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I am reading that a linear operator $A$ can be thought of as a (1,1) tensor [where $(r,s)$ corresponds to $r$ vectors and $s$ dual vectors]. This can be done by saying $$A(v,f) \equiv f(Av)$$ where $v …

results in $$\langle3|\hat a^2|3\rangle+\langle3|\hat a\hat a^\dagger|3\rangle+\langle3|\hat a^\dagger\hat a|3\rangle+\langle3|(\hat a^\dagger)^2|3\rangle$$ There are a few ways to think about the ladder operators … -i(\frac{1}{2 m \omega \hbar})^{1/2}P$$ When doing $\langle n|\hat{a}|n\rangle$ you are getting the answer you'd get by the integral using the above definition of $\hat{a}$, but the beauty of ladder operators …

$$\langle p \rangle = -i\hbar\int\psi^*\frac{d \psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\frac{d}{dt}\int\psi^*\frac{d\psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hb …

The book I'm reading (https://arxiv.org/abs/1508.02595 pages 108-109) is trying to demonstrate why a two qubit state $\rho_{AB}$ has a symmetric extension iff $$\operatorname{Tr}\left(\rho_{B}^{2}\rig …

If I have a Hamiltonian $$\mathcal{H} = \prod_j^N Z_j$$ where $j$'s are different sites on a lattice and $Z$'s are Pauli $Z$ operators does that mean that the Hamiltonian can also be written as $$\mathcal … {H} = Z_1 \otimes Z_2 \otimes \cdot \cdot \cdot Z_N$$ and if they are all Pauli operators could it just be $$\mathcal{H} = Z ^{\otimes N}$$ …