10 Angular Momentum II

10.1 Central force and pseudopotential Our goal is now to solve the Hydrogen atom completely at the nonrelativistic level (& for an inﬁnitely massive proton, an assumption we will soon relax), where the Hamiltonian is

h¯2∇2 e2 H = − − 2m 4π²0r Griﬃths follows a mathematical strategy in Ch. 4, using the tech- nique of separation of variables. He separates the variables by assuming ψ(r, θ, φ) = R(r)Y (θ, φ), and then after substituting in ﬁnds that the Schoedinger equation can be written in the form above Eq. (4.16) in his book. The angular momentum is then introduced purely mathematically, as a separation constant `(` + 1) which he pulls out of a hat, in (Eqs. 4.16-4.17). I would like to pursue a more physical strategy based on the angular momentum results we obtained last week. The idea is, after exploring properties of Lˆ, we now want to express H in terms of Lˆ in order to explicitly classify eigenfunctions by angular momentum eigenvalues. I will skip tedious calculation (however, it’s good for cultural background— you can read about it in Peebles, Quantum Mechanics, p. 141.). Remember what you are trying to do is express Lˆ2 2 ˆ2 ˆ2 ˆ2 ˆ2 in terms of ∇ , but L = Lx + Ly + Lz, and

Lˆx = ypˆz − zpˆy

Lˆy = zpˆx − xpˆz (1)

Lˆz = xpˆy − ypˆx, (2)

1 so in principle you can work it out. The ﬁnal result is −Lˆ2 1 ∂2 ∇2 = + r (3) h¯2r2 r ∂r2 (Note meaning of 2nd term—r−1(∂2/∂r2)rψ = r−1(∂2/∂r2)(rψ)). So kinetic energy operator in Hamiltonian is h¯2 h¯2 1 ∂2 Lˆ2 − ∇2 = − r + (4) 2m 2m r ∂r2 2mr2 and full Hamiltonian for central force problem is h¯2 1 ∂2 Lˆ2 H = − r + + V (r). (5) 2mr ∂r2 2mr2

1st want to solve t-ind. S.-eqn. Hψ = Eψ, have already argued should 2 be able to ﬁnd simultaneous e’fctns. of H, Lˆ , Lˆz. Label as ψn`m with

Hψn`m = Enψn`m, 2 2 Lˆ ψn`m =h ¯ `(` + 1)ψn`m

Lˆzψn`m = mhψ¯ n`m

2 S. eqn. is then (using fact that ψn`m are e’fctns of Lˆ ) h¯2 1 ∂2 h¯2`(` + 1)2 − rψ + ψ + V (r)ψ = E ψ (6) 2m r ∂r2 n`m 2mr2 n`m n`m n n`m

Convenient to multiply (4) by r, introduce radial wave fctn. un`m ≡ rψn`m. Then eqn. reads

  2 2 2 h¯ ∂ h¯ `(` + 1)  − u +  + V (r) u = E u (7) 2m ∂r2 n`m 2mr2 n`m n n`m | {z } | {z } 1D K.E. eﬀective 1D potential which looks like a 1D problem in the radial variable r for a particle moving in an eﬀective potential depending on `.

2 Reminder: similar equation found in classical mechanics in polar coor- dinates (example in 2D for illustration): 1 m H = mv2 + V (r) = (r ˙2 + r2θ˙2) + V 2 2   2 1  Lˆ  = mr˙2 +  + V  2 2mr2 

10.2 H-atom bound states

Electron sees potential V = e2/r from nucleus:   2 2 2 2 h¯ ∂ h¯ `(` + 1) e  − u +  −  u = Eu = −Eu (8) 2m ∂r2 2mr2 r with E = −E the binding energy of the electron as usual. Rewrite as ∂2u `(` + 1) 2me2 2mE − u + u = u (9) ∂r2 r2 h¯2r h¯2 This is just “radial S.-eqn.” we wrote down before assuming spherically symmetric solutions (Eq. 4-15 et seq.) only now including `(` + 1) term! Recall trick leading to solution back then was to write √ 2 u(r) = f(r)e−r 2mE/h¯ (10) which we guessed because 1) we wanted solns. localized around the origin, and 2) could get some terms to cancel...

Substitute, ﬁnd v u 2 2 u ∂ f `(` + 1) 2me u2mE df − f + f − 2t = 0 (11) ∂r2 r2 h¯2r h¯2 dr Now look for series soln.: X p f = Apr (12) p

3 Substitute, require coeﬃcient of each power of r vanish, ﬁnd  √  2 2p 2mE 2me  [p(p + 1) − `(` + 1)]A =  −  A (13) p+1 h¯ h¯2 p

A` is zero, since factor on lhs vanishes and [ ] on right is nonzero. Then all Ap are zero for p ≤ `. (Take p + 1 = `, then you ﬁnd A`−1 = 0, iterate again...) If series does not terminate, asymptotic value of ratio √ A 2 2mE 1 p+1 ∼ (14) Ap h¯ p √ p which has soln. Ap+1 ∼ (2 2mE/h¯) /p!, meaning √ X p 2 2mE r f ∼ Apr ∼ e ¯h (15) p √ − 2mE r which is no good since u ≡ fe ¯h blows up at r → ∞.

But can escape this conundrum by assuming termination if for some in- teger n > ` √ 2n 2mE 2me2 = (16) h¯ h¯2

( Then all Ap with p > n will vanish due to (11).) This gives energy eigenvalues as before: e4m 1 E = −E = , n = 1, 2, 3 ··· > ` (17) n 2¯h2 n2 and eigenfunctions (neglect angular part for now!) √ 2m|E | un`m ` n−1 − n r ψn`m ∼ = [A`+1r + ··· Anr ] e ¯h (18) r | {z } related to Laguerre polynomial (19)

Radial part of eigenfunction Rn`(r) diﬀerent for each `!

4 Now for each ` there are 2` + 1 degenerate orthogonal eigenfctns. corre- sponding to diﬀerent m’s, i.e. ∝ eimφ. Examples:

−r/a0 ψ100 ∼ e (20) −r/2a0 ψ200 ∼ (1 − r/2a0)e (21) iφ −r/2a0 ψ211 ∼ r sin θe e (22) −r/2a0 ψ210 ∼ r cos θe (23) −iφ −r/2a0 ψ21−1 ∼ r sin θe e (24)

Remarks:

1. Eigenfunctions of Lˆ2 are Legendre functions (sometimes called asso- ciated Leg. polynomials) P`m(cos θ). For m = 0, reduce to familiar Legendre polynomials P`(cos θ). Won’t belabor deﬁnitions now, can look up anywhere.

imφ 2. Recall e’fctns of Lˆz are ∼ e . 2 3. Simultaneous e’fctns of both Lˆ and Lˆz easy to construct, just multiply imφ two together, get spherical harmonics Y`m ≡ P`m(cos θ)e . Note orthonormality: Z ∗ dΩ Ylm(θ, φ)Yl0m0(θ, φ) = δ``0δm,m0 (25)

4. 1st few:

1 Y = √ 00 4π √ √ √ 3 −iφ 3 3 iφ Y1,−1 = − sin θe ,Y1,0 = − cos θ, Y1,1 = − sin θe 8√π 4π √ 8π 15 2iφ 2 15 iφ Y22 = e sin θ, Y21 = − e sin θ cos θ, √ 32π 8π 5 2 Y20 = 16π (3 cos θ − 1), ···

5 5. Full normalized e’fctns ψn`m for H-atom problem are radial fctns. (see Eq. 17) times spherical harmonics,

ψn`m(r, θ, φ) = Rn`(r)Y`m(θ, φ) (26)

Recap: nonrelativistic hydrogen energy levels • 1s. (Ground state) n=1,`=0, E=-13.6 eV spectroscopic notation: (s=“sharp”) • 2s. (1st excited state) n=2, ` = 0, E=13.6 eV/(22) • 2p. n=2,` = 1. Degeneracy = 2`+1 = 3. • 3s. n=3, ` = 0 spectroscopic notation: (p=“principal”) • 3p. n=3, ` = 1 Degeneracy =3. • 3d. n=3, ` = 2 Degeneracy =5. spectroscopic notation: (d=“diﬀuse”). .

6 10.3 Two body problem Need to ask what happens if proton allowed to have ﬁnite mass, also consider other 2– or more–particle systems where masses are similar. For now stick to 2 particles, think of electron and proton in H atom. Write down single wave function to describe entire two-particle system, fctn of electron position re and proton position rp:

ψ = ψ(re, rp, t) (27)

3 related to joint probability of ﬁnding the electron in d re at re simultane- 3 ously with proton in d rp at rp: 2 3 3 dP = |ψ(re, rp, t)| d rpd re (28) with normalization condition Z 2 3 3 |ψ| d rpd re = 1 (29) which we again write as (ψ, ψ) = 1. Total momentum: ∂ ∂ P = pe + pp = −ih¯ − ih¯ (30) ∂re ∂rp Total energy: pˆ2 pˆ2 e2 H = e + p + − (31) 2me 2mp |rp − re|

10.4 Momentum conservation in 2-body system Check commutators: • ∂ [pe,H] = [pe,V ] = −ih¯ V ∂re

7 dV ∂ = −ih¯ r dr ∂re dV r − r = −ih¯ e p (32) dr r

and similarly • dV r − r [p ,H] = −ih¯ p e (33) p dr r so total momentum P = pe + pp commutes with H [P,H] = 0 (34) Consequences: 1. Can ﬁnd complete set of eigenstates of energy and total momentum. 2. Momentum conserved: if ψ is eigenfctn. of P at time t, still an eigenstate at later time t0.

3. Total p is generator of translations in same sense that total ang. 0 mom. was generator of rotations. If ψ (re, rp, t) is deﬁned to be the wave function of a system described by ψ shifted in space by the displacement ~δ, can show (See Peebles p. 121) ~ ψ0 = Uψ, U = e−iδ·p/h¯ (35)

where again operator U is unitary, the exponentiation of a Hermitian operator

10.5 Reduction from 2-body to 1-body problem. Deﬁne as in classical mechanics:

8 Total mass M = me + mp (36) m m Center of mass position R = e r + p r (37) M e M p Relative position r = re − rp (38) or invert: m r = R + p r (39) e M m r = R − e r (40) p M

Change variables in S.-eqn. (Hamiltonian Eq. (27)) from re, rp to R, r: ¯ ¯ ¯ ¯ ¯ ¯ ∂ψ ¯ ∂ψ ∂r ¯ ∂ψ ∂r ¯ ¯ e ¯ p ¯ ¯ = ¯ + ¯ (41) ¯ ¯ ¯ ∂R r ∂re ∂R r ∂rp ∂R r and note ∂re/∂R = ∂rp/∂R = 1. So

¯ ¯ ∂ψ ¯ ∂ψ ∂ψ ¯ ¯ = + , or (42) ¯ ∂R r ∂re ∂rp ∂ ∂ ∂ = + (43) ∂R ∂re ∂rp Similarly ∂ m ∂ m ∂ = p − e . (44) ∂r M ∂re M ∂rp Now put (mostly old deﬁnitions):

∂ ∂ pe = −ih¯ pp = −ih¯ (45) ∂re ∂rp ∂ ∂ P = −ih¯ p = −ih¯ (46) ∂R ∂r

9 So have same relations as in classical mechanics:

P = pe + pp Center of mass momentum (47) m m p = p p − e p relative momentum (48) M e M p or inverted m p = e P + p (49) e M m p = p P − p (50) p M Rewrite kinetic energy operator in terms of center of mass and relative momentum:   2 2 ˆ2 pˆe pˆp P 1  1 1  2 + = +  +  pˆ 2me 2mp 2M 2 me mp Pˆ2 pˆ2 = + (51) 2M 2µ where µ is the reduced mass m m µ = e p (52) me + mp Since we normally are not interested in the center of mass motion, discard, and write for 2-body problem (e.g. H-atom with ﬁnite proton mass): pˆ2 H = + V (r) (53) 2µ exactly as in classical mechanics.

10.6 Addition of 2 angular momenta. We now ask the question, if we have two particles with diﬀerent angular momenta, what are the allowed eigenvalues of the total angular momentum? This situation might arise for two particles with intrinsic angular

10 momentum (spin), such as the two quarks in a pion, or it might be a question we would ask about two electrons in diﬀerent angular momentum states of a multielectron atom. In any case, suppose there are two particles, one with angular momentum operator L(1) and one with L(2). The total angular momentum operator is L = L(1) + L(2), and we assume that since particles 1 and 2 are distinct particles, any operator referring to 1 commutes with any operator referring to 2. So for example we have [Lx(1),Ly(1)] = ihL¯ z(1), but [Lx(1),Ly(2)] = 0. Then what are the possible sets of commuting observables we can construct whose eigenvalues we can use to label our states with? One obvious choice is 2 2 L (1),Lz(1),L (2),Lz(2), whose eigenvalues we would call `1, m1, `2, m2, as usual. But we can also show that

2 [L ,Lz] = 0 · ¸ L2,L2(α) = 0 · ¸ 2 Lz,L (α) = 0 but · ¸ 2 L ,Lz(α) 6= 0, α = 1, 2, which suggests that there is another set of four commuting oper- 2 2 ators we could chose with which to label our states, namely L ,Lz,L (1) and L2(2). Note that the total angular momentum operators for each particle L2(α) are members of both sets. For a given pair of quantum numbers `1, `2, however, the two sets differ, because using the first set, the states can be classified in addition by m1, m2, and we will write |m1m2`1`2i, whereas using the second set, states can be classified according to `, m, where `, m are the total angular momentum quantum number and total 2 z-component quantum number, i.e. the operators L and Lz have the following effect on the states |`m`1`2i:

11 2 2 L |`m`1`2i =h ¯ `(` + 1)|`m`1`2i

Lz|`m`1`2i =hm ¯ |`m`1`2i, and of course still 2 2 L (1)|`m`1`2i =h ¯ `1(`1 + 1)|`m`1`2i 2 2 L (2)|`m`1`2i =h ¯ `2(`2 + 1)|`m`1`2i. 2 Note however that |`m`1`2i is not an eigenstate of Lz(α), because L doesn’t commute with it. We have two diﬀerent ways of labelling the eigenstates of angular momentum, neither of which is preferred. How are the “new” quantum numbers `, m related to `1, `2, m1, m2? Well we have Lz = Lz(1) + Lz(2), so certainly m = m1 + m2. The relation between ` and `1 and `2 is not so simple, but the answer is contained in what is sometimes called the triangle rule:

|`1 − `2| ≤ ` ≤ `1 + `2. I will not derive this here, but you will derive it in part in the homework.

Since both the |`m`1`2i and the |m1m2`1`2i represent a complete set of states, it must be that we can always express any given state in either representation in terms of linear combinations of the other. For example, we can write

`1X+`2 |m1m2`1`2i = h`m1 + m2`1`2|m1m2`1`2i|`m1 + m2`1`2i `=|`1−`2| or X |`m`1`2i = hm1m2`1`2|`m`1`2i|m1m2`1`2i. m1+m2=m

The amplitudes (just numbers) hm1m2`1`2|`m`1`2i are called Clebsch- Gordon coeﬃcients, and are tabulated in Griﬃths on p. 188. A concrete example is for `1 = 2, `2 = 1,

12 v u u 1 u3 1 |3 0 2 1i = √ |1 − 1 2 1i + t |0 0 2 1i + √ | − 1 1 2 1i, 5 5 5 where (careful!) the left side of the equation is written in `m`1`2 notation, whereas the kets on the right are in the m1m2`1`2 notation. Compare directly with Griiﬁths’ Eq. below (4.185). It’s the same equation in diﬀer- ent notation. A third notation is given to you in the homework! Choose whichever seems most natural to you.