# Tag Info

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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 ...

2

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 $\psi(\mathbf ... 2 The claim is that$|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$, and ... 2 No. For example,$f(x, y) = xy$gives$f(x, p) = xp$which is not an observable since$(xp)^\dagger = px \neq xp$. 1 Yes. If$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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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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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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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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