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Write $$U = \frac{1}{2}(I_1 + Z_1) \otimes U'_{00} + \frac{1}{2}(I_1 - Z_1) \otimes U'_{11} + (X_1 + iY_1) \otimes U'_{01} + (X_1 - iY_1) \otimes U'_{10} = \left(\begin{array}{cc} U'_{00} &U'_{01} \\U'_{10}& U'_{11}\end{array}\right)$$ where $U'_{ij} \in G_n$. In terms of the latter, $$U' |\Psi\rangle = \sqrt{2} \;\langle 0 | U | 0\otimes \Psi \... 2 Your second equation isn't quite right. If you have a continuous complete set of states \{|p⟩\}, then the correct expansion of a given arbitrary state |P⟩ in that basis is of the form$$ |P⟩ = \int\mathrm dp \: f(p)|p⟩, \tag 1 $$with a single arbitrary function f(p) over the indexing variable p as a (continuous) coefficient. Here the dp denotes ... 4 There's no analytic proof, but numerical evidence suggests that if you know that the Hamiltonian is local, and it satisfies the Eigenstate Thermalization Hypothesis (which most local Hamiltonians do), then you can extract the entire Hamiltonian from a single excited eigenstate, though not from the ground state: https://arxiv.org/abs/1503.00729. 15 IF you know that your Hamiltonian is of the form$$ \hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r) \tag 1 $$for a single massive, spinless particle, then yes, you can reconstruct the potential and from it the Hamiltonian, up to a few constants, given any eigenstate. To be more specific, the ground state \Psi_0(\mathbf r) obeys$$ \hat H\Psi_0(\mathbf r) =...

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If unknown part of Hamiltonian is potential $V({\bf{r}})$, then you can write a stationary Schrodinger equation and figure out what the potential should be.

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Assume for simplicity that all the operators are bounded. If you know the wave function $\psi$ associated with the ground state of the unknown Hamiltonian $H$, then $H$ has the form $$H = E_0|\psi\rangle\langle\psi| \oplus K$$ where $K$ is another Hamiltonian defined on a subspace of the original Hilbert space of co-dimension 1, and $E_0$ is the energy of ...

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I will try to answer the integral question. A continuous function on a bounded interval is bounded. Throughout this answer I will use Riemann's definition for integrability. Using Cauchy-Schwarz Inequality- \int_{z\in\mathbb{C}}|F_{\phi}(z)|^2 e^{-|z|^2}\leq\Bigg|\int_{z\in\mathbb{C}}|F_{\phi}(z)|^2 e^{-|z|^2}\Bigg|\leq\sqrt{\int_{z\in\...

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I wanted this post to be a comment, but made it an answer instead. I cant delete it yet, so i'll extend it a bit. You might want to check Quantum mechanics for mathematicians by L. Takhtajan; the result is also in this book, but I don't remember if he proves it. In any case, you are right in that the uniqueness only holds for antiholomorphic functions. ...

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Spin operator of the total 2 electron system is tricky: the statistical requirement reduces the Hilbert space to a 3-d rather than 4-d version. Like a spin-1 system, the $S_z$, as represents in basis $|S_z=1,0,-1\rangle$ is (see http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html, for example) S_z=\hbar\left(\begin{matrix}1&0&0\\0&0&...

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Quantum states are rays in the Hilbert space. i.e. $e^{i\alpha}|\psi\rangle$ is the same as $|\psi\rangle$. The set of all $e^{i\alpha}|\psi\rangle$'s with $\alpha\in[0,2\pi)$ is said to form a ray in the Hilbert space. A point on the Bloch sphere represents a ray and not a state. If you want you can think of the point $(\theta,\phi)$ on the Bloch sphere as ...

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I will answer with an example on how eigen functions of momentum do not exist in a Hilbert space in general. When they do not, we say the momentum operator is not self-adjoint. Consider a one dimensional infinite potential well. Lets us place the walls at $x=0$ and $x=L$. For $x\ge L$ and $x\le 0$, the potential $V(x)=\infty$ and therefore we put the ...

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The boundary conditions determine whether an operator is Hermitian or not. Once you know your operator is Hermitian, you have the results on the spectrum. Without boundary conditions the momentum operator need not be Hermitian, hence its spectrum can have non-real values (here I am assuming that by Hermitian you actually mean self-adjoint). As an example, ...

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The extra one simply reflects the commutation relationship $[a,\,a^\dagger] = \mathrm{id}$ of the quantum mechanical harmonic oscillator. If each annihilation operator in the expression for $E^+$ acts only on its corresponding mode, and likewise for the creation operators in $E^-$, then the action of $E^-\,E^+ -E^+\,E^-$ on a state involving only excitations ...

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Sometimes a picture can help. Its not hugely rigorous (especially for a physics site) but using the vector model, pictures can be constructed to illustrate the states that two spins $\alpha$ and $\beta$ can form.

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In reality, you need the quantization field to prepare your atom in a given spin state and have it stay in that state- at zero field, any small perturbation could change it. But in a thought experiment that is not a problem. Still, you have to specify your atom as starting in some initial spin, and if it is starting as usual in an eigenstate of $S_z,$ you ...

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General wave functions can be expressed in terms of any set of eigenfunctions. But for bound systems, the energy eigenfunctions have a couple of appealing properties that make them popular: The energy eigenfunctions are the solutions to the time-independent problem, so you can work on a steady-state system. This often makes the math a lot easier. The ...

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Because it doesn't have total spin $s=0$ - it has total spin $s=1$, with the spin component parallel to the $z$-axis being zero. If you looked at that state in a different basis (e.g. the $x-$ or $y-$ basis) it would very clearly not have spin 0.

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Let me actually collect most of my comments in an answer attempting to be more coherent than they, or your labile question. In fact, you are piling up three different questions, logically distinct, but with strong and natural connections, so it might be worth splitting them apart, before bringing them back together in the final coda. First, there is plain ...

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Just to add to what's been said and to address the comment by @LightnessRacesinOrbit, you can indeed have perfectly good functions that are only defined on some subset of the real numbers. The reason the Dirac delta is not a function is not that it is only well-defined for some real numbers and not others; in fact, it is not defined as a function anywhere! ...

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Physicists usually generously relax the condition that the norm should be finite and they sometimes say that $|\vec r\rangle,|\vec p\rangle$ belong to the "Hilbert space". It's exactly the same "generous" language that allows physicists to say that $\delta(x)$ is a "function", the delta-function, even though its values around $x=0$ are infinite or "ill-...

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Here is what I understand: if you have a particle at state $|x \rangle$, active translating it by $a$ means moving the particle to state $| x + a \rangle$. Passive transformation means you keep the particle in the same place, and change the coordinate by new variable $x = x' + a$ (note that the coordinate system is translated backwards $-a$). I am not very ...

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This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to ...

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As commentators have indicated Hilbert space is a vector space. A manifold is a space with an atlas-chart construction with maps on overlapping regions that define connection coefficients and ultimately curvature. It is certainly possible to think of a finite dimensional complex vector space that is a locally flat region in an otherwise curved space. This ...

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Hilbert space is a complex normed vector space equipped with an inner product, where this inner product comes from the norm on the space (the norm of a vector on the hilbert space is the square root of the inner product of the vector with itself), but for a curved space like the minkowski space we will use a minkowski metric that differs from the usual ...

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Hilbert spaces are vectorspaces by definition. If you interpret a vector space as a manifold (which you can do) then it's a flat manifold.

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