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    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    @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...
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    A TD perturbation - state initially in continuous part

    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...
  5. N

    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
  10. N

    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    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...
  14. N

    A TD perturbation - state initially in continuous part

    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...
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    A TD perturbation - state initially in continuous part

    I can be very well wrong, but from a purely theoretical point of view, why cannot the transition happen to a single state in the continuous spectrum? In that case the weight within the integral is still a density of states in respect to energy, just given as a delta function. This is used again...
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    A TD perturbation - state initially in continuous part

    Also, on link Landau and Lifshitz discuss in detail on how to change the normalization condition from momentum to energy. I really think there is not much to be misunderstood there, but still it's either that or that the problem is in communication between us or in the end that the book is not...
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    A TD perturbation - state initially in continuous part

    Thanks. For my own clarifications may I ask you the following: On the link, equation (5.4) states what I have written in my previous post. It is either that I misunderstood something there or that the statement in that link is wrong? (PS I have no problem when pointed of being wrong, to me...
  18. N

    A TD perturbation - state initially in continuous part

    But still, having on one side a delta function with properties that \int\delta(x-x')dx=1 and \delta(x)=0 for x\neq x' and on the other \int|a_E|^2dE=1 and |a_E|^2=0 for E\neq E' doesn't tell you that |a_E|^2=\delta(E-E')? And again, can you answer the second question in my post...
  19. N

    A TD perturbation - state initially in continuous part

    Ok, may I ask you for your opinion on how would you describe a_E(0), if you know that \int |a_E(0)|^2dE=1 and for all E except E=E_0, a_E(0)=0? PS Can you answer the second question in my post #8?
  20. N

    A TD perturbation - state initially in continuous part

    PS are you aware of how the delta function is changed if its argument is a function of another quantity?
  21. N

    A TD perturbation - state initially in continuous part

    Thank for the answer. But could you please explain this in greater detail? I could think of an electron starting in a discrete state and later being excited into one of the states of the continuous spectrum. The time after the excitation would be something I consider.
  22. N

    A TD perturbation - state initially in continuous part

    For the simplicity and trying to avoid a chaos or confusion let's go one by one: Any physical quantity f, characterized with the operator \hat{f} has eigenvalues f and eigenfunction \Psi_f . Let's assume it can take continuous values. Then we can represent any wave function \Psi as \Psi=\int...
  23. N

    A TD perturbation - state initially in continuous part

    Thanks for the corrections! You are absolutely right about both of them :) I was not rigorous enough, although meant the same as you. As you pointed, there is no problem with the phase. The problem that I have is that in the later calculations I will have to use a_E(0) instead of |a_E(0)|^2...
  24. N

    A TD perturbation - state initially in continuous part

    Hi everyone, I am doing a time dependent perturbation theory, in a case when the electron is prepared in a state of the continuous part of the energy spectrum. Existence of the discrete part and the degeneracy of the continuous part is irrelevant at the moment and will not be considered...
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    A Transition dipole moment - polarized absorption

    Hi again, To me it seems that there is a problem with the previous derivation. Let us for simplicity consider a 2D case. B_{01}=\frac{\pi e^2}{\epsilon\hbar^2\omega^2_{01} m^2}|\langle\Psi_{S_1}|\vec{e}\cdot\vec{p}|\Psi_{S_0}\rangle|^2 and define \begin{split} D_r &...
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    A Numerical solution to SE - variational method, many electrons

    Hi everyone, I am trying to find electron wavefunction of a system I am working in. Numerical method I choose is the Variational method (VM). This method is convenient to find the ground state of the system. More details are available here. Problem I have can be explained on a very simple...
  27. N

    A Transition dipole moment - polarized absorption

    Thank you for your reply, I really appreciate it. In that case both expressions are equivalent: \vec{D}=\vec{D}_r+i \vec{D}_i with \vec{D}_r and \vec{D}_i being defined as: \vec{D}=( |D_x|e^{i\alpha_1},|D_y|e^{i\beta_1} ,|D_z|e^{i\gamma_1})= ( |D_x|\cos\alpha_1,|D_y|\cos\beta_1...
  28. N

    A Transition dipole moment - polarized absorption

    Hi everyone, I am interested how is polarized light absorbed by a molecule or an atom. Unfortunately, I come to a problem in the derivation where a complex vector in a real space appears. This is something I never seen before and I do not know how to interpret it. Therefore I would like to ask...
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