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Angular

We review some basic classical . Consider a classical point of µ whose at t is denoted by ~r(t) moving in a speciﬁed central potential V (r) (the potential depends only on r = |~r|) in three . The is given by

1 d~r d~r E = µ · + V (r) . (1) 2 dt dt

dV The is given by F~ = −∇~ V = −rˆ and points along the radial direction since V dr depends only on r; its magnitude depends on the magnitude of ~r. This is a . Clearly the ~r × F~ vanishes and hence, is conserved. Since the angular momentum, ~r × ~p, is perpendicular to ~r, the constancy of the angular momentum implies that the radius vector of the particle, ~r lies in a plane perpendicular to the angular momentum. Thus angular momentum conservation reduces three-dimensional to motion in a plane. For the , closed are .

Choose the plane of motion to be the xy-plane and use polar coordinates. 1 E = µ (r ˙2 + r2φ˙2) + V (r) (2) 2 Note that in elementary parlance φ˙ is the angular ω and the angular part of the 1 2 2 1 2 is 2 µr ω = 2 Iω where we have used the deﬁnition of of . We re-express the angular part of the kinetic energy in terms of the , the orbital angular momentum. The magnitude of the angular momentum is given by µr2 φ˙ (or Iω). More explicitly1

~r × ~p = µ ~r × ~v = µ rrˆ × (r ˙rˆ + rφ˙φˆ)=zµr ˆ 2φ˙ i.e., the angular momentum is orthogonal to the plane.

Since L = µr2φ˙ the angular part of the kinetic energy can be rewritten2 as

1 L2 E = µ r˙2 + + V (r) . (3) 2 2µr2 This separation important result and occurs in Hamiltonian in mechanics. The energy describes a particle moving in one (with the one - dimensional coordinate r restricted to non-negative values) in an eﬀective potential given

1The velocity consists of the radial velocityr ˙ alongr ˆ plus a tangential velocity rφ˙ along φˆ. 2Substituting for φ˙ from the expression for L

2 1 1 L L2 µr2φ˙2 = µr2 = . 2 2 µr2  2µr2

1 by L2 V (r) = V (r) + . eff 2µr2 This allows us to provide a qualitative description of the motion. Draw ﬁgures in class.

Angular momentum commutation relations

We will need the commutation relations for coordinates and momenta of a particle in three dimensions. Consider a particle in three dimensions with the coordinate and momentum operators along the three directions denoted by rl and pj = − i~∂/∂rj . The notation r1 = x, r2 = y, and r3 = z allows a compact way of representing the relations. The commutation relations are

[ri, rj] = 0 , [, pj] = 0 and [ri, pj] = i~δij (4) which state that the coordinates can be measured simultaneously, the momenta along the three directions can be measured simultaneously, so can a coordinate and a momentum component along a diﬀerent direction. We will use x, y, and z or 1 and 2 and 3 for the three directions.

We wish to ﬁnd the commutation relations for the angular momentum operators de- ﬁned in the text and in the lecture. . Since L~ = ~r × ~p, we have

Lx = ypz − zpy, Ly = zpx − xpz and Lz = xpy − ypx .

Note that the order of the operators does not matter since x commutes with py etc., but it is conventional to place the diﬀerential operators to the right. Check that these are Hermitian operators.

Finding the commutation relations is tedious; a little thought can reduce the labor. We obtain

[Lx, Ly] = i~Lz , [Ly, Lz] = i~Lx , and [Lz, Lx] = i~Ly .

Angular momentum in spherical coordinates

We wish to write Lx, Ly, and Lz in terms of spherical coordinates. Recall that the in spherical coordinates (See Griﬃths “Classical Electrodynamics”) is 1 1 ∇~ =r∂ ˆ + θˆ ∂ + φˆ ∂ . r r θ r sin θ φ

2 You should be able to write this down from a simple geometrical picture of spherical coordinates. Using L~ = ~r × ~p → −i~~r × ∇~ and orthonormality of the unit vectorsr ˆ, θˆ, and φˆ we ﬁnd 1 L~ = (−i~) φˆ ∂ − θˆ ∂ .  θ sin θ φ We use the expressions for the unit vectors (obtained from ) xˆ =r ˆ sin θ cos φ + θˆ cos θ cos φ − φˆ sin φ yˆ =r ˆ sin θ sin φ + θˆ cos θ sin φ + φˆ sin φ zˆ =r ˆ cos θ − θˆ sin θ .

Given L~ in spherical coordinates above and the cartesian unit vectors we obtain Lz using Lz =z ˆ· L~ . This is the simplest of the three components and only one term survives. We ﬁnd Lz = −i~ ∂/∂φ .

Give a simple geometrical interpretation: the angular momentum along z in classi- cal mechanics arises from about the z=axis, i.e., from temporal variations in φ. The linear momentum along x in arises from time variation of the position x. Since in px → −i~∂/∂x, one might naively expect Lz → −i~ ∂/∂φ and this expectation is checked by the calculations outlined above.

Computing the scalar products yields

~ ∂ ∂ Lx = −i − sin φ ∂θ − cos φ cot θ ∂φ ,   ~ ∂ ∂ Ly = −i cos φ ∂θ − sin φ cot θ ∂φ , ~ ∂  Lz = −i ∂φ .

A very useful operator is the total angular deﬁned by 2 ~ ~ 2 2 2 L = L · L = Lx + Ly + Lz . One can verify that

2 2 2 [L , Lx] = [L , Ly] = [L , Lz]=0 . (5) 2 The commutation relations imply that only two of the set of four operators{Lx, Ly, Lz, L } can have deﬁnite values (with no uncertainty) in a quantum mechanical state simultane- 2 ously; conventionally they are chosen to be L and Lz.

Given the cartesian components we can compute an explicit expression for L2 in spher- ical polar coordinates. A straightforward calculation yields ∂2 ∂ 1 ∂2 L~ · L~ = −~2 + cot θ + (6) ∂θ2 ∂θ sin2 θ ∂φ2  1 ∂ ∂ 1 ∂2 = −~2 sin θ + . (7)  sin θ ∂θ  ∂θ  sin2 θ ∂φ2 

3 This is precisely the combinations that appear in the Laplacian in spherical coordinates,

We recall that the Laplacian in spherical polar coordinates is given by 1 ∂ ∂ 1 ∂ ∂ 1 ∂2 ∇2 = r2 + sin θ + . (8) r2 ∂r  ∂r  r2 sin θ ∂θ  ∂θ  r2 sin2 θ ∂φ2

Thus the kinetic (denoted by H0) in three-dimensional, single-particle quantum mechanics can be written as follows: Note that

2 ~p · ~p ~ 2 H0 = = − ∇ 2µ 2µ Substituting for the Laplacian we have

2 2 2 ~ 1 ∂ 2 ∂ ~ 1 ∂ ∂ 1 ∂ H0 = − r − sin θ + (9) 2µ r2 ∂r  ∂r  2µ r2 sin θ ∂θ  ∂θ  r2 sin2 θ ∂φ2  ~2 1 ∂ ∂ Lˆ2 = r2 + . (10) 2µ r2 ∂r  ∂r  2µr2 We have restored the hat on L to remind you that it is an operator.

This is an important identity and you should be familiar with it. The angular part Lˆ2 of the kinetic energy operator is . The classical calculation would have naturally 2µr2 suggested this form.

2 Since L and Lz commute they share a set of eigenvectors, i.e., we can ﬁnd a set of functions f(θ,φ) such that L2 f(θ, φ) = λ ~2 f(θ, φ)

Lz f(θ, φ) = ν ~ f(θ, φ) . Of course, the eigenvalues are not the same; they do not even have the same dimensions. We have denoted the corresponding eigenvalues by λ ~2 and ν ~. This way the dimensions are taken care of and λ and ν are numbers. We have to determine the eigenvalues and eigenvectors.

Now it is easy to obtain the eigenfunctions of Lz:

Lz fθ,φ) = ν ~f(r,θ,φ) .

Since Lz = −i~ ∂/∂φ (as you remember!) θ is eﬀectively a constant and we consider functions of φ only and set f(θ,φ) = g(φ). Canceling a factor of ~ we have d −i g(φ) = νφ ⇒ g(φ) ∝ eiνφ . dφ

4 Since under φ → φ + 2π the system returns back to the original position and we expect nothing to be changed. Therefore, we demand that ν is an integer which will be denoted by m. (This is a slippery argument and I will just let it slip by.) Thus the eigenfunctions of Lz are exp(imφ) with eigenvalue m~. Please remember this!

We determine the eigenvalues and the eigenfunctions of L2 next. Canceling the con- stant factor ~2 the eigenvalue equation reads

1 ∂ ∂ 1 ∂2 − sin θ + f(θ,φ) = λf(θ,φ) .  sin θ ∂θ  ∂θ  sin2 θ ∂φ2 

m First the result: the eigenfunctions are the spherical harmonics denoted by Y` (θ,φ) and the eigenvalues are `(` + 1) ~2 where ` is a non-negative integer. That is to say λ = `(` + 1). The superscript m indicates the eigenvalue of Lz, m~.

We use separation of variables to solve the equation. Let f(θ,φ) = Θ(θ)Φ(φ) and multiply the equation by sin2 θ and divide by ΘΦ and obtain

1 d2Φ sin θ d dΘ − = sin θ + λ sin2 θ . Φ dφ2 Θ dθ dθ The left-hand side and right-hand side are respectively functions of φ and θ only and so they have to be equal to the same constant for the equation to be satisﬁed for all values of the variables. We denote the separation constant by m2 we obtain Φ(φ) = eimφ. This is consistent with our result for the eigenfunctions of Lz. The other equation becomes d dΘ sin θ sin θ + `(` + 1) sin2 θ Θ − m2Θ=0 . (11) dθ dθ We have used the notation λ = `(` + 1) anticipating the result. Note that there is no loss of generality since one can always deﬁne ` thus. The diﬃculty is to show that ` is a non-negative integer. We let z = cos θ and write the equation as

d2Θ dΘ m2 (1 − z2) − 2z + `(` + 1) − Θ=0 . (12) dz2 dz  1 − z2 

This is known as the associated Legendre equation. If m = 0 this is known as the Legendre (diﬀerential) equation. If P`(z) the Legendre polynomial of order ` is a solution to the Legendre equation we can verify that dm P m(z) = (1 − z2)m/2 P (z) ` dzm ` is a solution to the associated Legendre equation. The spherical harmonic is deﬁned by m imφ |m| Y` ∝ e P` (cos θ) with a proportionality constant given in the text.

5 m In Dirac notation we can write the normalized ket |`mi → Y` (θ,φ) in spherical coordinates.

L2|`mi = `(` + 1)~2|`mi (13)

Lz|`mi = m~|`mi (14)

A very important constraint can be deduced as follows: The norm of Lx|`mi determined 2 2 by h`m| Lx|`mi is non-negative. Similarly, h`m| Ly|`mi ≥ 0. Thus we can write

2 2 2 2 h`m| (Lx + Ly) |`mi = h`m| (L − Lz) |`mi ≥ 0. (15)

Now we know 2 ~2 2 2~2 L |`mi = `(` + 1) |`mi and Lz|`mi = m |`mi. Using the fact that |`mi is normalized we have m2 ≤ `(` + 1) . This implies that for a given value of ` the magnitude of m (recall that m can be a positive or a negative integer while ` is a non-negative integer) has a maximum possible value. With a little bit more eﬀort (see Griﬃths) we can deduce that for a given value of ` (this determines the magnitude of the total angular momentum to be `(` + 1)~2) m can only assume integer values ranging from −` up to +`. You must remember that for a given `, m assumes one of 2` +1 values, ranging from −`, −` +1 ··· , −1, 0, +1, ··· ,` − 1, ` . The spherical harmonics form a complete orthonormal set. What precisely does this mean? Any reasonable function of θ and φ can be expanded in terms of these:

∞ ` m f(θ,φ) = c`m Y` (θ,φ) (16) X`=0 mX=−` and as before we have m ∗ c`m = dΩ[Y (θ,φ)] f(θ,φ) . (17) Z ` We have used the abbreviated notation

π 2π dΩ = dθ sin θ dφ . Z Z0 Z0 This follows from ‘orthonormality’:

π 2π 0 ∗ m m 0 0 dθ sin θ dφ Y`0 (θ,φ) Y` (θ,φ) = δ`` δmm . (18) Z0 Z0 h i

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