Frequency of light calculated from wavefunctions

Click For Summary
The discussion focuses on how to derive the frequency of emitted light from wavefunctions during an electron's transition between orbitals. It emphasizes that while wavefunctions themselves do not have a frequency, they can produce observables that relate to physical quantities, including energy. The frequency of emitted light is directly linked to the energy difference between the initial and final states, expressed by the equation E = ħω. To calculate this, one must find the energy eigenvalues of the respective wavefunctions and subtract them, keeping in mind that not all transitions are permissible due to selection rules. Understanding the relationship between wavefunctions and energy is crucial for determining the emitted light's frequency.
granpa
Messages
2,268
Reaction score
7
if you have a wavefunction for an electron in one orbital and another wavefunction for the same electron in another orbital and assuming that the electron transitions from the one to the other orbital how would you derive the frequency of the emitted light from the wavefunctions themselves. (without just calculating the total energy released)

does the wavefunction have a frequency?
 
Physics news on Phys.org
granpa said:
if you have a wavefunction for an electron in one orbital and another wavefunction for the same electron in another orbital and assuming that the electron transitions from the one to the other orbital how would you derive the frequency of the emitted light from the wavefunctions themselves. (without just calculating the total energy released)

does the wavefunction have a frequency?

This is a very ODD request.

You already know that one needs to find the energy eigenvalue to calculate such a thing. Yet, you want to find this another way and use the "wavefunction". That's like asking that you know a wrench can be used to tighten a bolt, but can we eat it as a cake?

The wavefunction is only meaningful in the sense that it can produce "observables" that have physical meanings. It is these observables that give you the physical quantities associated with various parameters that we know of. That is why the nature of these Hermitian operators (the observables) are as important as the wavefunction themselves.

Zz.
 
the frequency of the emitted light IS related to the energy in the transition, the relation is just E = hbar * omega, so the energy eigenvalues of w.f's is the same as the freq. eigenvalue.

to first order approximation in explanation details, just take the energy eigenvalue of initial state w.f and the energy eigenvalue of the final state and subtract. But not all transitions are possible, but must obey certain selection rules /symmetries.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 44 ·
2
Replies
44
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K