# Tag Info

1

My PhD thesis (section 2.2) has an introduction to the Schwinger-Keldysh "closed-time-path" (CTP) formalism. https://arxiv.org/abs/gr-qc/0209010 CTP is a way of computing dynamical equations for expectation values of operators for quantum fields that are in disequilibrium. The basic idea is that the usual way of defining the amplitude for a quantum ...

1

By definition: $$G^T(t,t') = G^>(t,t')\theta(t-t') + G^<(t,t')\theta(t'-t),\\ G^{\tilde{T}}(t,t') = G^>(t,t')\theta(t'-t) + G^<(t,t')\theta(t-t').$$ By summing these two equations we readily obtain: $$G^T(t,t') + G^{\tilde{T}}(t,t') = G^>(t,t') + G^<(t,t').$$ This equality is thus the consequence of the definitions of the Green's ...

2

In general you may choose whatever boundary conditions you like for your fields. The condition you mention is standard, but certainly not the only kind of condition that appears in application. It really depends on what you're looking to do. To compute $D$, the standard procedure is to being Fourier transforming everything into momentum space. Once you do so,...

4

Yes. In something like: $$e^-+\mu^-\rightarrow e^-+\mu^-$$ you start with two free particle states, and end with 2 free particle states. In between, the electron and electromagnetic fields (and perhaps the weak field) go through all possible configurations that conserve energy an momentum. That is something that cannot be solved exactly, so it is ...

3

If you mean that the factor of $i$ is inconsistent, note that $$\det (i M) = i^d \det M$$ where $d = \dim M$ is the number of degrees of freedom that we integrate over. Hence, $i^d$ is a constant. With path integrals, $d = \infty$ and it is not clear how to make sense of $i^d$. This problem can be traced back to the definition of the path integral. In ...

4

Let me first state that functional determinants are not the same as determinants. So point A is wrong. About B, there is no single best way to do it, it is generally a hard task. Let me know elaborate. So a functional determinant is the determinant of a differential operators, that is the determinant of of a linear functional in a function space, that means ...

3

You might have also come across this problem when trying to invert $k^2P^{\mu\nu} \rightarrow (1/k^2)P_{\mu\nu}$. The nullspace equation for the projector, $P^{\mu\nu}(k)k_\nu = 0$, is a linear algebra equation rather than something that stems from the equations of motion. Generally, whenever an equation involves a Fourier transform from position space to ...

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I think Marek Danielewski may just have answered this quesion in this paper from Dec 2020: Foundations of the Quaternion Quantum Mechanics https://www.mdpi.com/1099-4300/22/12/1424# In summary: quaternions can be viewed as representing compression (the real part) and torsion (the three imaginary parts). They are used in condensed matter physics to model ...

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