Graduate Quantum Mechanics II
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Time-ยญโDependent Perturbation theory continued
For a Hamiltonian ๐ป! ๐ก = โ!๐!!"# we can write,
๐! ๐ก =๐ โ! ๐๐โ
๐!(!!"!!!)!! ๐๐ก!
!
which leads to the probability of going from ๐ โ ๐ :
๐ ๐ โ ๐; ๐ก =๐ โ! ๐
!
โ!sin ๐!" โ ๐ ๐ก
2๐!" โ ๐ ๐ก
2
!
๐ก!
This is under the assumption that the system is in state |๐ at ๐ก = 0. Otherwise, we have to replace t with (t โ t0)
Assume we prepare |๐ ๐๐ก ๐ก! = โ !!
What is ๐ ๐ โ ๐ ๐๐ก ๐ก = !! ? Simply replace t with ๐
For large ๐ โ โ we can use Fermiโs golden rule
๐!! = !!
โ๐ ๐ป! ๐
! ๐ ๐ฟ ๐ธ! โ ๐ธ! โ โ๐! , meaning we have a constant rate of transition per unit time.
How can we get these two contradictory results to agree for the same situation?
Note: Fermiโs Golden Rule can only work if either the final state energy or the frequency distribution of the perturbing Hamiltonian are continuous functions โ otherwise the delta-ยญโfunction will make no sense.
So letโs study the case where the final state wave function is an eigenstate to the momentum operator (and therefore the unperturbed Hamiltonian H0 is the free particle Hamiltonian). What is the probability to find the momentum somewhere in a small volume of momentum space?
๐(โ!๐) = ๐ ฮจ!โ!๐ Where โ!๐ = ๐!โ๐โฮฉ
So we need to find the overlap ๐!! = ๐ ฮจ
! and multiply by the volume space.
๐ ๐ โ ๐ + โ!๐ = ๐!!๐โฒ!๐๐!
!!!!!
!!!!!
๐ฮฉโ!
= ฮฮฉ ๐!!๐โฒ!๐๐!
!!!!!
!!!!!
Since we are integrating over a very small volume,
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๐ ๐ โ ๐ + โ!๐ = ฮฮฉ ๐ ๐!!๐โฒ๐๐!
!!!!!
!!!!!
Applying what we got earlier for ๐!! in this equation,
๐ ๐ โ ๐ = ฮฮฉ ๐๐! 2๐โ
๐ ๐ป! ๐! ๐ ๐ฟ ๐ธ! โ ๐ธ! โ โ๐! ๐
๐!!
2๐
!!!!!
!!!!!
The only variable that depends on ๐! is,
๐ธ! =๐โฒ!
2๐
And the integral becomes,
๐ฟ ๐ธ! โ ๐ธ! โ โ๐! ๐๐ธ!
!(!!!!! )
!(!!!!! )
= 1 ; ๐ผ๐ ๐กโ๐ ๐ฃ๐๐๐ข๐ ๐๐ ๐๐๐ ๐๐๐ ๐กโ๐ ๐๐๐๐๐ก0 ; ๐๐ ๐๐ก ๐๐ ๐๐๐ก ๐๐๐ ๐๐๐ ๐กโ๐ ๐๐๐๐๐ก
The final solution then becomes,
๐ ๐ โ ๐ = ฮฮฉ m๐! 2๐โ
๐ ๐ป! ๐! ๐
But what about the other equation ? how are we going to integrate it to get a solution closer to this one?
๐ ๐ โ ๐ =๐ ๐ป! ๐
!
โ!ฮฮฉ
sin ๐!" โ ๐ ๐ก2
๐!" โ ๐ ๐ก2
!
๐ก!๐โฒ!๐๐!!!!!!
!!!!!
Where,
๐!" =๐ธ!โโ๐ธ!โ=
๐!!
2๐โโ๐ธ!โ
If ๐!" โ ๐ ๐ก < ๐
Sine of a small number divided by that small number is one. So,
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๐ ๐ โ ๐ =๐ ๐ป! ๐
!
โ!ฮฮฉ๐!!โP t!
Now we look at the opposite case where โtโ is large, and to do this we modify the integral as follows,
๐ ๐ โ ๐ =๐ ๐ป! ๐
!
โ!ฮฮฉ
2tโ๐๐!
sin ๐!" โ ๐ ๐ก2
๐!" โ ๐ ๐ก2
!
t! ๐ก2๐โฒ๐โ
๐๐!!!!!!
!!!!!
Let,
๐ = ๐!" โ ๐ ๐ก2 and ๐๐ =
!!!!!โ๐๐! which make the integral
๐ ๐ โ ๐ =2โ๐๐!๐ก
๐ ๐ป! ๐!
โ!ฮฮฉ t!
sin ๐๐
!
๐๐!!
!!
And the integral is equal to ๐
๐ ๐ โ ๐ =2โ๐๐!๐ก
๐ ๐ป! ๐!
โ!ฮฮฉ ๐t!
๐ ๐ โ ๐ =2๐๐๐!๐ก
โ๐ ๐ป! ๐
!ฮฮฉ
This is exactly the same as we got earlier.
(Remember: ๐ป!" = ๐ ๐ป! ๐ ; ๐ป! = ๐ป! ๐!!!!! ; ๐! =!!โโ !!
โ ; |๐ = |๐! )
Example: Hydrogen Atom in electromagnetic field of incoming light wave
Letโs consider the classical Hamiltonian,
๐ป!" =๐!" โ
๐๐ ๐ด
!
2๐+ ๐ฮฆ ๐
The vector and scalar potentials are defined through
๐ธ = โฮฆ โ1๐๐๐๐ก๐ด
๐ต = โร๐ด
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Maxwellโs equations then become (the other two are automatically fulfilled due to the definition)
โ!ฮฆ +1๐๐๐๐ก
โ โ ๐ด = โ4๐๐
โ!๐ด โ1๐!
๐!
๐๐ก!๐ด โ โ โ๐ด +
1๐๐๐๐กฮฆ = โ
4๐๐ฅ๐
For free space no charge or current, so ๐ = ๐ฅ = 0
From the previous semester,
Gauge transformation,
ฮฆ โ ฮฆ! = ฮฆ +1๐๐๐๐กฮ
๐ด โ ๐ด! = ๐ด โ โฮ
leaves all the physical observables (electric and magnetic field) unchanged. We can therefore adopt additional requirements on the potentials. In particular, in the absence of charges and currents, we can use the Coulomb Gauge:
โ๐ด = 0 ; ฮฆ = 0
So using the only remaining non-ยญโtrivial equation above,
โ!๐ด โ1๐!
๐!
๐๐ก!๐ด = 0
Which has a solution like,
๐ด ๐, ๐ก = ๐ด! cos ๐ โ ๐ โ ๐๐ก =12๐ด! ๐! !โ!!!" + ๐!! !โ!!!"
For this solution to satisfy the above equation,
๐! โ๐!
๐!= 0 โ ๐ =
๐๐
๐ด โ ๐ = 0 (Coulomb Gauge)
To find the electric field, using equation one,
๐ธ = โฮฆ โ1๐๐๐๐ก๐ด
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๐ธ =๐๐๐ด! sin ๐ โ ๐ โ ๐๐ก
And,
๐ต = ๐ร๐ด! sin ๐ โ ๐ โ ๐๐ก = ๐ร๐ธ
Using the classical Hamiltonian we now write the quantum-ยญโmechanical Hamilton operator for the wave
๐ป! =๐!
2๐โ
๐2๐๐
๐ โ ๐ด + ๐ด โ ๐ +๐ด!
2๐๐!โ๐!
๐
(the potential due to the nucleus is of course non-ยญโzero; only the one from the incoming wave is).
If we apply this to any wave function
โ!โ โ ๐ด ฮจ ๐ฅ = โ
!๐ด โ โ ฮจ ๐ฅ (because of the Coulomb Gauge condition)
So
๐ โ ๐ด = ๐ด โ ๐
Now we write the total Hamiltonian including the electrostatic field from the nucleus
๐ป =๐!
2๐โ
๐2๐๐
๐ โ ๐ด + ๐ด โ ๐ +๐ด!
2๐๐!โ๐!
๐= ๐ป! + ๐ป!
๐ป! =!
!!"๐ด! โ ๐ ๐! !โ!!!" (we leave out the ๐!! !โ!!!" term since it cannot fulfill the requirement of
the delta-ยญโfunction in Fermiโs Golden Rule).
By comparing this with,
๐ป! = ๐ป! ๐!!!!!
We can get,
๐ป! =๐
2๐๐๐ด! โ ๐ ๐! !โ!
In the integral, r will be of order a0. Furthermore, for a photon of a few times 10 eV (which is what we
need to eject the electron), k is of order 0.1 nm-ยญโ1 and a0 = 0.05 nm. Therefore, we can take ๐ โ ๐~0.