Recent content by Ilmrak

I liked the view presented in these notes (p.96), where antiunitarity is derived from the definition of the effect of time reversal on asymptotic states. You are perfectly right, infact there are conservation laws that prevent particles to decay in their own antipartcles :smile: Ilm

That's nice to read that, I actually got scared when someone told me he had mainly to solve tremendous integrals with almost no clue about physics! I really hope it's more frequent your situation :) Ilm

From the first post it seems that dx^\mu is the light-like element of a basis of the one-forms on the null surface. This would be consistent with the rest of the argument I think. Ilm

Hello, remember that g_{\mu\nu} is not strictly speaking a metric, it has signature (-+++) so there are vectors orthogonal to themselves: light-like vectors. Ilm

Thank you very much for your reply, I saw in your profile you're working on a very interesting topic! If I understood right your work is almost completely independent, this certainly have some pros and cons. Is this common for who is working in your field? Do you think more...

Hello, You simply have to consider that \int_{\Omega}\mathrm{d}x f(x) \delta(x) = f(0) \; \mathrm{if} \; 0\in \Omega, \mathrm{or} =0 \; \mathrm{if} \; 0\notin\Omega; no integration by parts is needed :) Ilm

Yes it is easy but no, it isn't faster. And if it were a bigger matrix (or worst a differential operator) it wouldn't be so easy to solve that equation while it would still be easy to let a matrix (or a differential operator) act on a vector (or a function). To solve an equation is (almost)...

Hello! My suggestion is to try to explicitly add up the two states in matrix notation... The answer should then be obvious to you :) Edit: The answer you gave to the first question is right. Nevertheless it is more time consuming then it was necessary and time is precious during exams :)...

Hello everyone! It won't be long before I'll take my degree in theoretical Physics and then I'll have to apply for a Ph.D. I'd like to work on some fundamental Physics. The problem I have to face is that I actually know (almost) nothing about contemporary physics and so I don't have the...

In particular here we have E(k)= \frac{\hslash ^2}{2m} k^2, and then \mathrm{d}E=\frac{\hslash ^2}{m} k \, \mathrm{d}k. If you are not familiar with the Heaviside theta don't worry, it's very simple! \theta(t)= 0 \quad \mathrm{if} \quad t<0, \theta(t)= 1 \quad \mathrm{if} \quad t>0...

If I understand your problem you don't get the last step of that equation. Then the solution is very simple. Try with this expression for the states density g(E) = \frac{1}{\pi^2 \hslash ^3} \theta(E) \sqrt{2m^3 E} \; , You will see that, being E= \frac{\hslash ^2}{2m} k^2, then...

Hello, I's been a while since the last time I read that book and I don't have one here to read. That said, if I remember well what that formula was about, the motivation to introduce g(E) is that the last expression obtained in Eq. 2.60 is much more general then the one you started with...

I think the OP question is a bit more difficult to answer. If I understood well what he's asking, he wants to know what the wavefunction would be after a "partial" measurement of position, i.e his experimental apparatus can only say whether the particle is in some finite space region or not...

Your book is lying xD By definition a path is a continuos map \phi: \left[ 0, 1 \right] \rightarrow M\; . The integral is invariant under reparametrization, i.e. a change of the parameter s not of the metric. Ilm

I think it may be worth to remark that the bosonic or fermionic nature of a field is a symmetry property of that field. A field would have such a symmetry even if it was the only field of your theory, it doesn't depend on the other fields and in particular it doesn't depend on the interaction...