3

I hope that I do not misunderstand something about your question, but $e^{-ik}$ goes once around the unit circle, so $$0=\sum_k e^{-ik} = \sum_k (a_k a_k^\dagger-a_k^\dagger a_k) e^{-ik}=\sum_k a_k a_k^\dagger e^{-ik}-\sum_ka_k^\dagger a_k e^{-ik}$$ and thus, $$\sum_k a_k a_k^\dagger e^{-ik}=\sum_ka_k^\dagger a_k e^{-ik}$$


3

The commutator shown in the question is not zero, and it shouldn't be zero. Even if $f$ and $g$ both have compact support in spacetime, $\phi^-(g)$ and $\phi^+(f)$ are not local operators, so we do not expect them to commute when the supports of $f$ and $g$ are spacelike separated. Intuitively, they can't be local operators, because the definitions of the ...


2

Your calculations are all correct (from what I can tell) and completely applicable. Moreover, your conclusion is also correct: But this implies that if I "measure" the position of a particle, got the delta Dirac function for the position and then compute how it evolves from that point onward, the position of my particle would be unknown and ...


2

Use coordinate system with grid parallel to two hexagon sides (let coordinates of two vertices be $(1,0)$ and $(0,1)$): $$ \int_{\text{hex}}f(x,y)dxdy = \frac{A_{\text{hex}}}3\int_{\text{hex}}\tilde f(u,v)dudv. $$ then the domain of integration is the square $(0,0)-(2,2)$ minus two triangles $(1,0)-(2,0)-(2,1)$ and $(0,1)-(0,2)-(1,2)$


2

Producing the frequency spectrum of a time series is nothing but doing a Fourier transform, that is obtaining the amplitudes (and phases) of sinusoidal waves which result in the time series when all summed together. Given a time series $f(t)$ the Fourier amplitude in frequency space is $$F(\omega)=\int_{-\infty}^\infty dt f(t)\:e^{-i \omega t}$$ $F(\omega)$ ...


1

Given $$ \tag{1} C_{\ell}\left(z, z^{\prime}\right)=\int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right) P(k) $$ Question: how to invert the integral to find the function $P(k)$? The closure relation for spherical Bessel function: $$ \tag{2} \int_0^\infty x^2 j_n(xu) j_n(xv) dx = \frac{\pi}{2u^2} \delta(u-v). $$ Multipy Eq.(1) with $z^...


1

Hint: Use one of the definitions of the Dirac delta function $$\frac{1}{2\pi}\int_{-\infty}^\infty e^{ik(x-x')}dk=\delta(x-x').$$ It would be $$\int dx \phi^*_E(x)\phi_{E'}(x)=\left(\frac{m}{2}\right)^{1/2}\frac{1}{2\pi\hbar}\frac{1}{(E E')^{1/4}}\int dxe^{i\frac{\sqrt{2m}}{\hbar}x\left(\sqrt{E'}-\sqrt{E}\right)}=\\ =\left(\frac{m}{2}\right)^{1/2}\frac{1}{2\...


1

The impulse response is not "improperly" called impulse response, but the convolution is "improperly" called a convolution (by whoever). In general, the impulse response (the transfer function represented in the spatial domain) of a lens is not shift invariant. I.e. it does not only depend on the mutual distance of input and output as you ...


1

Your PDE: $$\frac{\partial \psi}{\partial z} = \frac{i}{2k} \frac{\partial^2 \psi}{\partial x^2}$$ is solvable by separation of variables. Assume ('Ansatz'): $$\psi(x,z)=X(x)Z(z)$$ Insert into the PDE: $$XZ'=\frac{i}{2k}ZX''$$ Divide through by $XZ$: $$\frac{Z'}{Z}=\frac{i}{2k}\frac{X''}{X}$$ $$\frac{2k}{i}\frac{Z'}{Z}=\frac{X''}{X}=-\lambda^2$$ where $\...


1

Sorry this answer is so long. I didn't have time to write a shorter one. Clarification of specific details Before answering in general terms, I'll clarify a few details from a book cited in the question (Peskin and Schroeder's An Introduction to Quantum Field Theory). In sections 2.3 and 2.4, where they discuss the free scalar field, their field $\phi(x)$ ...


Only top voted, non community-wiki answers of a minimum length are eligible