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Maybe this is an answer on the question you've asked. The main reason of applications of homotopy theory in QFT is the requirement of finiteness of an action. Suppose that we want to start from the field configuration for which an action is finite, and then to discuss perturbations near such configuration. The most simple way to satisfy the requirement of ...

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The simplification follows from the theorem which states that if such operator is conserved in Heisenberg sense, $$\frac{d\hat{Q}}{dt} = \frac{\partial \hat{Q}}{\partial t} - \frac{i}{\hbar}[\hat{Q}, \hat{H}] = 0,$$ than it commutes with S-operator: $$[\hat{Q}, \hat{S}] = 0$$ So that these two operators can be diagonalized simulatenously: in ...

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The action of the boost on momentum states specifies the matrix elements of $T_V$ in momentum representation, it does not "change the momentum representation of the state". In other words, from $$T_V\left|p_1,p_2,..,p_N\right\rangle=\left|p_1+m_1V,p_2+m_2V,..,p_N+m_NV\right\rangle$$ it just follows that $$\langle p'_1,p'_2,..,p'_N | T_V ... 1 A Schrödinger background is a background with Schrödinger symmetry, and that, in turn, is having the Schrödinger group, which is the central extension of the Galileo group by the non-relativistic mass operator, as a symmetry group. The relevance of the Schrödinger group in string theory and conformal field theory arises because the d-dimensional ... 0 opinion based question, so it may be closed. The author Vincent (family name) has a very good introduction to group theory for molecules. I like this book as it has questions for you to answer as you go along so you really learn it as you read. If you are interested in solid state then you will have to go further to space groups with another text - this ... 2 To construct a crystal you need a lattice and a basis. The lattice represents the translational symmetry of the system. Namely, graphene has a hexagonal lattice, meaning the two lattice vectors are 60 degress apart. Since the brillouin zone is constructed by inverting the lattice vectors, the brillouin zone is shaped based upon the lattice, but not the ... 0 In a relativistic field theory, a conserved quantity according to Noether's theorem is given by a conserved current (density) J^\mu, i.e. \partial_\mu J^\mu=0. Hence, the contradiction you suggest does not really exist. The symmetry corresponding to conservation of electric charge is indeed the global part of the U(1) gauge symmetry. 0 Thanks for the hint. This is what I got so far:$$\delta W=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x)$$Now since d^4x'=d^4x(1+(\delta x^\alpha)_{,\alpha}) this equation reduces to:$$\delta W=\int d^4x \mathcal{L}'(x')-\mathcal{L}(x)+(\delta x^\alpha)_{,\alpha}\mathcal{L'}(x')$$Whic is just:$$\delta W=\int d^4x \delta\mathcal{L}+(\delta ...

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In non relativistic quantum mechanics, referring to an irreducible projective unitary representation of Galileo group, up to a multiplicative factor (the mass) the position operator naturally arises as the generator of boost transformation. This is equivalent to the standard translation in the space of momenta. In relativistic QM the standard translation ...

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We can try to prove the relation ourselves: $$\mathrm e^{-iXq}|p\rangle$$ Insert a resolution of the identity: $$1=\int\mathrm dx\,|x\rangle\langle x|$$ between $\mathrm e^{-iXq}$ and the ket: $$\mathrm e^{-iXq}|p\rangle=\int\mathrm dx\ \mathrm e^{-iXq}|x\rangle\langle x|p\rangle=\int\mathrm dx\ \mathrm e^{-ixq}|x\rangle\frac{1}{\sqrt{2\pi}}\mathrm ... 2 All the alternatives to the Dirac Lagrangian are actually forbidden by the requirement of requiring the hamiltonian to be well behaved (bounded from below and unbounded from above) and hermiticity of the action. To see this most simply we write the Lagrangian in terms of the fundamental left and right handed fields,  \psi \equiv \left( \begin{array}{c} ... 1 Comments to the question (v3): OP's last formula is the standard expression$$ \delta W~=~ \int \! d^4x~ \left[ {\rm EL} \cdot\bar{\delta}u + d_{\mu} j^{\mu}\right], \tag{A} $$for the variation of the action W=\int\! d^4x~{\cal L} (= Wirkung in German). Here$$ j^{\mu} ~=~ p^{\mu} \cdot \bar{\delta}u + {\cal L} ~\delta x^{\mu}, \qquad ...

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In order to make the problem symmetric, consider this: What happens, when you take the dodecahedron and drive current $I$ into it from vertex $A$ and drive $I/20$ out from all every vertex (including $A$)? By Kirchhoff's laws and symmetry, there is a current $I_{A-out} = \frac{(I-I/20)}{3} = \frac{19}{60}I$ going from $A$ to neighbouring vertexes. Now ...

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Short answer: The Hartley transform is a subset of the results given by Fourier transforms, which is only the real part (assuming your signal is real, which is almost always the case in physics). Long answer: Practically, you need the amplitude and phase of the signal, and the Fourier transform gives you both amplitude and phase by taking the magnitude: ...

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