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In field theory, there are two vacua. The non-perturbative vacuum $|\Omega\rangle$ and the vacuum of the free theory $|0\rangle$. The wikipedia article makes reference to $|\Omega\rangle$ in terms of $|0\rangle$ and its excitations. The true vacuum is annihilated by the (dressed) annihilation operators, and can be thought of perturbatively in terms ...
Your idea of what $\psi(\vec x,t)$ is supposed to be is essentially correct. Given a space of states $\mathcal{H}$, the "Schrödinger state" is a map $$\psi : \mathbb{R}\to\mathcal{H}, t\mapsto\lvert\psi(t)\rangle$$ where $\lvert \psi(t)\rangle\in\mathcal{H}$ for every instant $t$. If $\mathcal{H}$ is a space of functions in a variable $\vec x$, then $\lvert ... 6 The operator$X^{-2}$does exist and is self-adjoint as it follows from standard spectral theory. Its domain is $$D(X^{-2}) := \left\{\psi \in L^2(\mathbb R, dx) \:\left|\: \int_{\mathbb R} x^{-4} |\psi(x)|^2 dx \right.< +\infty\right\}$$ and thereon $$(X^{-2}\psi)(x) := x^{-2}\psi(x)\:.$$ The eigenfunctions of the Hamiltonian operator of the harmonic ... 6 Dirac notation is ill-suited for non-self-adjoint operators. Here's why: Let$(-,-)$be the inner product on our Hilbert space. The expectation value of$AB$is then $$\langle AB \rangle_\psi = (\psi,AB\psi)$$ by definition, and Dirac notation writes$\langle \psi \vert AB \vert \psi \rangle$. for this. But, in this notation, it is no longer clear to which ... 5 Comments to the question (v5): In this quantum case the overline/bar notation$\bar{A}=\langle A\rangle$is borrowed from statistics and it denotes a quantum expectation value of a quantity$A$. See also Ehrenfest theorem. The problem from Ref. 1 considers a harmonic oscillator with Hamiltonian operator $$\tag{A} H~=~\frac{p^2}{2m} ... 3 (1) Yes, take {\cal H} = L^2(\mathbb R, dx)\oplus L^2(\mathbb R, dx) and thereon \left(X (\psi, \phi)\right)(x,y) := (x\psi(x),y\phi(y)). We have \sigma(X)=\sigma_c(X) and the degeneracy is just 2. (2) Yes, use the example (1) with a countably infinite copies of L^2(\mathbb R, dx) and use the Hilbertian direct sum of Hilbert spaces. (There are ... 3 I would say that you are suffering from a little bit of pop-science fatigue. You don't need to be convinced of these things because the great majority of working physicists don't think about quantum mechanics in this way. What if a brain dead person looks and doesn't comprehend? This is called "begging the question", I think. Except for a handful ... 3 Each separable infinite-dimensional Hilbert space carries both irreducible and reducible representations of any noncompact Lie groups you can name. But this information in itself is of little use. The Hilbert spaces in quantum mechanics always come with distinguished representations that give certain operators an interpretation as distinguished ... 3 I think the fundamental misunderstanding of superposition has a lot to do with the popular interpretation of quantum mechanics. That is, how Schrödinger's cat is portrayed in popular science. When a quantum system is in a state of superposition, it means that the outcome of a measurement of some property of that system is uncertain. The wacky thing about ... 3 The ladder operators satisfy: \bf{a^{\dagger}}$$|n>=\sqrt{n+1}|n+1>$ $\bf{a}$$|n>=\sqrt{n}|n-1> Taking into account <n|m>=\delta_{n,m} , you get the answer. 3 Comments to the question (v1): The operator f(\hat{p}) and the identity operator {\bf 1}= \int\!\frac{dp}{2\pi}|p\rangle\langle p| commute. The operator f(\hat{p}) and the integration \int\!\frac{dp}{2\pi} are independent of each other. The q-ket and the q-bra are independent of the integration \int\!\frac{dp}{2\pi}. Of course, if one ... 2 I know this question was asked a long time ago, but since I thought very hard about the same question today and didn't find the other answer very helpful, I decided to write my own. The problem is that OP is not really asking about why the product form used in the BO approximation is valid, but how the given expansion (which is claimed to be exact, see e.g. ... 2 Assuming everything is defined in the correct Hilbert spaces, project the decomposition$$ |\psi\rangle = \sum_n {c_n |n\rangle} $$onto the position kets ("states") |x\rangle and obtain$$ \psi(x) = \langle x |\psi\rangle = \sum_n {c_n \langle x |n\rangle} = \sum_n {c_n u_n(x)} $$where the u_n(x) = \langle x |n\rangle are the wavefunctions ... 2 Your confusion stems from the fact that \lvert x \rangle is not inside the Hilbert space of states. It cannot be because \langle x \vert x \rangle = \delta(x-x) = \delta(0) is not an allowed value for an inner product in a Hilbert space to have. There are several things to say about \lvert x \rangle: If you want to make precise what kind of objects ... 2 Just regarding question number 3: 1) If your friend knows anything at all about quantum mechanics, he will not say "Both", because quantum mechanics does not allow anything to be in two states at once, ever. 2) If your friend knows a small amount about quantum mechanics, he might make the mistake of saying "Neither", because, after all, most coins ... 2 The process of "collapse" can almost entirely be handled just by including your measurement apparatus in a quantum description. It's the quantum interaction between observer and observed that causes collapse. For the mathematics of this process, you might want to look at my answer to Particle interactions which are NOT considered observations? for more on ... 2 Born calculated solution to Schroedinger's equation corresponding to electron scattering experiment and what he got was continuous function of scattering angles measured with respect to the original direction of propagation of electrons. However, in experiment electrons are always detected at definite points of a screen. Clearly, there is no direct match ... 2 You can think of the spin state of an electron as represented by a vector (\alpha,\beta). Depending on how you set things up, "Up" might be represented by (1,0), "Down" by (0,1), "Left" by (1,1), and "Right" by (-1,1). Up is orthogonal to Down, and Left is orthogonal to Right, but Up is not orthogonal to Left. 1 when we define X=a+ a^\dagger, \frac{1}{X} and any power thereof do not exist. Proof: Consider the state \left|x\right\rangle\propto e^{-a^{\dagger 2} + 2 x\, a^\dagger}\left| 0\right\rangle, you can show that X\left|x\right\rangle = x \left|x\right\rangle, and in particular if you choose x=0, then X\left|x=0\right\rangle = 0. Now from ... 1 Angular momentum in quantum mechanics in general works like this: the total is measured by L^2 = \hbar^2 \ell (\ell+1) whereas the projection along any axis is measured by L_z = \hbar~m between -\ell \le m \le \ell. Both \ell and m are simultaneously measurable (i.e. the L^2 and L_z operators commute), and they must be spaced by integers but ... 1 First get the idea of kets as some component in \mathbb{R}^n out of your mind. kets are elements of a complex vector space, in this case a finite dimensional one. Yes the space will be isomorphic to \mathbb{R}^n, or rather \mathbb{C}^n, but by assuming they are \mathbb{C}^n, you may imbue them with properties of \mathbb{C}^n that are not true of a ... 1 Under conformal mapping z=>w(z) and \bar{z}=>\bar{w}(\bar{z}) a field of conformal dimension(h,\bar{h}) transforms as \tilde{\phi}(w,\bar{w})=(\frac{\partial w}{\partial{z}})^{-h}(\frac{\partial \bar{w}}{\partial\bar{z}})^{-\bar{h}}\phi(z,\bar{z}).. 1 The situation is impossible and therefore doesn't happen. When you claim to have two states \psi_1 and \psi_2 then I'll assume they are linearly independent, otherwise they aren't really two different states. Then you claim you have an Operator O such that POP^{-1} = \epsilon_3 P where P is the parity operator and further that P \psi_1 = ... 1 In the context of quantum mechanics, Hilbert spaces usually refer to the infinite-dimensional space of solutions to the time-dependent Schrodinger equation$$ i\frac{d}{dt} \left|\psi (t)\right\rangle = H(t) \left|\psi(t)\right\rangle$$for the state vector$\left|\psi (t)\right\rangle\$. This space is completely determined by the (in general) time-dependent ...