Bohr's electronic theory vs. Schroedinger equation?

  • Context: Graduate 
  • Thread starter Thread starter loom91
  • Start date Start date
  • Tags Tags
    Electronic Theory
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 5K views
loom91
Messages
404
Reaction score
0
Hi,

I was wondering, if the Schroedinger equation implies a continuous evolution of the state vector, then how does Bohr's theory of electron's discontinuous jumps between energy states fit in? Conversely, if an electron 'jumps' from one state to another with nothing intermediate, then how can its wave-function undergo continuous evolution? Thanks.

Molu
 
Physics news on Phys.org
Hi,

First of all, you cannot compare the validity of Bohr's theory with that of Schroder since the first one is partly incorrect. To see why, check out our FAQ (the first post to be exact : Why Don’t Electrons Crash Into The Nucleus In Atoms?) where we have answered this question.

Enjoy

regards
marlon
 
Yes, Bohr's concept of discreet orbits have been superceded by wavefunctions. But energy and angular momentum is still quantised, no? Their evolution is discontinuous? How does this fit into continuous unitary evolution?

Molu
 
loom91 said:
But energy and angular momentum is still quantised, no? Their evolution is discontinuous?

No, their evolution is not discontinuous. You can model a transition between two energy levels by constructing a linear superposition of the two time-dependent wave functions, in which the coefficients are time dependent:

[tex]\Psi(x,t) = a_1(t) \Psi_1(x,t) + a_2(t) \Psi_2(x,t)[/itex]<br /> <br /> If the wave functions are normalized properly, then at any time, [itex]a_1^*a_1[/itex] gives the probability that the system is in state 1, with energy [itex]E_1[/itex]; and [itex]a_2^*a_2[/itex] gives the probability that the system is in state 2, with energy [itex]E_2[/itex].<br /> <br /> Before the transition begins, [itex]a_1 = 1[/itex] and [itex]a_2 = 0[/itex]. At some later time, after the transition has finished, [itex]a_1 = 0[/itex] and [itex]a_2 = 1[/itex].<br /> <br /> During the transition, both [itex]a_1[/itex] and [itex]a_2[/itex] are nonzero, and the system does not have a definite energy. If you measure the energy partway through the transition, you may get either [itex]E_1[/itex] or [itex]E_2[/itex]. During the transition, the probability distribution [itex]\Psi^* \Psi[/itex] oscillates (sloshes around or pulsates) with frequency [itex]f = (E_2 - E_1) / h[/itex].[/tex]
 
Last edited:
Thank you, I think I understand. The wavefunction itself evolves continuously, but as measurement collapses the wavefunction, thus observations will seem to indicate discreet jumps. This is another manifestation of the measurement problem, a transition from the quantum world to the classical one.
 
loom91 said:
Thank you, I think I understand. The wavefunction itself evolves continuously, but as measurement collapses the wavefunction, thus observations will seem to indicate discreet jumps. This is another manifestation of the measurement problem, a transition from the quantum world to the classical one.


We have a number of threads going on different aspects of the measurement problem. Perhaps you would want to continue on one of them, rather than starting yet another new one?
 
I don't think I'm qualified enough to discuss the measurement problem seriously. I was just clearing a confusion brought about by the outdated quantum mechanics syllabus that includes wave mechanics without correcting Bohr's theory in our high school.