- #1
Nemanja989
- 79
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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.
First, the perturbed wave function is given as a superposition of the unperturbed wave functions:
[tex] \Psi(t) = \int a_E(t) \Psi_E dE [/tex]
where [itex] \Psi_E=\psi_E e^{-i\omega_Et}[/itex] and the integral is done over the whole spectrum of energies. The unperturbed wave functions are normalized to the Dirac delta with energy as an argument: [itex] \int \Psi_{E_i}^*\Psi_{E_j}dx=\delta(E_i-E_j)[/itex] and have units of [itex] \frac{1}{\sqrt{m⋅J}}[/itex], while the perturbated wave function has the units of [itex] \frac{1}{\sqrt{m}}[/itex]. The unit of [itex] a_E(t)[/itex] is [itex] \frac{1}{\sqrt{J}}[/itex], while the product [itex]|a_{E_0}(t)|^2dE[/itex] is the probability that the electron has the energy [itex] E_0[/itex] at the time [itex] t [/itex], which implies [itex] \int|a_{E_0}(t)|^2dE=1[/itex]. For more details one can look in the Landau Lifshitz course - https://archive.org/stream/QuantumMechanics_104/LandauLifshitz-QuantumMechanics#page/n29/mode/2up, introductory chapter about operators.
Now comes the issue:
If the electron is initially in the state with the energy [itex] E_0[/itex] , then it has to be that [itex]|a_{E_0}(t)|^2=\delta(E-E_0)[/itex], while it also holds that [itex] a_{E_0}=\int \Psi_{E_0}^*\Psi dx[/itex]. My question is, if one can write any expression for the value of [itex] a_{E_0}[/itex] only? The problem for me is that the [itex]\sqrt{\delta(E-E_0)}[/itex] does not seem to be well defined.
This would be very important as it would be used in the first order of the perturbation theory.
In other case, when the electron is initially in a discrete state, this problem is rather trivial and that's why I think I am missing something obvious here.
Thanks for your help!
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.
First, the perturbed wave function is given as a superposition of the unperturbed wave functions:
[tex] \Psi(t) = \int a_E(t) \Psi_E dE [/tex]
where [itex] \Psi_E=\psi_E e^{-i\omega_Et}[/itex] and the integral is done over the whole spectrum of energies. The unperturbed wave functions are normalized to the Dirac delta with energy as an argument: [itex] \int \Psi_{E_i}^*\Psi_{E_j}dx=\delta(E_i-E_j)[/itex] and have units of [itex] \frac{1}{\sqrt{m⋅J}}[/itex], while the perturbated wave function has the units of [itex] \frac{1}{\sqrt{m}}[/itex]. The unit of [itex] a_E(t)[/itex] is [itex] \frac{1}{\sqrt{J}}[/itex], while the product [itex]|a_{E_0}(t)|^2dE[/itex] is the probability that the electron has the energy [itex] E_0[/itex] at the time [itex] t [/itex], which implies [itex] \int|a_{E_0}(t)|^2dE=1[/itex]. For more details one can look in the Landau Lifshitz course - https://archive.org/stream/QuantumMechanics_104/LandauLifshitz-QuantumMechanics#page/n29/mode/2up, introductory chapter about operators.
Now comes the issue:
If the electron is initially in the state with the energy [itex] E_0[/itex] , then it has to be that [itex]|a_{E_0}(t)|^2=\delta(E-E_0)[/itex], while it also holds that [itex] a_{E_0}=\int \Psi_{E_0}^*\Psi dx[/itex]. My question is, if one can write any expression for the value of [itex] a_{E_0}[/itex] only? The problem for me is that the [itex]\sqrt{\delta(E-E_0)}[/itex] does not seem to be well defined.
This would be very important as it would be used in the first order of the perturbation theory.
In other case, when the electron is initially in a discrete state, this problem is rather trivial and that's why I think I am missing something obvious here.
Thanks for your help!
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