Transcript Document

Transition Probabilities of Atoms and Molecules
Spontaneous and stimulated emission
Einstein’s analysis:
 Consider transitions between two molecular states with energies E1
and E2 (where E1 < E2).
 Eph is an energy of either emission or absorption.
 f is a frequency where Eph = hf = E2 − E1.
If stimulated emission occurs:
 The number of molecules in the higher state (N2)
 The energy density of the incoming radiation (u(f))
the rate at which stimulated transitions from E2 to E1 is
B21N2u(f) (where B21 is a proportional constant)
 The probability that a molecule at E1 will absorb a photon is B12N1u(f)
 The rate of spontaneous emission will occur is AN2 (where A is a
constant)
Stimulated Emission and Lasers

Once the system has reached equilibrium with the incoming radiation,
the total number of downward and upward transitions must be equal.

In the thermal equilibrium each of Ni are proportional to their
Boltzmann factor
.

In the classical time limit T → ∞. Then
becomes very large.
and u(f)
The probability of stimulated emission is approximately equal
to the probability of absorption.
Stimulated Emission and Lasers

Solve for u(f),
or, use Eq. (10.12),

This closely resembles the Planck radiation law, but Planck law is
expressed in terms of frequency.

Eqs.(10.13) and (10.14) are required:

The probability of spontaneous emission (A) is proportional to the
probability of stimulated emission (B) in equilibrium.
Stimulated Emission and Lasers
Laser:
 An acronym for “light amplification by the stimulated emission of
radiation”
Masers:
 Microwaves are used instead of visible light.

The first working laser by Theodore H. Maiman in 1960
helium-neon laser
Stimulated Emission and Lasers



The body of the laser is a closed tube, filled with about a 9/1 ratio
of helium and neon.
Photons bouncing back and forth between two mirrors are used to
stimulate the transitions in neon.
Photons produced by stimulated emission will be coherent, and the
photons that escape through the silvered mirror will be a coherent
beam.
How are atoms put into the excited state?
We cannot rely on the photons in the tube; if we did:
1) Any photon produced by stimulated emission would have to be
“used up” to excite another atom.
2) There may be nothing to prevent spontaneous emission from
atoms in the excited state.
The beam would not be coherent.
Stimulated Emission and Lasers
Use a multilevel atomic system to see those problems.

Three-level system
1)
2)
3)
Atoms in the ground state are pumped to a higher state by some
external energy.
The atom decays quickly to E2.
The transition from E2 to E1 is forbidden by a Δℓ = ±1 selection rule.
E2 is said to be metastable.
Population inversion: more atoms are in the metastable than in the
ground state
Stimulated Emission and Lasers



After an atom has been returned to the ground state from E2, we
want the external power supply to return it immediately to E3, but
it may take some time for this to happen.
A photon with energy E2 − E1 can be absorbed.
result would be a much weaker beam
This is undesirable because the absorbed photon is unavailable
for stimulating another transition.
Stimulated Emission and Lasers

Four-level system
1)
Atoms are pumped from the ground state to E4.
They decay quickly to the metastable state E3.
The stimulated emission takes atoms from E3 to E2.
The spontaneous transition from E2 to E1 is not forbidden, so E2 will
not exist long enough for a photon to be kicked from E2 to E3.
 Lasing process can proceed efficiently.
2)
3)
4)
Stimulated Emission and Lasers

The red helium-neon laser uses transitions between energy
levels in both helium and neon.
Selection rules
electric dipole selection rules for a single electron: (1) ∆L =
±1, ∆M = 0, ±1; (2) ∆S = 0, ∆MS = 0.
electric dipole selection rules for many electron atoms are,
then: (1) Only one electron changes its nl state; (2) Parity
must change; (3) ∆J = 0, ±1; (4) ∆MJ = 0, ±1; (5) J = 0 ↔
0 is not allowed; (6) ∆L = 0, ±1; (7) L = 0 ↔ 0 is not
allowed; (8) ∆S = 0; where J ≡ L+S is the total orbital plus
spin angular momentum
The magnetic dipole selection rules are, then: (1) No
change in electronic configuration; (2) Parity is
unchanged; (3) ∆J = 0, ±1; (4) ∆MJ = 0, ±1; (5) ∆J = 0
together with ∆MJ = 0 is not allowed; in particular, J = 0 ↔
0 is not allowed; (6) ∆L = 0; (7) ∆S = 0.
Oxygen spectrum
Selection
rules for vibrational versus rotational-vibrational Raman spectra
Q-branch:Weak and for diatomic molecule not
allowed
Q-branch:allowed
Influence of nuclear spins on the rotational structure
HFS is not treated here
In thermal equilibrium a hydrogen molecule gas is a mixture of para to ortho in the ratio 1:3
The rotational spectrum can have no transitions with ΔJ= ±1and
therefore no allowed transitions at all
In contrast rotational Raman transitions with ΔJ= ±2 are allowed
They belong alternatively to para and ortho states
Nuclear statistics
symmetric with exchange of the nuclei(nuclear spins)
Antisymmetric with exchange of the nuclei(nuclear spins)
The odd rotational eigenfuctions with J=1,3,5…change their sign. Negative parity, antisymmetric
The even rotational eigenfuctions with J=0,2,4…do not change their sign.Positive parity,symmetric
Figure 9-16 p333
Why does Bose-Einstein Condensation of Atoms Occur?
Rb atom
Na atom
Eric Cornell and Carl Wieman
Wolfgang Ketterle______
Nobel Price 2001
Consider boson and fermion wave functions of two identical particles labeled “1” and “2”.
For now they can be either fermions or bosons:
:Solutions:
Ψ(1,2)
2
= Ψ(2,1)
Ψ 1,2 = ±Ψ(2,1)
2
Identical
probability density the same
+ symmetric =boson
- antisymmetric=fermion
Composite boson
1
Electrons S= 2
3
Ψ 1,2 = Ψ 1 Ψ 2 overall wavefunction of two noninteracting identical particles
Rb87 I= 2
Net wavefunction of two particles in different states is the linear combination
∑=S+I = 2 integer
1
Ψ𝑠 =
Ψ𝑎 1 Ψ𝑏 2 + Ψ𝑎 2 Ψ𝑏 1
Boson
2
1
Ψ𝐴 =
Ψ𝑎 1 Ψ𝑏 2 − Ψ𝑎 2 Ψ𝑏 1
2
For fermions in the same state a=b and Ψ 𝐴 =0 and Ψ 𝐴
For Boson a=b
Proof:
Ψ𝑠 =
∗
1
2
Ψ𝐴
2
Ψ𝑠 Ψ𝑠 =(
= 0 due to Pauli Exclusion Principle
≠0
Ψ𝑎 1 Ψ𝑎 2 + Ψ𝑎 2 Ψ𝑎 1
2
2
=
)2 Ψ𝑎 ∗ 1 Ψ𝑎 ∗ 2 Ψ𝑎 1 Ψ𝑎 2
2
2
2
Ψ𝑎 1 Ψ𝑎 2
= nonzero probability occupying the same state favors to be
in the lower states for Bose-Einstein Conclusion