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The simplest way is to exploit the symmetries here. So, instead of mindlessly going through the algebra, solving equations etc. you just use the fact that you are free to call any direction the x-direction, and set up a right handed coordinate system. In particular, this means that you are free to cyclically permute x, y, z in the equations. The problem of ...

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This gives a slightly more general formulation of Hindsight's answer. The most straightforward interpretation of superpositions in Quantum Mechanics is usually given in the context of orthogonal bases, which already involve more structure than the Hamel bases mentioned in the comments. For the purpose of keeping things simple, let us assume a countable basis ...

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I had the same conceptual problems you are having when I was studying QM. Let me spray a bit of philosophy which allowed me to finally move on and concentrate on the mathematical stuff which actually matters :) Quantum states are rays in the Hilbert space (or points in the projective Hilbert space, or etc.). All states have the same physical meaning: they ...

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"What really is X" is a tricky question in physics, especially when dealing with such a fundamental and abstract theory as quantum mechanics. So I can give a mathematical definition: Let $\mathcal{H}$ be a complex Hilbert space, that is, a complete vector space over $\mathbb{C}$ equipped with a positive-definite inner product $\langle\, \cdot\, |\, \cdot\, ... 4 A state is something that encodes our knowledge about the system. And that's it. There are many ways to encode a state in quantum mechanics. As a wavefunction ("Schrödinger representation"), as Fock momentum states ("Fock representation"), as a density matrix, as a ray in a Hilbert space, as a linear functional on the$C^*$-algebra of observables, as a ... 3 I) Let us for clarity use a subscript "$S$" (and "$H$") to denote the Schrödinger (Heisenberg) picture, where bras and kets evolve (are unchanged) and operators are unchanged (evolve), respectively. Moreover, let us assume that the two pictures coincide at the instant$t_0$(which Ref. 1 assumes is$t_0=0$). II) Recall first of all the possibly confusing ... 3 In a certain sense,$\psi(\mathbf x)$is to$|\psi\rangle$as$v^i$is to$\mathbf v$. We say that$\mathbf v$is vector in a vector space while$v^i$are the components of$\mathbf v$on some basis: $$\mathbf v = \sum_i v^i\;\vec e_i$$ where $$v^i = \mathbf v \cdot \vec e_i$$ Analogously,$|\psi\rangle$is a ket in a vector space while$\psi(\mathbf ...

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$\mathcal{E}$ is just a separable Hilbert space. Since all separable Hilbert spaces are isomorphic (non-canonically, alas), nothing further has to be specified about this. The elements of this abstract space are written as kets. Since $L^2(X)$ is a separable Hilbert space for $X=\mathbb{R}^n$ (it is for a larger class of spaces, but that's not relevant ...

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Lets put it another way. Every physical observable corresponds to a quantum mechanical (mathematical) operator, i.e. a (usually) differential that acts on the wave function. It is part of the postulates of quantum mechanics (page 2 in link) as wave mechanics given by the solutions of the Schrodinger equation. It is necessary that every physical observable ...

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Only the full hamiltonian is observable, as in "corresponds to a physical quantity that can be observed." The "free" and "perturbed" parts are a convenient split when we do a calculation but are not separately observable. In fact it would be surprising if they were, since the split between free and perturbed was arbitrary, it is our choice how to make the ...

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A real linear combination of any two observables is an observable. Suppose $A,B$ are observables, and you have a Hamiltonian $H$. Let $C=A+B$, and note: $C^\dagger=A^\dagger+B^\dagger=A+B=C$ $[H,C]=[H,A+B]=[H,A]+[H,B]=0$ So theoretically speaking, yes, it would be. (The result simply extends to multiplication by a real constant)

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Yes. If $A$ and $B$ are observables, then so is $A+B$. In this case, since $H_0$ and $H$ are observables, so is $H-H_0 = W$.

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No. For example, $f(x, y) = xy$ gives $f(x, p) = xp$ which is not an observable since $(xp)^\dagger = px \neq xp$.

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The claim is that $|0(\theta)>$ lies outside the Hilbert space built on the original vacuum |0>. To check that this true, consider the overlap of the new vacuum $|0(\theta)>$ and the (unnormalized) basis states $(a_k^\dagger)^n|0>$ generated from |0>, taking into account that $a_k(\theta) = a_k + \theta_k$, $a_k = a_k(\theta) - \theta_k$, and ...

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Observables (I refer here to Hermitian operators) are confusingly named since they are not observable. What is observable, as you noted, is the eigenvalues that represent the outcomes. So what are observables for? The clearest way to understand the issue is to look at Heisenberg picture observables. These observables change over time but the state does not ...

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Observables are measurable quantities represented by certain mathematical structures dependent on the theory. In classical mechanics, measurable quantities are represented by functions on a phase space. In quantum mechanics, they are represented by operators on a Hilbert space. In classical mechanics, a measurement is equivalent to evaluating the function ...

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In non relativistic QM observables are represented by hermitian operators $\hat{O}$ acting on the states $\psi$, they represent measurements of some physical quantity ($\hat{O}$ represents say angular momentum, or spin). The result of the measurement will be an eigenvalue $o$ of the operator $\hat{O}$. The eigenvalues represent the actual measured value.

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Once you admit the tunnelling the "topology" of the problem changes to "particle in a circle with an infinite barrier". In fact, to avoid arguments with energy when crossing the barrier, it can be described as "particle in a circle with an infinite barrier in one point". So physically it can be argued to be a different problem.

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You need an infinite dimensional Hilbert space to represent a wavefunction of any continuous observable (like position for example). Wavefunctions map real numbers, which correspond to classical observables like spin or position, to complex coefficients of some set of basis kets in a Hilbert space. That basis and those coefficients define a ket that can be ...

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