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Quantum entanglement is the property of two objects $A,B$ – more precisely two subsystems – or a relationship between these two objects whose quantities or observables aren't independent of each other. It means that there exist some quantities $a_j$ and $b_k$ describing $A,B$, respectively, such that the probability distribution for these observations ...


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Essentially you are asking why only scalars are allowed to develop a vacuum expectation value (VEV). A scalar (as the name suggests) does not point to any direction -it has spin 0- therefore it can have a VEV without breaking the Lorentz symmetry. On the other hand, a boson with higher spin, e.g. a vector (spin 1) would spontaneously break Lorentz by ...


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Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent. A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard ...


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Of course the SM Higgs gives mass to both fermions as well as the gauge bosons. However, the latter is much more fundamental and predictive than the former. Point is, it is enough for a scalar to transform non-trivially under a gauge symmetry to contribute to its associated gauge bosons masses (after taking vev). This contribution is constrained by the ...


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1) Universal covering groups are groups with the property of being simply connected. Each algebra has a unique covering group. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way $$G=\frac{\tilde G}{Ker(\rho)},$$ where $Ker(\rho)$ is the kernel of the group homomorphism $\rho:\tilde ...


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The general problem that the Higgs mechanism solves is giving mass to spin-one particles. It turns out that finding relativistic, unitary theories of spin-one massive particles is non-trivial. There are a few known ways of doing it (this paper has a pretty good list of sources), but the oldest and easiest is probably the Higgs mechanism. In contrast, there ...


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Every regularization scheme is somewhat arbitrary. There are three popular regularization schemes when it comes to path integrals and their associated perturbative divergent integrals: time slicing, mode regularization, and dimensional regularization. Time slicing is the usual procedure used to derive the path integral, and it is the discretization of time ...


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Hints: The starting point is the 2-point relation $$T(\phi(x)\phi(y)) ~-~:\phi(x)\phi(y): ~=~ C(x,y)~{\bf 1}, \qquad C(x,y)~\equiv~\langle 0 | T(\phi(x)\phi(y))|0\rangle,\tag{1} $$ cf. this Phys.SE post. The relevant Wick's theorem is a nested Wick's theorem $$ T(:\phi(x)^n::\phi(y)^m:)~=~\exp\left( ...


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There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient $7.181\times 10^{-16}$, when converted from MeV to eV, should give $7.181\times 10^{-46}$. Mega means a million, and it to the fifth power gives $10^{30}$, not just $10^{15}$. In this way, the OP has to add a $10^{-15}$ factor to his result. That makes his result $10^{-3}$ ...


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Most people prefer to do research in the Path Integral Formalism (PIF), instead of the Operator Formalism (OF), because it is "easier." Easier means that whatever property you have in the OF, if you have a PIF for that theory, you can have more properties in the PIF. For example, Noether's theorem is only valid, as far as I know, in a Lagrangian formulation. ...


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Particles are described by quantum fields, and the quantum field determines the mass, spin and charge. So for example all electrons (and positrons) have the same mass, spin and (magnitude of) charge because they are all excitations of the electron quantum field. Individual electrons can have different energy and momenta, but I'm guessing you wouldn't ...



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