- #1
jonas_nilsson
- 29
- 0
Hi!
I'm doing a basic course in atomic physics, and right now I'm looking at quantum beats. I've found several sources for basic descriptions, but there are some things that confuse me. I'm going to present some of the material in short here, and hope for comments that might make things more clear!
---
In Haken&Wolf's "the physics of atoms and quanta" p.387 , the description basically goes as follows:
An electron is excited to a state that is a superposition of two stationary states:
[tex] \psi(r,0) = \alpha_1 \phi (r) + \alpha _2 \phi _2 (r) [/tex]
in time the electron will make a decay transition to the ground state (wavefunction [tex] \phi _0 [/tex]). The occupation probability of the ground state decreases exponentially with the decay constant [tex]2\Gamma[/tex].
Here's what I oppose:
"The total wavefunction therefore takes the form
[tex] \psi(r,0) = \alpha_1 exp(-i E_1 t /\hbar - \Gamma t) \phi_1 (r) + \alpha _2 exp(-i E_2 t/ \hbar - \Gamma t) \phi _2 (r) + \alpha _0(t) \phi _0 (r) [/tex]"
This must be a statistical mixed state appropriate for a big bunch of atoms, or? I mean after a decay, the wavefunction should collapse to the ground state wavefunction.
---
I slightly different approach, that makes more sense to me, is given in Wolfgang Demtröder's "Laser spectroscopy" p. 661ff:
As above, a superposition of states is assumed at t=0, here just a bit more general with an unspecified number of states. At time t the wavefunction has evolved into
[tex] \psi (t) = \sum c_k \psi _k (0) exp((-i \omega _{km} + \gamma _k/2)t)[/tex]
with
[tex]\omega _{km} = (E_k - E_m)/\hbar[/tex]
Here the psi_k are stationary states (the excited ones) and m denotes a lower state to which we observe decay.
Looking at the simple special case used in Haken&Wolf with just a superposition of two states (1,2) and decay to 0, we get:
[tex] \psi (t) = \left[ c_1 \psi_1(0) exp(-i E_1t/\hbar) + c_2 \psi_2(0) exp(-i E_2t/\hbar) \right] exp(i E_0t/\hbar - \gamma/2) [/tex]
Here's what I don't get: how does the last exponent with E_0 and gamma end up there? How do you motivate that? And how is the norm preserved?
Then, Demtröder looks at the matrix elements of the dipole operator to get the transition probabilities to the ground state. The presented end-results (the emission intensity beating) in the two books are similiar.
I'm doing a basic course in atomic physics, and right now I'm looking at quantum beats. I've found several sources for basic descriptions, but there are some things that confuse me. I'm going to present some of the material in short here, and hope for comments that might make things more clear!
---
In Haken&Wolf's "the physics of atoms and quanta" p.387 , the description basically goes as follows:
An electron is excited to a state that is a superposition of two stationary states:
[tex] \psi(r,0) = \alpha_1 \phi (r) + \alpha _2 \phi _2 (r) [/tex]
in time the electron will make a decay transition to the ground state (wavefunction [tex] \phi _0 [/tex]). The occupation probability of the ground state decreases exponentially with the decay constant [tex]2\Gamma[/tex].
Here's what I oppose:
"The total wavefunction therefore takes the form
[tex] \psi(r,0) = \alpha_1 exp(-i E_1 t /\hbar - \Gamma t) \phi_1 (r) + \alpha _2 exp(-i E_2 t/ \hbar - \Gamma t) \phi _2 (r) + \alpha _0(t) \phi _0 (r) [/tex]"
This must be a statistical mixed state appropriate for a big bunch of atoms, or? I mean after a decay, the wavefunction should collapse to the ground state wavefunction.
---
I slightly different approach, that makes more sense to me, is given in Wolfgang Demtröder's "Laser spectroscopy" p. 661ff:
As above, a superposition of states is assumed at t=0, here just a bit more general with an unspecified number of states. At time t the wavefunction has evolved into
[tex] \psi (t) = \sum c_k \psi _k (0) exp((-i \omega _{km} + \gamma _k/2)t)[/tex]
with
[tex]\omega _{km} = (E_k - E_m)/\hbar[/tex]
Here the psi_k are stationary states (the excited ones) and m denotes a lower state to which we observe decay.
Looking at the simple special case used in Haken&Wolf with just a superposition of two states (1,2) and decay to 0, we get:
[tex] \psi (t) = \left[ c_1 \psi_1(0) exp(-i E_1t/\hbar) + c_2 \psi_2(0) exp(-i E_2t/\hbar) \right] exp(i E_0t/\hbar - \gamma/2) [/tex]
Here's what I don't get: how does the last exponent with E_0 and gamma end up there? How do you motivate that? And how is the norm preserved?
Then, Demtröder looks at the matrix elements of the dipole operator to get the transition probabilities to the ground state. The presented end-results (the emission intensity beating) in the two books are similiar.