Recent content by SisypheanZealot

Thank you, I had written things down incorrectly in my notes. The proper relationship was ##a^{\dagger}_{p}|0>=\frac{1}{\sqrt{2\omega_p}}|p>## which when applied does reconstruct the proper measure of ##\int\frac{d{3}p}{2\omega_p}##.

So if I make the appropriate changes to account for the relativistic realm. We get $$|p>=\frac{1}{\sqrt{2\omega_p}}\hat{a}^{\dagger}_{p}|0>$$ and $$\phi(\vec{x},t)=\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(\hat{a}_{p}e^{-ipx}+\hat{a}_{p}^{\dagger}e^{ipx})$$ Applying the field operator...

I think you guys have gotten to the crux of what WWCY was trying to talk about, but I would like to add my two cents. QFT does actually allow for the measurement of the exact position of a particle, and it is in such a way that does not violate the Uncertainty Principle. Two good theoretical...

So, for the interested. I think I finally got this thing figured out. The first thing that we are going to do is transition to the complex Fourier series definitions of ##f(x)## and ##\delta(x-y)## and set ##L=\pi##. i.e. $$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{inx}$$ and...

No problem

So, in the majority of cases with Hamiltonians like this the 1/2 term is to deal with the double counting of states. Look at the original summation term ##\sum_{ijkl\sigma\sigma'}##. Based on the anticommutation relations $$\lbrace c^{\dagger}_{i},c^{\dagger}_{j}\rbrace=0$$ $$\lbrace...

The power radiated is going to be equal to the power incident. Then think of how the normal vectors will behave when a single face of the cube, two faces, and three faces are pointing in the direction of the light. Also, it looks that the answers presented are for a single face of the cube and...

New thought. What if I work backwards on this. so start with $$\sum^{\infty}_{n=-\infty}f(2nL+y)$$ Then set this equal to $$\int dx f(x)\sum^{\infty}_{n=-\infty}\delta(x-(2nL+y))$$ Now look at the delta term as a periodic term, and look at the Fourier transform definition of the Dirac Delta...

I know it is something simple that I am missing, but for the life of me I am stuck. So, I used the identity ##sin(a)sin(b)+cos(a)cos(b)=cos(a-b)## which gives me $$\int^{\infty}_{-\infty}dx\:f(x)\delta(x-y)=\int^{\infty}_{-\infty}dx\:f(x)\frac{1}{2L}\sum^{\infty}_{n=-\infty}\lbrace...

Hello my people, I am passionate about learning and researching in the areas of: QM, E&M, QED, QFT, and Statistical Physics. I mainly focus on foundations and theoretical calculations.