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3

As the other poster said, the thing for which you use $E$ in your expression is not the eigenvalue of the Hamiltonian, but rather the eigenvalue of the kinetic part of the Hamiltonian. The Hamiltonian is: $$\hat H = \frac{\hat p^2}{2m} \otimes \mathbb{1} + \mathbb{1} \otimes h_z\sigma_z$$ and the eigenvalues are: $$E_\pm(p) = \frac{p^2}{2m}\pm h_z$$ ...

3

I slightly deviate from your notation and use $\phi$ to denote the scalar field as its more standard. Also I should point out that quantum fields are operators and thus under a transformation they get acted on from both the left and the right. The complex scalar field is given by, \phi (x) = \int \frac{ \,d^3p }{ (2\pi)^3 } \frac{1}{ ...

2

Okay, I don't quite get the details of what you are doing, but since this is linear algebra, I'd advise you to use linear algebra. You can then easily transfer between bra-ket notation and matrices. First, let's fix what we are talking about: You have one system $A$ containing one spin, so the system is a space $\mathbb{C}^2$ with basis states ...

2

The hamiltonian $$H=AJ_z^2=A\left(\sum_j S_z^{(j)}\right)^2\tag1$$ is a function of the individual $z$ spin projections $S_z^{(j)}$, and all of those commute. Therefore, the eigenstates will be product states of the form $$|\Psi\rangle=|a_1\rangle\otimes\cdots\otimes|a_N\rangle,\tag2$$ where each $|a_j\rangle$ is either $|\!\uparrow\rangle$ or ...

2

"If the particles are distinguishable, does that entail that one is a fermion whereas the other is a boson?" No, the particles being indistinguishable from each other does not imply that one is a fermion and the other a boson, they can be any types of particles as long as they aren't the same type. Eg: One could be an electron and the other a muon (both ...

2

I would argue that while the expression is indeed a solution of the Schrödinger-equation for the given Hamiltonian, the interpretation of $E$ as the eigenenergy of $H$ is not quite correct. The general solution for a time-independant Hamiltonian is of the form $$\Psi(t) = \exp\left(-it\hat{H}\right)\Psi(t=0)=\sum_nc_n \exp\left(-itE_n\right)\Psi_n$$ The ...

1

Thanks to @user12262 for pointing me in the direction of the KWW function. After perusing that link and searching SciFinder for stretched and compressed exponential functions in relation to NMR, I ran across this paper (subscription required, sorry). To (briefly) summarize the paper, the compressed exponential function, $e^{-kt^q}$, with $1 < q < 2$ ...

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