Physics 131: Quantum Mechanics

Roger Blandford, Stanford University

Disclaimer

This course covers Chapters 4(most of) - 8 of Griffiths. What follows are highly condensed notes covering these chapters plus some routines that complete and display calculations that are encountered in the course and homeworks. The derivations are mostly more direct, though less didactic than those found in Griffiths. To learn the material you should attend lectures and section, study the book and do as many problems as you can. These notes may then be useful for reference and revision purposes.

1: Wave Function

Time - dependent Schrödinger Equation

iℏ((∂Ψ)/(∂t)) = Hψ;  Copenhagen interpretation

Born Interpretation

ρ=|Ψ((|)^2) ; (d/dt)∫dxρ=0; ∫dxρ=1

Momentum Operator

<p>=m((d<x>)/(dt  ))=∫(SuperStar[dxΨ])(ℏ/i)(∂/(∂x))Ψ; Ehrenfest theorem

Uncertainty Principle

(Subscript[σ, x])(Subscript[σ, p])≥(ℏ/2)

2: Time-independent Schrödinger Equation

Stationary States

V(x)=>separable: Ψ=(e^(-(iEt/ℏ)))ψ(x);  <p>=0, (OverHat[H])ψ=Eψ;  <H>=E; <(H^2)>=(E^2); =>(Subscript[σ, H])=0. Ψ=∑(Subscript[c, n])(Subscript[ψ, n])(e^(-(((Subscript[iE, n])t)/ℏ)));

Infinite Square Well

(Subscript[ψ, n])=((((2/a)))^(1/2))sin(((nπx)/a)); (Subscript[E, n])=(1/(2m))(((((nπℏ)/a)))^2); ∫(Subscript[dxψ, m]^( * ))(Subscript[ψ, n])=(Subscript[δ, mn]).

ψ=Table[(2^(1/2))Sin[n π x],{n,1,5}];en=Table[(((n^2)(π^2))/2),{n,1,5}];
Plot[ψ,{x,0,1},Frame->True,FrameLabel->{x,"(Subscript[ψ, n])","Infinite Square Well",""},RotateLabel->False];
MatrixForm[Integrate[Table[ψ\[Conjugate][[m]] ψ[[n]],{m,1,5},{n,1,5}],{x,0,1}]];

Parabolic initial wave function

Expand in eigenfunctions and compute evolution.

Ψ=x-(x^2);A=(((Integrate[Ψ\[Conjugate] Ψ,{x,0,1}]))^(-1/2));Ψ=A Ψ;
c=Table[Integrate[ψ[[n]]\[Conjugate]Ψ,{x,0,1}],{n,1,5}];
Plot[{Ψ,Sum[c[[n]]ψ[[n]],{n,1,5}]},{x,0,1}]

Harmonic Oscillator

V=(1/2)((mω)^2)(x^2); [x,p]=iℏ; H=ℏω((SubPlus[a])(SubMinus[a])+(1/2)); (Subscript[E, n])=(n+(1/2))ℏω.
(Subscript[ψ, n])~ (Subscript[H, n])(x)(e^(-(x^2)/2)) in scaled units.

Free Particles

Fourier transform Ψ(x)=(((2π))^(-1/2))(∫_(-∞)^∞)(dke^ikx)ϕ(k), wavefunction evolution.

Scattering from Simple Potentials

Delta function, square well...
Reflection, transmission

3: Formalism

Hilbert Space

square integrability, orthonormality, completeness.

Observables and Hermitian Operators

Observables associated with Hermitian operators; eigenfunctions are determinate states with real eigenvalues
Discrete, continuous spectra.

Generalized Uncertainty Principle

(Subscript[σ, A]^2)(Subscript[σ, B]^2)≥((((1/(2i))<[A,B]>))^2). 
ΔEΔt≥(ℏ/2)
(d/dt)<Q≥<((∂Q)/(∂t))>+((i )/ℏ)<[H,q]> 
(Subscript[σ, H])(Subscript[σ, Q])≥(ℏ/2)|((d<Q>)/dt)|.

Dirac Notation

<ψ|≡∫(SuperStar[dxψ]) . relationship to vector space, matrices.

4:Three Dimensions

Needs["VectorAnalysis`"];

Spherical Potential

V(r)=>ψ=(u/r)Y; Y=Θ(μ)(e^(imϕ)); ((1 - (μ^2))Θ')'-(((m^2)Θ)/(1-(μ^2)))+l(l+1)Θ=0; u''+((2m)/(ℏ^2))(E-V)u-((l(l+1))/(r^2))u=0. ∫dr|u((|)^2)=1;∫dΩ|Y((|)^2)=1

Hydrogen

V=-((e^2)/(4(Subscript[πϵ, 0])r)); length in a=((4(Subscript[πϵ, 0])(ℏ^2))/(me^2))=53pm, Energy in Ry=(1/2)(α^2)(mc^2)=13.6eV. ψ∝((((r/n)))^l)(e^(-r/n))(Subscript[L, n-l-1]^(2l+1))(((2r)/n))(Subscript[Y, lm])(θ,ϕ).

n=3;l=2;m=-2;
ψ=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,ϕ]
Integrate[((Abs[ψ])^2)(r^2)Sin[θ],{ϕ,0,2π},{θ,0,π},{r,0,∞}]
ContourPlot[2 π  R (ψ^2)/.{r->((((R^2)+(z^2)))^(1/2)),θ->ArcTan[z,R],ϕ->0},{R,0,5n},{z,0,5n},Contours->{.001,.002,.003,.04,.05},ContourShading->None,FrameLabel->{"R","z","2π R|ψ((|)^2)",""}]

Angular Momentum

Ladder Operators

[(Subscript[L, i]),(Subscript[L, j])]=(Subscript[iℏϵ, ijk])(Subscript[L, k]) => [(L^2),L]=0 
(Subscript[L, ±]) = (Subscript[L, x])±(Subscript[iL, y])=> [(Subscript[L, z]), (Subscript[L, ±])] = ±(Subscript[ℏL, ±]), (L^2)f=l(l+1)(ℏ^2)f,  (Subscript[L, z])f=mℏf, |m|≤l,l=N/2, (Subscript[L, ±]) (Subscript[f, m]^l) = ((ℏ(l(l + 1) - m(m ± 1)))^(1/2)) (Subscript[f, m+1]^l)

Spin[1]

Generalities

Ladder operators describe more general relationships than the specific case of orbital angular momentum described using (Subscript[Y, l]^m)(θ,ϕ) 
Theorist Pauli (1924) and spectroscopists Uhlenbeck & Goudsmit (1925) proposed that electrons could take two forms 
Turns out to be an angular momentum satisfying same algebra as orbital angular momentum
Known as spin but not like rotational angular momentum of earth; electron has this intrinsic property despite being effectively a point
(Suppose that it were like classical spin, then (2/5)mrv ~(1/2)ℏ =>r~(ℏ/mc)(c/v)≳400fm which is much larger than measured)
Need relativistic quantum mechanics to understand - later cf understanding about rest mass and saying that E=(mc^2)+(1/2)(mv^2)instead of recognizing ((γmc)^2)as one component of a four vector
Know that it is a different object than a vector in euclidean space because you have to rotate through 4π not 2π to leave it unchanged 
Use |sm> to represent simultaneous eigenvectors of (S^2), (Subscript[S, z]) with eigenvalues s(s+1)(ℏ^2), mℏ etc; nb m refers to a specific z direction
Boundary conditions on solutions to Legendre’s equation restricts l to integers; no such restriction with spin so s can be half integral as well. 
Identical half integral spin particles - fermions - cannot occupy same state (Pauli Exclusion Principle) e.g. electrons, quarks, neutrinos have s=1/2
Identical integral spin particles - bosons - can occupy same state leading to Bose-Einstein condensates e.g. π, K have s=0; γ, (Z^0),(W^±) have s=1

Matrix approach

Specialize to s=(1/2) (e.g. electrons)
Eigenvectors: (SubPlus[χ])≡|(1/2)(1/2)>≡↑, (SubMinus[χ])≡|(1/2)-(1/2)>≡↓; orthonormal basis vectors; general vector (called a spinor) is χ=(SubPlus[aχ])+(SubMinus[bχ]); write its components in this basis as (({{a}, {b}})) where |a((|)^2)+|b((|)^2)=1
Likewise for any operator (OverHat[O]) write a matrix O with components in this basis (Subscript[O, 12])=<(SubPlus[χ])|(OverHat[O])|(SubMinus[χ])> etc; in this case (S^2)=(3/4)(ℏ^2)(({{1, 0}, {0, 1}})); (Subscript[S, z])=(1/2)ℏ(({{1, 0}, {0, -1}}))
We can also use the ladder operators (Subscript[(OverHat[S]), ±]) | sm >=(Subscript[(OverHat[(S]), x]) ±(Subscript[(OverHat[iS]), y]))| sm >= (([s(s + 1) - m(m ± 1)])^(1/2))|sm±1> to recover (Subscript[S, x])=(1/2)ℏ(({{0, 1}, {1, 0}})),(Subscript[S, y])=(1/2)ℏ(({{0, -i}, {i, 0}})) 
These are hermitian matrices, of course, with eigenvalues ±(1/2)ℏ and eigenvectors (({{(2^(-1/2))}, {±(2^(-1/2))}})), (({{(2^(-1/2))}, {±(2^(-1/2))i}})), respectively
So, if a measurement of (Subscript[S, z])is followed by one of (Subscript[S, x]), the two possible outcomes will have equal probability of (1/2)
Common to remove ℏ/2 prefactor and use Pauli spin matrices arranged as a vector σ={(({{0, 1}, {1, 0}})),(({{0, -i}, {i, 0}})),(({{1, 0}, {0, -1}}))}
The relations (Subscript[S, x]^2)=(1/4)(ℏ^2)I, [(Subscript[S, x]), (Subscript[S, y])]=(Subscript[iℏS, z]) etc => (Subscript[σ, i])(Subscript[σ, j])=(Subscript[δ, ij])+(Subscript[iϵ, ijk])(Subscript[σ, k])

Adding angular momenta

Two distinguishable particles

Four eigenstates of (Subscript[(OverHat[S]), z]) = (Subscript[(OverHat[S]), z]^1)+(Subscript[(OverHat[S]), z]^2) can be written ↑↑,↑↓,↓↑,↓↓.However, two middle states have same m=(Subscript[m, 1])+(Subscript[m, 2])=0. 
Get right linear combination by applying lowering operator to top state. ((Subscript[(OverHat[S]), x])-i(Subscript[(OverHat[S]), y])). Get symmetric combination of two states. Orthogonal state is antisymmetric combination. 
After normalization, states are:|1,1>=↑↑,|10>=(2^(-1/2))( ↑↓+↓↑), |1,-1>=↓↓, |00>=(2^(-1/2))(↑↓-↓↑). Use ((OverHat[S])^2)to confirm values of s. The s=0,1 states are called singlet and triplet.

Generalizations

With two spins, s ranges from (Subscript[s, 1])-(Subscript[s, 2]) to (Subscript[s, 1])+(Subscript[s, 2]), assuming (Subscript[s, 1])≥(Subscript[s, 2]). We have the same number of states because 2((Subscript[s, 1])-(Subscript[s, 2]))+1 +...+2((Subscript[s, 1])+(Subscript[s, 2]))+1=(2(Subscript[s, 1])+1)(2(Subscript[s, 2])+1)
We write | sm >= ∑(Subscript[C, (Subscript[m, 1])(Subscript[m, 2])m]^((Subscript[s, 1])(Subscript[s, 2])s)) |(Subscript[s, 1])(Subscript[m, 1])>|(Subscript[s, 2])(Subscript[m, 2])>.  Note that the only non-zero Clebsch-Gordan coefficients have (Subscript[m, 1])+(Subscript[m, 2])=m. 
We can calculate more economically using group theory. We can also generalize to adding more spins.

Magnetic Dipole[1]

Classical electromagnetism

Gyro frequency qvB = m ((v^2)/r)=>(Subscript[ω, G])=-((qB)/m) 
Magnetic moment of ring is current times area  μ =q(ω/(2π))  ((πr)^2)=>μ= (q/(2m)) L. Also for any axisymmetric “top” with charge ∝ mass
Potential energy V =-μ·B (Let dipole be μ=(Subscript[q, m])d and then let d->0)
Force F=-∇V=(μ·∇)B 
Torque N=μ×B=Ω×L=>Ω=-(q/(2m))B gives steady precession at Larmor frequency.

Quantum mechanics

Electron orbit: q=-e, μ=γL=-(Subscript[μ, B])(L/ℏ) where γ =-(e/(2m))is the gyromagnetic ratio and (Subscript[μ, B]) =((eℏ)/(2m))= 9.3 (yJT^(-1))=58μeV (T^(-1))is the Bohr magneton (for electrons)
Electron spin: μ =γS=-(ge/(2m))S where g is measured to be ~2 as predicted by relativistic theory.
Actually g=−2.002 319 304 362±.000 000 000 001which agrees with QED theory => trust quantum mechanics!
Muon, (use (Subscript[m, μ])~200(Subscript[m, e])) g~2; proton ((Subscript[m, p])~2000(Subscript[m, e]),q~+e) g~5.6; neutron g~-3.8 (Bloch)!
Hamiltonian (isolated at rest): (OverHat[H])=-γS·B +... =>  H=(Subscript[μ, B])B(({{1, 0}, {0, -1}})) with eigenvectors (Subscript[χ, ±])

Temporal evolution

TDSE=> we can choose an α in [0,π] and find an origin of time where we can write: a(t)=cos(α/2)(e^(iγℏBt/2)), b(t)=sin(α/2)(e^(-iγℏBt/2)). 
Hence, <(Subscript[S, x])>= ((SuperStar[a])(SuperStar[b]))(1/2)ℏ(({{0, 1}, {1, 0}}))(({{a}, {b}})) =(1/2)ℏsinα cosγℏBt and<(Subscript[S, y])≥-(1/2)ℏsinαsinγℏBt, <(Subscript[S, z])≥(1/2)ℏcosα. 
So α is the inclination angle and <S> precesses about B at the Larmor frequency just like the classical case. cf Ehrenfest th.
Note also that if we rotate through angle α =2π, spinor changes sign. Need to rotate through 4π to recover original spinor.

Stern-Gerlach

Thought experiment of few electrons in inhomogeneous field - dilute enough so that can ignore Coulomb repulsion. Switch on field and spin up electrons will accelerate along B and vice versa.
(In practice use neutral atoms with spin in a beam. Stern and Gerlach used silver atoms in 1922.)
Can use to prepare electrons in a specific spin state or, conversely, to measure the spin state as long as we perform the separation in many Larmor periods.
If B is vertical, and we separate spin up electrons, a subsequent measurement of these electrons will confirm that all electrons are spin up. However a horizontal field will send half to the left and half to the right. If we then take the spin left electrons we will find half of these are now spin down and so on.

Problems

4.27

Set ℏ=1

χ=Normalize[{3I,4}];mb[m_]:=χ\[Conjugate].m.χ
sx=(1/2){{0,1},{1,0}};sy=(1/2){{0,-I},{I,0}};sz=(1/2){{1,0},{0,-1}};{mb[sx],mb[sy],mb[sz]};
σsx=Sqrt[mb[sx.sx]-((mb[sx])^2)];σsy=Sqrt[mb[sy.sy]-((mb[sy])^2)];σsz=Sqrt[mb[sz.sz]-((mb[sz])^2)];
{σsy σsz-(1/2)Abs[mb[sx]],σsz σsx-(1/2)Abs[mb[sy]],σsx σsy-(1/2)Abs[mb[sz]]}

4.33

(({{(OverDot[a])}, {(OverDot[b])}}))=-i((Subscript[eB, 0])/(2m))cos ωt(({{1, 0}, {0, -1}}))(({{a}, {b}}))=> a=(2^(-1/2))(e^(-(((Subscript[ieB, 0])sin ωt)/(2mω)))), b=(2^(-1/2))(e^(((Subscript[ieB, 0])sin ωt)/(2mω))) as in (Subscript[χ, +]^((x)))  state at t=0. 
χ=(((a+b)/(2^(1/2))))((((SubPlus[χ])+(SubMinus[χ]))/(2^(1/2))))+(((a-b)/(2^(1/2))))((((SubPlus[χ])-(SubMinus[χ]))/(2^(1/2)))) =>(Subscript[P, -]^((x)))=(1/2)|a-b((|)^2)=(sin^2)(((Subscript[eB, 0])sin ωt)/(2mω)); complete flip if (Subscript[B, 0])≥((πmω)/e).

4.51

ℏ=1;(Subscript[s, 1])=(1/2), (Subscript[s, 2])=(3/2).
|22 >= |(1/2) ,(3/2)> in obvious notation
(SubMinus[S])| 22>= 2 | 21 >= | -(1/2), (3/2)>+(3^(1/2))|(1/2),(1/2)>=> | 21 >= (1/2)| -(1/2), (3/2)>+((3^(1/2))/2)|(1/2),(1/2)>
(SubMinus[S])| 21>= (6^(1/2)) | 20 >=(( 1)/2)((3^(1/2))| -(1/2), (1/2)>)+((3^(1/2))/2)(1|-(1/2),(1/2)>+2|(1/2),-(1/2)>)=> | 20 >= (1/(2^(1/2)))| -(1/2), (1/2)>+(1/(2^(1/2)))|(1/2),-(1/2)>; extend by symmetry
|11>=A|-(1/2),(3/2)>+B|(1/2),(1/2)>
(SubPlus[S])| 11>= 0=(1A+(3^(1/2))B )| (1/2), (3/2)>=>|11≥-((3^(1/2))/2)|-(1/2),(3/2)>+(1/2)|(1/2),(1/2)> using normalization and choosing overall sign
(SubMinus[S])| 11>= (2^(1/2)) | 10 >=-(( (3^(1/2)))/2)((3^(1/2))| -(1/2), (1/2)>)+(1/2)(1|-(1/2),(1/2)>+2|(1/2),-(1/2)>)=> | 10 >=- (1/(2^(1/2)))| -(1/2), (1/2)>+(1/(2^(1/2)))|(1/2),-(1/2)>; extend by symmetry

4.41

∇·J =((iℏ)/(2m)) [((2mΨ)/(ℏ^2))((SuperStar[VΨ]) + (Subscript[iℏΨ, ,t]^( * )))-((2(SuperStar[mΨ]))/(ℏ^2))(VΨ + (Subscript[iℏΨ, ,t])) ]=-|Ψ(Subscript[|, ,t]^2)
Set ℏ=m=a=1

n=2;l=1;m=1;ψ=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,ϕ]
j=(I/2)Simplify[Grad[ψ]\[Conjugate]ψ-Grad[ψ]ψ\[Conjugate],Assumptions->{r,θ,ϕ}∈Reals]
Integrate[Det[{{Cos[θ],-Sin[θ],0},{r,0,0},j}](r^2)Sin[θ],{r,0,∞},{θ,0,π},{ϕ,0,2π}]

5:Identical Particles

Exchange and symmetrization[2]

Two body problem

If the interaction is a potential that just depends upon the separation r=(Subscript[r, 1])-(Subscript[r, 2]), 
[-((ℏ^2)/(2(Subscript[m, 1])))(Subscript[∇, 1]^2)-((ℏ^2)/(2(Subscript[m, 2])))(Subscript[∇, 2]^2)+V]ψ=[-((ℏ^2)/(2m))(Subscript[∇, R]^2)-((ℏ^2)/(2μ))(Subscript[∇, r]^2)+V(r)]ψ=Eψ, where ψ((Subscript[r, 1]),(Subscript[r, 2]))=(Subscript[ψ, R])(R)(Subscript[ψ, r])(r), R=(((Subscript[m, 1])(Subscript[r, 1])+(Subscript[m, 2])(Subscript[r, 2]))/m) ,m=(Subscript[m, 1])+(Subscript[m, 2]), μ=(((Subscript[m, 1])(Subscript[m, 2]))/m),.  
Separate variables and find E=(Subscript[E, R])+(Subscript[E, r]) where the latter is what is relevant. 
Apply to H, D, positronium, muonic atoms...

Two Particles

For two spinless, identical but distinguishable particles in two (separate and unentangled)states, a, b, we write  ψ((Subscript[r, 1]),(Subscript[r, 2]))=(Subscript[ψ, a])((Subscript[r, 1]))(Subscript[ψ, b])((Subscript[r, 2])). 
However, two identical, quantum mechanical particles - e.g. electrons - are indistinguishable
We cannot detect a swap, we can only say we have two electrons on a combined state and (Subscript[ψ, b])((Subscript[r, 1]))(Subscript[ψ, a])((Subscript[r, 2])) should have been just as good.
There are two ways to combine these two possibilities so that the observable probability density |(ψ^2)|  is unaffected: 
 (Subscript[ψ, ±])=A[(Subscript[ψ, a])((Subscript[r, 1]))(Subscript[ψ, b])((Subscript[r, 2]))±(Subscript[ψ, b])((Subscript[r, 1]))(Subscript[ψ, a])((Subscript[r, 2]))]  where + is symmetric associated with integer spin bosons,  and - antisymmetric with half integer spin fermions. 
For fermions this expresses the Pauli Exclusion Principle.
If the states are orthogonal, A=(2^(-1/2)) , if the particles are bosons and the states the same, A=(1/2).
Formally, we introduce a mathematical exchange operator P which interchanges the two particles.
(P^2)=1, [H,P]=0, eigenvalues are ±1, eigenfunctions are either symmetric or antisymmetric with ψ((Subscript[r, 1]),(Subscript[r, 2]))=±ψ((Subscript[r, 2]),(Subscript[r, 1])) and the states are stationary.

Spin

If there is spin, the combined wavefunction, say Φ=ψχ, must obey these rules. 
For electrons an antisymmetric spin singlet state must be combined with a symmetric spatial state and vice versa.
For bosons both must be symmetric or both antisymmetric.

Einstein, Podolsky, Rosen and Bell

Consider (π^0)->(SuperPlus[e])+(SuperMinus[e]) in singlet state moving in antiparallel directions. Causally separate detectors measure opposite spins decay by decay. 
Orient the detector directions as a, b. and measure the expectation P(a,b)=<00|(Subscript[S, a]^1)(Subscript[S, b]^2)|00>=-a.b for QM (HW2). 
If we have a hidden variable, λ, we introduce the separate outcome functions A(a,λ)=±1, B(b,λ)=±1 for electrons and positrons. 
Now A(a,λ)=-B(a,λ) and P(a,b)=∫dλρ(λ)A(a,λ)B(b,λ)=-∫dλρ(λ)A(a,λ)A(b,λ)
Hence P(a,b)-P(a,c)=-∫dλρ(λ)[A(a,λ)A(b,λ)-A(a,λ)A(c,λ)]=-∫dλρ(λ)[1-A(b,λ)A(c,λ)]A(a,λ)A(b,λ) and |P(a,b)-P(a,c)|≤∫dλρ(λ)[1-A(b,λ)A(c,λ)]=1+P(b,c)
However, if we assume quantum mechanics with a·b=cosθ, a·c=cosϕ, this inequality cannot be satisfied for 0<ϕ<θ<π or π<θ<ϕ<2π. 
Abandon locality; need entanglement.
Experiments performed using photons, result as expected for QM.

Exchange Forces

Consider two distinguishable particles 1,2  in separate states a,b. ψ=(Subscript[ψ, a])((Subscript[x, 1]))(Subscript[ψ, b])((Subscript[x, 2])). <((((Subscript[x, 1])-(Subscript[x, 2])))^2)>= <(x^2)(Subscript[>, a])+ <(x^2)(Subscript[>, b])-2<x(Subscript[>, a])<x(Subscript[>, b])
where <(x^2)≥∫dx (x^2)(Subscript[ψ, a]^( * ))(Subscript[ψ, a])
For indistinguishable particles ψ=(2^(-1/2))[(Subscript[ψ, a])((Subscript[x, 1]))(Subscript[ψ, b])((Subscript[x, 2]))±(Subscript[ψ, b])((Subscript[x, 1]))(Subscript[ψ, a])((Subscript[x, 2]))], and there is an extra term involving an overlap integral  ∓2|∫(Subscript[dxxψ, a]^( * ))(Subscript[ψ, b])((|)^2)
Electrons in symmetric spatial, spin singlet state are closer => (misnamed) exchange force. 
This state has higher energy than having two electrons in an antisymmetric spatial state.

Covalent Bond

Consider a hydrogen molecule. When the two electrons are in a symmetric spatial state, the repulsive between the protons is screened out and the protons are attracted. 
This is an example of a covalent bond.

Helium[2]

Ground State

Ignore electron repulsion. Ground state will have two electrons in symmetric combination with n=1, l=0  and an antisymmetric spin singlet state. 
a∝(1/(Ze^2))=26pm and energy E∝-2(Z^2)(e^2)=-8Ry=-109eV. However, E=-79eV due to electron repulsion in state with large overlap integral.

Electron-electron Energy

Estimate using the electron unperturbed (symmetric) wave function ψ=(Subscript[ψ, 1])(Subscript[ψ, 2]), and  (Subscript[V, ee])=∫(Subscript[dV, 1])(Subscript[dV, 2])((e^2)/(4(Subscript[πϵ, 0])(Subscript[r, 12])))|ψ((|)^2). 
If the two electron radii are r,s and the angle between the two is θ, (Subscript[r, 12])=((((r^2)+(s^2)+2rscosθ))^(1/2)).  Scale radii to a((SuperPlus[He])) and energies to E((SuperPlus[He])). Substitute x to perform integral

ψ=((E^(-(r+s)))/π);
vee=Simplify[8(π^2)(r^2)(s^2)(ψ^2)((((r^2)+(s^2)))^(-1/2))Integrate[(((1-2x Cos[θ]))^(-1/2))Sin[θ],{θ,0,π}]/.x->((r s)/((r^2)+(s^2)))]
vee=Integrate[vee,{s,0,∞},Assumptions->r>0]
vee=Integrate[vee,{r,0,∞}]

The answer is (Subscript[V, ee])= (5/8) x 4Ry=2.5Ry=34eV. So the revised estimate is -109+34=-75eV which is quite close to -79eV. 
We will soon see how to formalize this approach and improve upon the answer.

Excited States

Again start with hydrogenic wavefunctions and ignore e-e repulsion. 
Promote one electron to n=2, E=-((4/(1^2))+(4/(2^2)))Ry=-68eV. 
Conventionally, we measure energies relative to that of ionized helium with E=-4Ry=-54.4eV so the actual ground state is E=-79.0-(-54.4)=-24.6eV and our estimate of the first excited state is E=-68-(-54.4)=-13.6eV with this convention. The correct answer is -4.0eV, a large discrepancy.

Combined States

We count states by assigning the two electrons to hydrogenic states. We combine them in two ways.
S=0 - parahelium, S=1 - orthohelium. (The ground state must be S=0.) Sometimes “parallel”, “anti-parallel” spins but singlet and triplet more accurate.

Electron-electron Energy

Start with one excited electron in the n=2, l=0 singlet state and a combined symmetric spatial wavefunction ψ=(2^(-1/2))((Subscript[ψ, a]^1)(Subscript[ψ, b]^2)+(Subscript[ψ, b]^1)(Subscript[ψ, a]^2))

n=1;l=0;m=0;ψa=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,ϕ]
n=2;l=0;m=0;ψb=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,ϕ]
Plot[{ψa,ψb/.{θ->0,ϕ->0}},{r,0,8}, Frame->True,PlotRange->{-.1,.6},FrameLabel->{"r","ψ"},RotateLabel->False]

The e-e Coulomb energy is now (Subscript[V, ee])=∫(Subscript[dV, 1])(Subscript[dV, 2])((e^2)/(4(Subscript[πϵ, 0])(Subscript[r, 12])))|ψ((|)^2) = ∫(Subscript[dV, 1])(Subscript[dV, 2])((e^2)/(4(Subscript[πϵ, 0])(Subscript[r, 12])))|(Subscript[ψ, a])((|)^2)|(Subscript[ψ, b])((|)^2)+∫(Subscript[dV, 1])(Subscript[dV, 2])((e^2)/(4(Subscript[πϵ, 0])(Subscript[r, 12])))|(Subscript[ψ, a])*(Subscript[ψ, b])((|)^2)

vee1=Simplify[2π (r^2)(s^2)((((r^2)+(s^2)))^(-1/2))Integrate[ψa ψa\[Conjugate](ψb ψb\[Conjugate]/.r->s)(((1-2x Cos[θ]))^(-1/2))Sin[θ],{ϕ,0,2π},{θ,0,π},Assumptions->{θ∈Reals,ϕ∈Reals}]/.x->((r s)/((r^2)+(s^2))),Assumptions->{r>0,s>0}]
vee1=Integrate[vee1,{s,0,∞},Assumptions->r>0]
vee1=Integrate[vee1,{r,0,∞}]
vee2=Simplify[2π (r^2)(s^2) ((((r^2)+(s^2)))^(-1/2))Integrate[ψa\[Conjugate]ψb (ψa ψb\[Conjugate]/.r->s)(((1-2x Cos[θ]))^(-1/2))Sin[θ],{ϕ,0,2π},{θ,0,π},Assumptions->{θ∈Reals,ϕ∈Reals}]/.x->((r s)/((r^2)+(s^2))),Assumptions->{r>0,s>0}]
vee2=Integrate[vee2,{s,0,∞},Assumptions->r>0]
vee2=Integrate[vee2,{r,0,∞}]
vees=N[vee1+vee2]
54.4vees

The revised estimate for the singlet state is  (Subscript[E, s])=-13.6+6.3eV=-7.3eV which is closer to -4eV.
For the triplet state, ψ=(2^(-1/2))((Subscript[ψ, a]^1)(Subscript[ψ, b]^2)-(Subscript[ψ, b]^1)(Subscript[ψ, a]^2))

veet=N[vee1-vee2]
54.4veet

(Subscript[E, t])=-13.6+5.1=-8.5eV instead of -4.8eV for the triplet state.
Despite the fact that there is no overlap integral ∫(Subscript[dVψ, a]^( * ))(Subscript[ψ, b])r, there is an exchange e-e Coulomb energy term. 
Note that the triplet state with S=1, L=0 is more bound as the two electrons are further apart
This is still very approximate but we have seen how exchange breaks the degeneracy of these two energy states. We will make better approximations soon.

Periodic Table [2]

Hamiltonian

For Z electrons, H is sum of N single electron Hamiltonians  with (Subscript[V, i])=-((Ze^2)/(4(Subscript[πϵ, 0])|(Subscript[r, i])|)) and (Subscript[V, ee])=∑((e^2)/(4(Subscript[πϵ, 0])|(Subscript[r, ij])|))  summing over all pairs. States must be totally antisymmetric.

Ground States

One particle hydrogenic states are orbitals and they are filled in an order corresponding to the lowest energy state for a given Atomic Number Z.
The complete set of orbitals with a given n is a shell and contains 2(n^2)electrons when full. (The shells are sometimes denoted K, L, M, N... for n=1,2,3,4...)
The complete set of orbitals with a given n, l is called a subshell, Filled subshells have 2(2l+1) electrons. They are denoted s,p,d,f. 
The order in which subshells are filled is:
1s: (Subscript[, 1])H - (Subscript[, 2])He non-metal, noble gas
2s: (Subscript[, 3])Li - (Subscript[, 4])Be alkali, alkaline earth
2p: (Subscript[, 5])B - (Subscript[, 10])Ne metals,  non-metals, halogen, noble gas,
3s: (Subscript[, 11])Na - (Subscript[, 12])Mg alkali, alkaline earth
3p: (Subscript[, 13])Al - (Subscript[, 18])Ar metals,  non-metals, halogen, noble gas,
4s: (Subscript[, 19])K - (Subscript[, 20])Ca alkali, alkaline earth
3d: (Subscript[, 21])Sc - (Subscript[, 30])Zn Transition metals
4p: (Subscript[, 31])Ga - (Subscript[, 36])Kr metals,  non-metals, halogen, noble gas,
5s: (Subscript[, 37])Rb - (Subscript[, 38])Sr alkali, alkaline earth
4d: (Subscript[, 39])Y - (Subscript[, 48])Cd Transition metals
5p: (Subscript[, 49])In - (Subscript[, 54])Xe metals,  non-metals, halogen, noble gas,
6s: (Subscript[, 55])Cs - (Subscript[, 56])Ba alkali, alkaline earth
4f: (Subscript[, 57])La - (Subscript[, 71])Lu Lanthanides nb the filling of subshells is not always completed from here on.
5d: (Subscript[, 72])Hf - (Subscript[, 80])Hg Transition metals
6p: (Subscript[, 81])Tl - (Subscript[, 86])Rn metals,  non-metals, halogen, noble gas,
7s: (Subscript[, 87])Fr - (Subscript[, 88])Ra alkali, alkaline earth
5f: (Subscript[, 89])Ac - (Subscript[, 103])Lr Actinides
6d: (Subscript[, 104])Rf - (Subscript[, 112])Cn Transition metals
7p: (Subscript[, 113])Uut - (Subscript[, 118])Uuo metals,  non-metals, halogen, noble gas; unstable!

Spectroscopic Notation

Designate multi-electron states 2S+1(Subscript[L, J]) where: 
S refers to the combined electron spin 
L=S,P,D,F...  (Sharp, Principal, Diffuse, Fundamental...) measures the combined orbital angular momentum L=0,1,2,3ℏ
J=L+S, the combined total angular momentum. See Fig 5.2.

Hund's Rules

Combine hydrogenic wavefunctions in a partially-filled subshell to give stationary states with designated S, L, J. Filled subshells are  1(Subscript[S, 0]).
Empirical, though plausible rules. 
H1: Maximimize S (to make spatial part as antisymmetric as possible and reduce electron repulsion energy)
H2: Maximize L consistent with antisymmetrization (pushes electrons out to larger distance)
H3: J=|L-S| if less than or half full; J=L+S if more than half full (TBD)

Examples

(Subscript[, 1])H: (((1s))^1)  ->2(Subscript[S, 1/2])only choice
(Subscript[, 2])He: (((1s))^2)  ->1(Subscript[S, 0]) filled shell
(Subscript[, 3])Li: (((1 s))^2)(((2s))^1) -> 2(Subscript[S, 1/2])only choice
(Subscript[, 4])Be: (((1 s))^2)(((2s))^2) -> 1(Subscript[S, 0]) filled subshell
(Subscript[, 5])B: (((1 s))^2)(((2s))^2)(((2p))^1) -> 2(Subscript[P, 1/2])only S choice; maximize L; minimize J
(Subscript[, 6])C: (((1 s))^2)(((2s))^2)(((2p))^2) -> 3(Subscript[P, 0]) maximize S requiring antisymmetric spatial wavefunction; L=2 is symmetric and so L=1;minimize J 
(Subscript[, 7])N: (((1 s))^2)(((2s))^2)(((2p))^3) -> 4(Subscript[S, 3/2]) maximize S; three p electrons must have m=-1,0,1 so M=0 is required which is only true for L=0 despite existence of antisymmetric L=2 state; only J choice
(Subscript[, 8])O: (((1 s))^2)(((2s))^2)(((2p))^4) -> 3(Subscript[P, 2]) S, L same possibilities as (Subscript[, 6])C and same choices; maximize J
(Subscript[, 9])F: (((1 s))^2)(((2s))^2)(((2p))^5)-> 2(Subscript[P, 3/2]) S, L same possibilities as (Subscript[, 5])B and same choices; maximize J

The Elements[Live] 2:18 Tom Lehrer Tom Lehrer In Concert Spoken& Audio

Metals[2]

Free Electron Gas

Metals release loosely bound electrons from outer shells to roam free and conduct electricity. Density n per unit volume. 
Simplest approximation (Sommerfeld)  is to ignore the charges on the ions they leave behind - free electron gas. 
Consider in 1D with ∞ well of size L>>a. (Subscript[ψ, n])=((((2/L)))^(1/2))sin ((nπx)/L)=((((2/L)))^(1/2))sin (Subscript[k, n])x; n≥1. 
Each spatial eigenstate represented by a point in 1D k (momentum)-space, with k>0 and separation Δk=(π/L) 
Combine in 2D phase space with axes x, k. The number of states per unit area of phase space  is (1/π) independent of L. 
However sin contains ±k. As an accounting trick we will allow k to be negative so the density becomes (1/(2π)). This allows the electrons to have momentum both ways.
Combine in 3 space - k-space dimensions - 6D phase space with state density (1/(((2π))^3)); equivalently k-space density is (Subscript[n, k])=(1/((((2π))^3)n))
Exclusion Principle mandates ≤2 electrons per state with E = (((ℏ^2)(k^2))/(2m)). 
Lowest energy state is a filled sphere in k-space with radius (Subscript[k, F]) (Fermi wavevector) satisfying 2(Subscript[n, k])((4 π)/3)(Subscript[k, F]^3)=1 or (Subscript[k, F])=(((3(π^2)n))^(1/3)). Fermi momentum, energy (Subscript[p, F])=(Subscript[ℏk, F]), (Subscript[E, F])=((Subscript[p, F]^2)/(2m))
Average energy (Overscript[E, -]) =(∫_o^(Subscript[k, F]))8(Subscript[πn, k])(k^2)dk(((ℏ^2)(k^2))/(2m))=(3/5)(Subscript[E, F]). “Exclusion” or “degeneracy” energy density is U = n (OverBar[E])∝(n^(5/3)); the pressure P=(2/3)U as usual. (Alternatively P=-((d(U/n))/(d(1/n))).) 
The surface of the sphere in k-space is the Fermi surface. When we heat the metal, the electrons at the Fermi surface are excited into states with more momentum.

Band Structure

Ionic potential is periodic: V(x+a)=V(x) in 1D model. Displacement (shift, translation) operator commutes with H => stationary state=>ψ(x+a)=λψ(x) for some λ. 
Impose periodic boundary conditions  in lattice of length N=>(λ^N)=1=>λ=(e^(2πin/N)); 0≤n<N; N states in total. |ψ((|)^2) is periodic.
Introduce k=((2πn)/Na) and write ψ=(e^ikx)(Subscript[u, k])(x) => (Subscript[u, k])(x+a)=(Subscript[u, k])(x),∀x - periodic with period a (Bloch theorem). 
Energies etc unchanged by increasing k by ((2π)/a) reciprocal lattice vector. Generalizes to 3D.
e.g. δ-function potential V=αδ(x)...  (repulsive for α>0). 
ψ=A sinkx +coskx, 0<x<a,
 =(e^(-iKa))[A sink(x+a) +cosk(x+a)], -a<x<0. 
Imposing [ψ]=0, [ψ’]=((2mα)/(ℏ^2))ψ=((2β)/a)ψ at x=0 =>cosKa=coska+βsinc ka. 
Solve for allowed momenta k. a=1

β=1;
Plot[{1,-1,Cos[k]+β Sinc[k]},{k,0,6π},Frame->True,PlotRange->{-4,4},FrameLabel->{"(2mE)^(1/2)/ℏ","cos(K)"},RotateLabel->False]

Observe that there are intervals (gaps) of E for which no states are possible. For each interval (band) where states are possible, there is one spatial state containing two electrons per ion. Fill a single band if each ion donates two electrons. 
Filled band - insulator as cannot excite electrons; partially filled band - conductor; conductor with small number of electron donors - semiconductor. 
Plot the solution for a given value of k.

k=1.5;kk=ArcCos[Cos[k]+β Sinc[k]];aa=(((E^(I kk))-Cos[k])/(Sin[k]));
Plot[{Re[If[x>0,aa Sin[k x]+Cos[k x],(E^(-I kk))(aa Sin[k(x+1)]+Cos[k(x+1)])]],Im[If[x>0, aa Sin[k x]+Cos[k x],(E^(-I kk))(aa Sin[k(x+1)]+Cos[k(x+1)])]],
((Abs[If[x>0,aa Sin[k x]+Cos[k x],(E^(-I kk))(aa Sin[k(x+1)]+Cos[k(x+1)])]])^2)},{x,-1,1},Frame->True,PlotRange->{-2,2},FrameLabel->{"x","ψ, |ψ((|)^2)"},RotateLabel->False]

Note that the electron probability is reduced around the (repulsive) potential region. The opposite is true for an attractive potential.
We can consider the general case. 
Write (Subscript[u, k])=ζ+iη, 
ζ'' + [((2 m)/(ℏ^2))(E-V)-(k^2)]ζ-2kη'=0
η'' + [((2 m)/(ℏ^2))(E-V)-(k^2)]η+2kζ'=0
wlog, set ζ(0)=1. There are four parameters, p= {E,ζ’(0),η(0),η’(0)} which should be adjusted so that q={ζ(1)-1,ζ’(1)-ζ’(0),η(1)-η(0),η’(1)-η’(0)} vanish

Statistical Mechanics[2]

Classical Statistics

QM, (d^6)N=(((d^3)(xd^3)k )/(((2π))^3))=(((d^3)(xd^3)p )/(h^3)). 
Classically, (d^3)(xd^3)p is conserved along Hamiltonian trajectories. If mix thoroughly, so that phase space smoothly populated, number of states also proportional to volume in phase space.
Consider N classical, distinguishable particles partitioned among k energy levels. Number of states (Subscript[Q, c])=N! (∏_(n=1)^k)(1/((Subscript[N, n])!)). This is a multinomial distribution. 
Subject to the constraint: N=(∑_(n=1)^k)(Subscript[N, n])=const. What is (Subscript[N, n])?
Thermal equilibrium assumes enough mixing that every state of a given energy is equally likely. (Assumption which can be partially justified.)
Need to calculate subject to second constraint E=(∑_(n=1)^k)(Subscript[N, n])(Subscript[E, n])=const. We assume N is very large and calculate the most probable distribution - that with the most combinations.
Easier to use lnQ.

Stirling’s approximation

N!~(N^N)(e^(-N)).  Asymptotic formula, better as N->∞. (Prove by integrating lnx between 1 and N). Also lnN!~NlnN-N. Improvements are unnecessary

{N[n!],N[(n^n)(E^(-n))]}/.n->10.

Lagrange Multipliers

Maximize a function subject to constraints. 
Write G = lnQ +α (N-(∑_(n=1)^k)(Subscript[N, n]))+β(E-(∑_(n=1)^k)(Subscript[N, n])(Subscript[E, n])). 
G = (∑_(n=1)^k)((Subscript[N, n])-(Subscript[N, n])(Subscript[lnN, n])-(Subscript[αN, n])-(Subscript[βE, n])(Subscript[N, n])) +const.
(Subscript[G, ,(Subscript[N, n])])=0 =>(Subscript[N, n])=(e^(-(α+(Subscript[βE, n]))))Maxwell-Boltzmann distribution

Degeneracy

If (Subscript[d, n]) states in energy level n, (Subscript[N, n])=(Subscript[d, n])(e^(-(α+(Subscript[βE, n]))))  
For free particles, (Subscript[d, n])∝ (p^2)dp∝(E^(1/2))dE

Mean energy of free particles.

<E> = (((∫_0^∞)(dEE^(3/2))(E^(-βE)))/((∫_0^∞)(dEE^(1/2))(e^(-βE))))=(3/(2β)) =(3/2)kT, associating β with temperature T using kinetic theory. This defines T more generally. Also define chemical potential μ=-(α/β) = (Subscript[(((∂U)/(∂N))), S,V])

Identical Fermions

Large number of states (Subscript[d, n]) in the n’th level
(Subscript[Q, f]) = (∏_(n=1)^k)(((Subscript[d, n])!)/((Subscript[N, n])!((Subscript[d, n])-(Subscript[N, n]))!)) => G = (∑_(n=1)^(k))(Subscript[d, n])(Subscript[lnd, n]) -(Subscript[N, n]) (Subscript[lnN, n])-((Subscript[d, n])-(Subscript[N, n]))ln((Subscript[d, n])-(Subscript[N, n]))+(Subscript[βμN, n])-(Subscript[βE, n])(Subscript[N, n]). 
G(Subscript[,, (Subscript[N, n])])=0=>-(Subscript[lnN, n])+ln((Subscript[d, n])-(Subscript[N, n]))  +β(μ-(Subscript[E, n]))=0
  (Subscript[N, n])=((Subscript[d, n])/(((e^(((Subscript[E, n])-μ)/kT))+1))) Fermi-Dirac distribution

β=10;
Plot[(x^3)((((E^( x))-1))^(-1)),{x,0,10},Frame->True,PlotRange->{0,2}]

NIntegrate[(x^(5/2))((((E^( x))+1))^(-1)),{x,0,∞}]

Identical Bosons

(Subscript[N, n]) particles in level n with degeneracy (Subscript[d, n ])>>1. Equivalent to  (Subscript[d, n  ])-1~(Subscript[d, n]) (identical) partitions
(Subscript[Q, b]) = (∏_(n=1)^k)((((Subscript[N, n])+(Subscript[d, n]))!)/((Subscript[N, n])!(Subscript[d, n])!)) => G = (∑_(n=1)^(k))((Subscript[N, n])+(Subscript[d, n]))ln((Subscript[N, n])+(Subscript[d, n])) -(Subscript[N, n]) (Subscript[lnN, n])-(Subscript[d, n])(Subscript[lnd, n])+(Subscript[βμN, n])-(Subscript[βE, n])(Subscript[N, n]).
G(Subscript[,, (Subscript[N, n])])=0=>ln((Subscript[N, n])+(Subscript[d, n]))-(Subscript[lnN, n])  +β(μ-(Subscript[E, n]))=0
  (Subscript[N, n])=((Subscript[d, n])/(((e^(((Subscript[E, n])-μ)/kT))-1))) Bose-Einstein distribution. nb μ<(Subscript[E, min]).

Black Body Radiation

Photons not conserved so drop α. Two polarizations. 
  (dN/(dω))=((2×4((πω)^2))/((((2πc))^3)((e^(ℏω/kT))-1))) ;Energy density  (Subscript[ρ, ω])=(((ℏω)^3)/((π^2)(c^3)((e^(ℏω/kT))-1))) ~ (((ω^2)kT)/((π^2)(c^3))) for ω<<(kT/ℏ); ρ=(((k^4)(T^4))/((π^2)(ℏ^3)(c^3)))(∫_0^∞)dx ((x^3)/((e^x)-1))= (((π^2)(k^4)(T^4))/(15(ℏ^3)(c^3))) =(aT^4)=7.6×(10^(-16))(T^4)J (m^(-3)); n=(((k^3)(T^3))/((π^2)(ℏ^3)(c^3)))(∫_0^∞)dx ((x^2)/((e^x)-1))=2.0 (10^7) (T^(3 ))(m^(-3))
Relate flux to energy density using distant spherical emitter subtending solid angle Ω=(((πR)^2)/(D^2))
L=((σT)^4) 4((πR)^2) = (aT^4)(Ω/(4π)) c 4((πD)^2)=> σ=(ac/4)

7.6 (10^(-16)) (2.73^4)
2 (10^7)(2.73^3)

Integrate[(x^3)((((E^x)-1))^(-1)),{x,0,∞}]
Integrate[(x^2)((((E^x)-1))^(-1)),{x,0,∞}]

(((π^2)(((1.38 (10^(-23))))^4))/(15(((1.05 (10^(-34))))^3)(((3 (10^8)))^3)))
((3 (10^8)%)/4)
(((((1.38 (10^(-23))))^3)2.40)/((π^2)(((1.05 (10^(-34))))^3)(((3 (10^8)))^3)))

Problems

4.50

Wlog (Subscript[S, a]^1)(Subscript[S, b]^2)=cosθ (Subscript[S, z]^1)(Subscript[S, z]^2)+(1/2)sinθ (Subscript[S, z]^1)((Subscript[S, +]^2)+(Subscript[S, -]^2)),  <00|(Subscript[S, a]^1)(Subscript[S, b]^2)|00>=((ℏ^2)/8)[cosθ < ↑↓-↓↑ | (Subscript[σ, z]^1)(Subscript[σ, z]^2) | ↑↓-↓↑ >+(1/2)sinθ<↑↓-↓↑|(Subscript[σ, z]^1)((Subscript[σ, +]^2)+(Subscript[σ, -]^2))|↑↓-↓↑>]= -((ℏ^2)/4)cosθ

4.55

a) Linear combination of two eigenstates of (L^2)both with l=1 so <(L^2)≥2(ℏ^2)with P=1
b) P(m=0)=(1/3),P(m=1)=(2/3).
c) Two eigenstates of (S^2)with same s, so <(S^2)>=((3(ℏ^2))/4)with P=1
d) (Subscript[S, z])also an eigenstate P((ℏ/2)) = (1/3); P(-((ℏ)/2))=(2/3)
e)|10>|(1/2)(1/2)>=((((2/3)))^(1/2))|(3/2)(1/2)>-((((1/3)))^(1/2))|(1/2)(1/2)>;|11>|(1/2)-(1/2)>=((((1/3)))^(1/2))|(3/2)(1/2)>+((((2/3)))^(1/2))|(1/2)(1/2)> from C-G table; P((15/4) (ℏ^2)) = (((((((1/3)))^(1/2)) ((((2/3)))^(1/2))+ ((((2/3)))^(1/2)) ((((1/3)))^(1/2))))^2)=(8/9); P((3/4)(ℏ^2))=(1/9)
f) P((ℏ/2))=1 
g)

n=2;l=1;m=0;ψa=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,0]
n=2;l=1;m=1;ψb=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,0]
Simplify[(1/3)((ψa)^2)+(2/3)((ψb)^2)]
Simplify[Integrate[((2 π)/3)((ψa)^2)(r^2)Sin[θ],{θ,0,π}]]

4.56

a) f(ϕ+φ)=(∑_n)(((f^((n)))(ϕ)(φ^n))/(n!)) =(e^((∂_ϕ^(()^(^)))φ))f(ϕ)=(e^((Subscript[(OverHat[iL]), z])φ/ℏ))f(ϕ). Generalizing, to spin angular momentum resolved along an arbitrary direction n, f becomes the spin wavefunction/spinor, χ, and (Subscript[L, z])->ℏσ·n/2 . 
b) σ·n=(Subscript[σ, x]), so (e^((Subscript[iσ, x])φ/2)) = 1 +(( ((((Subscript[iσ, x])φ/2))^2))/(2!)) + (((((Subscript[iσ, x])φ/2))^4)/(4!))+...+((Subscript[iσ, x])φ/2)+(((((Subscript[iσ, x])φ/2))^3)/(3!))+…=1 -(( (((φ/2))^2))/(2!)) + ((((φ/2))^4)/(4!))+...+(Subscript[iσ, x])[(φ/2)-((((φ/2))^3)/(3!))+…] using (Subscript[σ, x]^2)=1.
(e^((Subscript[iσ, x])φ/2))= cosφ/2 +(Subscript[iσ, x])sinφ/2=(Subscript[iσ, x])for φ=π =(({{0, i}, {i, 0}})), converting (SubPlus[χ])->(SubMinus[iχ]). 
c) Likewise (e^((Subscript[iσ, y])π/4))  = (2^(-1/2))((({{1, 0}, {0, 1}}))+(({{0, 1}, {-1, 0}}))) = (2^(-1/2))(({{1, 1}, {-1, 1}})). (SubPlus[χ])->2((SubPlus[χ])-(SubMinus[χ]))=(Subscript[χ, -]^x)
d) (e^((Subscript[iσ, z])π))=-1 cf earlier discussion.

5.4

∫ (Subscript[dV, 1])(Subscript[dV, 2])(A^2) | (Subscript[ψ, a]^1)(Subscript[ψ, b]^2) ± (Subscript[ψ, b]^1)(Subscript[ψ, a]^2) ((|)^2) = 2 (A^2)=1=>A=(2^(-1/2)) using orthogonalityof (Subscript[ψ, a]),(Subscript[ψ, b]).; ∫ (Subscript[dV, 1])(Subscript[dV, 2])(A^2) | (Subscript[ψ, a]^1)(Subscript[ψ, a]^2) + (Subscript[ψ, a]^1)(Subscript[ψ, a]^2) ((|)^2) = 4 (A^2)=1=>A=(1/2)

5.6

Width =1; (Subscript[ψ, n]) = (2^(1/2))sin nπx. Follow 5.1.2. Only have overlap integral if one is even and other odd.

Clear[l,n];ψl=(2^(1/2))Sin[l π x];ψn=(2^(1/2))Sin[n π x];
exl=Simplify[Integrate[x ((ψl)^2),{x,0,1}],Assumptions->{l∈Integers,l>0}];
exn=Simplify[Integrate[x ((ψn)^2),{x,0,1}],Assumptions->{n∈Integers,n>0}];
exl2=Simplify[Integrate[(x^2)((ψl)^2),{x,0,1}],Assumptions->{l∈Integers,l>0}];
exn2=Simplify[Integrate[(x^2)((ψn)^2),{x,0,1}],Assumptions->{n∈Integers,n>0}];
ov=Simplify[Integrate[x ψl ψn,{x,0,1}],Assumptions->{l∈Integers,l>0,n∈Integers,n≥0}]
msd=Simplify[exl2+exn2-2exl exn]
msb=Simplify[msd-((128 (l^2) (n^2))/(((((l^2)-(n^2)))^4) (π^4)))]
msf=Simplify[msd+((128 (l^2) (n^2))/(((((l^2)-(n^2)))^4)  (π^4)))]

n=2;l=1;m=0;ψa=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,ϕ]

5.9

(2^2) 13.6((3/4)-(1/4))
(26^2) 13.6((3/4)-(1/4))

54.4((1/(Subscript[n, 1]^2))-(1/(Subscript[n, 2]^2)))

Midterm

1. a)4((πK)^2)((((a/2)))^3)2=1.=>K=(((((πa)^3)))^(-1/2)).
    b) (∫_0^∞)(drr^2)(1-cr/a)(e^(-3r/2a)) = 0  => 2 - ((2c)/3) 6 = 0 => c =(1/2)   Check: (1/(2 ))(2-6+(1/4)24)=1  
    c) a->a/2
    d)(Ke^(-r/a))=∑(Subscript[c, n])(Subscript[ψ, n]^He)  =>(Subscript[c, n])=∫((dr4πr)^2)(Ke^(-r/a))(Subscript[ψ, n]^He)
    e) (Subscript[c, 2]) = ∫((dr4πr)^2) (((Ke^(-r/a))( ((πa)^3)))^(-1/2)) (1 - (r/a))(e^(-r/a)) =(1/2)(2-(1/2)6) =-(1/2) =>|(Subscript[c, 2])((|)^2)=(1/4). 
    f) ns states
2. a) (Subscript[χ, ±])=(({{1}, {0}})) , (({{0}, {1}})) are basis vectors
   b)(Subscript[S, z])= (ℏ/2)(({{1, 0}, {0, -1}}));  ((±ℏ)/2)
   c) Cannot be simultaneously eigenstates of /measure accurately (Subscript[S, x])(Subscript[S, y]) 
   d) (SubPlus[S])(SubMinus[χ])~(SubPlus[χ]), (SuperPlus[S])(SubPlus[χ])=0. etc  (Subscript[S, x])=K(({{0, 1}, {1, 0}}))ℏ,  (Subscript[S, y])=K(({{0, -i}, {i, 0}}))ℏ  => (K^2)(({{2i, 0}, {0, -2i}}))(ℏ^2)=i(ℏ/2)(({{1, 0}, {0, -1}}))  =>K=(1/2)
   e) (χ^x)=(({{(2^(-1/2))}, {(2^(-1/2))}})), (({{(2^(-1/2))}, {-(2^(-1/2))}})) with (λ^x)=(1/2)ℏ, -(1/2)ℏ.
   f) (S^2)=((ℏ^2)/4) (({{3, 0}, {0, 3}})), λ=((3(ℏ^2))/4)
3. a) (P^2)=I =>λ=±1  [P,H]=0 =ψ((Subscript[r, 1]),(Subscript[r, 2]))=±ψ((Subscript[r, 2]),(Subscript[r, 1])) 
    b) s - integer, symmetric, half integer, antisymmetric ψ=0 if same state
    c) (SubPlus[χ])(SubPlus[χ])is unique uppermost state. Down operator -> (2^(-1/2)) ((SubPlus[χ])(SubMinus[χ])+(SubMinus[χ])(SubPlus[χ])), (SubMinus[χ])(SubMinus[χ]). Orthogonality -> (2^(-1/2)) ((SubPlus[χ])(SubMinus[χ])-(SubMinus[χ])(SubPlus[χ]))
    d) ((Subscript[n, 1]),(Subscript[n, 2]))=(1,1), (0,2) including ZPE. Electrons: first has 1 (spin antisymmetric) state, second has 3+1 states - total 5
    e) 4+(4+4) = 12 states
    f) 2 x s=1: (2 2 +1) +(2 0 +1) = 6 spin symmetric states (2 1 +1)=3 antisymmetric states. So, 6+ (6+3)=15 states

5.26

F (x, y) = 4 xy + λ[((((x/a)))^2) + ((((y/b)))^2)- 1]; (Subscript[F, ,x])=4y+2((λx)/(a^2))=0, (Subscript[F, ,y])=4x+2((λy)/(b^2))=0=>λ=±2ab, bx=ay in first quadrant. x=(2^(-1/2))a, y=(2^(-1/2))b, (Subscript[A, max])=2ab

5.29

a) N>0 foE>μ,∀E
b) We showed that (N/V)=(1/(2(π^2)))(∫_0^∞)((dkk^2)/((e^((E-μ)/(Subscript[k, B])T))-1))=const; E=(((ℏ^2)(k^2))/(2m)), μ<0. μ becomes less negative as T decreases.
c) For μ=0,  x=(((ℏ^2)(k^2))/(2(Subscript[mk, B])T)), (N/V)=(1/(4(π^2)))(((((2(Subscript[mk, B])T)/(ℏ^2))))^(3/2)) ((π^(1/2))/2)ζ((3/2)).  ζ (((3)/2)) = 2.61 =>(Subscript[T, BEC])= (((2 ((πℏ)^2))/(Subscript[mk, B])))((((N/(2.61V))))^(2/2))

5.30

a) ((dρ)/(dλ))=(Subscript[ρ, ω])|((dω)/(dλ))|=((((((hc/λ)))^3)(((2πc)/(λ^2))))/((π^2)(c^3)((e^(hc/kTλ))-1))) =((16(π^2)ℏc)/((λ^5)((e^(hc/kTλ))-1)))
b) x=(hc/(kTλ))  (Subscript[ρ, λ])~(((x^5)((e^x)-1))^(-1))peaks at x=4.97=> T(Subscript[λ, max])=2.9mmK

Plot[(x^5)((((E^x)-1))^(-1)),{x,4.9,5}]

5.31

ρ=(((k^4)(T^4))/((π^2)(ℏ^3)(c^3)))(∫_0^∞)dx ((x^3)/((e^x)-1))= (((π^2)(k^4)(T^4))/(15(ℏ^3)(c^3))) =7.6×(10^(-16))(T^4)(Jm^(-3))

6:Time-independent Perturbation Theory

Perturbations [1]

General issues

Perturbation theory is widespread approach to dealing with broader range of problems and organizing general solutions
Keeping equations linear still valuable despite ability to compute nonlinear problems
Convergence is necessary!

Series Expansion

(H^0)(Subscript[ψ, n]^0)=(Subscript[E, n]^0)(Subscript[ψ, n]^0); ((H^0)+((λH)^1))((Subscript[ψ, n]^0)+λ(Subscript[ψ, n]^1))=((Subscript[E, n]^0)+λ(Subscript[E, n]^1)) ((Subscript[ψ, n]^0)+λ(Subscript[ψ, n]^1))
Introduce expansion parameter λ eg strength of interaction which we set to unity eventually
Consider terms ∝λ => (H^1)(Subscript[ψ, n]^0)+(H^0)(Subscript[ψ, n]^1)= (Subscript[E, n]^1)(Subscript[ψ, n]^0)+(Subscript[E, n]^0)(Subscript[ψ, n]^(1 )). 
Use Hermitian character of H, <(Subscript[ψ, n]^0)|(H^0)(Subscript[ψ, n]^1)> =< (H^0) (Subscript[ψ, n]^0) | (Subscript[ψ, n]^1)>=(Subscript[E, n]^0)<(Subscript[ψ, n]^0) | (Subscript[ψ, n]^1)> to conclude that  (Subscript[E, n]^1)=<(Subscript[ψ, n]^0)|(H^1)|(Subscript[ψ, n]^0)>.  
Expand the perturbation to the wavefunction (Subscript[ψ, n]^1)=(∑_(m≠n))(Subscript[c, m]^n)(Subscript[ψ, m]^0), we get ((H^0)-(Subscript[E, n]^0))(Subscript[ψ, n]^1)=-((H^1)-(Subscript[E, n]^1))(Subscript[ψ, n]^0) 
=> (Subscript[c, m]^n)=((Subscript[W, mn])/((Subscript[E, n]^0)-(Subscript[E, m]^0))),  where (Subscript[W, mn])=<(Subscript[ψ, m]^0)|(H^1)|(Subscript[ψ, n]^0)> is matrix element as unperturbed states non-degenerate. Expanding to second order - terms ∝(λ^2)
(H^0)(Subscript[ψ, n]^2)+(H^1)(Subscript[ψ, n]^1) =(Subscript[E, n]^0)(Subscript[ψ, n]^2)+(Subscript[E, n]^1)(Subscript[ψ, n]^1)+(Subscript[E, n]^2)(Subscript[ψ, n]^0) giving (Subscript[E, n]^2) =(∑_(m≠n))((|(Subscript[W, mn])((|)^2))/((Subscript[E, n]^0)-(Subscript[E, m]^0))) and so on.

Degeneracy

Choose two orthonormal eigenfunctions, (Subscript[ψ, a,b]^0) of (H^0) with common eigenenergy. 
We seek the linear combination (ψ^0)=(Subscript[αψ, a]^0)+(Subscript[βψ, b]^0) that is an eigenfunction of the perturbation (H^1)and breaks the degeneracy.
Taking first order perturbation to Schrödinger equation, we obtain
(({{(Subscript[W, aa]), (Subscript[W, ab])}, {(Subscript[W, ba]), (Subscript[W, bb])}}))(({{α}, {β}}))=(E^1)(({{α}, {β}})) ; nb W is Hermitian, (Subscript[W, ab])=(Subscript[W, ba]^( * )). 
We simply solve this as an eigenvalue equation; the relevant eigenfunctions of the combined Hamiltonian are the eigenvectors of this matrix equation. 
The generalization to triple etc degeneracy is clear.
The eigenfunctions we seek will also be eigenfunctions of any Hermitian operator that simultaneously commutes with (H^0)and (H^1).

1D SHO

V0=(1/2)((mω)^2)(x^2); (Subscript[E, n])=(n+(1/2))ℏω.  
Let ω->(1+ϵ)ω. (H^1)=2(V^0)ϵ. 
However<T>=<V>=>(Subscript[E, n]^1)=(n+1/2)ℏωϵ
Add electric field H'=-qEx (Subscript[E, n]^1)=0 (Subscript[E, n]^2)= (((qE))^2)∑((|<m|x|n>((|)^2))/((n-m)ℏω))
Now x=((((ℏ/(2mω))))^(1/2))((SubPlus[a])+(SubMinus[a])) etc...

Fine and Hyperfine Structure[1]

Relativity

Spectroscopy is accurate and corrections O((α^2)).
SR => T = ((p^2)/(2m))-((p^4)/(8(m^3)(c^2)))+...   identifying H’
Note that perturbation commutes with (L^2) so automatically diagonalize pertubation matrix if use standard hydrogen wavefunctions. Some legerdemain  because (p^(4 )) not totally Hermitian
(Subscript[E, r]^('))=-(1/(2(mc^2)))<(((E-V))^2)> = -((1/(2(mc^3))))[(Subscript[E, n]^2)+2(Subscript[E, n])(((e^2)/(4(Subscript[πϵ, 0])))) <(1/r)>+(((((e^2)/(4(Subscript[πϵ, 0])))))^2)<(1/(r^2))>] =-(((α^4)(mc^2))/(8(n^4)))[((4n)/(l+(1/2)))-3]

Spin - Orbit Coupling

Work in non-inertial frame with orbiting proton. The Biot-Savart field at the electron is B = ((Subscript[μ, 0])/(2r))(ev/(2πr))=(e/(4(Subscript[πϵ, 0])(mc^2)(r^3)))L
The magnetic moment using g=2 is μ=-(e/m)S.  Therefore, the perturbation to the Hamiltonian is H'= ((e^2)/(4(Subscript[πϵ, 0])(m^2)(c^2)))((L·S)/(r^3))
This is double the actual value due to the Thomas correction. A heuristic way to derive this is to observe that the radius vector connecting the electron to the proton rotates at the orbital angular frequency ω. This involves improper time intervals and so the precessional angular frequency in the frame of the electron is γω ~(1+((v^2)/(2(c^2))))ω. The inertial frame tied to the electron therefore precesses relative to that tied to the proton at a rate (((ωv)^2)/(2(c^2))). This must be subtracted from the precessional frequency of the dipole given by μ×B=Ω×S or Ω=(((ωv)^2)/((c^2) )). So the precessional frequency relative to the electron inertial frame is halved. LIkewise the energy. So, H'= ((e^2)/(8(Subscript[πϵ, 0])(m^2)(c^2)))((L·S)/(r^3)). Classically, L, S are not conserved and precess around the total angular momentum J=L+S. Hence L·S=(1/2)((J^2)-(L^2)-(S^2)) . In QM language, (J^2),(L^2), (S^2) commute with the Hamiltonian. The associated eigenvalues are 
((ℏ^2)/2)[j(j+1)-l(l+1)-s(s+1)] where s=(1/2) for a single electron. (This carries over to multi-electron atoms where L,S are the total orbital and spin angular momentum.)
For Hydrogen we need <(1/(r^3))>=(2/(l(l+1)(2l+1)(n^3)(a^3))).
The spin orbit energy shift is then (Subscript[E, SO]^1)= (((α^4) (mc^2))/(2(n^3))) (( [j(j + 1) - l(l + 1) -s(s+1)])/(l(l+1)(2l+1)))

Fine Structure

Adding the relativistic and spin orbit terms and using s=(1/2),  gives a fine structure energy shift
(Subscript[ΔE, FS]) = (((α^4) (mc^2))/(n^4))((3/8)-(n/(2j+1)))  which must be negative.The energy shift breaks the degeneracy in l and depends upon admissible j for given n
The approximations that we have made break down for large values of Z when L-S coupling is replaced by the j-j coupling approximation.

efs=((5.11 (10^11))/((137^4)(n^4)))(.375-(n/(2j+1)))/.{n->2,j->(3/2)}

Hyperfine Structure

Proton dipole moment (Subscript[μ, p])= (Subscript[g, p]) (e/(2(Subscript[m, p])))(Subscript[S, p]), (Subscript[g, p])=5.6.  B=((3((Subscript[μ, p])·r)r-(r^2)(Subscript[μ, p]))/(4(Subscript[πϵ, 0])(c^2)(r^5))) + ((Subscript[μ, p])/(6(Subscript[πϵ, 0])(c^2)))(δ^3)(r). Second term represents internal structure of proton in a model-free way. (A subtle point!)
H'=-(Subscript[μ, e])·B => (Subscript[E, hfs]^1)=(((Subscript[g, p])(α^2)a)/(2(Subscript[m, p]))) [< ((3 ((Subscript[S, p])·r)((Subscript[S, e])·r)  -  (Subscript[S, p])·(Subscript[S, e])(r^2))/(r^5))>+(8/3)<(Subscript[S, p])·(Subscript[S, e])>|ψ(0)((|)^2)]. The first term vanishes by inspection for an s state, and (Subscript[ψ, 100])(0)=(π^(-1/2))(a^(-3/2)), leaving
(Subscript[E, hfs]^1)=((4(Subscript[g, p]))/3)(((Subscript[m, e])/(Subscript[m, p])))(α^4)(Subscript[m, e])(c^2)(((Subscript[S, p])·(Subscript[S, e]))/(ℏ^2))  =((2(Subscript[g, p]))/3)(((Subscript[m, e])/(Subscript[m, p])))(α^4)(Subscript[m, e])(c^2)[s(s+1)-(3/4)-(3/4)].  where s=0,1 and []=(1/2), -(3/2) for triplet and singlet states.is  (Subscript[ΔE, hfs])=5.9μeV ≡1.4GHz≡21cm. 
This can be measured by a hydrogen maser to mHz accuracy.

Zeeman[1]

Weak Field

Field due to orbiting proton (Subscript[B, SO])~(((α^4)(mc^2))/(Subscript[μ, B])) ~1T. For B <(Subscript[B, SO]),  L, S precess around J  and the average value of S is that projected along J=L+S, just like in classical mechanics.
So,  l,s,m,(Subscript[m, j])are conserved quantum numbers and H'=(e/(2m))(L+2S)·B  =   (e/(2m))(1+((S·J)/(J^2)))J·B 
Using (J^2) = (L^2)+(S^2)+2(J-S)·S , S·J=(1/2)((J^2)+(S^2)-(L^2)), (Subscript[E, WZ]^1)=(Subscript[μ, B])(Subscript[g, J])(Subscript[m, j])B, where (Subscript[g, J])=1+((j(j+1)-l(l+1)+s(s+1))/(2j(j+1))) Lande g-factor

Strong Field

B>(Subscript[B, SO]), S,L precess around B conserved quantum numbers are l,s, m,(Subscript[m, s]). H'=(e/(2m))((Subscript[L, z])+2(Subscript[S, z]))B => (Subscript[E, SZ]^1)=(Subscript[μ, B])B(m+2(Subscript[m, s])).  
Include SO correction using <S·L>=(ℏ^2)(Subscript[mm, s]) .This leads to (Subscript[E, fs]^1)=(((α^4)(Subscript[m, e])(c^2))/(2(n^4)))[(3/4)-2n(((l(l+1)-(Subscript[mm, s]))/(l(l+1)(2l+1) )))].Paschen-Back effect

Intermediate Field

Expand weak field eigenfunctions as linear combination of strong field eigenfunctions using Clebsch-Gordan coefficients. 
Compute combined fine structure-Zeeman perturbation to Hamiltonian perturbation matrix in strong field basis. 
It has off diagonal elements but can be diagonalized to give eigenenergies and associated eigenfiunctions

Ultra Strong

The Coulomb field is a perturbation to the magnetic field ie (Subscript[ℏω, G])> (α^2)(mc^2)~4kT. Neutron stars have magnetic field up to 30 GT. Atoms are elongated along field.

Problems

6.5

H'=-qEx; (Subscript[E, n]^1)=0; (Subscript[E, n]^2)= (((qE))^2)∑((|<m|x|n>((|)^2))/((n-m)ℏω)), where x=((((ℏ/(2mω))))^(1/2))((SubPlus[a])+(SubMinus[a])) ; (SubPlus[a])|n> =(((n+1))^(1/2))|n+1>;  (SubMinus[a])|n> =(n^(1/2))|n-1>.  
So  (Subscript[E, n]^2)= (((qE))^2)((ℏ/(2mω)))∑((|<m|x|n>((|)^2))/((n-m)ℏω))= (((q^2) (E^2))/(2((mω)^2)  ))(n-n-1)=-(((q^2) (E^2))/(2((mω)^2)  )). 
The potential energy can be written as (1/2)((mω)^2)(x^2)-qEx=(1/2)((((mω)^2)(x-(qE/((mω)^2))))^2)-(((q^2)(E^2))/(2((mω)^2))) and we recover the energy perturbation.

6.12

Virial theorem: 2<T>=<r·∇V>=-V. =>V=2E=><(1/r)≥(1/((n^2)a))

6.32 a

Use 1st order perturbation theory. ((∂(Subscript[E, n]))/(∂λ))δλ=<(Subscript[ψ, n])|((∂H)/(∂λ))|(Subscript[ψ, n])>λ+O((λ^2)) and take limit

6.33

(Subscript[E, n]) = -((me^4)/(32(π^2)(Subscript[ϵ, 0]^2)(ℏ^2)(n^2)))for some integer l. Hence: ((-e)/(2(Subscript[πϵ, 0])))<(1/r)>= - ((-(me^3))/(8(π^2)(Subscript[ϵ, 0]^2)(ℏ^2)(n^2)))=><(1/r)>=(1/((n^2)a))
Write n=k+l for some integer k. (((2 l + 1) (ℏ^2))/(2m))< (1/(r^2)) >= ((me^4)/(16(π^2)(Subscript[ϵ, 0]^2)(ℏ^2)(n^3)))=><(1/(r^2))> =(2/((a^2)(2l+1)(n^3)))

6.36

a) H’=eErcosθ. => (Subscript[E, 1]^1)=<(Subscript[ψ, 100]^( * ))H'(Subscript[ψ, 100])>=0

Clear[ψ]
ψ[l_,m_]:=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,ϕ]/.n->2;
w=Table[Integrate[ψ[If[i==1,0,1],If[i==1,0,3-i]]ψ[If[j==1,0,1],If[j==1,0,j-3]](r^3)Sin[θ]Cos[θ],{ϕ,0,2π},{θ,0,π},{r,0,∞}],{i,1,4},{j,1,4}];
Eigensystem[w]

b) 3 levels
c) (2^(-1/2))((Subscript[ψ, 00])±(Subscript[ψ, 10])),(Subscript[ψ, 1-1]),(Subscript[ψ, 11]). (Subscript[E, 2]^1)=0,0,∓3eaE;<p>=0,0±3ea

6.37

Clear[ψ];n=3;
ψ[l_,m_]:=(2/(n^2))((((((n-l-1)!)/((n+l)!))))^(1/2))(((((2r)/n)))^l)(E^(-r/n))LaguerreL[n-l-1,2l+1,((2r)/n)]SphericalHarmonicY[l,m,θ,ϕ]
Eigenvalues[Table[Integrate[ψ[If[i==1,0,If[i<5,1,2]],If[i==1,0,If[i<5,3-i,7-i]]]   ψ[If[j==1,0,If[j<5,1,2]],If[j==1,0,If[j<5,j-3,j-7]]]
(r^3)Sin[θ]Cos[θ],{ϕ,0,2π},{θ,0,π},{r,0,∞}],{i,1,9},{j,1,9}]]

(Subscript[E, 3]^1)=0,0,0,±(9/2)eaE,±(9/2)eaE,±9eaE

6.22

(Subscript[E, r]^1)=-((Subscript[E, n]^2)/(2(mc^2)))[((4n)/(l+1/2))-3], (Subscript[E, SO]^1)=<ψ|((e^2)/(8(Subscript[πϵ, 0])(m^2)(c^2)(r^3)))S·L|ψ>=(((e^2)(ℏ^2)(Subscript[mm, s]))/(4(Subscript[πϵ, 0])(m^2)(c^2)l(l+1)(2l+1)(n^3)(a^3)))
Hence (Subscript[E, fs]^1)=(Subscript[E, r]^1)+(Subscript[E, SO]^1)=(((α^4) (mc^2))/(2(n^3))) [((2(Subscript[mm, s]))/(l (l + 1) (2 l + 1))) + (3/(4n)) -(2/(2l+1))]

Lyman β

Ground state: n=1, l=0, j=(1/2). Excited state:n=3, l=0,1,2, s=(1/2)=>j=(1/2), (3/2),(5/2).  (1/2)->(1/2); ν=2.924 PHz, (3/2)->(1/2), δν=3.24 GHz, (5/2)->(1/(2 )),δν=4.33 GHz

α=(1/137.036);m=9.10936 (10^(-31));mp=1.67 (10^(-27));c=2.99792 (10^8);h=2π 1.05457 (10^(-34));e=1.60218 (10^(-19));μB=9.3 (10^(-24));(((α^4)m (c^2))/(μB));
en1=-(((α^2)m (c^2))/(2  h));en3=-(((α^2)m (c^2))/(18  h));en1fs=-((α^2)/4)en1;en3fs=-((α^2)/9)en3{2.25,.75,.25};
en3+en3fs[[1]]-en1-en1fs
en3fs[[1]]-en3fs[[2]]
en3fs[[1]]-en3fs[[3]]

Lyman α

B=0. Excited state n=2, l=0,1, s = 1/2=>j=(1/2),(3/2). (1/2)->(1/2); ν=2.467 PHz, (3/2)->(1/2), δν=10.95 GHz.
(Subscript[B, SO])~25T so weak field regime. 
B=.01T. 
n=1 δν=±140MHz. n=2,  l=0,j=(1/2),δν=±140MHz,l=1,j=(1/2), δν=±47MHz, j=(3/2), δν=±94,±281MHz
(1/2)->(1/2) ν=2.467 PHz δν=0,±280MHz; (3/2)->(1/2) ν=2.467 PHz+10.95GHz δν=±140±47,94,281=±187,93,234,46,321,140MHz. 
14 lines in total

Clear[j];en2=-(((α^2)m (c^2))/(8  h));en2fs=-((α^2)/4)en2{1.25,.25};
en2+en2fs[[1]]-en1-en1fs
en2fs[[1]]-en2fs[[2]];
gj=1+((j(j+1)+s(s+1)-l(l+1))/(2j(j+1)))/.{s->(1/2),l->1,j->(3/2)};
((μB  .01)/h){2,2,(2/3),(4/3)}

21 cm line

For an electron in the 1s the fine structure (combined relativistic and fine structure) perturbation to the Hamiltonian is more subtle than given in the book involving the “Darwin” term (footnote 14). The associated energy perturbation is given by Eq. 6.66, (Subscript[E, fs]^1) =-(((α^4)(mc^2))/8)=-180μeV. 
The weak field Zeeman perturbation is ±(1/2)(Subscript[μ, B])B=±2.9μeV, comfortably less than the fine structure pertubation. 
The hyperfine splitting is 5.9μeV (Eq. 6.93) which is similar to the Zeeman splitting. 
The combined perturbation to the Hamiltonian due Zeeman and hyperfine lines can be written 
H'=(((Subscript[g, p])(e^2))/(3(Subscript[ϵ, 0])(Subscript[m, p])(mc^2))) (Subscript[S, p])·S(δ^3)(r)-(e/m)S·B 
The procedure is to use the eigenfunctions of the combined 1s fine structure-proton spin state and evaluate the elements of the perturbation matrix (Subscript[W, mn])=<(Subscript[ψ, m])|H|(Subscript[ψ, n])>. 
Diagonalize this matrix to find the eigenstates.

6.31

H = (H^0) + H'=(1/(2m))((Subscript[p, +]^2)+(Subscript[p, -]^2))+(k/2)((Subscript[x, +]^2)+(Subscript[x, -]^2)) -((e^2)/(4(Subscript[πϵ, 0])(R^3)))((Subscript[x, +]^2)-(Subscript[x, -]^2))+…, where (SubPlus[x])=(2^(-1/2))((Subscript[x, 1])+(Subscript[x, 2])) etc
(Subscript[ω, ±]^2)=(Subscript[ω, 0]^2)∓((e^2)/(2(Subscript[πϵ, 0])(R^3)))=(Subscript[ω, 0]^2)(1∓((e^2)/(2(Subscript[πϵ, 0])(Subscript[mω, 0]^2)(R^3)))), ΔV=(1/2)ℏ((SubPlus[ω])+(SubMinus[ω])-2(Subscript[ω, 0]))= -(((ℏe)^4)/(32(π^2)(Subscript[ϵ, 0]^2)(m^2)(Subscript[ω, 0]^3)(R^6)))
Using perturbation theory, (Subscript[E, 0]^1)=- ((e^2)/(2(Subscript[πϵ, 0])(R^3)))<(Subscript[ψ, 1]^0)(Subscript[ψ, 2]^0)|(Subscript[x, 1])(Subscript[x, 2])|(Subscript[ψ, 1]^0)(Subscript[ψ, 2]^0)>=0, (Subscript[E, 0]^2)=(((((e^2)/(2(Subscript[πϵ, 0])(R^3)))))^2)(∑_(Subscript[n, 1]))(∑_(Subscript[n, 2]))  ((|<(Subscript[ψ, 1]^(Subscript[n, 1]))(Subscript[ψ, 2]^(Subscript[n, 2]))|(Subscript[x, 1])(Subscript[x, 2])|(Subscript[ψ, 1]^0)(Subscript[ψ, 2]^0)>((|)^2))/((Subscript[ℏω, 0])(1-1-(Subscript[n, 1])-(Subscript[n, 2]))))=-(((((e^2)/(2(Subscript[πϵ, 0])(R^3)))))^2)  ((|<(Subscript[ψ, 1]^1)(Subscript[ψ, 2]^1)|(Subscript[x, 1])(Subscript[x, 2])|(Subscript[ψ, 1]^0)(Subscript[ψ, 2]^0)>((|)^2))/(2(Subscript[ℏω, 0])))
Now x=((((ℏ/(2(Subscript[mω, 0])))))^(1/2))((SubPlus[a])+(SubMinus[a]))  and  (SubPlus[a])(Subscript[ψ, 0])=(Subscript[ψ, 1]) => (Subscript[E, 0]^2) =- (((((e^2)/(2(Subscript[πϵ, 0])(R^3)))))^2) ((((ℏ/(2(Subscript[mω, 0])))))^2)(1/(2(Subscript[ℏω, 0]))) =-(((ℏe)^4)/(32(π^2)(Subscript[ϵ, 0]^2)(m^2)(Subscript[ω, 0]^3)(R^6)))

Series[1-(((1-x1))^(-1))-(((1+x2))^(-1))+(((1-x1+x2))^(-1)),{x1,0,1},{x2,0,1}]
Series[(((1-x))^(1/2))+(((1+x))^(1/2)),{x,0,2}]

7:Variational Principle

Trial wave functions[1]

ψ=(Subscript[c, n])(Subscript[ψ, n]); <H>=<ψ|H|ψ>=(∑_n)|(Subscript[c, n]^2)|(Subscript[E, n]) >(Subscript[E, 0]). Any estimate is an overestimate.
Get lowest upper bound by minimizing estimate.
e.g. Hψ=(1/2)(-((ℏ^2)/m)ψ''+((mω)^2)(x^2)ψ). <H>  =(1/2)(∫_(-∞)^∞)dx((∂_x)(ψ^2)+(x^2)(ψ^2))
Try ψ=((((15/(16a))))^(1/2))(1-((((x/a)))^2)), <H>=(5/(4(a^2)))+((a^2)/14), minimized when a = ((((35/2)))^(1/4))=><H>=((((5/14)))^(1/2))=0.6. Get exact answer if we use a Gaussian as the trial wavefunction!

ψ=((((15/(16a))))^(1/2))(1-(((x/a))^2));
Integrate[(ψ^2),{x,-a,a}]
en=Integrate[(1/2)(((((∂_x)ψ))^2)+(x^2)(ψ^2)),{x,-a,a}]
Solve[(∂_a)en==0]
Simplify[en/.Solve[(∂_a)en==0][[4]]]

First order perturbations overestimate energy of ground state. Can sometimes get first excited state  when have odd/even eigenfunctions.

He[1]

Understand why what we did by guesswork turned out to be good. Recall that if we ignore the ee repulsion (Subscript[E, 0])=-8×13.6=-109eV. 
We improved this estimate using the electron unperturbed (symmetric) wave function ψ=(Subscript[ψ, 1])(Subscript[ψ, 2])=((e^(-(r+s)))/π), measuring lengths in terms of the Helium Bohr radius. 
So,  (Subscript[V, ee])=∫(Subscript[dV, 1])(Subscript[dV, 2])((e^2)/(4(Subscript[πϵ, 0])(Subscript[r, 12])))|ψ((|)^2)=∫4((πr)^2)dr∫2((πs)^2)ds(∫_0^π)dθsinθ((e^2)/(4(Subscript[πϵ, 0])(Subscript[r, 12])))|ψ((|)^2)where the two electron radii are r,s and the angle between them is θ, and (Subscript[r, 12])=((((r^2)+(s^2)+2rscosθ))^(1/2))

ψ=((E^(-(r+s)))/π);
vee=Simplify[8(π^2) (r^2) (s^2) (ψ^2) ((((r^2)+(s^2)))^(-1/2)) Integrate[(((1-2x Cos[θ]))^(-1/2))Sin[θ],{θ,0,π}]/.x->(r s)/((r^2)+(s^2))]
vee=Integrate[vee,{s,0,∞},Assumptions->r>0]
vee=Integrate[vee,{r,0,∞}]

The answer is (5/8)×((e^2)/(4(Subscript[πϵ, 0])a(He)))=(5/2)×13.6=34 eV giving a total estimate of -75eV, 4 eV above the measured value of -79eV. What we actually did was to use a variational calcualtion with the full Hamiltonian and the unperturbed wavefunctions as trial functions. 
We can improve on this using ψ=(Subscript[ψ, 1])(Subscript[ψ, 2])=(((k^3)(e^(-k(r+s))))/π), varying the effective charge. 
The full Hamiltonian remains: H=-(1/2)((Subscript[∇, 1]^2)+(Subscript[∇, 2]^2))ψ+(-(2/(Subscript[r, 1]))-(2/(Subscript[r, 2]))+(1/(Subscript[r, 12])))ψ =-(1/2)((Subscript[∇, 1]^2)+(Subscript[∇, 2]^2))ψ+(-((2k)/(Subscript[r, 1]))-((2k)/(Subscript[r, 2]))-((2(1-k))/(Subscript[r, 1]))-((2(1-k))/(Subscript[r, 2]))+(1/(Subscript[r, 12])))ψ,  in units of 4Ry=54.4eV
Using the scaling that lengths ∝(k^(-1)),
< H >= 2× (k^2)-2×2 (k^2)-2×2(1-k) k +(5/8)k =2(k^2)-(( 27k)/8). This has a minimum value of -((3^6)/(2^9))≡-77.5eV when k=(27/32), just 1.5eV high.

(Subscript[H, 2]^(+))[1]

Normalization

H = -(1/2)(∇^2)+2(-(1/(Subscript[r, 1]))-(1/( (Subscript[r, 2])))) where (Subscript[r, 2])= (((((Subscript[r, 1])^2) +(R^2)-2(Subscript[r, 1])Rcosθ))^(1/2))) and we measure lengths in a and energy in Ry , 
Introduce Linear Combination of Atomic Orbitals: ψ=A[(Subscript[ψ, 0])((Subscript[r, 1]))+(Subscript[ψ, 0])( (Subscript[r, 2]))], where (Subscript[ψ, 0])(r)=(π^(-1/2))(e^(-r)). with (Subscript[E, 1])=-1
∫dV|ψ((|)^2)=2(A^2)(1+I) where the overlap integral is I(R)=2(∫_0^∞)(drr^2)(e^(-r))(∫_0^π)((sinθdθe)^(- ((((r^2) +(R^2)-2rRcosθ))^(1/2)))) =(e^(-R))(1+R+((R^2)/3)), so A=(([2(1+I)])^(-1/2)).

Integrate[Sin[θ](E^(-((((r^2)+(R^2)-2r R Cos[θ]))^(1/2)))),{θ,0,π},Assumptions->{r>0,r<R}];
Integrate[Sin[θ](E^(-((((r^2)+(R^2)-2r R Cos[θ]))^(1/2)))),{θ,0,π},Assumptions->{r>R,R>0}];
Simplify[Integrate[((4r (E^(-(r+R)))(-r Cosh[r]+(1+R) Sinh[r]))/R),{r,0,R}]+Integrate[((4r (E^(-2r)) (-R Cosh[R]+(1+r) Sinh[R]) )/R),{r,R,∞}]]

Electron energy

Hψ = - ψ-2A((((Subscript[ψ, 0])((Subscript[r, 2])))/(Subscript[r, 1]))+(((Subscript[ψ, 0])((Subscript[r, 1])))/(Subscript[r, 2]))),  so <H>=-1-4(A^2)(D+X) =-1-2((D+X)/(1+I)),
where the direct integral is D=<(Subscript[ψ, 0])((Subscript[r, 2]))|(1/(Subscript[r, 1]))|(Subscript[ψ, 0])((Subscript[r, 2]))> =(R^(-1))-(1+(R^(-1)))(e^(-2R))  and the exchange integral is X=<(Subscript[ψ, 0])((Subscript[r, 1])) | (1/(Subscript[r, 1]))| (Subscript[ψ, 0])((Subscript[r, 2])) >=(1+R)(e^(-R))

Integrate[Sin[θ](E^(-2((((r^2)+(R^2)-2r R Cos[θ]))^(1/2)))),{θ,0,π},Assumptions->{r>0,r<R}];
Integrate[Sin[θ](E^(-2((((r^2)+(R^2)-2r R Cos[θ]))^(1/2)))),{θ,0,π},Assumptions->{r>R,R>0}];
Simplify[Integrate[((  (E^(-2 (r+R))) (-1-2 r-2 R+(E^(4 r)) (1-2 r+2 R)))/(2R)),{r,0,R}]+Integrate[(( (E^(-2 r)) (-2 R Cosh[2 R]+(1+2 r) Sinh[2 R]))/R),{r,R,∞}]]
Integrate[Sin[θ](E^(-((((r^2)+(R^2)-2r R Cos[θ]))^(1/2)))),{θ,0,π},Assumptions->{r>0,r<R}];
Integrate[Sin[θ](E^(-((((r^2)+(R^2)-2r R Cos[θ]))^(1/2)))),{θ,0,π},Assumptions->{r>R,R>0}];
Simplify[Integrate[((  4 (E^(-(R+r))) (-r Cosh[r]+(1+R) Sinh[r]) )/R),{r,0,R}]+Integrate[(( 4(E^(-2 r))  (-R Cosh[R]+(1+r) Sinh[R]))/R),{r,R,∞}]]

Molecular energy

Including proton-proton energy, E = < H >+ (2/R)=(2/R)[(((1-2(R^2)/3)(e^(-R))+(!+R)(e^(-2R)))/(1+(1+R+(R^2)/3)(e^(-R))))]-1.

Plot[(2/x)((((1-2(x^2)/3)(E^(-x))+(1+x)(E^(-2x)))/(1+(1+x+(x^2)/3)(E^(-x)))))-1,{x,0,6},Frame->True,PlotRange->{-1.5,0}]

Upper bound

E is minimized when R=2.4, E=-1.14 which is bound relative to the dissociated ion. The measured values are 2, -1.2, which is lower, as it shoud be.
Repeating with an antisymmetric combination leads to an unbound state.

Thomas - Fermi Model

A simple approximate statistical technique that is used for high Z atoms is to suppose that the density of electron is just like it is in a metal, and the Fermi sea extends up to the continuum.
i.e. n =((8((π[-2 mV])^(3/2)) )/(3(h^3))) We then add Poisson’s equation (1/(r^2))(d/dr)(r^2)(dV/dr) = -((ne^2)/(Subscript[ϵ, 0])) to generate a nonlinear ODE. The BCs are V->-((Ze^2)/(4(Subscript[πϵ, 0])r)), as r->0, rV->0 as r->∞.
Introduce χ=-((4(Subscript[πϵ, 0])rV)/(Ze^2)), x=((((4/(3π))))^(2/3))((2(Z^(1/3))(me^2))/(4(Subscript[πϵ, 0])(ℏ^2)))r and (x^(1/2))χ''=(χ^(3/2)); χ(0)=1; χ(∞)=0.This model is quite good in practice for large Z.

xmin=0.001;xmax=10.;
Off[NDSolve::ndsz]
nds=NDSolve[{χ''[x]==(x^(-.5))((χ[x])^1.5),χ[xmin]==1,χ[xmax]==.025},χ[x],{x,xmin,xmax}];
On[NDSolve::ndsz]
Plot[-((χ[x])/x)/.nds[[1]],{x,xmin,xmax},Frame->True, PlotRange->{-3,0}]

Problems

7.6

-4×13.6+79=24.6eV

7.7

In units of ((e^2)/(4(Subscript[πϵ, 0])a(Z)))=2Z Ry,< H >= 2×[(1/2) (Zk^2)-(Zk^2)-Zk(1-k)]+(5/8)k =(Zk^2)-(2Z-(5/8))k, which has a minimum when k=1-(5/(16Z))of E=-((((16Z-5))^2)/(256Z))≡-((((16Z-5))^2)/128)Ry
For (SuperMinus[H]), k=(11/16),E=-(121/128)Ry; for (SuperPlus[Li]), k=(43/48), E=-(1849/128)Ry.

Yukawa

< H >= 4 π (∫_0^∞)(drr^2)(e^(-βr))H (e^(-βr))] = 4 π((1/(8β))-((Subscript[V, 0])/(((1+2β))^2))). This has a negative minimum (bound state) for (Subscript[V, 0])>1. For (Subscript[V, 0])=1, β=0.5. For (Subscript[V, 0])=1.35, minimum has β=.32, <H>=-1.4. 
Give general expressions solving general cubic.

V0=1.35;h=4π Integrate[(r^2)(E^(-2β r))(-((β^2)/2)+(β/r)-((V0 (E^(-r)))/r)),{r,0,∞},Assumptions->β>0]
b=NSolve[(∂_β)h==0,β]
Simplify[h/.b]

Double Oscillator

ψ = (Subscript[αψ, n])(x-a) + (Subscript[βψ, n])(x+a)
∫dx|ψ|=(α^2)+2(Subscript[αβC, n])+(β^2)=1, where (Subscript[C, n]) = <(Subscript[ψ, n])(x-a)(Subscript[ψ, n])(x+a)> is the overlap integral. 
< H >= ((α^2)+(β^2))(Subscript[D, n])+2 (Subscript[αβX, n]), where (Subscript[D, n])=<(Subscript[ψ, n])(x-a)(Subscript[Hψ, n])(x-a)>is the direct integral and (Subscript[X, n])=<(Subscript[ψ, n])(x-a)(Subscript[Hψ, n])(x+a)> is the exchange integral.  
The extreme values of <H> subject to the normalizaton constraint are (((Subscript[D, n])±(Subscript[X, n]))/(1±(Subscript[C, n]))). (Use Lagrange multiplier method.)
For ground states, (Subscript[ψ, 0]) =(π^(-1/4))(e^(-(x^2)/2)). (Subscript[C, 0])=(e^(-(a^2))), (Subscript[D, 0])=(1/2)(1+(a^2)),  (Subscript[X, 0])=(1/2)(1-(a^2))(e^(-(a^2)))=><H(SubMinus[>])-<H(SubPlus[>])~2(a^2)(e^(-(a^2))), for a>>1.

c0=Integrate[(π^(-1/4))(E^(-(((x-a))^2)/2))(π^(-1/4))(E^(-(((x+a))^2)/2)),{x,-∞,∞}]
d0=Integrate[(π^(-1/4))(E^(-(((x-a))^2)/2))(-(1/2)D[(π^(-1/4))(E^(-(((x-a))^2)/2)),{x,2}]+((π^(-1/4))/2)(x^2)(E^(-(((x-a))^2)/2))),{x,-∞,∞}]
x0=Integrate[(π^(-1/4))(E^(-(((x-a))^2)/2))(-(1/2)D[(π^(-1/4))(E^(-(((x+a))^2)/2)),{x,2}]+((π^(-1/4))/2)(x^2)(E^(-(((x+a))^2)/2))),{x,-∞,∞}]

7.20

a) For a particle in arm as x->∞, gd. state has ψ=cos((πy)/2), E=((π^2)/8).
b) Divide into octants. 
∫dxdy|ψ((|)^2)=((2(e^(-2α))(6+11α))/(9α)).
< H >= -(1/2)∫dxdyψ(∇^2)ψ=-(1/2)∮ dsψn·∇ψ+(1/2) ∫dxdy ∇(ψ^2) = 1/2 ∫dxdy ∇(ψ^2)=((3(3+2α+(α^2)))/(6+11α)),  which minimizes at α = 0.94 with<H(Subscript[>, gd])<1.058<((π^2)/8).

ψ=If[x<1,If[y<1,(1-x y)(E^(-α)),(1-x)(E^(-α y))],If[y<1,(1-y)(E^(-α x)),-1]];
n=8Integrate[(ψ^2),{x,0,∞},{y,0,Min[x,1]},Assumptions->α>0]
h=(8/n)(1/2)Integrate[ ((Norm[{D[ψ,x],D[ψ,y]}])^2),{x,0,∞},{y,0,Min[x,1]},Assumptions->α>0]
Plot[h,{α,.8,1},Frame->True,FrameLabel->{"α","<H>"},RotateLabel->False]
ContourPlot[ψ/.α->.94,{x,0,2},{y,0,2},Contours->{0,.1,.2,.3},FrameLabel->{"x","y"},RotateLabel->False]

8:WKB

Classical[1]

Background

“(J)WKB” (after Rayleigh!) method used for approximate solutions to second order differential equations including TISE. Use when need to know ψ, not just E.

1D Formalism

In 1D,ψ'' + (k^2)ψ = 0, where k = ((([2 m(E-V)])^(1/2))/ℏ).   If k const, ψ~(e^(±ikx)) for (k^2)>0 and ~ (e^(±|k|x)) , if (k^2)<0.  
If k varies, try ψ=(Ae^(iϕ)).   A''+2iA'ϕ'+iAϕ''-((Aϕ')^2)+(k^2)A=0.
Hence, (A^2)ϕ'=(C^2);  If amplitude varies slowly relative to oscillation, drop A'' so that ϕ’=±k => ϕ=±∫dxk or ψ=(Ck^(-1/2))(e^(±i∫dxk)).

Conservation

Note that |ψ((|)^(2 ))k∝ |ψ((|)^(2 ))v is const. Conservation of flux. 
Consider as a sequence of small discontinuities with r=(((SubMinus[k])-(SubPlus[k]))/((SubMinus[k])+(SubPlus[k]))),  t=((2(SubMinus[k]))/((SubMinus[k])+(SubPlus[k]))),    so that  (1-|r((|)^2))(SubMinus[k])=|t((|)^2)(SubPlus[k])- flux conservation. WKB is neglect of reflections.

Semi - infinite  Wells

wlog, consider a stationary state with V(x) in [0,1] and ψ=0 at the walls =>(Subscript[ψ, WKB])=(Ak^(-1/2))(Subscript[sinϕ, n]), where (Subscript[ϕ, n])(1)=nπ. Solve by Fourier series or special functions.
Example V=x(1-x). Closed form is Parabolic Cylinder functions. Use numerical shooting method instead with (Subscript[ψ, WKB]^('))(0)=(((2E))^(1/4))A=ψ’(0).

v=x(1-x);e=5.15;k=(((2(e-v)))^(1/2));
Plot[{e,k,v},{x,0,1},Frame->True,FrameLabel->{"x","E, k, V"},RotateLabel->False]
nds1=NDSolve[{ψ''[x]+(k^2)ψ[x]==0,ψ[0]==0,ψ'[0]==1},ψ,{x,0,1}];ψ[1]/.nds1[[1]]
nds2=NDSolve[{ϕ'[x]==k,ϕ[0]==0},ϕ,{x,0,1}];(π^(-1))ϕ[1]/.nds2[[1]]
Plot[{ψ[x]/.nds1,(((2e))^(-1/4))(k^(-1/2))Sin[ϕ[x]]/.nds2},{x,0,1},Frame->True,FrameLabel->{"x","ψ, (Subscript[ψ, WKB])"},RotateLabel->False]

Alternative Potential

v=(4.5/(1+100(((x-.5))^2)));e=6.9;k=(((2(e-v)))^(1/2));
Plot[{e,k,v},{x,0,1},Frame->True,FrameLabel->{"x","E, k, V"},RotateLabel->False]
nds1=NDSolve[{ψ''[x]+(k^2)ψ[x]==0,ψ[0]==0,ψ'[0]==1},ψ,{x,0,1}];ψ[1]/.nds1[[1]]
nds2=NDSolve[{ϕ'[x]==k,ϕ[0]==0},ϕ,{x,0,1}];(π^(-1))NIntegrate[k,{x,0,1}]
Plot[{ψ[x]/.nds1,(((2e))^(-1/4))(k^(-1/2))Sin[ϕ[x]]/.nds2},{x,0,1},Frame->True,FrameLabel->{"x","ψ, (Subscript[ψ, WKB])"},RotateLabel->False]

Tunneling[1]

Barrier transmission

Evanescent region with sharp boundaries. 
γ=∫dx|k|.  ψ~|k((|)^(-1/2))(e^(±γ)). If >>1, neglect backward wave. Flux transmission coefficient: T ~ (e^(-2γ)). Tunneling is QM and optics phenomenon, not classical mechanics.

Alpha decay

Observed that He nuclei (α particles) are emitted. Z=2,m=4(Subscript[m, p])=3.7GeV. 
Simple model brings out key point. Treat nuclear potential as square well with depth -(Subscript[V, 0]) relative to ∞ and width (Subscript[r, 1])~1.07(A^(1/3))fm 
Spherical TISE y=ψr satisfies 1D equation with combined nuclear and Coulomb potential.
Trapped particle has energy E>0. Classical escape (turning) point at (Subscript[r, 2])=((2(Ze^2))/(4(Subscript[πϵ, 0])E)).
γ =(( (((2 mE))^(1/2)))/ℏ)(∫_(Subscript[r, 1])^(Subscript[r, 2]))((dr(((Subscript[r, 2])/r)-1))^(1/2)) = ((2 (((Ze^2)(2 m))^(1/2)))/(4 (Subscript[πϵ, 0]) ((ℏE)^(1/2)))) (∫_((Subscript[r, 1])/(Subscript[r, 2]))^1)((dx((1/x)-1))^(1/2)) = ((2 (((Ze^2)(2 m))^(1/2)))/(4 (Subscript[πϵ, 0]) ((ℏE)^(1/2))))[(π/2) -  2 (((((Subscript[r, 1])/(Subscript[r, 2]))))^(1/2))]=2(Subscript[ZE, MeV]^(-1/2))-1.5(Z^(1/2))(Subscript[r, 1fm]^(1/2))
Flux transmission coefficient T=(((((y((Subscript[r, 2])))/(y((Subscript[r, 1]))))))^2) ~ (e^(-2γ)). 
Individual α has speed v = (((((2 (E+(Subscript[V, 0])))/m)))^(1/2))=> halflife (Subscript[τ, 1/2]) ~ ((2(Subscript[r, 1]))/v)(e^(2γ)). Insensitive to prefactor!
Get E from measured α energies, (Subscript[V, 0]) from nuclear theory.

Connection[1]

Airy Function

Tunneling is not always this simple. We need approximations that work near turning point x=0 with wave/evanescent solutions for x<0/x>0.  
Use V=E+xV’. ψ’’=zψ, z=(((((2mV')/(ℏ^2))))^(1/3))x where (Subscript[V, 0])=E, V’>0. ψ=aAi(z)+bBi(z) - related to (Subscript[J, 1/3]), (Subscript[Y, 1/3]).
Ai = (π^(-1)) (∫_0^∞)ds cos(sz+((s^3)/3))=Re[(π^(-1))(∫_0^∞)(dse^(i(sz+(s^3)/3)))]  which clearly satisfies the differential equation
For z<<0, Ai~Re [(π^(-1))(∫_(-∞)^∞)(dse^(2i|z((|)^(3/2))/3))(e^(((iz^(1/2))(s+(z^(1/2))))^2))]=Re [(e^(2i|z((|)^(3/2))/3))(((e^(3iπ/4))(((π(-z))^(1/2))))^(-1/2))]= (((π^(-1/2))(-z))^(-1/4)) sin[(2/3) (((-z))^(3/2))+(π/4)] using method of stationary phase.
Likewise for z>>1, Ai ~ (1/2)(π^(-1/2))(z^(-1/4))(e^(-(2/3)(z^(3/2)))). More properties in Table 8.1.

Plot[{AiryAi[z],AiryBi[z]},{z,-8,4},Frame->True,FrameLabel->{"z","Ai(z),Bi(z)"},RotateLabel->False]

Matching

For |x|, |z| ~0, ψ=a Ai(z).
For x,z<<0, ψ= a Ai(z)~((((aπ)^(-1/2))(-z))^(-1/4))sin((2/3)(((-z))^(3/2))+(π/4))=((((aπ)^(-1/2))(((2mV')/(ℏ^2))))^(1/6))(k^(-1/2))sin((∫_x^0)dx'k(x')+(π/4))=(Ak^(-1/2))sin((∫_x^0)dx'k(x')+(π/4)), wh a=((((Aπ)^(1/2))(((2mV')/(ℏ^2))))^(-1/6)).
For x,z>>0, ψ=aAi(z) ~ (a/2)(π^(-1/2))(z^(-1/4))(e^(-(2/3)(z^(3/2))))=(a/2)(((π^(-1/2))(((2mV')/(ℏ^2))))^(1/6))|k((|)^(-1/2))(e^(-(∫_0^x)dx'|k(x')|))=(A/2)|k((|)^(-1/2))(e^(-(∫_0^x)dx'|k(x')|)).
In words, we halve the amplitude and introduce (π/(4 ))phase relative to turning point. ψ singular at turning point but flux continuous as in α decay.

SHO

Eigenenergies

V = (1/2)(x^2);E=(1/2)(Subscript[x, t]^2); k=((((xt^2)-(x^2)))^(1/2))
 (∫_(-(Subscript[x, t]))^(Subscript[x, t]))dx k +2 ×(π/4)= πE +(π/2) = (n + 1) π=>E=(n+(1/2)), n≥0, which is correct.

k[x_]:=((((xt^2)-(x^2)))^(1/2));ak[x_]:=((((x^2)-(xt^2)))^(1/2));
ph=Integrate[k[xp],{xp,x,xt},Assumptions->x∈Reals&&x>0&&x<xt]
2ph/.x->0
la=Integrate[ak[xp],{xp,xt,x},Assumptions->x∈Reals&&x>0&&x>xt&&xt>0]
2la/.x->0

Eigenfunctions

ψ ~ (Subscript[H, n])(x)(e^(-(x^2)/2)). Normalize at turning point

n=5;xt=(((2n+1.))^(1/2));
ψ[x_]:=((HermiteH[n,x])/(HermiteH[n,xt]))(E^((1/2)((xt^2)-(x^2))))
Plot[{1,(((x/xt))^2),ψ[x]},{x,0,1.5xt},Frame->True,FrameLabel->{"x","V,ψ"},RotateLabel->False]

Airy Approximation

Valid near turning point, z, x ~ 0. a=(1/(Ai(0)))=(3^(2/3))Γ((2/3)).

ψA[x_]:=(3^(2/3))Gamma[2/3]AiryAi[(((2xt))^(1/3))(x-xt)]
Plot[{1,(((x/xt))^2),ψA[x]},{x,0,1.5xt},Frame->True,FrameLabel->{"x","V,(Subscript[ψ, A])"},RotateLabel->False]

WKB Approximation

Evaluate phase and decrement relative to turning point.
Use A=a (((π^(-1/2))(2(Subscript[x, t])))^(1/6)) to match amplitude

ϕ[x_]:=((xt^2)/2)ArcCos[(x/xt)]-(x/2)((((xt^2)-(x^2)))^(1/2))
λ[x_]:=(x/2)((((x^2)-(xt^2)))^(1/2))-((xt^2)/2)Log[((x+((((x^2)-(xt^2)))^(1/2)))/xt)];
ψW[x_]:=If[x<xt,(π^(-1/2))(((2xt))^(1/6))(3^(2/3))Gamma[(2/3)]((k[x])^(-1/2))Sin[ϕ[x]+(π/4)],(1/2)(π^(-1/2))(((2xt))^(1/6))(3^(2/3))Gamma[(2/3)]((Abs[k[x]])^(-1/2))(E^(-λ[x]))]
Plot[{1,(((x/xt))^2),ψW[x]},{x,0,1.5xt},Frame->True,FrameLabel->{"x","V,(Subscript[ψ, WKB])"},RotateLabel->False]

Comparison of Matched Solutions

We can improve on these excellent approximations by using additional terms from the asymptotic expansion (e.g. Matthews & Walker).

Plot[{1,(((x/xt))^2),ψ[x],ψA[x],ψW[x]},{x,0,1.5xt},Frame->True,FrameLabel->{"x","V,ψ,(Subscript[ψ, A]),(Subscript[ψ, WKB])"},RotateLabel->False]

Problems

8.4

(^238)U`(->^234)Th 
E=238.0508-234.0436-4.0026=.0046u =4.3 MeV (including electrons which cancel out) =>v ~ 14Mm (s^(-1)).
(Subscript[r, 1])=1.07*(234^(1/3))=6.6fm
Hence γ=((1.98×90)/(4.3^(1/2)))-1.485(((90×6.6))^(1/2))=50. τ ~ ((2r)/v)(e^(2γ)) ~ 8 ×(10^(14 ))yr (high) 
(^212)Po ->(^208)Pb
E=211.9889-207.9767-4.0026=.0096u =8.9 MeV  =>v ~ 20Mm (s^(-1)).
(Subscript[r, 1])=1.07*(208^(1/3))=6.3fm
Hence γ=((1.98×82)/(8.9^(1/2)))-1.485(((82×6.3))^(1/2))=20.7. τ ~ ((2r)/v)(e^(2γ)) ~ 600 μs (high)

8.15

(Subscript[ψ, i])=D|k((|)^(-1/2))(e^(-λ));  λ=(∫_(Subscript[x, 2])^x)dx'|k|
(Subscript[ψ, ii]) = 2 (Dk^(-1/2))sin(ϕ+(π/4))=2(Dk^(-1/2))sin(η+(π/4))sinθ+2 (Dk^(-1/2))cos(η + π/4)cosθ, where ϕ(x) = (∫_x^(Subscript[x, 2]))dx'k, θ=ϕ((Subscript[x, 1])); η=(∫_(Subscript[x, 1])^x)dx'k
(Subscript[ψ, iii])=Dsinθ|k((|)^(-1/2))(e^ν)+2Dcosθ|k((|)^(-1/2))(e^(-ν));ν= (∫_x^(Subscript[x, 1]))dx'|k|; ζ=ν(0).
ψ'(0)=0=>tanθ=2(e^(2ζ)); ψ(0)=0=>tanθ=-2(e^(2ζ))
ϕ>>1=>θ=(n+(1/2))π∓((e^(-2ζ))/2)=πE=>ΔE=((e^(-2ζ))/π)
Ψ(x,t)=(2^(-1/2))((SuperPlus[ψ])(e^(-(SuperPlus[iE])t))+(SuperMinus[ψ])(e^((SuperMinus[iE])t))) =>|Ψ((|)^2) =  (1/2)(| (SuperPlus[ψ])((|)^2)+| (SuperMinus[ψ])((|)^2))+(SuperPlus[ψ])(SuperMinus[ψ])cosΔEt. =>τ=2(π^2)(e^(2ζ))
2ζ=(a^2)+O(e)

Series[2Integrate[((((((x-a))^2)-2e))^(1/2)),{x,0,a-(((2e))^(1/2))},Assumptions->a>(((2e))^(1/2)) &&e>0],{e,0,1}]

8.17

m~0.5 kg, z~2cm, g~10(ms^(-2))=>V~0.1J. x ~3cm. γ~(( (((2 mV))^(1/2)))/ℏ)x~(10^32) ; rms speed ~ (((kT/m))^(1/2))  ~ 100pm (s^(-1)).  
=>(Subscript[t, 1/2])~ (10^11)(e^(2γ))s~ (10^(11+.87γ))~(10^(10^(32 )))s

Final