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The positive and negative frequency components are the energy-decreasing and -increasing parts (or conversely, depending on sign conventions), also called annihilation and creation operators. To deduce this, let $A(t)$ be any operator in the Heisenberg picture, with time-dependence given by $$i\dot A(t) = \big[A(t),\,H\big].$$ Consider the operator $A_\... 3 How do you want to define momentum to begin with? Perhaps the best way is to define the momentum operator$\hat p$to be the operator that satisfies the commutation relation $$[\hat x,\hat p]=i\hbar$$ with the position operator$\hat x$. You might like this definition because it has the same structure as the canonical Poisson bracket in classical mechanics$\...

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The represent the parts of the field that are responsible for emitting and absorbing photons. The usual convention (although this is not always followed), is that $E^{(+)}$ contains only annihilation operators, and $E^{(-)}$ contains only creation operators. So the quantum mechanical description of light intensity does not involve $[E(\vec{x})]^{2}=E^{(+)}... 2 The propagator - or any, arbitrary correlation function - depends strongly on the gauge of internal photons (the Ward identity deals with the variations of external photons' gauge). This was first noted by Landau and Khalatnikov (and around the same time by Fradkin) who basically analyse the quantised version of the gauge transformation field called$\alpha(...

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This is a topic with a long history. Even though the Ewald formula is just this year becoming a one-century old method to evaluate the electrostatic energy of periodic systems, it continues to motivate new studies. Interestingly, a diffuse confusion exists about the real role of the Ewald formula in connection with the conditional convergence of the direct ...

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In a compact space, the net charge must be zero, by a simple topological argument: the net charge is proportional to the volume integral of the divergence of the electric field. By Gauss' law, this is equal to the surface integral of the electric field over the boundary, but a compact (periodic) space has no boundary! Or you can also see it in coordinates: ...

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Consider a particle in some state $|\Psi\rangle$, and let's look at how its position and momentum space wavefunctions $\psi(x) = \langle x | \Psi \rangle$ and $\widetilde{\psi}(p) = \langle p | \Psi \rangle$ are related. The first step is to add insert the identity in between either of the equations, using the fact that: $$\mathbb{I} = \int_{-\infty}^\infty \... 1 The (not entirely) physical reason to try that ansatz is the fact that the Hamiltonian commutes with both p_x and p_z, and they obviously commute between themselves:$$ \begin{cases} [p_x, H]=0\\ [p_z, H]=0\\ [p_x, p_z]=0 \end{cases} $$hence \{p_x,p_z,H\} is a set of commuting observables. For what concerns \hat{x} and \hat{z} directions, we then ... 1 Intuitively there is a relationship between periodicity and being discrete between dual spaces (time vs frequency or space vs wavenumber). Periodicity in one space implies being discrete in the dual space and vice versa. There are four cases: The continuous Fourier transform is suited when representations in both dual spaces are non-periodic and are non-... 1 The general delta function in 3D curved space is$$ \frac{\delta^{3}({\bf x}-{\bf x}_c)}{\sqrt{ g }} $$where g is determinant of metric. For spherical coordinates, you should have \sqrt{g}=4\pi r^2, if you have spherical symmetry. Thus you should get$$ \frac{\delta( r- r_c)}{4\pi r^2} $$Edit: Delt function in spherical coordinates. Since \sqrt{g}=r^2\... 1 Thanks to user110971 in the comments, I think I've managed to find the solution. According to Wikipedia, the Convolution Theorem states that if f and g are two functions, then f \ast g denotes their convolution and$$ \mathfrak{F}[f \ast g] = \mathfrak{F}[f] \cdot \mathfrak{F}[g] where $\mathfrak{F}$ is the Fourier Transform and $\cdot$ is the ...

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