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One way to look at this is through the Schrodinger's equation: $i\hbar|\dot\psi(t)\rangle = H|\psi(t)\rangle$ Then a general solution to this equation is: $|\psi(t)\rangle = e^{-iHt/\hbar} |\psi(0) \rangle$ (Notice that $H$ is an operator here instead of a scalar. $H$ also has to be time-independent, as is usually the case for introductory quantum ...

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Consider for the shake of simplicity a free neutral scalar field $\phi$. Passing to the second quantization picture, it is a operator valued distribution $$C_0^\infty(M;\mathbb R) \ni f \mapsto \phi(f)$$ where $M$ is Minkowski spacetime and $\phi(f)$ is a densely defined symmetric operator on the Hilbert space $$F_+(\cal H) = \mathbb C \oplus \cal H ... 1 Let V be an (in)-finite dimensional vector space, V^* its dual and let$$ \sigma\colon V\times V\to \mathbb{C} $$be a scalar product, that is \sigma(v,u)\in\mathbb{C} on any pair of vectors in V. At this point notice the following: Proposition: every bilinear form \sigma on V induces a map \rho\colon V\to V^* as \rho_v(\cdot) = ... -1 U=e^{-iH(t_{1}-t_{0})} is indeed the formal time evolution operator in RQFT. However, typically, the U does not exist for a system with an infinite number of degrees of freedom and that is why one never sees a \psi(t) calculated explicitly. In Dirac's 1966 book "Lectures on Quantum Field Theory", section 7 on page 31 studies a model Hamiltonian for ... 0 I think this is saying much the same as Gennaro, but I'd phrase things slightly differently. In config \mathbf{1} your system has some eigenfunctions \psi, and you start with the system in one of these eigenfunctions \psi_n. When you change to config \mathbf{2} the function \psi_n is no longer an eigenfunction because your system now has a set of ... 1 It might help if we use a little mathematics to clear up what's happening here; I assume that you are familiar with vector spaces equipped with an inner product. These when first introduced are over the reals, and are finite-dimensional. In the standard presentation of QM, these two things changes: we no longer work over the real numbers, but the complex ... 1 I'm taking a QM course and I'm trying to make sense of why observables are sometimes conjugated for no apparent reason in their inner products. There is a lot of confusion even already. Let's start with the basics. There are vectors and operators and an inner product. If everything was finite dimensional the vectors would be complex column vectors, the ... 0 I did not get your question completely, but still would try to answer it. You see a bra and and a ket represent some state. Sometimes eigenkets are written as their eigen values like |x>. So if you are taking inner product of two states physically you are trying to find the probability amplitude (overlap) of one state collapsing to the other. Mathematically ... 2 A quantum system is described by a set of self-adjoint operators (A_1\ldots A_n, H) and a Hilbert space \mathcal{H}. The mentioned operators represent the observables that you can experimentally measure and their eigenvalues the possible outcomes. Among them there is a special one, the Hamiltonian H, describing the time evolution of the system. A state ... 0 There is no eigenvector corresponding to continuous spectrum. The formalism of Gel'fand triples does not give much help in solving your doubts either, and it has very few applications in my experience. One reason is that those "generalized eigenvectors" are not in the Hilbert space but on a bigger space, and what you can do with them is not much on a ... 1 This is possible iff the spectrum of the Hamiltonian (including multiplicities of eigenvalues) can be written as the sum of two other spectra. So it doesn't depend on a basis. 2 Another way to see this is to observe that any state |\psi⟩\in\mathcal H can be extended to an orthonormal basis of the Hilbert space, and in that basis the trace \operatorname{Tr}\left(|\psi⟩⟨\psi|\hat A\right) is exactly ⟨\psi|\hat A|\psi⟩. More explicitly, for any |\psi⟩\in\mathcal H there exists a sequence ... 8 Let |n'\rangle be a basis of the Hilbert space, then$$ \textrm{tr}\Big[|\alpha\rangle\langle\alpha|A\Big]=\sum_{n'}\langle n'|\alpha\rangle\langle\alpha|A|n'\rangle=\sum_{n'}\langle\alpha|A|n'\rangle\langle n'|\alpha\rangle = \langle\alpha|A\left(\sum_{n'}|n'\rangle \langle n'|\right)|\alpha\rangle=\langle\alpha|A|\alpha\rangle $$0 Choosing a Hilbert space is easy. It just needs to be mathematically convenient, while still completely characterising the system. (You could use l_2 to represent any quantum state, but you often don't because that is often mathematically very inconvenient.) As for defining operators, it is true one cannot explicitly define an operator without defining ... 3 There are many approaches. But first I want to make sure you know that when you have n spin zero particles in a 3d space and have a wavefunction that the function is from \mathbb R^{3n}, i.e. from configuration space. But also I want you to know when someone says infinite dimensional Hilbert Space, they mean the size of a set of mutually orthonormal ... 3 Separable Hilbert spaces are all isomorphic to one other, since they are all isomorphic to \ell_2, despite the dimensions (this holds even in the infinite dimensional case). No matter how you realise it, you may just choose one Hilbert space and its properties will exhaust all you need to perform the calculations and derive observable quantities (they do ... 0 So yeah, what you need to do is successfully integrate by parts. In spherical coordinates the integral is:$$\langle\phi|\hat B|\psi\rangle = \int_0^\infty dr~\int_0^\pi r~d\phi~\int_0^{2\pi}r\sin\phi~d\theta\;\phi^*(r,\phi,\theta) ~i\left(\frac{\partial\psi}{\partial r}\right)_{\phi,\theta}.$$The integration by parts on the variable r differentiates r^2 ... 0 Two main points are.... Generally \langle{x}|[X,P]|\alpha\rangle \not= \langle{x}|XP|\alpha\rangle-\langle{x}|PX|\alpha\rangle When [X,P]=XP-PX is an well-defined operator in a Hilbert space, H=L^2([a,b]), space of square-integrable functions in [a,b], the domain of definition of [X,P] is a set of functions |\alpha\rangle satisfying ... 0 I think you'd probably need to integrate that term by parts, lowering k = p'/\hbar to 1 while raising \exp[i~k~(x - x')]~dk into [-i\hbar/(x - x')]~\exp[i~k~(x - x')]. The result you get for the middle integral is then$$-2\pi i\hbar ~ \frac{\delta(x - x')}{x - x'}.$$If you hold off evaluating the integral further, the other integral will be ... 0 Let us first clarify the difference between a state and its wave function representation. A state |\psi\rangle is an element in a Hilbert (or equivalently) Fock space, whereas its wave function \psi(x)=\langle x |\psi\rangle is its representation onto the position (or momentum, respectively) basis. In quantum field theory things extend a little, ... 1 Hermitian conjugate (also called adjoint) of the operator A is the operator A^* satisfying$$\langle f,Ag\rangle\,=\,\langle A^* f,g\rangle \text{ for all }f,g\,\in H$$H is so-called Hilbert space and f,g are vectors. Since you are new to QM, you need not be confused with the word "Hilbert space". Just treat it as a special case of vector spaces. ... 1 Yes, this is something physicists like to sweep under the rug. \lvert x \rangle as an "eigenvector" of the position operator is not a vector in the physical Hilbert space of states \mathcal{H} - it's not normalizable, for one, as the odd "inner product"$$ \langle x' \vert x \rangle = \delta(x' - x)$$indicates. To deal with it mathematically, one has ... 0 We have \begin{eqnarray*} \psi (x,t) &=&<x|\exp [iH(t-t_{0})]\psi (t_{0})>=\int dy<x|\exp [iH(t-t_{0})]|y>\psi (y,t_{0}) \\ &=&\int dyG(x,t;y,t_{0})\psi (y,t_{0}) \end{eqnarray*} and in your case \psi (y,t_{0})=\delta (y-x_{0}). Here H is the harmonic oscillator Hamiltonian. For the Green's function I could not find an ... 0 Yes, it's not very satisfactory. The usual resolution is to modify the inner product.$$ \langle\langle A | B \rangle \rangle \equiv \langle A | c | B \rangle $$Then inserting \{ b, c \} = 1 into the norm no longer yields a vanishing result. Alternatively, as the questioner did, one can try to use the facts that |{\uparrow} \rangle = b | {\downarrow} ... 3 In QM the position operator and the Hilbert space of a particle are defined contextually: The Hilbert space is L^2(\mathbb R^3, d^3x) and the operator position along x_k is defined, in that space, as (X_k\psi)(x):= x_k \psi(x) with the obvious domain. You can adopt a more abstract viewpoint if you simultaneously define the momentum and the position ... 0 There is nothing circular. We assume that the physical reality of a particle corresponds to states in some vector space. At this point we don't know of a basis defined on this vector space because we haven't given it any structure. We give it structure by assuming the position operator is a Hermition operator in this vector space with eigenvalues on the ... 0 The states of string theory are quantum states. They represent a "vibration" of the string in the same sense that a particle in standard QFT represents a "vibration" of the quantum field. That is, they do not represent actual "physical" vibration at all. In particular, the states do not describe actual physical position, vibrations or whatever of the ... 0 Just to define the concepts: Given a Hilbert space {\mathcal H}, there always exists a complete set of distinct, but commuting observables \{ {\hat O}_1, {\hat O}_2, ...,{\hat O}_m \}, \left[{\hat O}_j, {\hat O}_k \right] = 0, whose common eigenstates define a basis of {\mathcal H} labeled exhaustively by the eigenvalues of each {\hat O}_k. This ... 4 Post measurement you want the eigenspaces to be orthogonal (and to have the projection onto the eigenspace to be entangled with a state of the measurement device). So you want the different eigenspaces to be orthogonal. And you want to be able to evolve to a post measurement state that has the right kinds of states. So really it is about the kinds of end ... 1 1) If all the eigenvalues of an operator are real, then it is Hermitian. You can see this by writing the operator (call it A) in the eigenvector basis. Then A has all real eigenvalues along its diagonal and zeros everywhere else. Therefore, A^\dagger = A which means it is Hermitian. 2) Many of the operators that we call "observables" are the generators ... 2 I) OP's question (v1) seems to be spurred by a common confusion: The Heisenberg instantaneous position eigenstate |x,t_0\rangle_H  does not evolve in time t but does depend on a time parameter t_0. In detail,$$\tag{1} | x, t_f \rangle_{H}~=~ e^{i\hat{H}\Delta t/\hbar} | x, t_i \rangle_{H}, \qquad\Delta t~:=~t_f-t_i, $$where we for simplicity have ... 2 The "time-independent Schrödinger equation" is just an equation for the eigenvalues and eigenvectors of the Hamiltonian operator on the Hilbert space of states (typically L^2(\mathbb{R}^3,\mathrm{d}x), the "space of wavefunctions") The spectral theorem tells us that the eigenvectors of any self-adjoint operator form a basis for the space the operator ... 2 It is a mathematical theorem that self-adjoint operators in Hilbert space have a complete spectrum. Note that "self-adjoint" has a special mathematical meaning. Not every Hermitian symmetric operator is self-adjoint. For example, the 1D free Schrodinger Hamiltonian on an open interval without boundary conditions is not self-adjoint. The reason is that we ... 2 The Hilbert space formulation, sometimes with the explanatory add-on: "where observables are represented by linear operators acting on Hilbert space". Both the wave and matrix sub-formulations are basically shadow-double formalisms for the very same structure. 2 Who says kets are not functions? A ket can be a function of time: |\psi(t)\rangle. Since they are elements of a vector space (with a complete inner product), you can define their derivative:$$\frac{d|\psi\rangle}{dt} =\lim_{h\rightarrow 0} \frac{|\psi(t+h)\rangle-|\psi(t)\rangle}{h}$$And given a Hamiltonian H : \mathcal{H}\to\mathcal{H}, which is a ... 3 The Schrödinger equation is not limited to any particular kind of Hilbert space. There's no problem with abstract kets. Given a space of states \mathcal{H}, a Schrödinger (or time-dependent) state is given by a (smooth) map$$ \lvert\psi(\dot{})\rangle : \mathbb{R} \to \mathcal{H}, t \mapsto \lvert \psi(t) \rangle$$so the Schrödinger state ... 0 Observables correspond to particular things you can do in the lab (or observe in nature). So let's first talk about something you can do in the lab. You can take a particle with spin and subject it to an inhomogeneous magnetic field. A particle with spin has a magnetic moment proportional to the spin and we know to Hamiltonian for a particle with a magnetic ... 3 To add to Yuggib's Answer, which I am in complete agreement with: I have never particularly liked the name "operator" for an "observable", because the former implies a mapping and, therefore, that the image \hat{A}\,\psi has a direct physical meaning. As in Yuggib's Answer, there is in general no direct physical meaning. Rather, an "observable", as I like ... 3 I do not think that the action A\psi has a direct physical meaning, when A is a generic observable. This is because the interpretation of a quantum system as a mathematical model yields the wavefunction and its corresponding Hilbert space as a sort of byproduct. In fact, the state may not always be a wavefunction: without entering too much into details, ... 1 Using the definition of the creation operator, a^\dagger = c(m\omega \hat x - i\hat p) where c is a constant, and \hat p = -i\hbar\partial_x, you can write the eigenvalue problem in the position representation as$$(m\omega x - \hbar\partial_x)\psi = \alpha\psi.$$You can solve this differential equation to find$$\psi = C\exp(m\omega x^2/\hbar - ...

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To add to Innisfree's correct answer, I'd like to emphasize something that the OP does not seem to know and that is that the creation operator has no eigenvectors (nor, therefore, eigenvalues). It is easy to see this: write a general state as a row vector $(\psi_0,\,\psi_1,\,\cdots)$ of superposition weights for the number states ...

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A coherent state is, amongst other interesting things, an eigenstate of the annihilation operator. It is not an eigenstate of the creation operator; hence, I'm not sure this "eigenvalue problem" makes much sense. This is easy to realize. You can quickly see that $\langle0|a^\dagger|\alpha\rangle=0$, whereas $\langle0|\alpha\rangle\neq0$. If you really want ...

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To get $\langle \chi|\chi \rangle =1$, you find that $A=\frac{1}{5}$. Then, when you calculate $\langle S_x \rangle$, you have to use that value for A and then you find the right result of $\hbar/50$. You if didn't get it, I'll do explicitly for you :)

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The norm of your spin wave function is 5 so they divided by 25 to normalize it, then multiplied it by hbar over 2 to get 1/50

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I did show that this equation works, but I want to know how to prove it just working with the fact that the Pauli matrices span a basis in 2x2 Hilbert space and that M is hermitian. You can do this if you can specify exactly what you mean by "span a basis in 2x2 Hilbert space," which sounds really convoluted and mathematically wrong for me. For ...

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First, check that the 2x2 hermitian matrices form a (finite dimensional) real vector space. Convince yourself, that the set $\{1,\sigma_i\}$ is linearly independent. You may now either directly expand a generic hermitian matrix in terms of $\{1,\sigma_i\}$, or note that the dimension of the aforementioned space is four, thereby proving that $\{1,\sigma_i\}$ ...

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Let $\chi$ be the spinor defined as follows:- $$\chi=\begin{pmatrix} a\\b\end{pmatrix}$$ then for measuring $S_x$ we need to find the eigenspinors of $S_x$ which are $$\chi_{+}^x= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \end{pmatrix} ,\hspace{1cm} \chi_{-}^x= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ -1 \end{pmatrix}$$ Now the spinor $\chi$ can be ...

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The adjoint eigenspinor you multiplied by was the unit length eigenvector of $\sigma_z$ with positive eigenvalue. If you you want a spin up result for the direction $(n_x,n_y,n_z)$ find a unit length eigenvector of $n_x\hat\sigma_x+n_y\hat\sigma_y+n_z\hat\sigma_z$ with positive eigenvalue. And use that instead. If you wanted to do an interaction in the x ...

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