PY 451 Notes — April 12-19, 2018 Mcintyre Chap. 7 Notes — Spectrum of Angular Momentum Professor Kenneth Lane1 A. Preliminar

PY 451 Notes — April 12-19, 2018 McIntyre Chap. 7 notes — Spectrum of Angular Momentum

Professor Kenneth Lane1

A. Preliminaries

1.) In the following I will use J = (Jx,Jy,Jz) = (J1,J2,J3) for a generic angular momentum, L = r × p for the orbital angular momentum of a particle whose position (operator) is r and momentum is p, and S for the spin operator of the particle. (For a massive particle, its spin is defined as its total angular momentum when it is at rest. The spin is therefore an intrinsic fundamental property of the particle, like its mass, electric charge, and other quantities which are not effected by the particle’s motion. Defining the spin for a massless particle like the photon is a little tricky, so we won’t be concerned with that here.) For a particle in motion, its total angular momentum (operator) is J = L + S.

2.) As for spin, angular momentum is a hermitian operator and its commutation relations with itself are [Jx,Jy] = i~Jz plus cyclic permutations. They are written compactly using the summation convention of summing over repeated indices:

3 X [Ja,Jb] = i~ abcJc ≡ i~ abcJc, (a, b, c = 1, 2, 3). (1) c=1

Here, abc is the completely antisymmetric Levi-Civita tensor introduced in the Chapter 2 notes. 3.) It is convenient to deﬁne ladder operators for angular momentum,

† J± = Jx ± iJy = J1 ± iJ2 = J∓. (2) These have the commutation relations (from Eq. (1))

[Jz,J±] = ±~J± , [J+,J−] = 2~Jz . (3) 4.) The square of the angular momentum operator is

2 2 2 2 1 2 J = JaJa ≡ Jx + Jy + Jz = 2 (J+J− + J−J+) + Jz . (4) This is also hermitian and its possible eigenvalues must be non-negative (≥ 0). It is easy to see 2 1 that [Ja, J ] = 0 for a = 1, 2, 3 (just as it was for spin- 2 ). Summing on repeated indices again and using [A, BC] = B[A, C] + [A, B]C, we have (by the antisymmetry of abc = −acb)

2 [Ja, J ] = [Ja,JbJb] = [Ja,Jb]Jb + Jb[Ja,Jb] = i~ abc(JcJb + JbJc) = 0. (5)

2 Thus, one component of the angular momentum, usually taken to be Jz, and J can be simultaneously 2 diagonalized. Considered by themselves, with no other operators, (Jz, J ) constitute a complete set of commuting observables (CSCO); see the Chapter 2 notes again.

1 c Kenneth Lane, 2018.

1 B. The Spectrum of Angular Momentum

2 1.) We denote by |jmi the simultaneous eigenstates of Jz and J , where j and m are real, dimen- sionless numbers we determine below. They label the total angular momentum and its z-component, respectively. The eigenvalue equations are

Jz|jmi = m~|jmi, (6) 2 2 J |jmi = j(j + 1) ~ |jmi. (7) Since J 2 is a positive semi-deﬁnite operator, j(j +1) ≥ 0. For simplicity in this derivation, I assume 2 that j ≥ 0. For convenience, I set ~ = 1 in this derivation. As usual, the eigenstates of Jz and J form a complete orthonormal set in the space in which these two operators act:

0 0 hjm|j m i = δjj0 δmm0 , (8) X |jmihjm| = 1 , (9) j,m where the sum is over all possible values of j and m.

2.) For fixed (j, m), consider the states obtained by applying the ladder operators J± to |jmi. We 2 act on these states with Jz and J to find their m and j-values. Using the commutation relations in Eqs. (3, 5) (fill in the missing steps yourself!)

Thus, J±|jmi is a state with total angular momentum j and z-component m + 1:

± J± = cjm|j, m + 1i. (12)

Hence the name “ladder operators” for J±.

± 3.) Now determine cjm:

± 2 2 2 |cjm| = hjm|J∓J±|jmi = hjm|Jx + Jy ± i[Jx,Jy]|jmi 2 2 = hjm|J − Jz ∓ Jz|jmi = j(j + 1) − m(m ± 1). (13)

± 2 Since the squared norm of J±|jmi must be non-negative, |cjm| ≥ 0, we must have (for j ≥ 0): j(j + 1) − m(m + 1) ≥ 0 =⇒ m ≤ j, (14) j(j + 1) − m(m − 1) ≥ 0 =⇒ m ≥ −j. (15) (16)

Choosing the positive square root (a simple choice of overall phase), we have

± p p cjm = j(j + 1) − m(m ± 1) = (j ∓ m)(j ± m ± 1). (17) This is a useful and important equation. You should try to memorize it.

4.) Finally: For ﬁxed j ≥ 0, each application of the ladder operator J± to the state |jmi raises or lowers m by one unit until m = j (for J+) or m = −j (for J−). The next application of J+ or J−

2 gives zero: J+|jji = J−|j, −ji = 0. Thus, there are 2j + 1 values of m = −j, −j + 1, . . . , j − 1, j. But, then 2j + 1 must be a non-negative integer!. (Had we assumed j ≤ −1 so that j(j + 1) is still ≥ 0, this shows that we would have obtained a contradiction.) We have now found the eigenvalue 2 2 spectrum of the Jz and J (or of J and any other component of J):

Jz|jmi = m~|jmi, (18) 2 2 J |jmi = j(j + 1) ~ |jmi; (19) 1 3 j = 0, 2 , 1, 2 ,..., (20) m = −j, −j + 1, . . . , j + 1, j. (21)

More usually, one writes −j ≤ m ≤ j.

A question: Can you prove that there are no states between |jmi and j, m ± 1i? Recall the similar question about the states of the harmonic oscillator.

5.) I hope you have enjoyed this derivation of the spectrum of angular momentum. With the ladder operators, it is somewhat similar to determining the spectrum of the harmonic oscillator and, indeed, many other diagonalization problems (including the hydrogen atom!). So, now we 1 know why the spin operators studied in Chapters 1-3 can describe spin- 2 , 1, etc. Everything is determined by the commutation relations of the operators and the fact that they are hermitian! As you know, the spectrum of orbital angular momentum contains only the integers: l = 0, 1, 2,... with −l ≤ ml ≤ l. I will explain the reason for that when we discuss orbital angular momentum — which is next.

C. Orbital Angular Momentum

As discussed in some detail in the discussion section on April 11, the orbital angular momentum operators L = r × p in the r representation are ∂ ∂ ∂ ∂ ∂ ∂ L = −i r × ∇ = −i y − z , z − x , x − y , (22) ~ ~ ∂z ∂y ∂x ∂z ∂y ∂x and they are easily (more easily than in McIntyre!) seen to satisfy the commutation relations of angular momentum: [La,Lb] = i~ abcLc. (23) For spherically symmetric potentials V (r) — which we shall focus on almost exclusively for the rest of the semester — the entire Hamiltonian is spherically symmetric (invariant under rotations), and so, as we have discussed many times now,

[H, L] = 0, (24) or, what is the same thing (as we have also discussed repeatedly), orbital angular momentum is 2 conserved: dhL(t)i/dt = 0. Thus, we can simultaneously diagonalize H, Lz and L , and the eigenstates of H, states with a particular energy eigenvalue En` and orbital angular momentum ` will be (2` + 1)-fold degenerate. These states are labeled |n`mi, and

3 The reason for labeling the energy with both a “principal” quantum number n and the orbital angular momentum ` will become clear presently. (Note: We are ignoring the possibility of spin here. In atoms like the hydrogen atom, spin-eﬀects are small and can be ignored to a good approximation.)

Now, it turns out that the only allowed values of ` for orbital angular momentum are ` = 0, 1, 2,... . 2 Why is that? The usual answer is that the wave function (here the eigenfunction of Lz and L imφ must single-valued, i.e. that the eigenfunction e of Lz = −i~ ∂/∂φ must have the same value at φ = 0 and φ = 2π; one complete revolution by 2π around the z-axis should give us back the same wave function. But, since the overall phase of the wave function has no physical meaning, that 1 3 argument cannot be right! Furthermore, it is not correct for spin 2 , 2 , etc. The real argument is more complicated; I’ll merely state the result below.

The components of L and L2 in the (θ, φ) representation are:

∂ ∂ L = i sin φ + cos φ cot θ , x ~ ∂θ ∂φ ∂ ∂ L = i − cos φ + sin φ cot θ , y ~ ∂θ ∂φ ∂ L = −i , (26) z ~∂φ 1 ∂ ∂ 1 ∂2 L2 = − 2 sin θ + . (27) ~ sin θ ∂θ ∂θ sin2 θ ∂φ2

This expression for L2 should look familiar. It appeared in the energy eigenvalue equation for the Hamiltonian in spherical coordinates on page 7-5 of the notes labeled “Chapter-7-prelims”. Using it here with our notation in Eqs. (25), that equation is

2 1 ∂ ∂ L2 − ~ r2 + + V (r) ψ (r, θ, φ) = E ψ (r, θ, φ) , (28) 2µ r2 ∂r ∂r 2µr2 n`m n` n`m where ψn`m(r, θ, φ) = hr|n`mi ≡ hr, θ, φ|n`mi. Note the separation of the kinetic energy (derivative) terms into a purely radial piece and a purely orbital piece. This is very important, because the 2 orbital term will now become part of the central potential. Since L |n`mi = `(` + 1)~2|n`mi, Eq. (28) becomes — in the spherical coordinate representation of operators and states:

2 1 ∂ ∂ `(` + 1) 2 − ~ r2 + ~ + V (r) ψ (r, θ, φ) = E ψ (r, θ, φ) . (29) 2µ r2 ∂r ∂r 2µr2 n`m n` n`m Thus the orbital kinetic energy has become part of an eﬀective central potential called the “centrifugal barrier”: `(` + 1) 2 V (r) = V (r) + ~ . (30) eﬀ. 2µr2 As we shall see, for ` 6= 0, it is a repulsive potential keeping the particle with reduced mass µ away from r = 0. This is consistent with classical mechanics: a classical particle will pass through the origin if and only if L = r × p = 0, i.e., if and only if r and p are parallel.

Think about what we just did and what we ended up with. We started with a Hamiltonian which includes an orbital kinetic energy term which is a diﬀerential operator L2 in the angles θ and φ. Using the fact that the wave function ψn`m is an eigenfunction of this operator with eigenvalue

4 `(` + 1)~2, we ended up with an energy eigenvalue equation in which the Hamiltonian is just a function of r, i.e., it’s one-dimensional! What is going on?

Let us try a solution of the energy eigenvalue equation in which we separate the radial variable r from the angular variables θ and φ. With knowledge aforethought, I write ψn`m as (using my more common notation for Y`m, not the book’s)

ψn`m(r, θ, φ) = Rn`(r)Y`m(θ, φ). (31) Using this in Eq. (28), it is easy to see that 1 ∂ ∂ 1 ∂2 L2Y (θ, φ) ≡ − 2 sin θ + Y (θ, φ) = `(` + 1) 2 Y (θ, φ). (32) `m ~ sin θ ∂θ ∂θ sin2 θ ∂φ2 `m ~ `m That is, Y`m(θ, φ) ≡ hθ, φ|`mi (33) is the eigenfunction of L2 in the (θ, φ) basis! I won’t go through the solution of the angular eigenvalue equations to obtain the Legendre polynomials, 1 d` P (z) = (z2 − 1)` , (34) ` 2` dz` (where −1 ≤ z ≤ 1 is the variable z ≡ cos θ), the associated Legendre polynomials, 1 dm+` P (z) = (1 − z2)m/2 (z2 − 1)` , (35) `m 2` dzm+` and the spherical harmonics, s (2` + 1) (` − |m|)! Y (θ, φ) ≡ (−1)mY ∗ (θ, φ) = (−1)(m+|m|)/2 P (cos θ) eimφ . (36) `m `,−m 4π (` + |m|)! `m

That is amply done in McIntyre’s book in Sects. 7.6.1–7.6.4. The derivations for P` and P`m follow the usual series method of solving ordinary diﬀerential equations, just as we did for the Hermite polynomials in the harmonic oscillator wave functions. And, as there, in order to make the series terminate and yield normalizable solutions — the boundary condition again! (see pages 230-231) — we ﬁnd that ` must be a non-negative integer! ` = 0, 1, 2, 3,... ; −` ≤ m ≤ m, (2` + 1 values). (37) That is the reason for integer-valued orbital angular momenta! Note that the only φ-dependence in imφ Y`m is the simple e , and that ∂ L |`mi = m |`mi =⇒ hθ, φ|L |`mi = −i hθ, φ|`mi = m hθ, φ|`mi ≡ m Y (θ, φ) . (38) z ~ z ~∂φ ~ ~ `m

Another expression of the fact that orbital angular momenta are integer-valued is that the spherical harmonics are a complete orthonormal set of functions in the space of functions of cos θ ∈ [−1, 1] and φ ∈ [0, 2π): Z 1 Z 2π ∗ 0 0 d(cos θ) dφ Y`m(θ, φ)Y`0m0 (θ, φ) ≡ h`m|` m i = δ``0 δmm0 , (39) −1 0 ∞ ` ∞ ` ! X X ∗ 0 0 X X 0 0 0 0 Y`m(θ, φ)Y`m(θ , φ ) ≡ hθ, φ| |`mih`m| |θ , φ i = δ(cos θ − cos θ )δ(φ − φ ) . (40) `=0 m=−` `=0 m=−`

5 Finally, using Eq. (31) in Eq.(29) and factoring out the Y`m, we get the one-dimensional radial equation, the energy eigenvalue equation for central potentials in three dimensions:

2 1 ∂ ∂ `(` + 1) 2 − ~ r2 + ~ + V (r) R (r)(r, θ, φ) = E R (r) . (41) 2µ r2 ∂r ∂r 2µr2 n` n` n`

I repeat: The eﬀective potential is the sum of the central potential and the centrifugal barrier term.