130_notes.dvi

13.3 Sample Test Problems A particle of massmin 3 dimensions is in a potentialV(x,y,z) =^12 k(x^2 + 2y^2 + 3z^2 ). Find the ene ...

14 Angular Momentum 14.1 Rotational Symmetry If the potential only depends on the distance between two particles, V(~r) =V(r) th ...

This leads to a great simplification of the 3D problem. It is possible toseparate the Schr ̈odinger equationsincerandL^2 appear ...

[Li,Lj] = i ̄hǫijkLk [L^2 ,Li] = 0. We have shown that angular momentum is quantized for a rotor with asingle angular variable. ...

whereℓis an integer. Note that we can easily write the components of angular momentum interms of the raising and lowering operat ...

Its easy to find functions that give the eigenvalue ofLz. Yℓm(θ,φ) = Θ(θ)Φ(φ) = Θ(θ)eimφ LzYℓm(θ,φ) = ̄h i ∂ ∂φ Θ(θ)eimφ= ̄h i i ...

∫ dΩYℓm∗Yℓ′m′=δℓℓ′δmm′ We will use theactual functionin some problems. Y 00 = 1 √ 4 π Y 11 = − √ 3 8 π eiφsinθ Y 10 = √ 3 4 π co ...

14.3.1 Parity of the Spherical Harmonics In spherical coordinates, theparity operationis r → r θ → π−θ φ → φ+π. The radial part ...

Now we Taylor expand this equation. Hψ(x,y,z) +Hdθ ( ∂ψ ∂x y− ∂ψ ∂y x ) =Eψ(x,y,z) +Edθ ( ∂ψ ∂x y− ∂ψ ∂y x ) Subtract off the or ...

whereǫijkis the completely antisymmetric tensor and we assume a sum over repeated indices. ǫijk=−ǫjik=−ǫikj=−ǫkji The tensor is ...

We expect to need to keep the radial derivatives so lets identify those by dotting~rinto~p. This will also make the units matchL ...

θ φ x y z r We now proceed to calculate the angular momentum operators in spherical coordinates. The first step is to write the∂ ...

Bringing together the above results, we have. ∂ ∂x = sinθcosφ ∂ ∂r + 1 r cosφcosθ ∂ ∂θ − 1 r sinφ sinθ ∂ ∂φ ∂ ∂y = sinθsinφ ∂ ∂r ...

The operatorsL+,L−andLZdo not changeℓ. That is, after we operate, the new state is still an eigenstate ofL^2 with the same eigen ...

The raising and lowering operators changemin integer steps, so, starting fromm=−ℓ, there will be states in integer steps up toℓ. ...

14.5.2 The Expectation Value ofLx What is the expectation value ofLxin the stateψ(~r) =R(r)( √ 2 3 Y^11 (θ,φ)− √ 1 3 Y^10 (θ,φ)) ...

An eigenvalue ofℓ(ℓ+ 1) ̄h^2 = 2 ̄h^2 impliesℓ= 1. We will need a linear combination of theY 1 m to get the eigenstate ofLx=L++ ...

b) Assuming thatR(r) is given, write down the energy eigenfunctions for the ground state and the first excited state. c) Assumin ...

Higher powers ofrare OK, but are not dominant. Plugging this into the equation we get [ s(s−1)rs−^2 + 2srs−^2 ] −ℓ(ℓ+ 1)rs−^2 = ...

The lowestℓNeumann functions(irregular at the origin) solutions are listed below. n 0 (ρ) =− cosρ ρ n 1 (ρ) =− cosρ ρ^2 − sinρ ρ ...