130_notes.dvi

doesn’t play a role, again, because only theℓ= 0 state is totally antisymmetric. Since the shell is just half full we couple to ...

26.6.4 Oxygen Ground State Oxygen, withZ= 8 has the 1S and 2S levels filled givingj= 0 as a base. It has four valence 2P electro ...

27 Molecular Physics In this section, we will study the binding and excitation of simple molecules. Atoms bind into molecules by ...

We can now compute the energy of these states. 〈H 0 〉± = 1 2[1±S(R)] 〈ψA±ψB|H 0 |ψA±ψB〉 = 1 2[1±S(R)] [〈ψA|H 0 |ψA〉+〈ψB|H 0 |ψB〉 ...

By settingd〈dyH〉= 0, we can get the distance between atoms and the energy. Distance Energy Calculated 1.3 ̊A -1.76 eV Actual 1.0 ...

where the spin singlet is required because the spatial wfn is symmetric under interchange. The space symmetric state will be the ...

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27.5 Vibrational States We have seen that the energy of a molecule has a minimum for some particular separation between atoms. T ...

27.6 Rotational States Molecules can rotate like classical rigid bodies subject to the constraint that angular momentum is quant ...

27.7 Examples 27.8 Derivations and Computations 27.9 Homework Problems In HCl, absorption lines with wave numbers in inverse ce ...

28 Time Dependent Perturbation Theory We have used time independent perturbation theory to find the energy shifts of states and ...

Now we want to calculate transition rates. To first order, all theck(t) are small compared to ci(t)≈= 1, so the sum can be negle ...

perturbation carefully. V(~r,t) = 2V(~r) cos(ωt)→ 2 Vcos(ωt) =V ( eiωt+e−iωt ) We have introduced the factor of 2 for later conv ...

= Vni i ̄h [ ei(ωni+ω)t ′ i(ωni+ω) ]t′=t t′=0 = Vni i ̄h [ ei(ωni+ω)t− 1 i(ωni+ω) ] = Vni i ̄h ei(ωni+ω)t/^2 [ ei(ωni+ω)t/^2 −e− ...

Γi→f= 2 πVni^2 ̄h ρf(E) whereρf(E) is the density of final states. When particles (like photons or electrons) are emitted, the f ...

As long as the E field is weak, the initial state will not be significantly depleted and the assumption we have made concerning ...

table of integrals), so the result is simple. ∫∞ −∞ d∆f(∆)g(∆) = f(∆ = 0) 2 π t g(∆) = 2 π t δ(∆) [ 4 sin^2 ((ωni+ω)t/2) (ωni+ω) ...

29 Radiation in Atoms Now we will go all the way back toPlankwho proposed that the emission of radiation be in quanta withE= ̄hω ...

vector~k). Using classical E&M to compute the energy in a field (See Section 29.14.1) represented by a vector potentialA~(~r ...

We have taken a step toward quantization of the EM field, at least when we emit or absorb a photon. With this step, we can corre ...