Transition Probabilities of Atoms and Molecules

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)

Spontaneous and stimulated emission

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 and u(f) becomes very large.

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

Stimulated Emission and LasersLaser: 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

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

Stimulated Emission and LasersUse a multilevel atomic system to see those problems. Three-level system

1) Atoms in the ground state are pumped to a higher state by some external energy.

2) 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.

3) 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.

2) They decay quickly to the metastable state E3.

3) The stimulated emission takes atoms from E3 to E2.

4) 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.

Stimulated Emission and Lasers The red helium-neon laser uses transitions between energy

levels in both helium and neon.

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.

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

Selection rules

Oxygen spectrum

Selection rules for vibrational versus rotational-vibrational Raman spectraQ-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

Antisymmetric with exchange of the nuclei(nuclear spins)

symmetric with exchange of the nuclei(nuclear spins)

The odd rotational eigenfuctions with J=1,3,5…change their sign. Negative parity, antisymmetricThe 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?

|Ψ (1,2)|2=|Ψ (2,1)|2

Ψ (1,2 )=±Ψ (2,1)overall wavefunction of two noninteracting identical particlesNet wavefunction of two particles in different states is the linear combinationΨ  𝑠=

1

√2 |Ψ 𝑎 (1 )Ψ𝑏 (2 )+Ψ 𝑎 (2 )Ψ𝑏 (1 )|Ψ  𝐴=

1

√2 |Ψ 𝑎 (1 )Ψ 𝑏 (2 )−Ψ 𝑎 (2 )Ψ 𝑏 (1 )|

Ψ  𝑠=1

√2 |Ψ 𝑎 (1 )Ψ 𝑎 (2 )+Ψ 𝑎 (2 )Ψ 𝑎 (1 )|= 2

√2|Ψ 𝑎 (1 )Ψ 𝑎 (2 )|

Ψ  𝑠∗Ψ  𝑠=¿

Rb atom Eric Cornell and Carl WiemanNa atom Wolfgang Ketterle______

Consider boson and fermion wave functions of two identical particles labeled “1” and “2”. For now they can be either fermions or bosons:

Nobel Price 2001

+ symmetric =boson - antisymmetric=fermion

Identical probability density the same

For fermions in the same state a=b and =0 and due to Pauli Exclusion Principle

For Boson a=b

= nonzero probability occupying the same state favors to be in the lower states for Bose-Einstein Conclusion

:Solutions:

Proof:

Composite boson

Electrons S= Rb87 I=

∑=S+I = 2 integer Boson