... = 1$ then $$P(\psi_1) = \left|a_1\right|^2, P(\psi_2) = \left|a_2\right|^2$$etc. Slightly more generally, if you know ...


0

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\ad}[0]{\hat{a}^\dagger}$ $\newcommand{\ao}[0]{\hat{a}}$ Firs of all I'd like to tell you that the position operator $x$ is given in terms of the ladder operator by the following relation: $$\hat x = d (\ao+\ad )$$, where $d=\sqrt{\frac{\hbar}{2m \omega}}$. Computing $\hat{x}^3$ gives then, as you ...


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There is indeed a way to construct the projection-operators, when you only know the operator itself and its eigenvalues. The derivation can be found in Julian Schwingers "Quantum Mechanics: Symbolism of Atomic Measurement" and leads to $$ P_j = \prod_{i\neq j} \frac{A-a_j}{a_i - a_j}, $$ where the product goes over all distinct eigenvalues $a_i$ of $A$. ...


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Eigenvectors are undetermined up to a scalar multiple. So for instance if c=1 then the first equation is already 0=0 (no work needed) and the second requires that y=0 which tells us that x can be anything whatsoever. This is fine, and correct, as $x=re^{i\theta}, y=0$ is a fine solution when c=1. If $c\neq 1$ then none of the equations are already 0=0 so ...



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