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As I am not allowed to comment for lack of reputation and cannot find a way to message I would like to point you to a reasonably recent source on the Kerr metric. I am by no means an expert but from what I have read and from what I understand the "Lines" of space time do twist and become unstable at the Cauchy Horizon. From what I gather the rotation isn't ...

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The assumption is, that the spin $S$ is a large parameter. A conjecture that is apparently not valid for $S=1/2$. The expansion is in $1/S$, which is assumend to be close to zero. $$S^+_j = \sqrt{2S-n_j}a^\dagger_j = \sqrt{2S}\sqrt{1-\frac{n_j}{2S}}a^\dagger_j\approx\sqrt{2S}\cdot\left(1-\frac{n_j}{4S}\right)a^\dagger_j$$ The second term, being of order ...

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Initially the pair is in state $(P\otimes M)+(M\otimes P)$. (I am writing "P" and "M" instead of "+" and "-" so as not to confuse the state "plus" with the addition operation in the state space.) You observe the first electron and happen to measure $P$. Now the pair is in the state $P\otimes M$. Note that this state is not entangled (i.e. it is a tensor ...

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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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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}{ ...

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Each free particle or field (each component of corresponding field or wave-function) must satisfy the Klein-Gordon equation, because it corresponds to relativistic energy-momentum relation (or, more formally, refers to the Casimir operator $p^{\mu}p_{\mu}$ of the Poincare group). But each free particle or field must satisfy relation $W^{\mu}W_{\mu} \Psi = ... 0 Following what I had been explained above, I think I might be able to solve my issue by writing it as such: Initially: $$\left|\psi\right> = \left|\psi_1\right> \otimes \left|\psi_2\right> = \left|\uparrow\right> \otimes \frac{1}{\sqrt{3}}\left( \left|-1\right> + \left|0\right>+ \left|1\right> \right)$$ ... 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 ... 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: $$E_\pm(p) = \frac{p^2}{2m}\pm h_z$$ ... 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 ... 0 An experimentalists answer: Why do same/opposite electric charges repel/attract each other, respectively? Because careful physicists have made an innumerable number of observations and have found that this is what nature does. There is a long history of observations before any theory could be solidified. They observed the behavior of attraction with ... 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 ...

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