Recent content by Kekeedme

Wow! Thank you so much for your help. I appreciate it, and am glad that you made me go through it step by step to I could see it myself. Thanks so very much indeed!

The terms in which the indices are the same go back to themselves when we do ##\vec k \rightarrow -\vec k##, so they are even in ##\vec{k}## and so times the ##-\vec k## will be odd and go to zero in the integral. The terms with the differing indices go back to one another ##\vec k \rightarrow...

Thank you so much! You have been a GREAT help. Can you give me a little clarification on the index relabeling in this particular case to make the argument work? I wouldn't want to do it for myself in a hand wavy way. Thank you so much

It does still work for ##\epsilon_{\sigma}(-\vec{k}) \cdot \epsilon_m(-\vec{k})## right? Just not ##\epsilon_{\sigma}(\vec{k}) \cdot \epsilon_{m}(-\vec{k})## that's what you mean right? So in this case, the sum does not collapse, but the resulting terms times ##-\vec{k}## are still odd and that...

Oh my days! I can't believe I performed an improper book keeping on taking the delta function, that completely changes the signs on the arguments on the "magnetic"side of the product. Thank you! As such, I get this: Attached. From which I see that 1 and 4 will go to zero in the integration...

Ok. It is probably starring at me or I am doing something wrong, but unfortunately, I am not seeing what else I can do. Attached you will see when I expand all four terms and do the triple product etc. The first and fourth results are almost complex conjugates of one another except for not being...

Right, and this is where I had gotten confused. I think I am not seeing what the symmetry argument should be, for instance, let us say I distribute the first a term, we'd get something like this: Attached and I would get another one like that with the complex conjugate

Ah, sorry, in my actual I made it a double sum not to prejudice the fact that the electric and magnetic field would be at same polarization all the time. So I actually had it as a sum of ##\sigma## and a variable ##m##. Sorry I didn't write that

Hi @Matterwave I am not making any real headway, would you mind taking a look please? I attached the part I am working with now after using the delta function. I have tried to use the fact that the ##\vec{A}(\vec{k},t)^*= \vec{A}(\vec{-k},t)## and see that since I am doing integrals over...

hmm, I will go back and check. I am not sure which symmetry properties are helpful here, but I might be overlooking.

Hi Matterwave, thank you for your reply. This is how I did it at first, plug and chug, used ##dˆ3k## variable for say ##\vec{E}## and ##dˆ3l## variable for ##\vec{B}## and then got an ##e^{i\vec{k}\cdot\vec{x}}## term and also an ##e^{i\vec{k}\cdot\vec{l}}## term, which gives me a delta...

I am sorry to hear that Bill. But thank you again for all your suggestions.

Hi Bill, I have been working on this paper for a while, and it is really helpful! Thank you! I get stuck on deriving the momentum field operator (equation 5.88 on page 73). I was able to get exactly what he got for the energy operator, but for the momentum part I tried a lot of different things...

This seems amazing! Thank you so much for this resource

Thank you so much for the reply. So at first, does it mean that we should indeed get this expression by ignoring zero-point energy and equation the classical field energy to that of the single-photon? For the V portion, that is very helpful, and also very unfortunate that they have this...