The following could come handy:
Let's consider the momentum operator \hat{p} , whose eigenfunctions are \Psi_p, which are further normalized as \int\Psi^*_{p}\Psi_{p\ '} dx=\delta(p-p'), where p is the corresponding eigenvalue.
We know that...
Really the center! I like this very much. As you say, one should consider \sqrt{\delta(x)} as something one layer more general than distributions and in order to make that generalization, we have to give up on something. As you say, not all the operations can be done as with distributions. Like...
@A. Neumaier
Have you seen from my first post that the wave function I started with is normalized to 1 and not to delta function?
I want to know properties of the wave function \Psi, for which I know i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\Psi, where \hat{H}=\hat{H_0}+\hat{H'} and I...
Because I got curious of the problem they didn't mention in the book. Namely, if you have the electron starting from the continuous spectrum, but not from discrete, and then if I do perturbative expansion and try to solve the equations the only coefficient in the zeroth order that is non zero is...
Exactly, the integral has the meaning. But I have nowhere said that |a_E|^2 is a probability. Please check my first post! From it is clear that I regard |a_E|^2 as a probability density and |a_E|^2 dE as probability for it to be in the energy interval E+dE (I originally said only E...
For me the problem is that PeterDonis and DrDru say that it cannot be done like in LL, I would be very happy if they could justify their statements. For me, I don't see any problem whatsoever. Maybe only as you say in some pathological cases, that are more relevant for a mathematician than for a...
Thanks! That sounds reasonable, although I still like rigor to a certain extent. There is a very good and serious video that at one point shows in a funny way a difference between a theoretical physicist and a mathematician - link. The part I am referring to begins at 28:04 and culminates at...
Thank you very much for your comment. I read your post and like the idea of what you have done and how you defined it. However, as I do not have a very good knowledge of distributions, I am not fully confident of it, specially because later in the same topic member samalkhaiat said that it was...
I am sorry if you felt like wasting your time, however I highly appreciate yours and anyone else's time spent on the topic. Also, it's hard to communicate through messages like here and one unfortunately may get a different impression of what was meant (there is no intonation or gesticulation as...
Yes, I asked about a_E, but my claim was about |a_E|^2. In the initial post, I said that it has to be |a_E|^2=\delta(E-E') and to this you said in the comment #4 that it makes no sense and that I am misunderstanding the source I am using. After your last post, I have impression that you agree...
And I would really like to hear a critical opinion of the others on the method that Landau and Lifshitz used for normalizing a continuous state, as this is something as far as I understood they were strongly opposing.
(In the end for me, this forum serves as a place for a constructive...
Because of your statement earlier. You said something that I found contradicting some other fact. Further, you said that very thing will solve all of my problems.
I am not playing any game, but honestly trying to understand something. I get your last comment as a bit sarcastic, but I still want...
No, they do not call the probability density as probability. What they did change is calling the rate of probability as probability.
Here it goes:
|a_{fi}|^2=(2\pi/\hbar)|F_{fi}|^2\delta(E_f-E^{0}_i-\hbar\omega)t
and then they say |a_{fi}|^2 d\nu_f is the probability of a transition from...
Sorry for looking as if I am stubborn, but I simply want to clarify things for myself.
The fact that it is not possible to induce a transition to the single state of the continuous spectrum is in the direct contradiction with, I quote from the LL book:
"If the energy levels of the continuous...