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On the set ${\cal L}_1(\mathbb{R})$ of integrable functions $f:\mathbb{R}\to \mathbb{C}$, we can define the sine and cosine transforms $$\tag{1}\left\{\begin{array}{ccc} ({\cal C}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~\cos(\omega t) f(t),\cr ({\cal S}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~\sin(\omega t) f(t). \end{array} \right.$$ ...

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Basically, the answer is yes: $H$ is TRI because it is real. Reality condition really means that the Hamiltonian obeys a certain anti-unitary symmetry. In this case, the time-reversal operation is simply $T=K$ where $K$ is the complex conjugation. It is not the usual one($T=K\prod_i i\sigma^y_i$), and in particular $T^2=1$, so there is no Kramers' theorem ...

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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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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 $$\langle p'_1,p'_2,..,p'_N | T_V ... 2 To construct a crystal you need a lattice and a basis. The lattice represents the translational symmetry of the system. Namely, graphene has a hexagonal lattice, meaning the two lattice vectors are 60 degress apart. Since the brillouin zone is constructed by inverting the lattice vectors, the brillouin zone is shaped based upon the lattice, but not the ... 2 I would argue that there are two different antiunitary operators that are both commonly called the "time-reversal operator," and you need to specify which one you mean. Under one definition (T = K \prod_i i\, \sigma_i^y), all three Pauli matrices change sign under T. This definition is more "physical" because spin is odd under time-reversal. (E.g. if ... 2 Note: All products between \chi's are to be understood as tensor products. Assuming the fermions are both spin 1/2 particles, one recalls that the spin-part of the total wavefunction in the spin 1 triplet is either one or a linear combination of$$ \chi(j=1,m=1) = \chi(1/2,1/2)\chi(1/2,1/2)  \chi(j=1,0) = \frac{1}{\sqrt{2}}\big( ...

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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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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 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: $$\mathrm e^{-iXq}|p\rangle=\int\mathrm dx\ \mathrm e^{-iXq}|x\rangle\langle x|p\rangle=\int\mathrm dx\ \mathrm e^{-ixq}|x\rangle\frac{1}{\sqrt{2\pi}}\mathrm ... 1 Comments to the question (v3): OP's last formula is the standard expression$$ \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 ... 1 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: ... 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 ...

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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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