J.D'Almbert

I've realized it's not. Thank you for your help. If I may ask one last question (I promise), is the energy lost by a photon the same as the energy lost by an EM wave when both are travelling through the same conductive material?

just to clarify, do you mean \hat{N} is the particle number operator? Also I still want to have E_p=f(\hat{H}) so could I say E_p=\frac{\frac{d}{dp_x} \frac{d}{dp_y} \frac{d}{dp_z} \hat{H}}{\sum_{\lambda} \hat{a}_{\lambda}^{\dagger}(\mathbf p) \hat{a}_{\lambda}(\mathbf p)} ?

Could you solve $$\hat{H}=\sum_{\lambda}\int d^{3}\mathbf p E_{\mathbf p}\left( \hat{a}_{\lambda}^{\dagger}(\mathbf p)\hat{a}_{\lambda}(\mathbf p) + \delta (0)\right)$$ for $E_{\mathbf p}$ (ignoring the constant)?

@AndrewMcAddams Last questions, is it possible to rearrange equation (6) for $E_{\mathbf p}$ and if so how?

@AndrewMcAddams Does a decrease in the energy of the electromagnetic wave mean a decrease in the energy of the photon or the number of photons?

@AndrewMcAddams if $\delta (0)$ is the dirac delta function wouldn't it mean the solution is infinite as it approaches infinity?

@AndrewMcAddams Could you explain the notation used in $$\sum_{\lambda}\int d^{3}\mathbf p E_{\mathbf p}\left( \hat{a}_{\lambda}^{\dagger}(\mathbf p)\hat{a}_{\lambda}(\mathbf p) + \delta (0)\right)$$?

@AndrewMcAddams Sorry that it took me this long to accept it my internet was disconnected and thank you for all the work that you put into your answer.

@AndrewMcAddams May I ask what $\mathbf r$ is?

@AndrewMcAddams Ok, so the question is now, how does one relate the energy of sets of photons to $E$?

Yes, could you explain some of the terms and more importantly you state that $$ E = \int T_{00}d^{3}\mathbf r = \int \frac{1}{8 \pi}\left( \hat{\mathbf B}^{2} + \hat{\mathbf E}^{2}\right)d^{3}\mathbf r. $$ but what does $E$ denote? I'm looking for $E_{p}=f(E_{w})$

Could you get it?

Its is said that " However even higher order algebraic equations can be solved explicitly if an associated Galois group is solvable ", can I use this idea?

k

Can't I approximate the aforementioned solution of x

W is the solution of the octive, or the root.

But u can see why that isn't really helpful because what that means is W is represented as a root object, and the variable of that root object can be represented as a root object

K so what your saying is that there is no representation for the second solution of x in terms of [L,A]

What other detail is needed?

I'm looking for a simple solution for W, which would be the second solution of x.

The only problem is I have been using a trial version of Mathematica so I would greatly appreciate it if you could apply N

Ok so that means the only solution I'm concerned with is the second one. Now, is it possible to represent this solution as just a f[A,L]?

When you simplify "sol" could you make A=2 and L=8.7532759347618317962647585592473591641754313583030763

It is important to add that L>>A, e.g. when W=3 and A=2, L=8.7532759347618317962647585592473591641754313583030763

Does this just proves that there are two positive real roots? How can I find these roots as a f[A,L]?

Thank you for your help

How did you come up with the 2 approximations?

Oh ok so the red is used for that whereas the blue is used for the other case

So the curves are used for when psi>>gamma or psi is close to gamma, but what about gamma>>psi?

So the blue can used for when psi>>gamma and psi\approx gamma (which seems to be the most accurate)? Also how do I increase the precision for these approximations so they aren't rough estimates?

Ok, but how much bigger?

very nice, I am just trying to follow you atm

How would I start approximating this into its closed form? Can I do what you have done for your older answer?

Ok so I have two cases, where psi>>gamma -> x is in the nbhd of 0 and the aforementioned

My point was $x$ is always in the nbhd of 1, regardless of how close or far apart $\psi$ and $\gamma$ are. I want to be able to do what you did for your older answer for this problem.

Sorry I misjudged the situation, $x$ ergo $E(x)$ must be closer to 1 since when $\psi$ and $\gamma$ are close together, for example when $\psi=1$ and $\gamma=2$ $x \approx 0.952890514$ and when $\psi$ and $\gamma$ are further apart, for example when $\psi=10000$ and $\gamma=2$ $x\approx 0.9999999995$

$x$ can range from 0 to 1 so how would I know how large my argument of $E(x)$ is? $4 E(x)=\int_0^{2 \pi } \sqrt{1-\frac{4 \pi ^2 \psi ^2 \left(\sin ^2(\theta)\right)}{\gamma ^2+4 \pi ^2 \psi ^2}} \, d\theta =4 \int_0^{\frac{\pi }{2}} \sqrt{1-x^2 \sin ^2(x)} \, d\theta$