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2

As suggested by WSA aka RV, I copy my comments into a (partial) answer. The key point is that the theorem says "for any given $\psi$ there exist two bases $\{i_A\}$ and $\{i_B\}$ such that...". This means that the choice of the bases depends on the vector $\psi$ we are considering. So there is not an $n$ dimensional common basis that span the whole ...


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Denote $|\psi\rangle = \sum\limits_{i = 1}^m \sum\limits_{j = 1}^n h_{ij} |ij\rangle$ as $|\psi\rangle \rightarrow H = (h_{ij})_{m \times n}$. Then we have the following lemma: Lemma: Define matrix $U$ (in the original basis) as a new setting for Alice and $V$ for Bob, then a state $|\psi\rangle$ in the basis of the new settings is $U^* H V^\dagger$. ...


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How can a scattering process have bound states? We are familiar with bound states from our everyday experience. For example, two hydrogen atoms interact through the Coulomb force. This leads to the formation of a bound state, namely, the hydrogen molecule. The most simple model of this situation is the square-well potential. This potential has ...


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DFT is exact concerning ground state properties. However, the bandgap is not a ground state property. Not sure, if this simple explanation is correct, but I find it somehpw intuitive: in order to speak about a bandgap, you either need a (at least fictitious) electron in the conduction band, which therefore is in an excited state, or you need a ...


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In the context of ultracold Fermi gases, a BEC-BCS crossover means that by tuning the interaction strength (the s-wave scattering length), one goes from a BEC state to a BCS state without encountering a phase transition (thus the word "crossover"). It is also useful to know that the BEC state is a Bose-Einstein condensate of two-atom molecules, while the ...


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In a system with non-interacting particles you may know that the right quantity to look at is $\rho \Lambda^3 \sim \left(\Lambda/l \right)^3$ where $\Lambda$ is the (thermal) de Broglie wavelength and $l = \rho^{-1/3}$ the typical inter-particle distance in the system. Basically if $\Lambda/l \ll 1$, then your system behaves classically i.e. you do not ...


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Is there a simple argument why the first version is right and the second one not? The correct form for a two-body interaction in second quantization is always: $$ V=\frac{1}{2}\int d^3r \int d^3r' V(|\vec r-\vec r'|)\hat \psi^\dagger(\vec r)\hat \psi^\dagger(\vec r')\hat \psi(\vec r')\hat \psi(\vec r) $$ Note carefully the order of the $\vec r$s and ...


3

For the Laplacian $$ \Delta ~:=~ -\frac{d^2}{dx^2} ~\geq~ 0, $$ the corresponding HS transformation reads $$\exp\left(-\frac{a}{2}\Delta\right) f(x)~=~\int_{\mathbb{R}} \!\frac{dy}{\sqrt{2\pi a}}\exp\left(-\frac{y^2}{2a}\right) f(x+y), \qquad a~>~0.$$ Proof: Use Fourier transformation.



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