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In your first paragraph you describe a Stern-Gerlach device as one with a magnetic field in the $\hat z$ direction. And then later you talk about having a large homogeneous component of the magnetic field. I'm not sure you have an accurate physical model of a Stern-Gerlach device. The Hamiltonian for a Stern-Gerlach has magnetic fields components combined ...

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Think about the kinetic energy observable $T$ for a spin-1/2 particle in 3D space. The particle's Hilbert space is technically ${\mathfrak H} = L^2({\mathbb R}^3) \otimes {\mathbb C}^2$, yet the kinetic energy operator $T$ is first defined on $L^2({\mathbb R}^3)$, where its eigenfunctions $\Psi_{\bf p}({\bf x})$ are easily found. The extension of $T$ to ...

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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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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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Let $\mathcal{H}$ be the space of states of our theory. Then, time evolution is given by a unitary operator $U(t_2,t_1) : \mathcal{H}\to\mathcal{H}$ that evolves "stuff" from time $t_1$ to time $t_2$. For time-independent Hamiltonians it is just $\mathrm{e}^{\mathrm{i}H(t_2 - t_1)}$. If we are in the Schrödinger picture, we say states "carry the time ...

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Let me try to use a programming analogy. A quantum mechanical object is like a class. It has certain attributes, which would be physical attributes, e.g. position, momentum. It also has certain methods, which are physical operations that can change or modify the object or modify the environment using the object. In physical terms these are the unitary ...

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It is worse than lazy evaluation. Haskell I can treat a lazy value as if it were already there and manipulate it as such. In quantum mechanics you can't do that. What you have is something that tells you the relative frequency of getting lots of results for different interactions if you did one of the various interaction first. And you can't do them ...

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Absolute four-momentum is not an observable. Relative four-momentum is. We cannot find the four-momentum of the lab itself, but we can (and do, regularly) measure the four-momentum of particles relative to a given lab, which then allows us to calculate the four-momentum of said particles relative to any frame you care to name. Whoever said that wasn't ...

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